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~J. Ba\a !

Er. Anurag Mishra
Main & Advanced
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Heat& Thermodynamics for JE E
(Mo'. & Adva.ced)
by:
Er. Anurag Mishra BTech (Mech. Engg.)
HBTI Kanpur
SHRI BALAJI PUBLICATIONS (EDUCATIONAL
PUBLISHERS
AN ISO 90012008
CERTIFIED
& DISTRIBUTORS}
ORGANIZATION
Muzaffarnagar (U.P.)  251001
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11 Published by:
SHRI BAlAJl PUBLICATIONS lEDUCATIONAL PUBLISHERS & DISTRIBUTORS)
6, Gulshan Vihar, Gali No.1, ,,
Opp. Mahalaxmi Enclave, Jansath Road, Muzaffarnagar (U.P.) Phone: 01312660440 (0), 2600503 (R) website: www.shribalajibooks.com email: [email protected]
11 First edition
, ,,,
: 2010
11 Third edition: 2013 11 Reprint
: 2018
, ,
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/
~ 275.00
11 Typeset by : Sun Creation Muzaffarnagar
III
All the rights reserved. No part of this publication
may be reproduced. stored in a retrieval system or transmitted, in any form or by any means, electronic. mechanical. photocopying, recording or otherwise, without the prior permission of the author and publisher. Any violation/breach
shall
be takAn into legal action,
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Preface I am very thankful to all those readers who gave warm welcome to Mechanicsl & II and Electricity & Magnetism
~ I & II. These books have got whelming
regular demand of the students which compelled short span of time. "Heat and Thermodynamics" IIT.JEE syllabus. This book will help the students addressing
key misconceptions
I sincerely
me to complete
response
to the
this volume in very
is one of the most scoring topic in the
in building analytical
and developing
confidence
and quantitative
skills,
in problem solving.
wish that this book will fulfill all the aspirations
of the readers.
Although utmost full care has been taken to make the book free from error but some
errors inadvertently may creep in. Author and Publisher shall be highly obliged if suggestions regarding improvement indebted
Neeraj
Ii for providing
and errors are pointed out by readers. I am me an opportunity
to write
a book
of this
magnitude. I am indebted
to my father Sh. Bhavesh Mishra, my mother
Mishra, my wife Manjari, my sister Parul, my little kids Vrishank their valuable Moradabad,
time which I utilized during the writing
who supported
I am also thankful Sunil Manohar, fortheir
me throughout
to Mr. T. Konda)a
Mr. S.P. Sharma,
valuable suggestions
Shri Balaji Publications
my career.
Mr. Sudhir
In the last, I also pay my sincere
and Ira for giving
of this book and people of
Rao, Mr. Abhishek
in improving
Smt. Priyamvada
Sharma
Sinha (Ranchi),
and Mr. P. Narendra
Mr.
Reddy
the book.
thanks
to all the esteemed
members
in bringing out this book in the presentform.
Anurag
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How to face the challenge?
Following are some doubts which arise in the mind of almost all the students but may face them by taking some care. 1.
/ can not solve numerical because my concepts are notc1ear. In fact numerical solving itselfis an exercise to learn concepts.
2.
/ can not study because I am in depression, I fell into it because 1 was not studying! Depression is escape mechanism of people afraid offacing failures. Failure is integral part oflearning. I understand everything in class but can not solve on my own. WRITI NG work is vital. It is a multiple activity. initially idea comes in mind then we put into language to express it, we are focussed in hand eye coordination, visual impression
eyes create
on brain which is recorded there. WRITING WORKS ARE
EMBOSSED ON BRAIN LIKE CARVINGS OF AJANTA CAVES. 4.
In exams my brain goes blank, but I can crack them at home. Home attempt is your second attempt! you are contemplating
about it while home back. You
do nDt behave differently in exam YDUreplicate your instincts. Once a fast bDwler was bDwling no balls. His cDach placed a stump Dn crease, in fear Df injury he got it right. CONCEPTUALIZATION, SOLVING, THEN PROBLEM GETS TO CONCLUSION!
WRITING
EQUATION,
S.
I am an average student. It is a ratiDnalizatiDn used by peDple afraid of hard wDrk. In their reference frame Newton's first law applies "if I have a misconceptiDn I will continue with it unless pushed by an external agent even I will surround him in my web Df misconceptiDn yielding zero resultant." AVERAGE IS NOT DUE TO CAPACITY LACUANE BUT DUE TO LACKOF DETERMINATION TO SHED INERTNESS.
6.
A famous cliche "I do not have luck in my favour" PRINCIPLE OF CAUSALITY: CAUSE OF AN EVENT OCCURS IN TIME BEFORE OCCURRENCE OF THAT EVENT i.e., cause Dccurs first then event Dccurs. SHINING OF LUCK IS NOT AN INSTANTANEOUS EVENT IT IS PRECEDED BY RELENTLESS HARD WORK. Sow a seed Dfaspiration in mind, water it with passion, dedication will bear fruit, luck can give you sweeter fruit .
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Useful Tips 1.
I
Do not take study as a burden actually its a skill like singing and dancing. It
has to be honed by proper devotion and dedication. 2.
Without strong sense of achievement you can't excel. Before entering the competitive field strong counselling by parents is must. Majority do not know what for they are here. No strategic planning, they behave like a tail ender batting in frontof5teyn's bouncers.
3.
Science is not a subject based on well laid down procedures or based on learning some facts, it involves very intuitive and exploratory approach.
Unless their is desire and passion to learn you can not discover new ideas. It requires patience and hard work, whose fruits may be tangible later on.
•
4.
Some students realize very late that they are studying for acquiring skills and honing them. Their is a feeling that they can ride at the back of instructor and achieve excellence. Study comes as torturous exercise enforced on them and theiris some mechanism that can take this burden of them.
S.
Science is not about gaining good marks, up to Xth by reading key points good marks are achieved but beyond that only those survive who have genuine interest in learning and exploring. Self study habit is must.
6.
IF YOUWANT TO GAIN LEAD START EARLY.Majority of successful students try to finish major portion elementary part of syllabus before they enter Coaching Institute. Due to this their maturity level as compared to others is more they get ample time to adjust with the fast pace. They are less traumatized by the scientific matter handed over. For those who enter fresh must be counselled to not get bullied by early starters but work harder initially within first two months initial edge is neutralized.
7.
Once a student lags behind due to some forced or unforced errors his mind begins to play rationalization remarks like I am an average student, my mind is not sharp enough, I have low IQ etc. These words are mechanisms used to avoid hard work. These words are relative terms a person who has early start may be intelligent relative to you. Intelligence means cumulative result of hard work of previous years, that hard work has eventually led to a development of instinct to crack things easily.
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"....;;;;

1. Temperature, Heat and the Eguation of State, Heat Transfer Thermal Equilibrium and the Zeroth Law of Thermodynamics
(1); Thermometers
and
Temperature Scales (2); The Triple Point of Water (2); ConstantVolume Gas Thermometer (3); Constantpressure Gas Thermometer (3); Thermal Expansion (4); Thermal Stress (18); Determination of the Coefficient of Expansion of Liquid (20); Calorimetry (25); IdealGases (33); The IdealGas Law (34); Gas Laws (35); Microscopic View of Thermodynamics (36); The Eqoipartition of Energy Theorem (38); Ideal Monoatomic Gas (39); Ideal Diatomic Gas (39); Specific Heat Capacity of Solids (41); Maxwells Law of Distribution of Velocities (42); HeatTr?nsfer (59); Conduction (59); Why is Conductive Heat Flux (60); Radiation (75); Kirchhoff's Law (77); Temperature of the Sun (79); Subjective Problems (100). Level I: Only One Alternative is Correct (II 4); Solutions (129). Level.2:
More than One Alternative is/are Correct (135); Solutions (139);
Leve!3: Comprehension Based Problems (140); Matching Type Problems (144); Assertion and Reason (147); Solutions (150).
~232Y
2. Thermodynamics
Thermodynamic Processes (153); Atoms and Moles (153); IdealGas Processes (154); Heat Transfer to a System (162); The First Law of Thermodynamics (162); Conversion of Graphs (167); Mixing of Gases (186); Condition for the Molar Heat Capacity is Negative (193); Analysis of Thermodynamic Cycles (213); The Second Law of Thermodynamics and Heat Engines (218); Absolute Zero and the Third Law of Thermodynamics (219). LevelI: Only One Alternative is Correct (227); Solutions (246). Level2: More than One Alternative is/are Correct (250); Solutions (259);
Level3: Comprehension Based Problems (260); Matching Type Problems (268); Assertion and Reason (271); Solutions (273). l!,'
Previous Year Problems with Solutions
•
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GROW
GReeN
Save NaTURe
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THERMAL EQUILIBRIUM AND THE ZEROTH LAW OF THERMODYNAMICS Two systems are said to be in thermal contact if heat transfer can take place between the systems. When two systems at the same temperature are placed in thermal contact with each other, there is zero heat transfer between them, Fig. 1.1 the systems are said to A cup of cooling coffee illustrates the be in thermal Idea of a thermodynamic system. equilibrium. When two systems in thermal equilibrium are put in contact through a diathermic wall, their variables of state do not change. A diathermic boundary between lWa systems allows thermal interaction, whereas an adiabatic wall inhibits thermal interaction. For a given process, system choice can became that allows the three different ways to describe a cup of coffee. (a) The system is the coffee. Its environment is the cup the room and everything else. (b) The system is the coffee and the cup. Its environment is the room and the rest of the world. (c) The system is the coffee, the cup, and the room. The environment is everything outside the room. 'TWosystems can be in thermal equilibrium even if they are not in direct contacr. If systems A and B each are in thennal equilibrium with a third system C, then A and Bare in thermal equilibrium \'iith each other. This result is called zeroth law of thermodynamics. Two systems in
thermal equilibrium with a third system are in thermal equilibrium with each other.
Example
1
A thennometer (system 3) consists of a quantity of alcohol in a glass tube. When the thermometer is in thennal contact with soft ice cream in a box (system 1), the length of the alcohol column is 5 em. When placed in a freezer (system 2), the alcohol column remains 5 cm long. What happens to the ice cream when it is put into the freezer 7
Solution: The length of the alcohol column measures temperature. When they are in thermal contact, the ice cream and thermometer come to thennal equilibrium. Similarly, the thermometer comes to thermal equilibrium with the freezer when placed inside. Since the alcohol column has the same length in the freezer and in the ice cream, the thermometer is simultaneously in thennal equilibrium with both ice cream and freezer. By the zeroth law, the ice cream and freezer are in equilibrium with each other. There is no net exchange of energy and the ice cream neither melts nor freezes hard. We may restate the zeroth laws as follows. 1\vo systems are in thermal equilibrium with each other if and only if they have the "ame temperature. In this form, the zeroth law defilles temperature as the quantity that must be the same in two systems in thermal equilibrium.
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THERMODYNAMICS
Thermometers
and Temperature Scales
A thermometer relies on some physical property of matter, known as a thermometric property that changes with temperature. Operation of the familiar mercuryinglass thermometer depends on the empirical observation that the volume occupied by a fixed mass of mercury vT) AR>SR
XX
'I
_,
Scale reading alr (r W.rr
Thus with the rise in tempertlture buoyant decreases and apparent weight will increase. Effect of Temperature on The Barometer with a Brass Scale
Reading
force
The clock wil1100se ilT =N(7' 1'0)
of a .
Let true height of mercury column and density of mercury be ho and Po respectively, and 11 and p be the corresponding values after temperature change t:..T. Neglecting any significant change in atmospheric pressure, we have
24 x 60)< 60
=
where
r 1 =(l +
Po
"fllg
_
h 0
Thus
=12x
T
=
2J't
d1'
(l + YIl~AT)
the
Time period of pendulum is =
211:
II;;
I~
\ g
Number of oscillations performed in nventyfour hours, N=24x60x60
2rr..•.jloig When the temperature changes by t degrees, the new length is L =Lo(1 + MT) Change in time period of oscillations of the pendulum is 1'1'0 =2n[~~ 27t =
fjJ
L ~Lo
become slow if T1i",,1 > Tim1i"I' i. e., ilT > 0, will loose time. The clock will bewme fast if 1"t;",,1 < T,mhal' i. e., ilT < 0, will gain time. The loss or gain of time is independent of time period
Example
will
4
_
A steel rod with a crosssectional
area A is set lengthwise between fwo rigidly secured massive steel plates. With what force wiII the rod press against each plate if the temperature of the rod is increased by ~T? The Young's modulus a/steel is Y and the coefficient of linear expansion of steel is a.
Solution: If the rod is free to expand, the change in its length is L  La = aLoilT , From the given conditions, boundary' of rod is rigid, it cannot expand. Hence the plates will exert compressive stress corresponding to a compression of L  La. From the definition of Young's modulus, y=
or
FL AM t.L F=YAL
=YAu..1T
fgc
_lILLu
g ..,I1 ,'g 0
I
clock
1'.
g (i: + if," \g
1 =  (1 + MT) 2
=x24x60x60x(l{ilT) 2
_
of a clock has length Land the clock then goes accurately. The coefficient of linear expansion of the pendulum material is a = 1.85x 105• How much the dock wil/loose or gain in nventyfour hours if the ambient temperature is 1O"C higher than To?
To
1
Time lost in [\'ientyfour hours 1
At a temperature 'fo, the pendulum
Solution:
(g is constant
1 ,\T = MT 2
11
~ The
3
= K ,'7
1 dl
=2 L
T
8.'[' )
If is coefficient of expansion of brass scale on, barometer,
Example
f!.
\g
=l' 2 I T'  T 1 L'  L

\ g
60x 60ut s
Method 2:
or
(l
F
\ g
=[~}
110
aloilT
r;; g
27t
hoPo =hp =p
or
:t
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THERMODYNAMICS d !. = (l +
Example
3a6T)
d, An optical engineering firm needs to ensure that the separation between two mirrors is unaffected by temperature changes. The mirrors are auached to the ends of two bars of
d _
= dj (1 + 3a6Tr'
different materiaLs chat are welded together at one end as shown in Fig. 1£.5. The surfaces of the bars in COntact are lubricated. Show chat the distance 1 does not change with temperature if aliI'" uzl2• where uj and u2 are the respective thennal
coefficients of temperature.
the required lengths IJ and fz, in terms
"d, (I  3aAT)
Example
Also detennine
OfU"U2
and 1.
Mirrors ~I
Lubricaling oil
d,
2  (l + 3M T)
An aluminiUm cup of volume V, is filled with a liquid at temperature T. How much liquid will spill out if the temperature of both, the cup and liquid, is increased by 6T? Assume coefficient of expansion of aluminium aa is less than coefficient of expansion of liquid y.
Solution:
The change in the volume of the cup, 6V, =3aaV,!J.T
'1.1, Fig_ 1E.5
Solution: At any temperature t, the lengths of the two bars are
The change in volume of the liquid, llVI = yV, !J.T Nare that the original volume of liquid is same as the original volume of cup. The volume of liquid that spills is
1'1 =1](1+u1t),
1'2 =12(1+(120
... (1)
From the given conditions. ['21\=[
and 1211=[
...(2)
From eqos. (1) and (2), 1'2
flVI 6.V, = (yV, 3aaV,)6.T •. Volume of liquid after temperature change, V'I = V(l + yl!J.T ) Volume of container after temperature change, V', = V(l + y,llT)
l'1 = (12 iI) + l20.2t I, o.lt 12u2 = 1,0.,
12 I,
... (3)
Change in volume of liquid relative to container, V'I  V', = V(YI  y,)llT
Ct.)
"
... (4) 0.2
YI  Yc is called apparent coefficient of expansion. Level of liquid in the container will rise or fall depends on whethet YI >Y, or 11 Ul) at (TJ > T2 Find the percentage error in the reading shown.
re.
Solution: When the scale itself expands, is smaller
than true reading.
~I ::: (t. dT is given by, therefore,
a
is increased
11 ~
A
reading
f. :::£. When the temperature
.E:> To). a AI and Y waltr are given.
Solution: Height of the water column inside [he
v
vessel = h =A errors and
Not only the linear dimensions but also the radius of gyration (k) has the same fractional increase elK
=
a. dT. Its
/(
area, being L:2 in its dimension will increase by a fraction y. dT = 20.' dT. Since the mass remains a constant, dm = a m
hence, the moment of inertia will also have proportional increase 2adT Since the angular momentum is conserved, dL dJ dw =0=+
L
dw
Hence, for
the

w
I
= 
dI I
rotational
= (20.. dT)
energy
for
instance,
L' Ek
.
An aluminium cube floats OT! mercury at temperature 300 K. How much farther will the block sink if temperature is raised to 350 K. PI/g =13.6 glee, PA]=2.7 glce, '(Ilg = 1.B>.1.T,
Where Ya is the coefficient of apparent expansion of the liquid. OT
:::::_ B ......
Yc(~~)
:: ~(~:J(~:J
or
la ='~r V, '1=Vc
Ifyp and
If the liquid fills the entire. vessel then" = 1 la =Y. Y,
If Ya is positive, upon heating the liquid seems to expand relative to the container. And, If Ya is negative upon heating, the liquid seems to contract relative to the container. If there is no apparent expansion of liquid relative to container, then 'I a "'" D V, 'Ie and 11 =~Ve Y,
'Ie
be the respective volume expansivities of the
liquid and the glass apparatus, then V2 =Vt(l+y,T)andpt
or
VI = Vl
or
and £.!. = 1 + 'I,T
1 l+y,T
P2
From Eq. (0 we get m
l+y,T
m2
l+y,T
l ~
=1+h,+'I,)T
m m
(Y,+Y,)T=
or
also
The apparatus has a glass bulb B with a narrow outlet pipe P. A vessel containing the liquid for coefficient of expansion is to be determined, is connected to bulb B. (Figure 1.16)
or
The bulb and the outlet pipe completely filled with the liquid is separately weighed at two different temperatures
I
m,
y,+y'=Y(J=
2
m m 1
m,
2
...
y,=[m\':2m2]+y,
or
DETERMINATION OF THE COEFFICIENT OF EXPANSION OF LIQUID 1. Weight. Thermometer Method
The glass bulb is cleaned and weighed. Let it be W. Next, the bulb alongwith the stem is completely filled with the liquid.
andp2 (l+yT)
where T =T2 TI
Anomalous Expansion 'of Water When pure water at DOCis heated slowly then initially, water is found to contract, till a temperature of 4°C is obtained. Beyond that water expands for a further rise in temperature. The density of water increases from DOCto 4°C and decreases, beyond 4°C. Such a behaviour of water is referred to as anomalous (abnormal) expansion of water.
...(0
T=T2T1=
...(iii)
n1jm2
m2(Yr
y,)
T =[n1:tY~~~cJ+TI 2
(ii)
...
(iv)
Equation (ii) yields the coefficient of apparent expansion of the liquid relative to the material of the apparatus (glass) . Equation (iii) gives the coefficient of real expansion of the liquid which is obtained by weighing the liquid. The method is known as weight thermometer method. The appartus is reffered as thermometer because knowing Y the coefficient of apparent expansion, any
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TEMPERATURE, HEAT AHD THE EQUATION OF STATE, HEAT TRANSFER unknown Ov).
(Tz)
temperature
can be obtained
hz hI (htTZ 1I2Tl)
from equation
Dulong and Petifs Method
y=
33
Example
Fig. 1.17 shows the experimental arrangement consisting of a glass U tube filled with the experimental
_
An isosceles triangle is formed with a rod of length II and coefficient of linear expansion aI for the base and nvo thin rods each of length 12 and coefficient of linear expansion (12 for the two pieces, if the distance benveen the apex and the midpoint of the base remain unchanged as the temperatures
!
varied show that .!.L= 2 a 2
'2
\a
.
J
"m" I"u,
Fig. 1E.33
Fig. 1.17
liquid to the same height in the two limbs. Both the limbs are enclosed in two different jackets, maintained at steam point (lOOCC) and ice point (aCC) by means of steam and ice cold
water circulated in the two jackets respectively. The two jackets contain accurate thermometers to give their respective temperatures. When the two thermometers yield thermal equilibrium
readings,
the levels of the liquid in
two limbs are recorded (say hI and hz). Since, the base of the liquid columns in the twO limbs are at the same horizontal level, pressure, at those points will be the same. If Po and Pwo be the densities of the liquid at DoC and lOooe and Po the atmospheric pressure, then Po + h1Pog or
hlpo
=
increment of length with increment in dt ~ temperature of system Given, Because I remains constant with rise in temperature dl~
therefore dl = 0 de So,
Po + hZPlOog
dll = IJaldt,
PIOO = l+y.l00 hz
or
y
==
12
==~~
Example
The experiment can be carried, even with temperatures other than the ice and the steam point. IfTI °C and T2 °C be the temperatures of the liquid columns in the two limbs, corresponding to heights h] and liz respectivel}~ then h2P2 1 + yT2 =
hIP I hI 01
= Po + hzp2g ==
li2
htpo l+yTI
=
dt
~=2/u.z
hI (l + lOOy) h2 iiI 100hl
Po + htPIg
0.2 x
11>< l}atdr = 412 >< lza2dr 112a} = 41;0.2
= hZPlOO
p, So,
dl2 = 12 x
,
\
0.1
34 V
A brass rod of mass m = 4.25 kg and a cross sectional area 5
cm2 increases its length by 0.3 mm What amount of heat is spent for coefficient of linear expansion for specific heat is 0.39 kJ/kgK and 8.5>< 103 kg/m3.
I +yTI
Solution:
hzpo 1+1T2
Given
III
=0.03cm,
m = 4.25 kg,
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upon heating from DoC. hearing the rod ? The brass is 2>< 1O5/K. Its the density of brass is
0.= 2>< 105,
A = 5cm2
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THERMOOYNAMICS Initial length of the rod is
p~.=
I = ~ =_ (rn_r_l ~ A
A
I~
_rn_
Ap
1'1\' =2x
4.25
4 10 /oC,'l
.• =3a.,
fJ.T = SoC,
5x 104 x 8.50x 103
lOs/oe,
1'$ =3x1.2x
1 5)(101)(2
~
mg =?
P,,'voO + ysfJ.T) = 109 + m (I + y",~J')
= 1 m =IOOem
So if increment in length is At = o.3mm ""D.03cm then temperature
PI\'
O+y",M)'
... (3)
Putting P,,'vo from eqn. (1) in (3), we get
rises by li/=/aAO All =
=
L\l
La
0.03 100)(2xlO$
= 150C
li(J = 15°C
So amount of heat supply to rise the [emperature from
Mg=109g=p'.V'og
acc to lS~C
Mg
Q = 4.25x 0.39)( 103 x IS J Q =
Fig. 1E.35 (b)
24.86x 103J= 24.86kJ =25kJ
11 x lOBx 0 +ysfJ.T) = 109 + m (I + yw"T)
35y
Example
A submarine made of steel weighing 109g has to take 10IJg of water in order to submerge when the temperature of the sea is woe. How much less water it will have to take in when the sea is at 15"C ? (Coefficient of cubic expansion of sea water = 2 x 104 joc, coefficient of linear expansion of steel
11 x lOB =>
[1 + 5x 2)( 1O4J
11 x lOB x (1+ 1.8 x 104)
~. 0+10x104)
8
= 10 g + 109g ... (1)
PwVo9
=
10' 10'
+m
+m
m = 99098901.1 gm So at this position it takes 99098901.1 mass and wiJI be 99098901.1 gm. So finally reduction in the weight will be =lOB 99098901.1 == 901098.9gm = 9.01 x lOSgm
or
v.
Example
36
A U.tube filled with a liquid of volumetric
coefficient of 1O5 lies in a venical plane. The height of liquid column in the left vertical limb is 100 em. The liquid in the left vertical limb is maintained at a temperature = DGe while the liquid in the right limb is maintained at a temperature = lODGe. Find the difference in levels in the two limbs.
1°c
Mg=PsVo9"'109g 8 M",g= 10 9 ',' Fig. 1E.35 (a) Note: Initially it takes in 108 g weight water then mass will be 108 in gm (gram)
Condition when temperature rised from 10°C to ISoC p;"VQg =p~VQg[Mg]+mg p;" Vog = 109 g + mg
=
104 (l0000  80.2) = m(1001)
Solution: p,,'vQ8" = M ",g + Mg[Mg = Ps Vo8"l P".vo = 11 x 108
[1+5)(3.6)(1OsJ
11x lOB + 18x 11x 104 = 109 + 106 + m + mx 103 1011 _ 80.2 x 10" = m(1 + 0.001)
~1.2xl05/oC)
P",.voS
x
Solution:
_ Pt"C 
... (2)
Po
(l + yfJ.O)
at point C the pressure will be equal, PI = Pz
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TEMPERATURE, HEAT AND THE EQUATlDN DF STATE, HEAT TRANSFER 1/'1/ = V' 1l~~VII~
OT 100cm
I
AV
100"C
AVUl'
=
~,,=V ~ 3u,;:/ ~ ~r = VH;: ~ YHg
[ by 4 kg] =
from DOCto 5.DOC
of water cooling from 20°C to SoC
mic"C;ccAT + m;c.Lj + mioeCwAT = nl",C",AT m(2100)(D"C  (lOOC)] + m(333) + m(4186)
L All ice 2. A mixture of ice and water at O'C
(5.0'C  0.0° C)
3. All water at 20"C
= 4(4186)[5.0'C
Energy release required to bring the 3.0 kg of water down to O"C, QI =m",C",(20"C0"C)
Method 2:
Heat required to raise temperature to DOC, QI
= 2S1 kJ to change the ice from 10'C
Qz =m;"cC,cc[O"C(lOOC)] = (0.50)(2100)(10)
 20.0'C]
m(21DO)(1O) + 333m + m(210) = 251D kJ from lO'C
= (3)(4186)(20) Energy required
T)
A calorimeter contains 4.00 kg of water at 20.0"C. What amount of ice at 1OOCmust be added to cause the resulting mixture to reach thernlal equilibrium at 5.0"C. Assume that heat transfer occurs only benveen the water and ice.
Solution:
~8_~
= (3)(41B6)(20'C
T = S.l"C
which on solving yields
Example
msC.,AT. = m",C", AT", + m,'alC"al~TC"1 (0.150X', (540  30.5) = (0040)(4186)(30.5  10)
Example
water cooling
Hence, 10.5 + 167 + 0.50(4186)T
76.4C,
]
from 20"C to T
write
.possibilities
[
900 Jlkg.
Solution:
10>[ by
3.0 kg of
200 g aluminium calorimeter cup. The final temperature of tile mixture is 3D.SOC. Calculate the specific heat of the alloy
given C ,at
]
::
mi""C
;ceAT
:: m(210D)[D"C  (lOOC)] to O'C,
= m(21DD) (lO) kJ/kg Hear required for phase change of ice, Qz :: m j""L j = m( 333) kJ/kg
= 10.5 kJ
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of ice
Study Anurag Mishra with www.puucho.com , THERMOOYNAMICS Heat required to raise water (formed from ice) to final temperature of S.O"C, Q3 = nJi....,Cw.6.T
"m(4186)(S"C  o.O"C) = 21m kJ/kg Heat lost by water in cooling from 20.0"C to S.O"C, Q, "mwCwH" (4)(4186)(S.0"C  20.0"C) ~2S1 k.J Since system is isolated from surroundings, Q1 +Q2 +Q3 +Q4 =OJ m(2100)(lO) + 333m + 210m  2510 = 0 Note that equations formed are mathematically same.
Thus nearly 87% of the original mass of the water can. be frozen by intensive evaporation.
Example An aluminium container of mass 100 gm contains 200 gm of ice at 20"C. Heat is added to the system at the rate of 100 call sec. Find the temperature of the system after 4 minutes. (specific heat of ice = 0.5 and L = 80 eal/gm, specific heat of Al = 0.2 cal/gm/oC)
Solution: Total heat supplied in 4 min = 240sec at the rate of 100cal/sec_ 100 x 240 ""24 x 103 cal and heat required to rise the temperature of ice from 200C to DoCand then melt it into warer
Example A vessel from which the air is rapidly being pumped out contains a small amount of water at oec. The intensive evaporation causes a gradl/a/freezing of the water. What part of the original water can be converted into ice by this method?
=
200x 0.5x 20 + 200x 80"" 18x 103cal
So out of 24 x 103 cal, 18 x 103 cal will consume and rest 6x 103 will utilize for increase the temperature of the system from DOC to ODe. 6x 103"" 100x0.2x.68
Solution:
At any temperature, a layer of vapour is formed over the surface of the liquid. An equilibrium exists between vapour and liquid. The vapour pressure at this stage is termed saturated vapour pressure. When air is pumped out, panial pressure of the vapour falls and funher liquid evaporates to saturate it. The rate of evaporation can be increased by anificially removing vapour from the surface of the liquid. The latent heat required for evaporation comes from the water itself thus freezing it. During freezing, latent heat of fusion is liberated which is used up for evaporation. Suppose there is initially m kg of water present out of which ml kg evaporates and m2 kg freezes. If Lll and Lf are respectively the latent heats of evaporation and freezing, then we have ' miL"
or Since ml + m2 finally be frozen is
m2 =
== m2Lf miLt, ==
3
6x 10
m
(220)tl.8
tl.0 =< 6000 220
=
0:25, ' = 5J/gmCC =
Example
500 J/kgOC
59
_
Toluene liquid of volume 300 cmJ at ooe is contained il1 a beaker an another quantity of toluene of volume 110 cmJ at 100°C i, in another beaker. (The combined volume i5 410 em] ). Determine the total l'olume of the mixture of the toluene liquid" when they are mixed together. Given the coefficient of volume expa1l$ion'f = 0.00 Ire and allforms of heat losses can be ignored. Also find the final temperature of the mixture.
Solution: If suppose density of toluene of O"C is Po' 3
Then volume of 110 cm toluene and become Vo at E)°C then
at 100°C will be change
110 = Vo(l + 'I x 100) = \"0[1 + 103 x: 100) =::> Vo = 100cm3
110 = Vo[l.l]
Total mass will be (300po + 100Po) = 400po at O°C and mass will remain constant with temperature So if (lOOpo) mass toluene from O°C to 100°C
\vill increase it's volume and it will
V2 = 300[1 =
3
final volume = 409.25cm
307.5c013 3
= VI
/,\V = (V,  Vi)
in the volume =
=
= Vf
= 41Ocm3
And total initial volume Net reduction
O)J
V = VJ + V2 = 409.25 em
And now net volume So
+ y(25 
300[1 + 25x 103]
409.25  410
Example
e
Given, Density of water, P ••. = 1 gm em
=>
Density of ice, Pi =0.9gm/cm Specific heat of water,
Sw
=
1callgmOC
Specific heat of ice si = 0.5 eal/gmOC Specific latent heat of ice, L = 80 callgm
Solution: When the \','ater at SOC is powered into the ice vessel then when the thennal equilibrium reach the temperature of wClter will down from ijQC and due to this temperature change if there is no change in density of water therefore \'olume of water remain same but at this stage total fall in height /'\11 =0.5cm it is only due to that some of the ice will melt and convert into water. So ice column height (h = IDem) will change to (h ""II) = 9.5 em and at this moment not the complete column of ice will melt but some column height (y) out of h will melt y =?
M icc
from
= 300po
AXYXPicc
40 = 100 =>
e = 25°C
Clrea is A then mass ofy height ice =
1\1 waler
=Ax(yo.5)xflw
pD.9oCyD..5)
~
x sx 0
y
=
5cm
Means 5 cm height ice will melt out of 10 cm height ice and it will convert into 4.5 em height water.
of [he mixture
So at equilibrium at 100°C will reduce
3,
J
=:>
e) = 30
_
lee at 20o is filled upto height h = 10 em in a unifonn cylindrical vessel. Water at temperature O°C is filled in another identical vessel upto the same height h = 10 em. Now, water from second vessel is poured into firs( vessel and it i.~ found that level of upper surface falls through /'\h = 0.5 em when thermal equilibrium is reached. Neglecting thermal capacity of vessels, change in density of water due to change in temperature and lOS$ of heat due to radiation. Calculate initial temperature 0 of water.
lfvessel crosssection
at mixing 100po x sx (l008)
60
will rise the temperature
Then mass will also remain (lOOpo) but volume 100c013 at O°C will change to 110c013 at 100°C.
(lOa 
and 300cm:1 toluene become
= 0.75cm3
O.2gmjcm3 x lOcm3
temperature
31
temperature 110 em 3 toluene which is it's volume at it will be
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Study Anurag Mishra with www.puucho.com THERMODYNAMICS If (A x 10 x P IV) gm water powered into ice finally it will come to DOC temperature and it will be equilibrium
temperature. total heat lost by ice = heat used by ice to temperature gain + to melt ice Axl0xp",x(OO) =Si x lOx A x Pi(O+ 20) + 5>< Ax PiX L IOAp",O =Si x lOx A xPi x 20+ 5x A xPi xL lOx 1 x 0 = 0.5x lOx 0.9x 20+ 5>< 80 tOn = 9>LO
To ),jind the constant heat flow through the jacket. Apply the heat conduction equation to steady state radial heat flow corresponding to cylindrical symmetry .
Total heat flowing per second is

105
The heat transferred
Fig.1E.104
K2.3rrR
K=K,+3K2 4
~ 87 W
The total heat flow is the sum of the heat flow through the wall and that through the \..•.. indow. =
2
L
0'
 (6.10)(30) (2.10)
dQ dt
K,rrR
L
TH T, wall
2
~+
R]
In [Ro)
of
r'dT Tl
= _ 2nLK (T2 Tj)
R, ... (2)
~_2m.K q
q
dQ 2trKL (T1  T2) q=~______ dt In(RoiRd
or
At a distance r from axis the Icmperature given by eqn. (I) dT
dr
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q_ 2rcrLK
= __
...(2) gradient
is
Study Anurag Mishra with www.puucho.com THERMODYNAMICS R R (T T2)
2 t 1 +(R2R1)r
Integrate this equation to obtain nr) ""(q)lnr
+ constant
2dk
Example
On substituting temperature values at the boundaries, T = T1 at r = R1 and T = T2 at r = R2, and value of heat flow per second, we obtain the temperature distribution T(r)=T
+ (T2 T))ln(rjR1) 1
InCRolRd
Consider two concentric spheres of radius R, at temperature T1 and radius R2 at temperature T2• A material of (hennal conductivity K is filled in the space between shells. Determine the steady state radial heat flow Qnd the temperature distribution in the material.
Solution: Consider a spherical shell of thickness dr and radius r. The heat flow is radial, therefore, it flows through area 4nr 2. The conducting material can be considered to consists of large number of such shells, like onion peels. The heat crossing each shell is same in the steady state. The heat transferred per second is dQ , dT q ~  = K(4nr )dt dr Rearranging the variables and noting that q is same for each spherical shell, we have Rz dr = _ 47tK T2 dT ... (I) Rl r2 q 1i.
f
f
11) ( R1 R2 or
4;d{ =(T 2
4;d{(T,  T,)
q = dt =
(.!  .!) R1
q
2KA
Heat per second flowing to B from E T2 T =3L 2KA
Heat travelling from B to A per second T T1 =LlKA
In steady state, from conservation of energy, net heal arriving at B must be equal to net heat flowing OUtfrom Bthus we have
... (2)
c
,
B
o
T,
gradient is A
Flg.1E.107
KA(T,T)
q(!) 4;d{
respectively.
Heat per second flowing to B from F,
+
Integrating this equation, we have =
at B be T. Total
thermal resistance from F to Band E to B is ~
... (3)
;I; =  4n:r2K Ter)
Let the temperature
R2
At a distance r from axis the temperature given byeqn. (1) dT
Solution:
Tt)
q
dQ
Four identical rods AB, CD, CF and DE are joined as shown in Fig. 1E.107. The length, crosssectional area and thermal conductivity a/each rod are L, A, K respectively. The ends A, E and F are maintained at temperatures Tj, T2 and T3 respectively. Assuming no heat loss ta surroundings, determine the temperature at B.
+ constant
On substituting the boundary values, T = T1 at r = R1 and T = T2 at r = R2, we can calculate the value of constant. On substituting constant and q, we get
KA(T,T)
~L
~L
2
2
KA(TT,)
 ~ L
which on solving for T we get T == Zf3 + Zf2 + 31'1 7
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Study Anurag Mishra with www.puucho.com TEMPERATURE, HEAT AND THE EQUATION Of STATE, HEATTRAl6fER

Example
108
_
I Example
rod of 50 em length and having lcm2
A cylindrical
65
cross
sectional area is used as a conducting material between an ice bath at DOCand a vacuum chamber at 27 "'c as shown in figure. The end of rod which is inside the VQCllUm chamber behaves like a black body and L~ at umperature 17 °C in :;teady state. Find the thermal conductivity of the material of rod and rate at which ice is melting in the ice bath. Given that latent heat of fllsion of ice is 3.35 x 105 J I kg.
109
The temperature drop through each layer of a two layer furnace wall is shown in Fig. lE.109. Assume that the external temperature TJ ond T3 are maintained constant and T1 > T3• If the thickness of the layers x I and x 2 are the same. which of the following statements ore correct. 11., 11.2
tEL
T,
T,
? .
_
)1.,)1.2
.

)I.
Flg.1E.109
H
k1 > k2
(a)
(b) kl < k2 Vacuum chamber
Ice Bath
atO'C
at2rC
Fig. 1 E.1 08
Solution: It is given that the system is in steady state. So heat absorbed by the end of the rod in vacuum chamber by radiation is conducted [0 the ice bath through the rod. Hence kAeTB TA) _ A(T' T') I a ",Jj or or
k(l7'C
 DOC) = 5.67 x 1O~8[(300)'(290)')
(c) kJ =" k2 but heat flow through material through (2) . (d) k1 = k2 but heat flow through that through (2) .
dT
q = KA 
= constant
1'1)' Find the rate of !leat
(,)
(b) Fig.1E.117
flow per unit area of rod. Area is decreasing
Solution:
1:1
In Fig. IE.II? (b),
T'~~~IT,
Example
dx
is decreasing; area is increasing
118 ~
Fig.1E.116
A composite body is made up of two cylindrical pipes (joined
Heat transfer across each differential element is same q= )' o
q
_(K.l
ta k ing
1 pace
and Rz will be into
Rj
I R1=
0
oc
supp I"ymg
heat
perpendicular to the interface then the series R.q = R1 + Rz , 21 R K,qxwxh  .q
4rtKRr
height ..
'.
h x2xwl
R1
',f
2
f
Example
K
Rfq
4rrKRr
If a shell of radius r and thickness dr freezes in time dt Here (dm)Lj = (q)dt
or
=
KBIVI
_(Rr)
lh 
~JR KT
eq
Rz =
1 1 =+
Solution:
R
h
h R1 = KAwl' R
of the two parallel
Req
Study Anurag Mishra with www.puucho.com ~O
THERMODYNAMICS So rate of heat retiucing by vessel (1) will conducted by .. 3 dT ro d In umt time =  nR 
':2~y
Example.
.
Two identical thermally insulated vessels, each containing n mole of an ideal monatomic gas, are interconnected by a rod a/length 1and crosssectional area A. Material o/the rod has .:.ermal conducrivily K and its lateral surface is thermally insulated. If, at initial moment (t = 0), temperature of ga.~ in twO vessels is TJ and T2 TJ), neglecting thermal capacity of
«
the rod, calculate difference between temperature of gas in two vessels as afunction a/time.
Solution: Suppose if initially at t "" 0 according (0 problem temperature of vessels TJ > Tz then temperature difference will be T1  Tz = To (Initial temperature difference at
t
= 0) (j)
4
and. in unit time from (t = 0) heat conducted by rod [beca~s.e initial temperature difference (T1  T2)} will be equal to the rate of heat reducing by vessels (1) so at'any time t if temperature difference is T then after unit . 3. dT KAT t1menR=4 dt L Here it is imagined that rate of heat supplying will remain same. After the unit time because after unit time temperature difference between vessels not much more changed. So it is taken same.
n, T,
I
3
dT
KA
4
T
L
f
Y=
nR=dt
(%J K. L,A
n. T2
T ljT2
Fig.1E.124
And in starting heat (internal) energy of two vessels U=f/2nRT UI
"" ~
2
nRTI
and
Uz = ~ nRT2 at this moment if (at
2
0) temperature of one vessel (1) Decreased by dT, due to dQ heat supplying in dt time then in same dt time dT 2 temperature of vessel (2) will rise because it will receive the dQ heat from vessel (1) by conduction through rod.
t =
So in dt time temperature difference will be reduced by (dT) dT =dT, +dT2 dQ1 = dQ2
and 3
= Example
2
dT
dT1 =dT2 =
2
dT2
. 3 And If heat dQ = dU1 =  nRdT1 ( ve sign because 2 temperature decreasing) reduced by vessel (1) in dt time. Then heat reduced in unit time
0
4KA dr 3nRL
4'0:( T = (T1  T2)e 311RL
125 y
surrounded by a coaxial layer of a material having thennal conducrh'ity K and negligible heat capacity. Temperature of surrounding space (out side the layer) is To, which is higher than temperature of the cylinder. If heat capacity per unit volume of cylinder material is s and outer radius of the layer is b, calculate rime required to increase temperature of the cylinder from T1 to T2. Assume end faces to be thennally insulated.
(To = outer temperature)
 nRdT1 =  nRdT2
=
f'
A highly conducting solid cylinder of radius a and length I is
3
dTl
dT
Solution: Given condition To > T2 > T,
dUt =dU2
2
dt
= _~ nR dTt 2
And dTI = dTI2 , then heat reduced 3 dT 1 nRxx(ate=0) 2 dt 2
dt
in unit time =
For cylinder suppose if at any instant temperature of the inner surface is T and this temperature T maintain for time period dt so we can apply the steadystate condition formula of conduction between outer temp, To and inner surface temperature T (at any instant). For small time dt
I = KA dT ,.
d,
A = 2itrl
Here area A is not constant it is changing radially. Take dr thickness element at distance r from centre. r is not changing in dr width in steady state. I will be same for every cross.section. dT 1= +KI2rr.r
d,
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71 mssp
/
'/
'
.•. .
Example
./
•
Fig.1E.125
dr is the drop in temperature in dr width. Here sign is +ve because if dT reducing then dr is also reducing with dT so (+ve), if in case only dr reducing but dr is not decrement in radial path (or radius) then sign (ve)
=f'
d,
126 __
A vertical brick duct (tube) is filled with cast iron. The lower end of the duet is maintained at a temperature T1 which is greater than the melting point Tm of cast iron and the upper end at a temperature T2 which is less than the temperature of the melting point of cast iron. It is given that the conductivity of liquid cast iron is equal to k times the conductivity of solid cast iron. Determine the fraction of the duet filled Wir/I molten metal.
Solution: Given TI >Tm >T2 both the ends are at temperature T) > T2 respectively so if upto LI height the cast iron melts and rest of L2 part has solid cast iron, And if conductivity of heat of L2 part is K 2 then for L1 part will be (KK 2) according to given condition at steady state condition the rate of heat flow from TI to T2 L
•
1
2~IK (ToT)
L,
L,
Liquid Cast Iron
I =
CIa 
2rriK
log,[~)
Fig.1E.126
Q=
KeqA(T1 T2) L K ~qA(Tl  T2)
L,
KK]A(T1 Tm)
=
L,
KIA(Tm T2) =~~~
(To  T)dt
L,
log,Cb'a) 21tlK ('['0 'f)d t ms, dT =p loge(b a) 2nlk
dt
ms.,p In(b.'a)
time taken to rise temperature from TI to T2
fT2 1j
dT (To T)
Iog [ToT,) , To  T
2

Jt
0
ms
sp
L
R=R]+R2,
dQ = mS,pdT
efT (To T)
27r./K d loge(b,'a) I,
2nlK
= , ms,p iogc(ba)
,L=LI+L2
Q=~
J = dQ = 2,,1K (To T) dt !oge(b'a) 2rr/K
Solid Casllron
T)
It is the rate of heat supplying to the inner most cylinder which is at temperature T for dt time and system is in the steady state condition for dt time. Then cotal heat supplied for dt time
dQ =
J
To T2
2.lK dT
To
IOge(~)=
[ToT ~~,
2K
,,"'
.••
br
a's ,=~iog~(bla)log~
d,
,
f'
" :rals
""
KeqA
L) L =+ 2 KIA
KKIA L
LI
(LLI)
K,'qA
KK]A
KIA
L
L
L
L
K.qA
KKIA
KIA
KIA
=+
I I =+
K [LJ(L
t
cq
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=lKKI
L, LJ K'; +K I
Study Anurag Mishra with www.puucho.com
:\,~,~,~"' ...
72 ACTl

T2) =
ACT}  T2)
(K~J [:;1 ~: =
+
extracted
:J
THERMODYNAMICS
per second by conduction
.
K. A(O) conduction APiCC'L = ,~ dt y A8 dt f oyyd =f' 0A.LK P,~ ice
y
KK1A(T} Tm)
L,
,
...
2
2Kice8t
y Ll(TlT2)=KKl(TlTm{~+!:....0.) KK1 K} LI(T1 T2)
KiCC'St =,y~ 2 PiCC'L
K1
=CTt Tm)[L1 +KLKLd
LITI L1T2 =L}Tt TmL1 +KLT1 KLTm  KLtT1 KLtTm TmL1 L1T2 +KL}Tj
KLtTm
=KLT} KLTm
L}[(T m  T2) + Key}  Tm)J = KLeTl L} = Key} Tm) L
[(Tm T2)+K(T1 =
then rate of heat
dy

Tm)
PkeL
So in time t the y height of water will convert into ice and when ice will formed then it's volume will be greater than the water So mass is constant of ice and water. Mice = M waler PwxAxYw=PicexAxy YiCC'> YI
Y••.= PiCC' y,
P.
Tm)]
!J.y =yYw increment in the column height after converting into ice from water
Key} Tm) [K(TtTm)+(TmT2)]
Example
!J.y=y _PiceY
P.
=Y(l:':)
Water is filled in a nonconducting cylindrical vessel of uniform crosssectional area. Height of water column is ho and temperature is DOC. If the vessel is exposed CO an atmosphere having constant temperature of a°e DOC)at t = 0, calculate coral height h of the column at time t. Assume
«
thennal
conductivity
afice to be equal to K. Density
atwater
=
So at time
t
=ho+
P••.
2K St(l_Pke) ice
PiceL
P ••.
A lagged stick of crosssection area 1 cm2 and length 1 m is initially at a temperature of DOC. It is then kept between 2 reservoirs of temperature IDDOC and DOC. Specific heat capacity is 10 Jjkg_oC and linear mass density is 2 kglm. Find:
Flg.1E.127
Solution: Suppose at any time t after t = 0, the height of the ice column isy and time taken to make ice ofy column from water suppose t then t =? At instant t dy layer of water convert into ice into the dt time then rate of forming ice? rf [(A x dy x Pice) x L] energy required to make ice of dy thick layer in dt time then rate of forming ice (Apice x dy jdt)
t XL)
ke
PiceL the total height
is P ••.•and that of ice is P let' Latent heat of fusion of ice is L.
and it will require ( APice x
2K 8t(1_Pice)
100'C
1:;::=. '.=::::;.11 ', Flg.1E.128
(a) temperature
environment at SoC temperature. So ( APice x
t XL)
(a)
gradient along the rod in steady state.
(b) total heat absorbed by the rod to reach steady state.
heat has to be extracted
from it (water) to convert into ice. So by conduction it will reduce this heat to the
O'C
Ii,
(
100'C
heat
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x
(Ix)
'hi
• dx
T Flg.1E.128 (b)
jl) O'C
Study Anurag Mishra with www.puucho.com TEMPERATURE, HEAT AND THE EQUATION OF STATE, HEAT TRAN.;,;S_FE_R •••• __
Solution: Suppose the total length of the rod i 1 and if linear mass density is 2kgjm, then mass of the dx element (2d.x"). lHere element dx is taken very small like that if in steady state temperature at a distance x from 100°C end is T Ca) Then temperature will not vary in dx length it will remain constant T. So in steady state KA(JOOT) KA(TO) take the element
(dx) from the hot end (lODoG) and
x
(Ix)
73
of heat supply by conduction tiT
dt
To = 400K]
dT
3Stl
1
300
2kg/m
x
=
KAt IllsL
T}t
H
dH =
I' 2000(1 xjl)dx 0
will be the heat
r
= 200~ x
x']'
2'/
0
=2000(1~)==1000J
E'lTHEI!'RMOrulillDYNAM!U'2 •• ,;;;cs;'! P = Pa,mos + Mg = constant
... (3)
A
This pressure is independent of the temperature of the gas or the height of the piston, so it stays constant as long as M is unchanged. If the cylinder is warmed, the gas will expand and push the piston up. But the pressure, determined by mass M, will not change. This process is shown on the p. V diagram of Fig. 2.5 (b) as the horizontal line 1 ~ 2. We call this an isobaric expansion. An isobaric compression occurs if the gas is cooled, lowering the piston. Any isobaric process appears on a P  V diagram as a horizontal line.
An isobaric process appears on a PV diagram as a horizontal line
.•.
,
v,
•
Ibl
Fig. 2.5 A constantpressure
Constant Temperature
2
v
v,
(isobaric) process
Process
A constant temperature process is called an isothermal process. An isothermal process is one for which Tf = Ti• Because PV = nRT, a constant temperature process in a closed system(constant n) is one for which p/V/
=P;Vi
.•• (4)
is an isothermal process. One possible isothermal process is illustrated in Fig. 2.6 (a). A piston being pushed down to compress a gas, but the gas cylinder is floating in a large container of liquid that is held at a constant temperature. If the piston is pushed Push
•• • ••
• • •• •
:(
p =
nRr
2
V constant
An isothennal expansion would move in the opposite direction along the hyperbola.
./ ••
.:
=V The inverse relationship between P and V causes the graph of an isothermal process to be a hyperbola. The process shown as 1 ~ 2 in F;g. 2.6 (b) represents the isothermal compression shown in Fig. 2.6 (a).
P P2
slowly, then heat energy transfer through the walls of the cylinder will keep the gas at the same temperature as the surrounding liquid. This would be an isothermal compression. The reverse process, with this piston slowly pulled out, would be isothermal expansion. We have the P relationship An Isothermal process appears on a PV diagram as a hyperbola
P,
v,
v,
v
Ibl
Increasing temperature
Isotherms
1'1
v
Fig. 2.6 A Constanttemperature (isothermal) process.
The location of the hyperbola depends on the value of T. A lowertemperature process is represented by a hyperbola closer to the origin than a higher temperature process. Fig. 2.6 (c) shows four hyperbolas representing the temperatures T1 to T4 where T4 > T3 > T2 > r}. These are called isotherms. Work Done in A Thermodynamic Process Fig. 2.7 shows a cylinder with a movable piston. By virtue of its pressure the enclosed gas has moved the piston through a small displacement dx.
...........
~
••••••••••••• / Fig. 2.6 (a)
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Area A dv=Adx
p
.....
For an expanding gas (V2>V 1) The area under the PV curve is positive (integration direction is to the right) Thus the environment does negaliv~ work on the gas, o ,,
..~'.
F =PA
o
, ,
•
0
1 ", ..•.. ::
. f
Fig. 2.7
=;
,
F dx = (PA) dx = P(A dx)
o o
>
~ o
i
The product A dx is change in the volume of the gas. The work done by a gas when its volume changes from Vi to Vj is
VI
o
.'
v
~. V2
Integration
direction
(.J
W=f1pdV v, on lt1e piston
p
with force Fgas outward To keep the piston in place. an external force must be Pressure P
o
f
=p dV
The gas p~hes
o o o o
o o o o o
The infinitesimal work dW is dW
'" "~.~~
For a compressed gi.'ls (V2>Vj) The area is negative because the integration direction is 10 the right Thus the environment does positive work. on the gas.
and :OPPOSile 10 Fgas
equal
F,., v
, o
• Integration
Piston area A
(bJ
(.)
Fig. 2.9 The work done on a gas is lhe negative of Ihe area under the curve.
As the piston moves OX. the external force does work Fga.dx on the gas "
\
direction
:o \~' ~:
d,"o
'.
, ,
0 0
,
0
: :
o
,
:, ::
The volume change by dV=Adx as the piston moves dx (bJ Fig. 2.8 The external force does work on the gas as the piston moves
.~
~ F
During a quasistatic process any of the thermodynamic variables may be considered to be a function of other, and the functional relation between them can be depicted as a curve on a twodimensional graph. A common choice is a p  V diagram. Fig. 2.10 (a) is a graph p of P versus V for several different T values. The PI form of the equation PV = nRT is a hyperbola. For larger values of PI constant the hyperbolas are farther from the origin. v, V, The constant temperature Fig. 2.10 ( 0 and the wor\l' done by the gas is positive. When the gas is compressed, V, < Vi so that L\V < 0, and the work done by the gas is negative. The work done by the gas is represented on the Pv diagram by the rectangular area under rhe isobaric path OJ) the diagram. Whether the area is positive or negative, depends on whether the gas expands or is compressed. Work Done by a Gas in an Isochoric Process In an isochoric process, volume of the gas remaim constant, it is represented by a vertical line on a p. v diagram. Since the volume does not change, For an isochoric process ••' the area under the PV ,." curve is zero. No worK ; is done.
P
V=(~)T
i
A graph of V versus T is a straight line with a slope nRjP. Small Pvalues yield steeper slopes. Constant P lines are called isobars. The work done by a gas is the area under the P versus V curve. There are two important points regarding the definition of work: 1. The work depends on the path between initial and final states. Note that area under the P versus V culVe depends on the shape of the culVe. 2. When a gas expands, the system does work on the environment; the work done is positive. When the environment does work on the system, the work done is negative. Work Done by the Gas in an Isobaric Process A horizontal path on PV diagram is an isobaric process. The work done is found from the equation
j
f
P,
~~v V1=V,
Fig. 2.12
dV = 0 W=!VlpdV=O v,
Whether the pressure change is positive or negative. tht: work done by the gas in an isochoric process is zero. Work done by a gas in a process in whicl PV n = constant, where n is a constant It can be considered to be a general process. P Isotherm for
P
P
v[
!v,
= Pi L\V
('j
(e)
(b)
V~(";)T
P2 >Pl
i:'
.
., ..,
For an isobaric process the area Is PAV. The worK done on the gas is PIN
/
I"m""",""
f
.. ..'
,, ,, ,, ,
_ •.•••••• ~ ••••• ,
: :
,,
v,
AV:
.
v,
DYNAMICS
,, Pf
v
Fig. 2.11
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 •••
r
T
_
,
v
Study Anurag Mishra with www.puucho.com THERMODYNAMICS
159 pv" = p;v;n
=
PI
V/
Pyt
PjV;
V"
Vn
= constant
Work Done by a Gas Taken Around a Cycle Some thermodynamic systems involve taking a gas around a closed path on a PV diagram, so that some useful work is done around the closed path in each repetition of tycle. A closed curve on a PV diagram is called a cycle; the gas returns to initial point on the PV diagram irrespective of where the process begins on the cycle.
P~~W~fVI
,, PdV.
P
Path 1
~
=
p;vt
(V)n _
,, ,,, ,
v/n)
Pfix! V
Vln

p.V"
v,
V'"
"xi'
'' ''
~
'
,
, :
' ' ' ''
, ,, ,
1 n
P
v
v,
,, :, ,
v,
Wz 0, otherwise for anticlockwise Wto1al < O.
Process A
The area under the process A ,.''curve is larger than the area ••• under the process B curve.
,,
Vf
Fig. 2.14
V
PM
e~=>ecr.P
RT
Process B
~ nRTln (VI ) ~ nRTlJ!l) Vi
'\pf
~ nRT 1J
2)
V Fig. 2.16 The work done during these two ideal gas processes is not the same.
'~ef
If Vj > Vi' the gas is expanded and the work done is .positive. If Vi < V" the gas is compressed and the work done is oIlcgative.
The work done by the gas is the area under the path on othe PV diagram and is positive or negative depending on whether the gas is expanded or compressed.
n\lo processes is not the same. The area under the curve of process A is larger than the area under the curve of process B, so IWAI>IWBI. As this example shows, the work done during an idealgas process depends on the path followed through the P.V diagram.
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, The path dependence of work has an important implication for multi step processes such as the one shown in Fig. 2.17. The total work done on the gas during the process 1 + 2 + 3
must
calculated
p
p
P=aV2
Vo
be
=
Flg.2E.2
, (V )'! w=fv/av dv=a""""3
Y
a "(V!
p
P(alm)
0 T
,, '
2.00
Vi
3
V)
'
"(~J
[(0,25)3  (0.10)3]" 9,75x 103 J
3 V
clockwise circular path of a gas on a PV
Q
diagram. Volumes and pressures are as shown. What is the total work done by the gas around one cycle?
~
3
3
Example Fig. 2£.1 shows
3
2
Fig. 2.17 The war\( done during the process1'lo2t3 must be calculated In two steps
and W2103 is negative. The initial and final states are the same, but the work is the same because work depends on the path followed through the PV diagram.
1.00
=
as
+ W2to3' In this case, WJ102 is positive
Wlto3 = W\to2
5,00
Work = Area y 0.10 m3, VI 0.25 m3
A thermodynamic system is taken from initial state A to final state C via an intermediate state B, as shown in Fig. 2£.4. Compute the cotal work done by the gas from A to B to C. P.Pa
Volume radius
:
,, .
8,00
• V (liter)
L
0
600
p""",, I radius
300
.'y
A
C~B
:, ,,
2.00
Flg.2E.1
' 5.00
Fig.2E.3
Solution:
Area of the cycle on the PV diagram represents total work done. Since the path is clockwise in P.V the total work is positive. Area = 1t (pressure radius)(volume radius) = rt{2.00x 1.013x 105)(3.00x 103) = 1.91x 103 J We have expressed pressure in pascal and volume in cubic metre to get work in joule.
Pressure and volume of a gQ.';are related by the law P = aV2, where a = 2 N / m2. The gas expands from a volume of 100 to 250 1. How much work is done by the gas during the expansion?
Solution: The work is
Fig. 2E.2 shows a graph of the expansion.
Solution: The work performed by the gas is area bounded by the curve and V.axis. From A to B total area is the area of triangle ABC, and rectangle below CB. wAH • .!c [(5,00x 1O3)(2.00x 103)][600300] 2 + [(5 x 103)  (2 x 103 )](300) = 1.35 J Work done from B to C is negative because the gas is being compressed, volume is decreasing. Be is isobaric process. Thus work done is
Wac =P~V=PB(VC VB) = (300)[(2.00x 103)

(5.00x 103)]
"0,90J The negative work shows that work is done on the system. Change in volume, ~V, is always Vj  Vi' The net work done by the system is WIOta! =WAB +WBC =1.35J+ (0.90J) = 0.45 J
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P
: Determining work done in some, special
casesWork done when A P.V graph for a process in which pressures is a linear function a/volume. The net workdone/or the
~ ~
process 1 + 2 area of the graph between volume axis and the given process as shown shaded in the Fig. 2.18
v
L Fig. 2.20
P
P,
._...• ____ .2
P,
,, ,,
P,
P,
; 3
,
:5
4
.:~r
v
v, v,
IWAI
Q
= Q
An adiabat is the path/allowed by the system paim in the pv diagram during an adiabatic process. 2. In an adiabatic expansion, the system does work at the expense of its internal energy: ~U ::= W. The adiabat drops to lower isotherms as tile internal energy (and hence temperature) decrease. Adiabar.s are steeper than isotherms.
(Free expansion) ilU ::=0 In an adiabatic free expansion the internal energy of any gas (ideal or real) does not change. In the special case of an ideal gas, there is no change in temperature in an adiabatic free expansion. One can conclude that the imernal energy of an ideal gas depends only on temperature, not on pressure or volume. Sensitive experiments show a small temperature change for a real gas at high pressures and low temperatures. This indicates that the internal energy of a real gas is also a function of pressure or volume. Table 2.2 Summary of Thermodynamic Process and general result
Result for ideal gas
bochoric W =0 ilU::= nCv /1T Q=W
P
Processes
(True also for nonisochoric process)
Isobaric T,
+
dT
W '= PilV ilU ::=nCl, ilT::=Q  W Q::= nC p /1T
P, dV
Isothermal
w
=
J P dV
Q::=ilU+W
v,
V
W =Q = nRT
In(~o)
ilU =0
Fig. 2.26: A differential piece of a PVdiagram showing two neighboring isotherms and an adiabal crossing them.
Adiabatic W :/1U
When the gas is compressed along an adiabat, its pressure rises from PI to PI + dP. Work is done on the gas but no heat transfer occurs. According to the first law of thermodynamics, the internal energy and temperatu,e of the gas increase. Thus point E is on a higher isotherm than A Adiabats are steeper than isotherms.
QO
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::=nCv I1T
PVY::=constant(y::=Cp/C,.)
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mJ~6!!6=:;::;:=: v
Graphs: p.v, V.T and poT
 
v,
1. Isochorlc process:

P
P,
v
]8
,'
. ••
....
.'
... . :
T
• • •
A
...
•• •
v,
Fig. 2.29 (e)
v
T
I')
(b)
Fig. 2.28
. therefore
V = constant
(a)
v, •••... /
.,B
A, Pl __._ ••
8
dV nR (e) slope= =dT Po 3. Isothermal
F F ..1. = ...l. and Pl < P2 T1 T2
dF
process:
p
p A
dV dT
slope =  = 00 dV
(b) slope=.0
f
P ~8 P, ••.... ; ..• ~
P1
A
8
v
i
.~
," : ... ...
T
(.)
(b) Fig. 2.30
.'
11
1 To = constant; P:x: 
T
T2
V To = constant
FIg. 2.28 (c)
dF
dF
(a) slope = ; PV = nRT dV => PdV + VelP = a
dF nR (c) slope = =dT V 2. Isobaric process: p
;>
p
dF
(b) slope =dT
p
=dV V
v 8
•
Po
A m
••••••
•
:
A
8
I
v
8
(.)
Ib) T
Fig. 2.29
P = constant,
V
therefore...!. = l. and V2 > VI T1 T2
dF (a) slope==O
dV
v
Fig. 2.30 (e)
dV
ee) slope =  = 00 dT
dF
(b) slope:::  = 0 dT
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CONVERSION
Adiabatic Graph
Example I.
P P,
P
~
P
L
A/
,A ,
,,
v
(.J
,,
P A
B
P,
OF GRAPHS
,, ,
.Ll
B
c T
V
(OJ
... c ....
T
v V,
v,
::::~B A
V
,, ,
Tt
..
T
T2
('J StutcPl,Vl,TlP2,V2,T2;
Qc
Molar heat capacity of A is maximum. (4) The plot a,b, ande represent adiabatic process for H2, He and H20 identify them: p
Fig. 2.45
The sand particle are remove very slowly and the gas expand adiabatically. In the case of mono atomic gas sample. The final pressure will be minimum and the removal of sand will be maximum. The final temperature 'A'ill be minimum in the case of monoatomic gas sample. First Law of Thermodynamics In differential form
,
dQ::= dU + dW
b
dQ
,
b = H2
(di)
c = He
(mono)
+
PdV ;
dQ dU dW _=_+_ dt dt dt
Fig. 2.42
(trO
nC"dT
; dW ::=PdV
for a differential time dt V
a = H2O
=
; dU::= IlC1,dT
... (1)
dQ + rate at which heat is supplied i.e., Thermal de power supplied. dU +
rate
0f.mcrease 0 f.Imerna
de
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~0
.;;:==:;;::::===\~R~e~la~t?,o=n:t Connecting process \
dW ~ me ch'lamea power output. dt
Isothermal
For a dT change in temperature dQ dU dW =+dT dT dT
Adiabatic
constant = constant
Y =
1 I
pOY)i1T
=constant
C. Use TableJ to calculate ~U,Q and W for each path.
Example
(2)
Problem.Solving
PV
TV
dQ = C proc~""dT dU dU =Cl,dT; dT =Cv• dW =PdV.
Cprocess
V  = constant T
Isobaric
dW PdV =dT dT from eqn.
constant
T
'." a Process
To find the molar specific heat of any process in general, let C proceS!J be the molar specific heat for the given process taking one mole of gas. The amount of heat dQ supplied to increase its temperature by dT is
dT =Cproces,;
=
P  = constant
Isochoric ,.... (2)
How to Determine Specific Heat of
dQ
PV
PdV
=Cv +dT
Strategy
for
Thermodynamic
Cycles A thermodynamic cycle may involve unknown parameters. The following general strategy may help in determining them: I. Sketch the process in the P.V plane. Make a table of t.V, Q and W for each segment of the cycle. A. Fill in the given informations in the appropriate box of the table. B. Apply first law,.1.U =Q  W for each segment. C. At the end of a complete cycle all the state variables return to their original values, including the internal energy. For a complete cycle, 6U cycle Q Therefore, Q  W 0, i. e., work obtained from a reversible cyclic process is equal to the heat transferred into the system. We reemphasize the fact that irrespective of the process performed bet'Neen t'NO states with temperatures Ti and Tf, the change in internal energy is
=
=
~gm; cylinder hm; O.6mole of an ideal gm;. It is taken through the process ABC as shown in Fig. 2E.4. What is the work perfonned by the gas, the heat framferred. and the change in internal energy for each of the processes A ~ Band B ~ C. Specific heats at comtant volume and constant pressure of the gas are 12.5 JLmol!K and 20.8 J/moVK respectively.
Solution: Process A ~ B is isobaric, and B ~ C is isothermal. We construct a table and fill up flU = 0 for B~C, because there is no change in internal energy along a n Isot . heern 0f'd an leal I •••. as.
i'
.u
Process
w
Q
A>B B>C
0
Total Process A ~ B is isobaric, so the work done Is WJ\B =
v,
f v, Pdv
P, Pa Isotherms
100
= P(Vf  Vi)
6U =nCv(Tf Ti)
= (100)(2 1)
If all the boxes are not filled by using I, apply the strategy in II below:
= 100 J
II. Determine the temperature, pressure and volume at each vertex of the cycle.
The change in internal energy from A to
A. If two state variables are known, the equation of state can be used to find the third.
B is
B. If one variable is known at a given vertex, the second can be determined by relating the values at neighbouring vertex by knowing the process connecting the two vertex,
V, '" V2 = V3 123
V,m3
Flg.2E.4
tJ.U = nCv~T = (0.06)(12.5)(400  200) = ISO J The heat transferred can be computed in two ways. 1. Q = nG pliT = (0.06)(20.8)(400  200) =250J
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THERMODYNAMICS 2. From first law, Q
== E,U
+ W
Example
+ 100) J
== (150
A cylinder witll a movable piston encloses one mole of helium. The cylinder is placed in contact with various reservoirs and insulated when required; the helium peiform.~ the cycle as shown in Fig. 2£.5. Compute the internal energy change, hear transferred and work peifonnedfor each segment of the cycle and the toral amount of each of these quantities for the entire cycle. Assume helium to be an ideal gas, with Ct, = 12.5 j/moVK, C p = 20.81/moVK.
~2S0J Process B ) C is isothermal; so the work done is
wBe
~Jv1v lpdV V,=V2
=nRT
Vf =
J
v3dV 
V
Vi" V2
~ nRTln[ ~:
J
~ (0.06)(8.314)(400)
In(~)
Solution: First we calculate temperatures TA =TB, Te, TD• the volume VB' and the pressure PRo We can determine temperatures from the ideal gas equation = PAVA =(8.00)( 104)(2.00)( nR (1)(8.314)
T
:: 80.9 J
A
Since 6U = a for an isothermal process,
= 19.2 K
QBC == Wile == 80.9 J
We can now complete our table. We use the law for adiabatic process PV"' , for the process B ) C to compute pressure and volume at point B. PeVJ=PIJVJ
... (1)
For helium, we have
Cp =(20.8)= 1.66
'{ ==
C,.
(aBC" 0)
TA"'Ts"'19.2K
12.5
10J
(.J
However, we have two unknowns, both Pfl and VB' so we need another equation, which is PBVB =nRTB
r
... (2)
Eliminating Pjj from eqos. (1) and (2), we have PeVZ =PBV;
r1
~P '" 0
=PSVB(VB
=nRTB(vB)r1
Po
On solving for VB' we have VB =
.
400 K
=Wt+WZW3W4
V2
) (_1_
C
Taking sign of heat transferred and work, we get
)0
3
T
QH +QL =W1 +W2 +W3 +W4 +OJ
)0
)1'('" V
'Fig. 2£.9 shows a cyclic process perfonned on one mole of an ideal gas. A total of1000) of heat is withdrawn from the gas in a complete cycle. Find the work done by the gas during the process B )0 C. •
tOlal
(b) For the adiabatic process 2
3,
)0
By substituting these values in eqn. (12), and doing some algebraic manipulation, we obtain
L\Utotal =L\Ut +L\U2 +L\U3 +L\U4
For the adiabatic process 4
... (12)
VI
nR=PtV1
For complete cycle, the gas returns to its initial state; the total change in internal energy of the gas is zero. i. e.,
Net heat tiansk""d to the gas
V,
In
From ideal gas equation,
... (6)
Since W" is negative (compression work), tJ.U4 is positive, indicating that temperature of the gas increases.
QHQL,
minimum
V2
= nR(TH  TL)
Wlotal
.'
=nCv(TH TLl
QtOlal = Wtotal + L\V
0_
V4
The process is adiabatic, Q = 0; the first law becomes OJ=.6.U4 +W4
the
V2 V4 =nRTH In + nRTL 10VI V3 Vt V2 =nRTH In + nRTL InV2 VI
(5)
=nRTL
at
QeyB
., =
= 831 J
... (3)
W A ./l is positive; expansion of gas takes place.
Heat is added to the system. Process B + C is isothermaL From first law of thermodynamics,
a
...•c = 1831 J
Wu
(600 300)
= ISOOR
From eqns. 0), (2) and (3),
1000 = 831 + W/l_.e +
2(~R)
QBC =WBC'
Negative work dO:le implies that compression of the gas takes place. .
= nRTe
Example
as
I:1U
In(~:) In(4V2V
A)
= (2)(R)(600) Two moles of a monatomic
ideal gas is taken through a cyclic
A
process starting from A as shown in Fig. 2£.10. The volume ratios are VB = 2 and Vv = 4. If the temperature TA at point VA
= (l200R) In (2) = (1200R)(0.693)
VA
~ 83L6R
A is 27"C, calculate:
v
Heat is added to the system.
v, ODe
Process C + D is isochoric. From first law of thermodynamics,
Vs ••••
QCD = nC,.(TD Tc)
B
A VA ~ ...•.. TA
'
=n(%R
i,
Ts
)(TA T,)
T =2(%R )(300600)
Flg.2E.10
(a) the temperature
=0
of the gas at point B,
= 900R
(b) heat absorbed or released by the gas in each process,
Negative sign implies heat is rejected by the system.
(c)the total work done by the gas during the complete cycle.
Process D + A is isothermal.
Express your answer in terms of the gas' constant R.
Solution:
QDA =WDA
Gas is monatomic; hence 3 5 C =R and C =R u 2 p 2
= nRT,
= (2)(R)300
n=2 TA =ZrC=300K (a) Process A + B is a straight line passing through origin; hence it is an isobaric process. Number of moles
VA
VB
A
B
=600R
In(~)
ilU=O
=2x 300K=600K is isobaric.
Qp =nCp(TB
4VA
Negative sign implies heat is rejected by the system. (c) In a cyclic process,
Ts =(~:)TA A + B
In ( VA )
= 83L6R
=T T
(b) Process
In(~; )
Wne1 =
rAJ
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QAB + QBC + QCD + QOA
=
ISOOR + 83l.6R  900R  831.6R
=
600R.
•
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An ideal gas having initial pressure P, voll1~e, V and temperature T is allowed to expand adiabatically until its volume becomes 5.66 V while its temperature/ails to Tj2 (a) How many degrees of freedom do the gas molecules have? (b'. Obtain the work done by the gas during expansion as a function
of initial pressure and volume.
The pf;fSSures and temperatures at A, B, etc_, are denoted by PA, TA, PB, TH, etc., respectively. Given TA = 1000 K, PB = (2/3)PA and Pc = (lj3)Pk Calculate (a) the work done by the gas in the process A ) B, (b) the heat lost by the gas in the process B } C, and (c) temperature TDGil'en (2/3)(2'5) = 0.850_
Solution:
Solution: (a) For an adiabatic process, temperature and volume are related as T Vr1  2 2
y1 TV 1 1
1Vy1
(a) For an adiabatic process, Tr pl r = constant Trp1r
_Trpr1
AA
BB
=(~)(5.66V)Yl T•• logt2 =0.6931=0.4 loge 5.66 1.7334
yl=
r
where y = 5/3 for a monatomic gas.
1.4 The relation between y and degrees of freedom is 2
TA
(~:
.1000G)
".•
850 K
y =
y=1+
1. 8.31 (1000 850) • (5/3) 1
/.5 (b) Work done in an adiabatic process is given by = PI VI  P2V2
yl
1869.75 J (b) Process B } C is isochoric. Hence W =a From first Jaw of thermodynamics, Q =.t!.U + W
PIV} =nRT=PV
T
PV
2
2
=nCvt1T+O
3 2
PlVl =nR  =
Thus
.nR(Te
1 (PV _ PV) = 1.25PV (lAI) 2
W =
= nR(TA TB)
yl
/
7
as
w
2
1.4.1+
w
Work done in an adiabatic process is given by
 350)
For an isochoric process, p  = Constant
T
I:::xam~le
.Te T Pc
One mole of a monatomic shown in Fig. 2E.12.
ideal gas is taken through the cycle
PB B
Pe
Te=.T,
T
P,
A
t:::J
c
'_v Flg.2E.12
A } B adiabatic expansion B } C cooling at conMant volume C } D adiabatic compression D } A heating at constant volume
_ (I/3)PA (2/3)PA'
.T• .
T
2 850 2 = 425 K 3 Hence Q = 1 x  x 8.31(425  850) 2 = 5297.625J Negative sign implies that the system has lost heat.
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Study Anurag Mishra with www.puucho.com IHfRMODYNAMICS ee) Process C
Pressure are known.
D is adiabatic.
+4
TA
at D. PD is unknown,
but pressures
:=lOx 10" x 10 =240.66K 500 x B.31
T
=
at A and C
Process D + A is isochoric; hence PD Tv =0'
lOx 10" x 20 =481.32 K 500x 8.31 4
Tn :=5x 10 x 20 =240.66 500 x 8.31
TA
K
(b) Change in internal energy is path independent; hence we cannot predict the path taken. (c) Process A B C consists of isochoric process A B and isobaric process B C.
Therefore,
}>
}>
}>
rl,'! D
=T
~ [ cPT
C A
A
D
A
=
'3 ['21(2)"
}>
H'
T" =(T,2" )[ (1'3)P P
x
WA>s ....•c :=WA ....•B + Wn ....•c :=0+P(V2  VI)
]
:=
0 + 105(20 10)
:=
106 J
]"
1000
]( 3 )"
x 1000
Change in internal energy,
1000
h.U:=nC,,(TcTA)
=(500)(~
Tv =500K
R)( 4~0
_1~0)
:=2.25x106J
E':>
}>
D and
}>
p
= P(V2 
VI) + 0
:=5x 10"(2010) =0.5x
10
Change in internal energy aU :=2.25 x 106 J; there is no need to calculate h.U once again. Internal energy is path independent; therefore,
20
Flg.2E.13
Solution:
106 J
tJ.UA ....•B ....•C :=f,UA
Ca) Number of moles of helium,
....•D ....•C
From first law of thermodynamics,
n := mass :=500 mol molecular weight
Q=au+w '" (2.25x 106) + 0.5x 106 =2.75xI06J
From ideal gas equation. T= PV nR
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Example
[c
Three moles of an ideal gas
p
RJ
=~
at pressure P and
temperature T is isothermally apanded twice to its initial volume. It is then compressed at constant pressure to its original volume. Finally the gas is compressed at constan~ volume to its original pre.ssure P. (a) Sketch p.v ~nd p.T. diagram/or the process. (b) Calculate the net work done by' the gas and heat supp'lied to the gas during the comp'/ete cx.cle.
Solution: (a) The initial state A is characterized by state variables: pressure P, volume V, temperatureT. Process A . B is isothermal expansion. PAVA =PaVs B
B +
V
T =a
Example One mole of an ideal monatomic gas is taken around the cyclic process ABCA as shown in Fig. 2E.15; calculate: (a) the work done by the gas, (b) the heat rejected by the gas in the path absorbed by the gas in the path B 4 e,
2
Area of triangle ABe
P
PI2
P/2
(P/2, 2V)
V
TI2
T
C p "R 2 T
From ideal gas equation, nR •
T
== l:!.U + W cycle
Clcrcle
== W cycle
B
== 2PoYo nR
2
W.''\~B ==nRT Jog~v
,
,,
Vo
2Vo
Fig. 2E.15 (a)
nR
fiR
S
A
=PoYo 2
""3RT logt 2 ""2.08RT Isobaric process: WB....• C ""P(VcVB)=nR(Tc
0
,
"n(~ R)(POVO _ ZPoVo)
VB
""
3::_1~~c
Qp ""nCpCTA Te)
Isothermal process:
WC_.A
(2Yo  Yo)
P
gas
TA == PoVo
Qcycle
»(
}o
From first law of thermodynamics,
Isochoric process:
e,
(b) Process C A is isobaric, For an ideal monatomic
Flg.2E.15
",,3R(~T
}o
(Base»( (Height)
1 2
C (Pf2, T/2) 8 (Pf2. T)
V
2V
""..!.
== (3Po Po
s
C~B (PI2, V)
A and the heat
""PoYo
LJ(P'TI
P
4
2
(b) For a cyclic process l:!.U = 0, as the gas returns to its initial state at the end of each complete cycle. P
e
_
Process C + A is isochoric process.
P
O.58RT
Solution: (a) Work done in a cyclic process is area under PY diagram of the cycle. Work done is taken positive if the cycle is clockwise.
c
Vc Tc =Ta VB
""
Cd) the maximum temperature attained by the gas during the
c "T T B
Qcytle
W cycle and Qcycle are positive, hence heat is added to system, while net expansion takes place.
f'Y,c1"e,
C is isobaric process. VB
W cycle""
(c) the net heat absorbed by the gas in the path B
= PA VA = PA VB 2
F
Process
+ WB~C + WC~A ""2.0BRT  1.5RT + 0 " O.5SRT
Wcycle ""WA~B
To)
Negative sign implies that heat is rejected by the system to surroundings. Process A
) ""I.5RT
}o
B is isochoric.
For an ideal monatomic gas, C v "C R"R3 p 2 From ideal gas equation.
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Q" =/le,,(ra TA)
_PoVo)
=n~R(3PoVo 2 nR
nR
= 3PoVo
Positive sign implies that heat is added to the system.
Temperarure versus volume curve is a parabola as shown in Fig. 2E.1S(b). During expansion, temperature will increase and af::er reaching a maximum will decrease. For maximum temperature,
(e) Out ofthe three processes A ) B, B + C, C + A, we know the heat involved in the A + Band C + A. We also know the oct work done in the cycle. Since !'J.U = 0 for a cyclic process, two processes
+
QA>B
QB ....•C =W..yde
Vo~V 4 0
2Vo
v
Fig. 2E.15 (b)
dV
(4
Po  V R
V+ 5
o
)
=
0
S
+
QB ....•C
QC
....•A
V = Vo 4
= W(:rde
QA ....•1l
S = PoVo +  PoVo  3PoVo 2
Hence
Tmax
Po [
=R" 
2 2S 2 16 Vo o
V
2S
+4
Vo
]
25 PoVo =
1 =  PoVo
Cd) Along
P
dT = 0
= W cycle
Qcycle QC ....•A
T 3PoVl)
8
2 process A .......• B,
temperature
increases
from PoVo 3PoVo TA =toT n ==3TA• nR
R
This is the maximum temperature entire cycle.
attained during the
Examp~le 16y
nR
Along process C + A, temperature decreases from T = ZPoVo to TA = poVo. c oR nR The process B ) C is a general process whose equation is nm known. Since it is a straight Hne with negative slope and positive intercept with axis, its equation is
Two moles of an ideal monatomic ga.~is taken tllrough a cycle ABCA as shown in the PT diagram Fig. 2E.16. During the process AB, pressure and temperature of the gill vary, such that PT = constant. 1fT) = 300 K, calculate: P 2P,
c
B
P =aV + b
where a and b are slope of straight line and P intercept respectively. The straight line passes through points B(3Po, Yo) and C (Po, 2Vo); so we have
P, ••• T1
3Po =aVo + b
and b = SPo So the equation of the process is 2Po P=V+SPo From ideal gas equation, T = PV for one mole of gas, R
Vo
T
Solution:
Vo
2
2T1
(a) the work done on the gas in the process A ) B, and (b) the heat absorbed or released by the gas in each of the processes. Give answers in terms of the gas constant R.
2Po a=Vo
R
,A
Fig.2E.16
and Po =2aVo T b On solving these equation, we get
T = Po [_
.
•••
v sv] 2
From ideal gas equation PV = nRT On differentiating the gas equation, pdV + vdp =nRdT pdV + (n~T)
+
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dP
=
nR dT
..,(I) .. ,(2) .. ' (3)
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Process A ) B follows the law
(atm)p 3P,
PT = constant
On differentiating we get PdT+TdP=O
",(4)
TdP=PdT
On substituting expression forT dP in eqn. (3), we hC5P,~ele
liU = nCvliT
The average degrees of freedom per molecu.lesfor a gas is 6. rhe gas perfonns 25 J of work when it expands at constant pressure. Find the heat absorbed by the gas, [Pi = P2 = P (at
=(2)(~R)(300600) =  900R
From first law of thermodynamics on process A ) B,
Pconstanl )},
Solution:
QAoB =WA ...•B +Ii.U
liQ = liW + liU ""P(V2
= 1200R  900R = 2100R
Negative sign implies that heat is rejected by the gas.
.3
Q..
P(V
,C :
liQ = P(V2
 VI
+'~2~.'V(lit)
= P(V, 
Fig. 2E.17 fa)
into two parts AB, Be. In A ) B compression and B ) C expansion Upto initial volume. But pressure is changing continuously and increase at every stage.
 VI)
Given and
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)[1 + _1_] '(I
1 •• ~ :A
Solution: Workdone in the process ABC is divided
(,1)
I
2 =P(V2 \'t)+~~~[':PI (y I)
"
"L" B
TI)
=P(V _V,)+(P2V2PIVI) 2 I (,1)
In the PV diagram shown in Fig. 2E.17. ABC is a semicirde. ;Pindthe workdone in the process ABC.
"
VI) + nCvdT
"Q = P(\' _ v ) + (nRT,  nRT, ) 2
s

nR =P(V2 Vl)+(T2 (y I)
v,[y] 1,1
Ii W = P(V2  VI) = 25, y =1 +
2/1 = 4/3.1 =6
4/3 tlQ = 25 x ' = 100J
1/3
=P2 =PJ
Study Anurag Mishra with www.puucho.com 181
[HERMODYNAMI(S 3P
Examplg. 1 mole of an ideal gas at initial temperature T was cooled isochorically till the gas pressure decreQ.';edn times. Then by an isobaric process, the gas was restored to the initial temperature T. Find the net heat absorbed by the gas in the whole process.
Solution: TI =T,
PI =P,
T
P
T2
=v
VI
=. Pz =, V =v n n
T3=T,
constant
pocT
P = constant
VocT
v
2
=
V2 T2 V =T
P PJ=,V3=? n
3
3
V V3
0_0
'
Tin T
so
(U) 6.U =
p_V diagram of monoatomic ideal gas is a straight line passing through origin. Find the molar heat capacity in the process. 1
Solution: Equation of the line is P = KVbeacuse P x V
J PdV (ndT)
0
n=l,
PlY] =nRTl•
~QoR1\l~J
6W=
,
K z VdV=(V
2
2 VI) 2
2
p
xVZZ_
6W='!(PZ
PV=RT'
2 Vz
1 =(PzVz 2
ndT = n(T2

\
V1
2 XV1 )
PIVl) nR
Tl) = (Tz R
 Tt)
1
=(nRTz ~nRTI) R 1 =(P2VZ P1VI) R C =C •.
+B.x 2
:
,, , •
V2 VI
1
7"
p __.0j),:
J
pdV=K
=  (KV2 KV\ ) 2
Pressure versus temperature graph of an ideal gas is shown. Density a/gas at point A is Po' Find the density of gas at B.
3P ••
J"
VI
.. ~\...> P
nCdT = PdV + nCvdT,
C=Cv+
=PV(ll/n)
Example
f
6Q = 6W + 6U,
n
.1.Q=6.W P P c.Q= hoW=(V3 V2) =(nVV) n n
and
...(2)
From (1) and (2)
V3 = nV Total change in
=Pn x(~)x ZTo
(PzVz ptV\) (P2V2 PlVt)
oC, +Rj203Rj2+Rj2oZR T
C oZR
.~ampLe .... ...""""""' ..=. Solution:General equation for an ideal gas PV = nRr
An empty pressure cooker of volume 10 litres contains air at atmospheric pressure 105 Pa and temperature of 27°C. It contains a whistle which has area of 0.1 ern 2 and weight of 100 gm. What should be the temperature of air inside so that the whistle is just lifted up ?
P'= m RT M V
m
R
V
M
p=_x_xT
p=px(~)xT P =Pox(~ }To
...(1)
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Flg.2E.22
Study Anurag Mishra with www.puucho.com THEIMOIllIWIJCS Solution: When the whistle is just lifted up, then pressure inside of the cooker
+7)
P2 = Pi" =(Po
One mole of a gas mixture is heated under constant pressure, and heat required t.Q is plotted against temperature difference acquired. Find the value 011 for mixture.
105 + 100x 103 x 10 0.1 X 104 = 2x 105 N/mZ =
V2 = V = constant,
Example
2':O~~
Tz =?
Pi" V =nRT2
o
T _Pi"V 2  nR
PIV} = 105 n =RT}
lOx 103_ Rx300 X
Solution:
""
600K,
or
t.Q =nCpt.T
"Q
=nCp
.•. _ 2xlOs x lOx 103 xRx 300 " 105 x lOx 103 x R T2
100K AT
,_'.'9••.2E.24 "1' =c "Q
p
"Tn
327°C
yR =(y1)
2500
yR =100 (y1)
Example
YR~25(J)
V.T cu"'e for 2 moles of a gas is straight line as shown in the graph here. Find the pressure of gas at A.
(y1)
V(lil(~
(~~ +
Kmol
1)=y
y=(~~Y+1)
T(K)
1=1(1 ~:)
Flg.2E.23
Solution:
V"" tan 53°T
(.: n = 2)
1 = y(25;5&3) =
4 V=T
3 4 VA =TA 3 PA VA = 2R.TA
25 16.7 = Y
1.499~1.5=y
PA =2RX(~:) 3 4
PA = 2R.
x_
PA =.:?R 2
Jjlitre
P = 3/2x &31 J/m3 A
ye::)
Exam..R.'e Ideal diatomic gas is taken through a process t.Q = 2t.U. Find the molar heat capacity for the process. (where t.Q is the heat supp..lied and is cha[!ge in internal energy)
t.u
Solution: Given,
"Q = I>W + I>U ~Q= 2liU UU =t.W +t.U
103
~w=t.U
PA =1.25x 104 N/m2
I'~Q= nCt,dT
t.u
= nCvdT
nCdT = 2ne ,.dT
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=~R)
C =2x~R(fOrdiatomicgasc,.
aV2 == nRT
T _ uV~ a(nVo)2
T2 ==
and
Exampl.e
nR
A gas is undergoing an adiabatic process. At a certain stage A, the values of volume and temperature =: (Vo,To) and the magnitude of the slope of VT curve is m. Find the value of C p and C",
Solution:
PVY
=
l'.U == nC
log.T+(yl)log.
R
1
Of
l'.U == aV~ (n 2 1) (yI)
V=log.K1
dV =0 V dT (yI) dV 1 =V dT T dV __ Vx _= dT
T
(ii)
__
2:[n2Vi
1
W
 Vi]
a~,2
1]
2 (iii) Consider one mole of gas undergoing process. P == HV
dU ==C"dT:
dQ ==Cprocc"dT;
1
dQ == dU
(yl)
I;:,:, HT~~I
c:=
by
dV
C,'
=mRTo Vo
__
dT =:>
R+
'\
+ PdV dT
C
R
_0
=(m~:'o), =( n~:o)
2a vdV PdV _0_ dT
2aV c:=
RdT R 2
R
Vo
,
C pro('ess == ,
Example
dW =PdV
dW =:> Cprocm
PV ==RT uV2 == RT
=:>
CI'
Cp=U(l+mTo)
+
"
because y > 1 and (y 1) )tvc C_yR C_R '(yI)' "(y1) Cp
2
0_[71
x (y~l)
V, To
0
J pdVo J::' aVdvoa[V:[
0
2
m=~x
(y 1)
W
==
()'1)
(~~)(VO.TO)= ~:
An
aVo2(n2 1)
R
K
.!.+ ("(1)
so
2 1:( a112V~RaVo )
o
1Vy1 =K1
T
nR
1 
CdR
R +2
How to Determine Equation of Process
ideal gas whose adiabatic exponent equals
y
is expanded
.E:pl!,,31\.7 Q joule of heat is given to n mole of monoatomic
=
nRT
dT
V
Ta
+ 
.I
~."'
PdV + Vdp = nRdT
••• .
,
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. ."..
dT
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THERMODYNAMICS •
l86
nPR p!!.
dQ = nCudY+dT V
Q
=
J dQ = J:"
dT
=
adV
(c,
+ :~
)adV
Mixing of Gases Consider mixture of three gasses "1 mole of monatomic, "2 mole diatomic'"3 mole triatomic, Heat supplied to raise temperature of gas by tIo T is given by ': P""Po
+~ V
PodV "nR
dU= PoC" R
nRT
=:
J
=PoCv (V2 VI)
V2
dV
R
VI
PoV + a
.: nRdT = PadV
One mole of an ideal gas with heat capacity at constant pressure C p undergoes a process given by the equation, T =To + aV, (i) Find out molar heat capacity as a function
of' v' .
Find out amount of heat transferred it the volume be increases f!..om VI to V2' (ii)
Solution: 0) dQ = nC"dT + PdV
'; T =To +aV
',' dT =u=> dV
dv
dT
o a)"!"a
';
= nC" + ••.• nJ;l(T \ V +
C
=..!.. dQ = c n dT
=..!...;
aV
P = nR V
(ii)
£\Q ). is the heat capacity
~T
Remaining
mole of
Diatomic
Remaining
mole of
Monatomic
a
>(nI
~
R
)+ n
2
n2

Kn2
=
n2(1
K)
= ZKn2 + n}
5: which is greater
if the heat capacity is increases Molar heat capacity =
than molar capacity
in dissociation
£\Q/£\T nl+n2+Kn2
The molar heat capacity is decreases.
In the compartment (1) shown in the Fig. 2£.36 a aissociation had not taken place monoatomic gas is present in the state shown. In the compartment (2). A diatomic gas has been enclosed in the state shown. Both the compartment are thennally, insulated.
aV
p
=

RTo +
aV dT ( C + RTo)
~Q,'~T
volume =
Heat capacity ~Q 3 SR =(n1 +2Kn2)R+(1K)n2tIoT 2 2
dQ =CdT =
of the gaseous mixture
(To + aV)
RTo =Cp R++R aV C =Cp
~Q 3 5 6 = "}R + n2 R + n3R tIoT 2 2 2
PV=nRT
a
+ RTo + Ra tt
T
Suppose we have n2 mole of diatomic gas and n1 mole of monoaromic inside a closed insulated box after some time we find that K fraction of diatomic gas molecule dissociate
Example
dT
"3 + Tri
Dissociation
V,
dT
"2 + Dia
6 +n3MT.
"l+n2+n3
=Po(V2. V1)+ulnVI
dT +PdV
n, + Mon
2
Molar heat capacity at constant
W = J dw = J (Po + ~ )dV
v
= constant
Volume
2
dW=PdV=(Po+~}V
dQ =nC dT
2
•
dU =nCudY dU = nC
..
Heat supplied 3 5 ilQ = n1 R6.T + n2 R6.T
dT
=a
dV
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Mixture of Gases T
mono
Oi P2,V2.T2
Pl' V,. T1
P2V2 n2=
P1V,
n1= RT
,
RT,
j
Flg.2E.36
(a) Find out the final pressure, volume and temperature
of the
gas mixture. (b) Find out the CVmix. and C Prnl~_
Solution:
(a)
Vm•
us assume that
Let
the
final
temperature of the gas mixture is T Initial internal energy of the mixture 3 5 ="1 RT, +"2 RTz 2 2 Final internal energy of the mixture 3 5 "'", RT + n2 RT 2 2 Since the internal energy of the gas mixtures can not change therefore. 3 5 RT njRT1 +nzRTz =(3n. +5n2) 2 2 2 2
(n
. ~RTI
t
T=
2
2

".m..,.,....RJ"
=.'.,.,.RT
('(mix 1)
(y\1)
+Un
+ f' _,RT
~ll
+ ".3R_T_+ ..... + _f,~"_R_T_
(Y2 1)
+
(y\1)
(1'31)
+ n,Cp'
of the mixture). )..12
+
(Y21)
~
)..11
"3 ~
+"3C" 2
+
+ .... +
)..In
(Yn1)
"2Cp2
+
"1+"2+"3
)..12
(Y2 1)
+
+.....
)..13
with the help of above expression for a mixture of gases. Cp and Cv for a mixture
'(nix,
+
lin
(rn1)
(Y31)
tri
(C¥)mi~. _Pl(Cv)l
can be obtained
"3Cp3
+Un
+)..I2(C¥}z +113(Cvh+ PI +P2
3
"I +"2 +"3 "" "ICPl
+
(y\ 1)
Umix =UI +U2 +U3+
di,
j
C"mhr. ""
)..13 (131)
For a mixture
,,'
") + "2 + "3
n2
(Yn1)
)Jmix.
(Ymix1)
n}+n2
"ICl' +"2C"
_
Umax, ""UI +U2 +U3+
)JI +)..12 +)J3+ ... )Jn
+nz)CVtnix.tJ.T
n} ~ mono,
'l
Internal energy of the mixture Internal energy is an extensive property. It is additive in nature i.e ..
= "IG" mono L1T + T!2C"A_. ul /),T = rt1CL'mono +n2C"df
_ n,C"".ono
C Pmhr.
+)..I2M2 +1l3M3+ ....
"" ~~~~~~~~~
SO.
Q, is the heat given to 1 mole of the gas mix to increase its temperature by 1"C at constant volume. Prm.
Weight of the mixture total no. of moles
= ~
~tl+~t2+P3
Where
liU = AU, + .1.uz
C
IlIMl
M mixture
)..Irnix.
VI + V1
C"mlx
. Ma 1ar mass a f mixture
('(mix1)
The final volume of the gas mixture VI + Vz p= nRT
Q1 =(nl
Molar Mass of the Mixture
(where T: temperature
+ n2 .~RT2)
R(3n] +5n2)
(b)
Consider a homogeneous mixture of 'n' inert (nonreacting) ideal gases (G1 ,G2 •... Gn). Let theirmoles be ~ll' )..I2')..I3.... Pn respectively. Molar masses be M},M 2,M3 .... M n. Molar heat capacities at constant volume be C Y ,C v2 ,C v), .... Cyn. Under the ideal gas approximation, the properties of a gas are not influenced by the presence of other gases, and each gas component in the mixture behaves as if it exists alone at the mixture temperature T m and mixture volume
+~13+ ... +Pn
As Cp Cy =R So, from the above expression, (CP)mix =R+(C~')mix
CPmi>.
'ml" C~
,~
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.... lln(Cv)n
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lij188g:::::lZ== R + PI (CV)1
h + 1l3(CV
+ }1z(Cv
h+"'lln(CV)n
Vo
III +112 +fl3+ ... lln
.
'.'
1l2(CV}z +J.L3(Cvh+ .....+lln(CV)n ~+III
+J.12+J.I3+ ...•• +j..ln
IlI[R + (Cv)d + )..l2[R + (Cv hl + 1l3[R + (Cv hl + ...•• +lln[R
~ III
Ilt(Cp)l
B ,
A
no
PoVo
2PoVo
nR
nR
Temperature (T)
Fig. 2E.37 (b)
+ (CV)n]
+Ilz +J.l3+ .... +J.ln
+j..l2(Cph +1J3(Cph+
.....+lln(Cp)n Draw the P.T and VT diagrams for an isobaric process at. p:pansion, corresponding to n moles of an ideal gas at a pressure Po> ftom Vo to 2Vo'
III +J.l2 +Jl3+ ....• +lln

Draw the PT and VT diagrarru of an isochoric process of n moles of an ideal gas from pressure Po volume Vo to pressure 2Po indicating the pressures, and temperatures, of the gas, in the inieial and the final states.
Solution. From the equation state PV = nRT
.,
Po
A
u;
B
•
€
Solution. From the equation of state for an ideal gas PV
=
nRT for an isochoric process
i.e.,
P = T~T
(a)
P
oc T
PoVo
2PoVo
nR
nR
Temperature
For and at
p =po
and
PoV V=VnT= o "' nR
v=v"
P = 2Po•
T=2PoVo v'
V
At
V=Vo
or
T1 .= PoVo nR
T = 2PoVo 2 nR for the graph of P versus T the variation is a straight line the temperature vatying from T1 to T2 as shown in figure. and
B
PaVo
~(~)r v~r
We have,
nR
The graph of P vs T is a straight line passing through the origin, when interpolated.
(T)
Fig. 2E.38 (a)
V=2Vo
2PoVo
nR nR Temperature (T) Fig. 2E.37 (a) PoVo
PoVo 2PoVo increases fr am to __ nR ,nR volume remaining constant, so, the graph of V.T will be a straight line parallel to T axis as shown in figure.
(b) Since, temperature
2PoVo
nR nR Temperature (T)
Fig. 2E.38 (b)
For the graph .of V''Versus T the equation V = ( ~ )T is a straight line the volume varies directly, as the temperature (Charle's Law).
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189
Efficiency of a Cyclic Process Efficiency (11) of cycle is defined as the ratio of the net work performed to the total heat supplied to the system, per
Illustration: Find the efficiency of the thermodynamic cycle shown in Fig. 2.48 for an ideal diatomic gas.
cycle.
work done per cycle Th us 11 = total heat supplied per cycle
__ ~Lt
Vo
Fig. 2.48
.
:
• Vo
2Vo
Volume (V) Fig. 2.47
Illustration: Fig. 2.47 shows the pressure' versus volume graph corresponding to n moles of an ideal monatomic gas for a cyclic process ABCDA. Find the efficiency of the cycle.
Let n be the number of moles of the gas and the temperature be To in the state A. Net work done during the cycle is area under curve, 1 1 = W =  x (2Vo  Vo)(2Po Po) = PoVo 2 2 For the heat given during the process (6QI) A + B, we have ~Ql =~WAB +~UAB
~W AB = area under the straight line AB 1 =(Po 2Po)(2Vo Vol
Network done by the gas during the cyclic process is given by area enclosed by cyclic process W
=
(2Po ~Po)(2Vo
2
 Vol
~UAB=nCt,~T = {S:)c4To
= PaVo
For the isochoric process A to B the heat supplied is given by, = nC"t.T + 0
n(
3:)
Vo(2P~ :Po)
3R
=PoVo 2 Temperature
T
A
~ lSnRTo
in respective states A,B,C,D is given by
= PaVo TD = 2PaVo
nR'u
oR'
To)
_ lSPoVo
2 2 3 15 ~Ql =PoVo +PoVo =9PoVo 2 2 The processes B + C and C + A involve the rejection of heat by the gas. . ( ) Work done per cycle Effi Clency 11 = Total heat supplied per cycle
~Q=8.U+~W
=
2Vo
Volume (V)
T = 4PoVo T = 2PoVo c nR' D nR
1
i.e.,
for the isobaric process B to C the heat absorbed will be given by
2:PoVo 11 =9PoVo
1 =18
~Q2 =flCp~T
=
n( S:) 2Po(2~
Vo)
For the processes C to D and D to A, the heat is rejected from the system. . ( ) f h I Net work doen per cycle Erfi C1ency 11 ate eyc e = Tmal heat given per cycle PoVo 5P"
3PoVo + 2
oVo
...3.13
An isolating vertical cylinder fitted with an insulating frictionless piston contains n moles of an ideal diatomic gas. The initial volume and temperature of the gas are Vo and To respectively, with the piston fixed tightly. The atmospheric pressure is Po. if the piston be released, find the equilibrium temperature and volume of the gas, if the heat capacity of the cylinder and piston is negligibly small. Weight of the piston and cross sectional area of the cylinder are Wand A respectively.
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Solution. (a) Since the system is isolated from the surrounding, the work done by the gas will be at the cost of its own internal energy. Let the final temperature and volume be T and V respectively. Evidently, the final pressure P of the gas will be just enough to hold the piston in equilibrium against its weight and the atmospheric pressure. W
i.e.,
P = Po + 
...0)
A Change in internal energy of the gas,
nR8T
=
The process D + A is isobaric in nature, the gas getting compressed from a volume 2Vo to Vo
(yl)
nR(T To)
G 1)
=
Since P changes form Po to 2Po, so accordingly T changes from To to 2To' Next, the process C + D is an isothermal compression, the temperature remains constant throughout at Zfo state D.
tJ.U=nC tJ.T=II
The process B + C is isochoric in volume remains constant at 3Vo' From equation of state PV = nRT P oc T
From equation of state PV = nRT
SnR(TTo)
2
Work done by the gas against a constant
VxT the temperature varies linearly from Zfo to To. __ 3PoVo Here, To nR
external
pressure Po + W, will be A
EO
:> 2To
~ ~ To
Using equation of state PV = nRT, we have
(Po + :)v since =:>
(p
=
nRT
W}v 0+A

V)0
E
~
...(ii)
tJ.W +tJ.U = 0
[..."Q :
SnR(TTo) +~:O
'
'
:
V = 2Vo +
,
To
(.)
,,D
:,
Temperature
2To (T)
(b)
Fig. 2.50
Illustration: Fig. 2.50 (c) shows the V.T diagram for a cyclic process performed on n moles of an ideal gas.
T = ~o + (Po + :):~ and
C
 ••
A
,
2Vo 3Vo Volume (V)
OJ
Solving Eqs. (ii) and (iii), for V and T, we get
.:
, ,

Vo
... (iii)
2
B
• 20 ,
aW =pav=(po + :}vVo)
D
C
5nRTo
7 {Po+:)
Illustration: Fig. 2.49 shows the PV diagram for the cyclic process ABCDA performed on n moles of an ideal gas. Here we draw the T.V and PT diagrams for the process. Initially, if To be the temperature of the gas in state A, then from equation of state (3Po)(Vo) = nRTo
TO 2To 3To 4To Temperature (T)
Fig. 2.50 (c)
Here we draw PV and P.T diagrams. Let Po be the initial pressure of the gas, Po .: nRTo Vo
Vo
2Vo
Process A + B is isochoric the volume remains constant. If V is constant, then P IX T
3Vo
Since temperature increases from To to also increases linearly from Po to 2Po'
Volume (V)
____
'.19.2.49
To = 3PoVo nR the process A + B is isothermal expansion, so the temperature of the gas remains constant at state B it is To. 0'
BC
zro, the pressure
The process B + C, is isobaric, because the straight line passing through the origin, thus, V :x T.
The pressure remains at 2Po from state B to state C. For the process C + D, the volume remains constant,
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191
pxT
and
As the temperature decreases from 4To to ZTo the _pressure also gets halved to Po from 2Po' The process D 4 A is agains isobaric in nature as the .line DA, pusses through the origin.
~Ll :,
'
Vo
,
,
A
2Vo
'
,
2nyRTo (y1)
"W "=oQ
,D,
Temperature
in
= nRTo(31n4)
(T)
3nyRTo
(.)
("I
=
:. Efficiency 'r]' for the process is given by
,
To 2T03T04To
Volume (V)
0)
Total heat supplied, oQin = oQAll pO
Z0~,, ~,, " ,~ : ' .
=
=nRTo(31n4) '" 3nC T
,
Po
(:.oUC4
Total work done oW = llW AB + oWBC + oWC4 = 3nRTo + 0  nRTo In 4
g; 2Po ~
6.QJJC = oWC4
=[1_~)(1_1n4J 3
y
(y I)
Fig. 2.50
Illustration: Let us find the efficiency of the cycle shown in Fig. 2.51.
E>e is more steeper (y being greater than 1) than an isothermal one, Qoth passing through a common point.
7. For n = 1 or
= 1, the equation of curve is P = kV
(l
(,n)
C =Cy

1n
1 CV(Y+2 )
Putting n = 1; we have C
2. For a general polytropic process, we have
Pv"
1 =(yCv
= k where k i.~constant.
2 1
k
or
P=_
V"
=(Cp 2
+CvJ +Cv)
CP;Cv",c",Cv
Example
J ! c
Fig. 2E.44 shows a cycle performed on an ideal gas, referred to as Otto cycle. Show that its efficiency is given by
"
~=l(~:rl
where
VJ!V2
is called compression
ratio.
Solution: The heat exchanges take place at constant volume.
IQHI=nC,,(T3T,) IQLI=nCv(T4 p
T1)
Adiabatic
Volume (V)
v
Fig. 2.52
For all positive values of the index n (i.e., 0 < n < 00), the curves are hyperbola, which touch the coordinate axes at infinity. In particular for an isothermal process n = 1, the equation
reduces
to P
ox;
.!,
which
represents
V hyperbola and/or other process (where 0 < n < the curves are unequilateral hyperbolae.
= constant,
Efficiency of a heat engine is given by ~ = 1
rectangular. 'J)
J..g;J
IQH I
or n :f'I),
=1_[T.T,] T3  T2
3. For the isobar n = 0 and P = constant, C = C p and is represented by the horizontal line. 4. For the isochor 11 ~ ::!:(X) and V is represented by the vertical line.
Flg.2E.44
Process 1 ~ 2 is adiabatic. T, T)
C = C y and
5. For values of n such that y > n > 1, the value C (molar.. heat capacity is negarive[shown in figure). 6. For values of n which are negative, say n = a, the equation of the polyrrope can be rewritten as P = kVT =Q f +2
corresponds to C p of a monatomic gas and G of a diatomic gas. Slope of straight line 1 is maximum, i. e., it corresponds to minimum heat capacity, hence it corresponds to a monatomic gas in an isochoric process. Similarly, slope of straight line 3 is less, corresponding to greater heat capacity. Hence.it represents an isobaric process involving a diatomic gas.
For monatomic gas f = 3 and for diatomic gas f = S. Thus slope of the straight line representing the W versus Q is 2/5 for a monatomic gas and 2/7 for a diatomic gas. Straight line 1 corresponds to a monatomic gas and straight line 2 corresponds to a diatomic gas.
A gas is expanded from pressure PI' volume VI to final state volume V2 in two ways, isothermally and adiabatically. In
L.
p
where
f is the number of degrees of freedom. Slope ofw
versus Q line is W
2
In an isochoric process work done is, zero, the corresponding straight line coincides with the horizontal axis.
which of the two processes is the final pressure higher and in which is the work greater? p
p Compression
In an adiabatic process heat is not absorbed, the corresponding straight line coincides with the vertical axis.
Example ~"''._'=_..
.,
i~i
In Fig. 2£.47 are shown different processes for a monatomic and diatomic gas on a temperature versus heat supplied diagram. Which processes correspond to these straight lines? The graphs of what processes coincide with the coordinate axes? The initial states of the two gases are the same.
,, ,,, • ,: Isothermal. ,
.
.
v, (.J
v,
v
(') Fig.2E.48
Solution: An adiabatic curve is steeper than an isotherm. Hence for expansion to same final volume, the final pressure in the case of adiabatic process will be lower. The area under isotherm is greater than the area under adiabatic curve. The area under PV diagram represents work done.
oT
3
Fig.2E.47
In case of compression, from same initial volume to final volume, final pressure attained in case of adiabatic compression.
Solution: Along horizontal axis, I!T = 0, hence it represents an adiabatic process. Along vertical axis Q = 0, hence it represents an adiabatic process.
Fig. 2£.49 shows two adiabatic curves for two gases, helium and carbon dioxide. Which cu",e corresponds to which gas?
a
p
Heat absorbed in a process is given by Q =nGI!T
I!T =(n~)Q Hence a straight line on I!T Q diagram has slope (l/nG). Molar heat capacity of a monatomic gas in an isochoric process is, G" = (3/2)R and that of a diatomic gas is, Gv = (S/2)R. Molar heat capacity of a monatomic gas in an isobaric process is, G p = (S/2)R and that of a diatomic gas is, C, = (7/2)R. Since heat capacity Gv of a diatomic gas and the heat capacity Gp of a monatomic gas are same, the slope of tiT_Q curve (a straight line) is same. Therefore, the straight line 2
'_v Flg.2E.49
Solution: An adiabatic process is given by PVY = constant
IG
where y = G p \J = 1 + (2/ fl. f represents degrees of freedom. A helium molecule (monatomic) has three degrees
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On the loglog scale the pressure dependence of compressibility is shown by a straight line. The lines corresponding to isothermal process and adiabatic processes involving argon and carbon dioxide are shm ..•.. n in Fig. 2£.52.
Example
•
•.'c'
....' ...
51
'.
Find the /lumber of degrees of freedom of molecules of a gas: (a) whose molar heat capacity (C) = 29 l/molK Give expression for the compressibility of an ideal gas as a function of pressure for two cases, one when the gas is
compressed isothermally and the other when the gas is compressed adiabatically. Draw a rough graph of compressibility
versus pressure in the two cases. Also draw on
(b) whose molar PT = constant.
N
U "" 
The
is given
by
when
for 0 moles of gas
oRT
~~dU~.!.3(N nRT)~NR
C,
compressibility
(C) ""29 llmolK
2
and carbon dioxide same?
ndT
the
relation
capacity
Solution: (a) According to eguipartition of energy theorem gas with N degrees of freedom has internal energy
the loglog scale the pressure dependence of compressibility for argon and carbon dioxide. [s the isothennal line for argon
Solution:
heat
ndT
2
... (1)
2
From the relation,
P
logP
Cp ""C"
+R
N
=R
+R
2
~
Adiabatic
~(N;2)R
Isothermal
,,,'"'~,, P'~" process
logP
P
... (2)
y=C =(N+2) p
Hence
C"
Fig.2E.50
... (3)
N
From egn. (2), I dV
~~For an iSOlhermal process, PV = constant
=
PV~ =
p
3diabatic
constant
=p y
For an adiabatic process, I3P =.!.. y
versU.'i
(1)
From ideal gas equation, PV = oRT Eliminating T from eqns. (1) and (2), we have p2V = constant or PVL2 = constant
(2)
This is a polytropic process, with n = 1/2. For a pOI}1rOpicprocess, molar specific heat is given by
For an isothemtal process,J3P = 1 ""constant
On all
N + 2
N =5
dV+VYdP""O (Il)
2
(b) The given process is PT = constant
I
H~nce,
N+2
which on solving for N, yields
For an adiabatic process,
'(PVy1
Cp
29
I
(f\)isOlh"rmaJ
2
8.314 ~
PdV+VciP=O
Hence,
R
~
VdP
R R C~yl /11
"" constant
P diagram isothermal as well as adiabatic
processes are represented by hyperbolas; the constant decides the nearness to origin. The constant is lesser for adiabatic process, hence it is nearer to origin.
If N
=
degree of freedom,
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N+ 2
With C = 29J/molK, R = 8.314J/molK,
Y = 
and
N
I
n =, we have 2 29
I
1
[(N;2)_I]
+ = ~~ &314 (1/2)
'.'
On solving for N, we get,
,
Example Consider an ideal gas with density p = 1.3 mglcm3 at standard pressure and temperature. If velocity of sound propagation in this gas is v = 330 mis, calculate the degrees of fr.eedom of gas molecules.
Solution: Speed of sound is given by
N = 3.
v=ff
COTL5idera mass m = 15gm of nitrogen enclosed in a vessel at temperature T = 300 K. Find the amount of heat required to double the root mean square velocity of. its molec!;IJes.
Solution: All the diatomic gases (H2•
°
2,
where p is density, P is pressure, and y is adiabatic constant for the gas. From the above equation. we can find y.
pu'
y=p
;J2• etc.)
exhibit 5 degrees of freedom between 1001000 K, 3 of translation and 2 of rotation. Hence, internal energy of n moles of N 2 at T = 300 K would be
Now, y depends on the degrees of freedom of gas molecules. If there are f degrees of freedom, then average inte~al
U =~ nRT 2
.
energy of each molecule is f>
0 __, ..=~ .
Vb :: Tb
n moles of an ideal gas (y = 1.5) is taken through the thermodynamic cycle shown in figure. Let the initial pressure Po and volume Vo at point a be given, and let Vb = 4Vo' Assume step a 4' b is an adiabatic process. Find the
V,
A system containing
pressure, volume and temperature of the system at points b and c in terms of the given quanti/)' and also the total work done by the system during one complete cycle.
Vo x PoVo _ PoVo
T::
,
T,
4Vo

2nR
8nR
T :: PoVo , 8nR
Work done during the complete cycle is area of the closed cycle
J
WI ::work from a to b ::[P1V1 P2V2 y1
Po • a
~nD(T, T'J~nD(T" TbJ~nR[!'9~ ~] '\
11
1.51
2nR
Fig.2E.92
Vo' P" = Po.y = 1.5 Vb =4VO.Pb =?
P" V}
'\
PoVo ::nRx 2 ::PoVo
v, Solution: Given Va
y1
=
= Pb V~~
Work from b to c (isobaric process) Po 3 W2 :: Fev,  Vb)::  [Vo 4Vo] ::  VoPo B 8
3
Wntl =PoVo PoVo B
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::PoVo B
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Example
(b) Examine the following
vL2:v~ TL2
A movable heavy piston is supported by a spring inside a vertical cylindrical container as shown. When all air is pumped out of the container, the piston is in equilibrium as shown in Diagram 1, with only Q tiny gap between the piston and the bottom of the container. When a portion of gas at temperature T is introduced under the piston, the later rises to a height h as shown in Diagram 2. What would be the height of the piston above the bottom of the container if the gas is then heated to a temperature Zf ? Assume that the piston moves without friction and that the spring obeys Hookes law.
•.••.
v
T
(I)
Flg.2E.9J
Solution: Initially there is no gap between piston and cylinder because
and fill in the table: > or
(b) In (i)VOCT,ButPV=:RT
V==(:~}T
2
1
~
P2
So slope of V  T plot is mR MP Here slope of 1 < slope of 2. mR mR => P2 P2 MP} MP
the number of moles PIAh P2Ah'
(h,)2
=>
is P2 and
P, h ==P2 h'
h2
=>
... (1)
P2.A==kh'
Conserving
PI
(ii)
mLim
P, . A ~ kh After raising the temperature new height h'
(i)
Solution: (a) We have m = VI 'PI and m  Lim = V2 ,p 2 where VI and V2 are volume of the vessels at t} °C and t 2°C and P I and P 2 are densities of {he liquid at ttOC andt2°C
force balance weight of piston. Now after introducing gas in container, the decrease in extension of spring is because of the pressure of gas because due to gas pressure spring begins to compress let the pressure of gas is PI' A l' Area of crosssection of piston
(iii)
Fig,2E.94
..
Diagram2
P
(II)
',':~ ,.
Diagram1
diagrams :
2
In (ii) we notice that corresponding P,V1 VI or VI < V2 mR mR
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THERMOOYNAMICS
Problem
1.
(e)
2.
5  To 3 4 To 3
(b) ~To
(c) W2
2
3::SR
4 '
unit
increment
specific heat for this process (a) 3R
(d) data is insufficient
2
is related to its pressure U (a)~3P V
U
(e) 
V per
w, =
5. The energy density U I V of an ideal monoatomic
(d)"T
in temperature.
The
~
2 (d) 9R
2
2
P as : U 3 (b)~P V 2
pUS
(d) 
V
3
H by three different
~P 2
processes:
p
(b) SR
(e) 7R
gas
6. One mole of an ideal gas is taken from state A to state
is :
2
work
1n(2)
A spherical soap bubble (surface tension = 5) encloses 11 moles a monoatomic ideal gas. The gas is heated slowly so that t..hc sUlface area of the hubble increases by
3.
isothermal. Let W] and W2 be the respective done, then: W, (b) W2 =(a)W2=Wjln2
If 2 moles of an idee'll monoatomic gas at temperature To is mixed with 4 moles of another ideal monoatomic gas at temperature 21'0, then the temperature of the mixture is : (a)
is Correct
Only One Alternative
ePeve I
Pressure versus temperature graphs of an ideal gas are as shown in Fig. Choose the wrong statement :
L,~,J~, (i)
(ii)
(iii)
Ca} Density of gas is increasing in graph
(i)
(b) Density of gas is decreasing in graph (ii) (c) Density of gas is constant in graph (iii) (d) None of the above 4. A gas is expanded different processes.
to double its volume by two On is isobaric and the OTher is
7.
A
v
co
ACB (ii) ADB (iii) AEB as shown
(a) (b) (e) (d)
diagram. The heat absorbed by the gas is : greater in process (ij) than in (i) the least in process (ii) the same in CO and (iii) less in (iii) than in (ii)
in the
PV
Sixty per cent of given sample of oxygen gas when raised to a higher temperature dissociates into atoms. Ration of its initial heat capacity (at constant volume) to the final heat capacity (at constant volume) will be :
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Study Anurag Mishra with www.puucho.com THERMODYNAMICS 8 (b) 25 (a) 7 28 (e) 10 (d) 25 7 27 8. In a cyclic process shown in the Fig. An ideal gas is adiabatically p taken from B (0 A, the work done on the gas during the process B + A is 30 J, when the gas is taken from A + B the heat v absorbed by the gas is 20 J. The change in internal energy of the gas in the process A+Bis: " (a) 20 J (b) 30 J (e) 50 J (d) 10 J 9. Three moles of an ideal monoatomic gas performs a cycle 1+2+3+4+1 shown. The gas temperatures in differem states are
T
the gas during the cycle is : (a) 1200R (b) 3600R (e) 2400R (d) 2000R 10. Three moles of an ideal monoaromic gas perform a cycle shown in Fig. The gas temperatures in different states are Tt = ZOOK, T2 = 400K. T3 = 1600K, and T4 = BOOK.The work done by the gas during the cycle is: (Take R = 25/3 J/molK)
(e) IS kJ
11.
12.
(a) 95
(b) 31
(e) 39 (d) 45 13. Pressure versus density graph of p an ideal gas is shown in Fig. (a) During the process AB work done by the gas is positive (b) During the process AB work .... done by the gas is negative 6n p (c) During the process BC internal energy of the gas is increasing. (d) None of the above 14. Ideal gas is taken through the process shown in the
~Jl.. ' .' I J /"A
Fig. p
T1 :=: 400K, T2 = 24DOK and T41200K. The work done by
(a) 5 kJ
water in a 0.45 kg copper cup from 298 to 373K ? esp. heat of copper = 389 J I kg K)
(b) 25 kJ (d) 20 kJ
Unit mass of liquid of volume VI completely turns into a gas of volume V2 at constant atmospheric pressure Po and temperature T. The latent heat of vaporization is L.Then the change in internal energy of the gas is : (a) L
(b) L+PO(V2
(c) L  PO(V2  VI)
(d) zero
VI)
If all the heat produced were used, how many Iitres of natural gas at NTP (heat of combustion of the gas 6 3 = 37.3 x 10 MJ 1m ) is needed to heat 4.54 kg of
(a) In process AB, work done by p system is positive (b) In process AB, heat is rejected (c) In process AB, internal energy increases (d) In process AB, internal energy decreases and in process BC, intef!lal energy increases 15. A process I + 2 using diatomic p gas is shown on the PV diagram below.
j: T
.;71
P2 =2P1 =106N/m2, V2=4\)=0.4m3. The
2
_.V
molar
:, ,
:3' v
heat capacity of the gas in this process will be : (a) 35R/12 (b) 25R/13 (c) 35R/ll (d) 22R17 16. Select the correct statement from among the following: (a) A monoatomic gas has three degrees of freedom because it undergoes translational as well as rotational matiGn (b) A diatomic molecule undergoes both translational and rotational motion and has six degrees of freedom (c) A diatomic molecule is capable of rotating energetically about each of three mutually perpendicular axes (d) The molecule of a polyatomic gas is capable of rotating energetically about each of three mutually perpendicular axes and undergoes both translational and rotational motion 17. A gas is expanded from volume Vo and 2Vo under three different processes. Process 1 is isobaric, process 2 is
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21. A cylinder of ideal gas is dosed by an 8
and AU3 be the change in internal energy of the gas in
kg movable piston (area 60cm2)
as
these processes. Then:
shown in Fig. Atmospheric pressure is Gas 100 kPa. When the gas is heated from 30°C, to 100~C the piston rises by 20cm. The piston is then fixed in its place and the gas is cooled back to 30°C. Let .6QI be the heat added to the gas in the heating process and I.6Q21 the heat lost during cooling. Then the value of [.6Q2 1l'.Q21l will
isothermal
and process 3 is adiabatic.
Let AU1•
P p
,.
1 ,
2 3
v
v, (a) AU) > ilUz (b) flU) < AU2 (e) D.U2 < AUl (d) IlU2 < tJ,U3
be:
> t.U3 < AV3 < L\U3 < L'!.U}
22.
(a) zero
(b) 136 J
(e) 136 J
(d) 68 J
PV diagram of an ideal gas is as shovvn in Fig. Work done by the gas in the process ABeD is : P
18, A cylindrical tube of uniform crosssectional area A is
fitted with n\'o air tight frictionless pistons. The pistons are connected to each other by a metallic wire. Initially the pressure of the gas is Po and temperature of To. Atmospheric pressure is also Po, Now the temperature of the gas is increased to Zfo• the tension in the wire will be :
_.:.:.:.:Gas::':' :::::::. :.:: :.:.:::::::::::::::::::::::::::: :::::.::: :: (a)
(b) P,A
2PoA
ee) PoA
(d) 4PoA
2
19. Let V denote the root mean square speed of the molecules in an ideal diatomic gas at absolute temperature T. The mass of a molecule is 'm'. Neglecting vibrational energy terms, which is false? (a) A molecule can have a speed greater than .../2 t' (b) V is proportional to JT (c) The average rotational kinetic energy of a molecule is mv 2 / 4 (d) The average kinetic energy of a molecule is
2P,
Hence,
P~I+[~r
LA, ~ , ,
SR (c) _ SPoVo 4R
•
:
,, ,,
,, ,
,
v,
V
3V,
(a) 4PoVo
(b) 2PoVo
(c) 3Po Vo
(d) PoVa
23. A vessel of volume 20 litres contains a mixture of hydrogen and helium at temperature of 2rc and pressure 2.0 atm. The mass of the mixture is 5g. Assuming the gases to be ideal, the ratio of the mass of hydrogen [0 that of helium in the given mixture will be: (a) 1 : 2 (c) 2: 1 24.
(b) 2 , 3
(d) 2 , 5
Pressure versus temperature graph of an ideal gas of equal number of moles of different volumes are plotted as shown in Fig. Choose the correct alternative: tan 9 =.nB.
Po
and
Vo
are
a process constants. T
Change in temperature of the gas when volume is changed from V = Vo to V = 2Vo is : (a) _ 2PoVo
D
V
One mole of an ideal gas undergoes P,
..._:r:
P,  ••• .8 ,, ,,
Smv2 / 6
20.
Piston
rz,";;~;;"zz1
b) llPoVo
(lOR (d) PoVo
(a) VI = V2, V3 = V4 and V2 > V3
(b) VI = V2, V3 = V4 and V2 < V3 (c) Vj =V2 =V3 =V4 Cd) V4 > V3 > V2 > VI Volume versus temperature graph of two moles of 25. helium gas is as shown in Fig. The ratio of heat shown
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in Fig. The ratio of heat absorbed and work done by the gas in process 12 is : V
30.
(a) 3
26.
27.
(b) 5/2 Ce) 5/3 Cd)7/2 The PV relation for a monatomic ideal gas undergoing an adiabatic change is : (a) PVS'3 "" Constant (b) PV2 "" Constant,' (c) PVZ.'3 = Constant (d) PV7/3 "" Constant ~,
fuT;]
(c)
2aTi
3 , 2 Cd) 3uTo2 (b) uTo
Pressure versus temperature . graph of an ideal gas is as shown in Fig. corresponding density (p) versus volume (\I) graph will be : t
,
p
C
B,C
Pressure versus temperature graph of an ide'll gas is as shown in Fig. Density of the gas at point A is Po' Denshy at B will be : P
(a)
(a)
.:./00
(b) ~A,O
v
V
,p
3Po
7
B
Po ••••••
'A
,
:
:,
Ce)
,, to
Cd)
B~A
v
t
"0
3 3 (b) "Po 4 4 (c) Po Cd) 2po 3 28. In the p.v diagram shown in Fig. ABC is a semicircle. The work done in the process ABC is : (a) Po
31.
'b: V
The Fig. shows the graph of logarithmic reading of pressure and volume for two ideal gases A and B undergoing adiabatic process. From Fig. it can be concluded that:
P(atm)
3 ..••
~.rc~.:c
,, ~ ••• t ••. 
~A
''
B
:A
loV
, ,,
,, V(L)
(a) zero
IT
(b) atm.L 2
(c) ~atm.L (d) 4 atm.L 2 29. Pressure P, volume V and temperature T of a cerrain
(a) gas B is diatomic (b) gas A and B both are diatomic (c) gas A is monoatomic (d) gas B is monoatomic and gas A is diatomic 32. A thennodynamic system undergoes cyclic process ABCDA as shown in Fig. The work done by the system is :
2
material are related by P = aT
•
Here, a is a constant.
V
The work done by the material when temperature changes from To to ZTo while pressure remains constant is :
p
3PJ 2P
P,
~, ,
, ...... ~_.
,, , A'
v,
•
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Study Anurag Mishra with www.puucho.com mRMODYNAMICS p
(b) 2PoVo
(a) Po Vo (e) PoVo
(d) zero 2 33. Certain amount of an ideal gas are contained in a dosed vessel. The vessel is moving with a constant velocity v. The molecular mass of gas is M. The rise in temperature of the gas when the vessel is suddenly stopped is ('f ==Cp ICv) Mv' Cb) Mv'C ..,I)
Ca) 2R(y I)
Ce)
2R 2
Mv'
Cd) 
Mv y
.
2R(y I)
34. A cylinder contains He and 02 of equal volume. They arc separated by massless freely moving piston as shown. If the gas is adiabatically compressed by moving the piston so that volume of He becomes half. Finally: t ••
L
L
v (a) >
1
(b) < 1 (e) I
(d) data insufficient A sample of an ideal gas is taken through a cycle as 37. shown in Fig. It absorbs 50 J of energy during the process AB, no heat during Be, rejects 70 J during 0\.40 J of work is done on the gas during BC. Internal energy of gas at A is 1500 J, the intemal energy at C would be : p
.1
v 0,
H.
Area A
Adiabatic wall
massless
massless piston
(a) pressure in He chamber will be equal to pressure in O2 chamber (b) pressure in He chamber will be less than pressure in 02 chamber (e) volume of He chamber will be will be equal to volume of 0 2 chamber (d) volume of O2 chamber will be LA I (2)25121 35. The ratio of specific heat of a gas at constant pressure to that at constant volume is y. The change in internal energy of a mass of gas when the volume changes from V to 2V at constant pressure P is : Ca) (e)
R
Ca) 1590 J (b) 1620 J (e) 1540 J Cd) 1570 J 38. A liquid of density 0.85 g I cm 3 flows through a calorimeter at the rate of 8.0 em 3/S. Heat is added by means of a 250 W electric heating coil and a temperature difference of 15"C is established in steadystate conditions benveen the inflow and the outflow points of the liquid. The specific heat for the liquid will be : (a) 0.6 kcaVkgK (b) 0.3 kcaVkgK (c) 0.5 kcaVkgK Cd) 0.4 kcallkgK 39. One mole of a monoatomic ideal gas udergoes the process A + B in the given PV diagram. The specific heat for this process is : p
(b) PV
6P o •.•....
Cd) yPV "f 1
3Po
y 1 PV 'f 1
/ .•••
.
•B •
36. The Fig. shows two paths for the change of state of a gas from A to B. The ratio of molar heat capacities in path 1 and path 2 is :
"".f... • ••
.
Vo
Ca) 3R (e)
:
'
v
5Vo
Cb) 13R
2
6
5R 2
Cd) 2R
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40. An ideal monoatomic gas is taken round the cycle ABCDA as shown in the PV diagram (Fig). The work done during the cycle is : p
i"o.V 2~.2V A
(a) W2 >W} >W3 (c) WI > W2 > Wt
(b) W2 >W3 >W1 (d) WI > W3 > W2
46. An ideal gas is initially at temperature T and volume V. It~ volume is increased by .6.V due to an increase in temperature liT, pressure remains constant. The quantity 8 = .6.V/V.6.T veries with temperature as :
,
,
D
P,2V
P,V
v
(a) PV
Cb) 2PV
ee) 112 PV
Cd) zero
Cb)
/ T
T+dT
T
41. If M1.M2,M3
•••••••• are the molecular weights of different gases, R1,R2,RJ ••••••••• are the gases constants respectively then which of the following is valid? (a) M1 +M2 +M3 + .•........ =1
R1 +R2 +R3 +
(b)
M1
=
M2
R1
=
R2
(e) MIR} =M2R2 Cd) none of these
MJ
=
T
=Constant
RJ
=MJRJ =..... =Constant
pistons contain the same ideal gas at the same temperature and the same volume V. The mass of the gas in A is rnA and that in B is mB" The gas in each cylinder is now allowed to expand isothermally to the same final volume 2V. The changes in the pressure in A and B are found to be /1P and 1.5/1P respectively, then:
=
(b) 2mA = 3mB (d) 9mA = 4mB
47. 1\vo monoatomic ideal gases 1 and 2 of molecular masses mt and m2 respectively are enclosed in separate containers kept at the same temperature. The ratio of the speed of sound in gas 1 to the gas 2 is given by: Ca)
r'
(b)
m2
(c) L2 / L}
(b) L1 / L2 (d) {L2 / L} )2/3
44. The C p IC v ratio for a gas mixture consisting of 4 gIns helium and 32 gms of oxygen is : (a) 1.45 Cb)1.6 (e) 1.5 Cd)1.66 45.
r'
mt
(d) m2
(c) ~
m2
ml
48. PV plots for two gases during adiabatic processes are shown in the Fig. Plots 1 and 2 should correspond respectively to :
43. A monoatomic ideal gas initially at temperature Tt, is enclosed in a cylinder fitted with a frictionless piston. The gas is allowed to expand adiabatically to a temperature T2 by releasing the piston suddenly. If L} and L2 are the lengths of the gas column bfare and after expansion respectively, then T1 / T2 is given by: (a) (LI / L2)2/3
~
"
42. 1\vo identical containers A and B with frictionless
(a) 4mA 9mB (c) 3mA = 2mB
(d)
p
~, 2
v (a) He and O2 (c) He and Ar
(b) 02 and He (d) 02 and N 2
49. An ideal gas is taken through the cycle A + B + C + A, as shown in the Figure. If the net heat supplied to the gas in the cycle is 5 J, the work done by the gas in the process C + A is :
Starting with the same initial conditions, an ideal gas expands from volume V} to V2 in three different ways, the work done by the gas is W} if the process is purely isothermal, W2 if purely isobaric and W3 if purely adiabatic, then:
V(m3)
2 1
Cf',t . . •
.
A
•
10
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Study Anurag Mishra with www.puucho.com TH!RMODYNAMICS (bl 10 J 5 J (d) 20 J (e) 15 J 50. Which of the following graphs correctly represent the (a)
variation of
P = dV I
dP with P for an ideal gas at
V
constant temperature: p
p
.,
233
(a)"Rf"tn 2
(b) ~ RT
(el RT
(d) ~RT
2 2
54. In the PV diagram of Fig. shown, work along isotherm ab and 4 J What is the change in the internal the gas traverse the straight path
the gas does 5 J of along adiabatic be. energy of the gas if form a to c ?
p
~
r
a
~, p
v ~ p
p
a process as shown. The total amount
of
heat imparted to the gas equals Q = 27.7kJ. Determine the ratio of molar specific heat capacities. T
/c 273K
A," '
.~ , .' , .'.
B
v (a) IAI
(b) 4 J J (d) 9 J (e) 5 J 55. An adiabatic change in a perfect gas in one: (a) which should be carried alit very rapidly III thinwalled, perfectlyconducting vessel (b) in which the temperature T and pressure pare related by the expression PyIT" constant (c) in which internal energy remains constant (d) in which no heat enters or leaves the system 56. For twO different gases X and Y, having degrees of freedom II and 12 and molar heat capacities at constant volume C VI and C V2 respectively, for adiabatic process, the In P versus In V graph is plotted as shown: (a) I
(d)
51. At a temperature of To = 273K, n\lo moles of an ideal gas undergoes
b
p
p
p
(el
\
(b)
(a)
;=
I,P
(b) 1.63
(e) 1.33 Cd)None of these An air bubble of volume V 52. o is released by a fish at a depth h in a lake. The bubble rises to the surface. Assume
constant
temperature
and
standard
atmospheric pressure above the lake. The volume of the bubble just before touching the surface will be (density) of water is p : (b) Vo(pgh/p) (a) Vo
ee)
t'o
(l+P~~)
(d)
I,V
(a) II >12 (e) CVz >CV1
(b)/2>fl (d)Cv1 >C"z
57. The VT diagram of an ideal gas for the process A ~ B ~ C is as shown in the Fig. Select the correct altemative{s) :
vo( I + P~h)
v
53. One mole of an ideal gas is contained within a cylinder
by a frictionless piston and is initially at temperature T. The pressure of the gas is kept constant while it is
heated and its volume doubles. If R is molar gas constant, the work done by the gas in increasing its volume is :
o
T
(a) Pressure of the gas first increases then remains constant (b) Pressure of the gas first decreases then remains constant
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Study Anurag Mishra with www.puucho.com 234 (c) Pressure of the gas remains constant throughout Cd) Nothing can be said about the pressure of the gas from this graph 58.
An ideal gas, contained in a cylinder by a frictionless piston, is allowed to expand from voltime VI at pressure PI to volume V2 at pressure P2' Its temperature is kept constant throughout. The work done by the gas is : (a) zero, because it obeys Boyles's law and therefore
62. One mole of an ideal gas with heat capacity at constant pressure Cp undergoes the process T = To + aV where To and a. are constants. If its volume increases from VI to V2 the amount of heat transferredto the gas is : (a)
CpRTo
11~~r
(b) a.CP(V2Vl)RToln~1
P2V2PIVl =0
(b) negative, because the pressure has decreased and so the force on the pistOn has been diminishing (c) zero, because it has been kept at constant temperature and so its internal energy is
unchanged
(c) aCp(V2 VI)+RTo (d) RTo
InIV,[
lV1
lnl~
aC p (V2  VI)
Cd) positive, because the volume has increased 59.
60.
The molar heat capacity of an ideal gas at constant pressure is greater than that at constant volume because (a) work has to be done against intermolecular forces as the gas expands (b) work has to be done against external pressure as the gas expands (c) the molecules gain rmational kinetic energy at the gas expands Cd) the molecules move faster when heat is supplied at constant pressure than when supplied at constant volume of a state of a gas is given by P(V b) = nRT. If 1 k mol of a gas is isothermally expanded from volume V to 2V, the work done during the process is : 2V V (a) RT (b) RTlnl ~ Vb
63. Fig. shows a parabolic graph between T and
V mixture of a gas undergoing an adiabatic process. What is the ratio of VmLS and speed of sound in the mixture? T o
2T
bl
I2Vb I
Vb (c)RTln
bl
.'
(1NoJ
(b)
T T+AT Temperature K
T T+AT Temperature K
, (e)
, ~
0
T T+t.T Temperature K
(4NoJ
(1N)
Ca) 13
(b)
.J2
f2
(d)
.fj
V~
(e)
V3
64. The density p versus temperature T graph of an ideal gas is shown in Fig. Choose the incorrect statement(s): p
v=bI
, 0
:
(d)RTlnl~1
61. An ideal gas has temperature T and volume V. The quantity is = dVjVdT varies with temperature under isobaric conditions as:
(al
7 ~
To •
The equation
Inl '1
!. for a
o
(a) The curve represents an isobaric compression (b) The internal energy of the gas decreases (c) The system absorbs heat during the process (d) The curve shifts upwards at higher pressures 65. fig. shows three isotherms at temperatures 300 K, sao K and 700 K respectively on PV diagram. lWo thermodynamic process (A and B) are performed on one mole of monoatomic ideal gas as shown in Fig.
~
(d) 0
T
T+AT
Temperature
T
K
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THERMODYNAMICS 68.
p
T, T, T
Liquid oxygen at 50 K is heated to 300 K at constant pressure of 1 atm. The rate of heating is constant. Which of the following graphs represents the variation of temperature with time?
Ca)
V
en
The gas in process A is heated while the gas in process B is cooled. Oi) The work done in process B is less than the work done in process A. (iii) The amount of heat is absorbed by the gas in both processes. Which of the above observations must be correct? (a) (ii) only (b) (ij) and (iii) only ec) (i) and (ii) only Cd) (i) and (iii) only 66. The poT diagram of thermodynamic heat engine cycle is shown in Fig. Two curved processes arc adiabatic. The work done for one mole of monoatomic gas in one cycle is given by : 8
69.
70.
C
U
(b)
(e) T
.
p
TL TEt Lt Lt
3T, ST'
Cd)T
When an ideal diatomic gas is heated at constant pressure the fraction of the heal energy supplied which increases the internal energy of the gas is : (a) (2/5)
(b) (3/5)
(e) (7/5)
(d) (517)
During the adiabatic expansion of 2 moles of a gas, the change in internal energy was found to be equal to (IOOJ). The work done during the process will be equal to : (a)
zero
(b) 200 J
(a) 1.25RTQ
(b) 2RTo
(e) 100J Cd)100J A container contains n moles of oxygen molecules (02), The molecules get dissociated if the temperature T of the gas is greater than some minimum temperature Td• The percentage dissociation into
(c) 2RTo
Cd) 1.25RT,
atomic oxygen is given by (~
To
71.
2T
:~,0
,>
0
T
a
:[78 P
c
ee)
c
v p
.LJ:
Cd)
v
PV
==
nRT
(b) PV
==
nRy
Ce) PV = 2nRT Cd) PV = nR(2T Td) In a thermodynamic process for an ideal gas, the system and the surrounding can influence each other only through work. Which of the following curves best represents this process? p
A~=J
Cb)
p
2
(a)
72. T
A'CJ
Ca)
The ideal gas
Td
... P
I).
equation can be written as (for Td ::::;T ::::;2Td):
67. A cyclic process ABeD is shown in a VI' diagram, the corresponding PV diagram is : V

'V
8~A
~I
v
c
III (isothermal) II
+_v (a) I
(b) II
(c) III
Cd) IV
v
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Study Anurag Mishra with www.puucho.com THERMODYNAMICS 73. PV curve for the process whose
vr
curve is:
T
adiabatic
C7' . ,' b
p~v Is~~vIt=" J;
(a) P~y
(e)
V
(b)
(d)
,
74. Find the approx. number of molecules contained in a vessel of volume 7 litres at O°C at 1.3 x lOs pascal: (a) 2Ax 1023 (b) 3.0x 1023 Ce) 6.0x 1023 Cd) 4.8 x 1023 75. One mole of an ideal gas is kept enclosed under a light piston (area : IO2m2) connected by a compressed spring (spring constant IOON/m). The volume of gas is a.83m3 and its temperature is lOOK. The
gas is heated so that it compresses the spring further by 0.1 m. The work done by the gas in the process is (Take R = &3J/mole and suppose there is no atmosphere) : (a) 3J (b) 6J (e) 9J (d) 1.5J 76. An ideal gas mixture filled inside a balloon expands according to the relation PV2l3 = constant. The temperature inside the balloon is: (a) increasing (b) decreasing (c) constant (d) can't be said 77. An ideal gas undergoes a thermodynamics cycle as shown in Fig. Which of the following graphs represents the same cycle?
p~"
(b)
(a) 5 kg (e) 20 kg
(a) 510 K (e) 100 K
(b) 200 K (d) 73 K
81. When 2 g of a gas are introduced into an evacuated flask kept at 25°C the pressure is found to be one atmosphere. If 3 g of another gas added to the same flask the pressure becomes 1.5 atmospheres. The ratio of the molecular weights of these gases will be: (,)1:3 (b)3:1 (e) 2 : 3 (d) 3 : 2 82. An ideal gas foHows a process PI' = constant. The correct graph between pressure and volume is:
t2
(a)
(b)
v
p~
v
p~ plL (d)
•••
1~1:."
(b) 10 kg (d) 40 kg
79. A perfect gas of a given mass is heated first in a small vessel and then in a large vessel, such that their volumes remain unchanged. The PT CUlVesare: (a) parabolic with same curvature (b) parabolic with different curvature (c) linear with same slopes (d) linear with different slopes 80. At a temperature 11A
(e)  25 kJ
B
A\]:
=sOOkJ
(b) +25 kJ
p
.' e
Q/JoC
(a)  20 kJ
T
Af7 .
and
The heat flow in the process
/'A
(a)
1.5 v(m3)
6.U.•._.H =+400kJ
V
p
,
2
(e) Q
5
(b) ~Q 5 (d) ~Q
3
130. One mole of an ideal monoatomic gas at temperature
To expands slowly according to the law P/V '" constant. If the final temperature is 2To• heat supplied to the gas
.e
is: T (a) 2RTo
3
(b) ~RT, 2
1 2 131. A diatomic gas follows equation PV'" '" constant, during a process. What should be the value of m such that its molar heat capacity during process'" R ? (c)
RTo
(a) 2/3 (e) 1.5
(d)~RTo
(b) 1 (d) 5/3
132. One mole of an ideal gas at temperature
T1 expends
according to the law ~ '" a (constant). The work V' done by the gas till temperature of gas becomes T2 is:
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(e) R(T2 TI)
4
133.2
mole of a diatomic gas undergoes the process: PT2 IV =constant. Then, the molar heat capacity of the gas during the process will be equal to : (a) sR/2 (b) 9R/2 (e) 3R
(d) 4R
134. A resistance coil connected to an ~ external battery is placed inside an adiabatic cylinder fitted with a v m frictionless piston and containing j . R" i an ideal gas. A current i flows , through the coil which has a resistance R. At what speed must the piston move upward in order that the temperature of the gas remains unchanged? Neglect atmospheric pressure .
twice its volume and then compressed at constant pressure to CVo/2) and the gas is brought to original state by a process in which P IX: V (Pressure is directly proportional to volume): The correct representation of process is :
I P (a)
Po
••••
 ••
..
IP, P
(b)
.,
(b) Rmg i'
(a) ~
Rg (e) mg i'
(d) i'R
mg
135. A long glass tube of length L, sealed at both ends, contains a small column of mercury (density =p) of length 'o'(a «L) at its middle and air at pressure P on both sides. The tube is fixed horizontally. If the mercury column gets a small displacement, the time period of its oscillations would be (assuming that the air on the sides undergoes isothermal expansion or compression) : (a) n[pLajP]"'
(b) 2n[p La/P]>2
(e) n[2pLajP]"'
(d) n[pLa/ZP]u,
136. Two pistons having low thermal conductivity divide an adiabatic container in three equal parts as shown. An ideal gas is present in the three parts A.B & C having initial pressures as shown and same temperatures. Now the pistons are released. Then the final equilibrium length of part A after long time will be: L
___ (a)
L/B
(e) L/6
B
c
2P
P
•
U3
(b) L/4 (d) L/s
137. One mole of an ideal gas at pressure Po and temperature To volume Vo is expanded isothermally to
.?
VQ
(d)
t
v 1 (a) 2Po
100 cm3
p
Po
P
..J! 2
3 Cd) 4Po 3
(e) Po
v
300 cm3
(a) heat rejected by the gas during the process is 10J (b) heat absorbed by the gas during the pTQcessis 10J (c) work done by the gas is 20 J (d) work done by the gas is 10J 145. Oce mole of an ideal monoatomic gas is taken from point A to C along the path ABC. The initial temperature at A is To. For the process A 1> B 1> C:
(b) Po
3
b
':
A B
. ~
,c
v
142. Cubical tanks X and Y have the same volumes and
share a common fixed perfectly conducting wall. There is 1 gm of helium in tank X and 2 gm of helium in tank Y. Which of the following is the same for both samples treating them as ideal gases? x
/
.
.:1gm . He
..
y
\
.
2gm' .. He'
".
(a) Heat absorbed"" RTo/2 (b) Heat liberated"" 2R'fo (c) Change in internal energy ""0 (d) Work done = 0 146. One of the straight lines in the Fig. depicts the dependence of the work done IWI on the temperature variations T for an isobaric process. The two of following are the adiabatic curves for argon and nitrogen. The isobaric process which depicts both the gases is :
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'.' process AB is two times the work done in the process CD then what is the value of T1/T2 ? T
.._~ .. (a) 1 (e) 3
•••• B
T''P~,' 2 " •• ' D
T
..
(b) 2 (d) 4
147. A same amount of same gas of temperature Tare enclosed in a three identical A, B & C. The temperature of wall of three container is TA,TB &Tc• (TA >Ta >TcJ respectively. The pressure on wall of vessel : (a) PA >PB >Pc I
.'.
(b) PA < 105N / m2
ndT
15. [dl
atm. pressure
7
nj x2+n2x4",S (nJ + 2n2)
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=
2.5
... (1)
...(2)
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(
Solving equation (1) and (2), "1
mH
",,(lx%+lX~)R=4R
2Cv
0.74, "2 '" 0.88 0.74)( 2
0:
Cp
100xo
105. [a]
1
2
100
V
z
[41]=1.5
207=CpxlO 6.Q=Cyx5 6.Q '" 207 x.s:.. = 103.5 2 Cp Y
dT +3:dV =0 T 3 V
Minimum
38. [aJ This is a problem of 'Flow calorimeter' used to measure specific heat of a liquid. Amount of heat supplied to the water per second by the heating coil", Q.• = 250 joule
y
=
Ymax
250
135. [a]
,
15°C
mst = 0.85 x 8x 103 x sx 15
•
I
•
p
l. ,'
•
I
250x 103 = 4186>< 0.85)( 8x 15
P1V1 "'PZV2
0.6 kcaVkg K
41. [c) The product of molecular weight and constant is Rw(universal gas constant)
,
(La)
(La)
Mass of this liquid = (0.85) x 1000 x 8 x lO6kg =
maximum
(61.5$ 6.Q< 103.5)
4186 the volume of liquid flowing out per second = 8.0em:'! = 8x IO6m3
Temperamres rise of this mass of liquid
value of 6.Q is for ="35 for monoatomlc. gas
~Q ~ 103.5 ~ 61.5 (5/3)
~kcal.
gas
Hence MjRj ""M2R2 ""Constant 44. [c] The molecule contains 1 mole of helium and one mole of oxygen 4 gm of helium (monatomic weight and molecular weight) corresponds to 1 mole 32 gm of oxygen (diatomic molecular weight 32) corresponds to one mole.
t
100
~_x_
~NkdT ~ _ NkT dV
=
n~T)x 1O~2
0.2 (Total compression) 1 2 1 Wgas =xl00x(0.2) xl00x(0.1) 2 2
dW ~PdV ~ NkTdV V
S
=(
Xl =
dE ~ ~NkdT 2 Moreover using the ideal gas law
Hence,
[xo is initial compression]
x ~~I_xlxa3xl00x_1_=o.l o 100 0.83 100
E ~~NkT 2
=
~1.5
4R
75. [d] kxo = PA
given by
Hence, 250 4186
~
Cv
26. [a] The first law of thermodynamics gives with E as internal energy, dE = dE for the adiabatic process. The energy of monoatomic ideal gas is
2
6R

1.48 """,*2: 5 3.52
=
Z)R ~ 6R 2
2C ~(IX~+ h p 2
~ H coce, mHe 0.88x 4
•
•
•
• •
p
•
I
la
,+' p
•
I
p(L;a)A ~P2(L;a X)A _ pela) P 2 
(La2x)
p(L;a)A =(L;a
+
X
)AP2
pela) P. _ (La+ 2x) 2 
_P' _ P22
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peL
a)
[(La)(2x)J
peL
a)
[(La)+2x]
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tV' = F
4P(L a) [(L _all _4x1]
= tV'A =
140. [bJ ' ~Q=dU+PdV
4PA(L  a) [(L_a)2
_4x2]
~dT=CvdT+PdV T
=>
~=~L:=2~L:
a RT dT=CvdT+dV
=>
T =~(L~r2
T
= ~
=>
PV
C = VTy1e
T = (Po + (1  a)V2)V nR
PI = Po/2 For isobaric process
2
dT = Po + 3V (Ia) dV nR nR
Vo VI = Vo/ 2, Tf = To. 2x 2Vo
For P
V
I
oc
Po = 3(a 1)V2
To =4 V process
T oc V
=;.
p2 ocT
=;.
PJ
V2 =
3(al) p
P=Po+(la)
V must be straight line V  T must be parabolic P  T must be parabolic
p 2
",,0
0
3(a 1)
Po 3
2P 3
=Po= o
138. [bl e.g .•
143. [eJ
u = nf
RT, where 2 internal energy.
V
v
=>
Po W = Po' Vo = 2" (VI  Vol VI
= 3Vo
=>
Since the total volume of the room is constant and pressure is also constant therefore energy of the room remains constant. Same molecule will leave the room.
T, = '3[0/2
TB > TA
144. [b] Work done = area under the CUlVe
139. [dl Asy>1 for and for
1Vy1 = constant
d'T dT < 0 >0 dV dV'
=!2 x 200x
T1 =KPyl
d'T
dT > 0  atmospheric pressure Po)' Spring of force constant k is initially unastretched. Piston of mass m and area S is frictionless. In equilibrium piston rises up a distance xo, then:
k m, ,
(c) Heat absorbed/released by the gas is maximum in path 1 (d) Temperature of the gas first increases and then decreases continuously in path 1 4. Two moles of a monoatomic
p 1 A~B
'..;.7 3
ideal gas undergoes a thermodynamic
(a) final pressure of the gas is Pa + kxo + mg S S (b) work done by the gas is .!. kx5 + mgxo 2 (e) decrease in internal energy of the gas is 1 ,  kxo + mgxo + PaSxo 2 Cd) work done by the gas is 12 kx5
'V
L
V'
process 
r'
=
constant, ifthe temperature is raised by 300 Kthen; (a) work done by the gas is 400R (b) change in internal energy is 900R (c) molar heat capacity of the gas for the process is 13(6 R Cd) molar heat capacity of the gas for the process is 3/2 R 5. During the process AB of an ideal gas: p
2 2. P.V diagram of a cyclic process ABC4, is as shown in
Fig. Choose the correct statement.(s)
:
p
T
v (a) .6.QA>B = negative (b) .6.UBoC = positive (e) tiUC ...•A = negative Cd) i'1WCA.B ••• negative 3. A gas undergoes the change in its state from position A to position B via three different paths as shown in Fig. select the correct alternative (s) : (a) Change in internal energy in all the three paths is equal (b) In all the three paths heat is absorbed by the gas
(a) Work done on the gas is zero (b) Density of the gas is constant (c) Slope of line AB from the Taxis is inversely proportional to the number of moles of the gas (d) Slope of line AB from the Taxis is directly proportional to the number of moles of the gas 6. Temperature versus pressure p graph of an ideal gas is shown in Fig. During the process AB : A08 (a) internal energy of the gas remains constant (b) volume of the gas is T increased
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251
(e) work done by the atmosphere
on the gas is
positive
(d) pressure is inversely proportional to volume 7. One mole of an ideal monoatomic gas is taken from A to C along the path ABC. The temperature of the gas at A is To' For the process ABC : p
'','V Vo 2V o
Ca) work done by the gas is RTo
Cb) change in internal energy of the gas is .!2 RTo 2 (e) heat absorbed by the gas is
2
(d) heat absorbed by the gas is 13 RYo 2 (R = universal gas constant) 8. Select the correct statements about ideal gas: (a) Molecules of a gas are in random motion colliding against one another and with the walls of the container (b) The gas is not isotropic and the constant 0/3) in equation p = .! pu2 r.m.s. is result of this properry 3
9. For two different gases X and Y, having degrees of freedom I, and 12 and molar heat capacities at constant volume Cv and C v2 respectively, the In P versus In V graph is plotted for adiabatic process, as shown: (a)
I, > 12
(c) CV2
>CVj
collision lasts is time of free path between molecules molecules and the
InP
'~I"
11. A barometer
is faulty. When the true barometer reading are 73 and 75 em of Hg, the faulty barometer reads 69 em and 70 em respectively: (a) The total length of the barometer tube is 74 em (b) The true reading when the faulty barometer reads 69.5 em is 73.94 em (c) The faulty barometer reading when the true barometer reads 74 em is 69.52 em (d) The total length of the barometer tube is 69.5 em 12. The density (p) of an ideal gas p varies with temperature T as B shown in Fig. Then: (a) the product of P& Vat A is equal to the product of P& V at B
.!...! RTo
(e) The time during which a negligible compared to the between collisions (d) There is no force of interaction among themselves or between wall except during collision
(c) C I' < 5/2 R for a diatomic gas (d) Cv >R for a diatomic gas
V
(b) 12 > /, (d) CYj >C1'2
10. At ordinary temperature, the molecules of an ideal gas have only translational and rotational kinetic energies. At high temperatures they may also have vibrational energy. As a result of this at higher temperatures: (C~, == molar heat capacity at constant volume) (a) C v = 312 R for a monoatomic gas (b) C Y == 312 R for a diatomic gas
(b) pressure at B is greater than A the pressure at A T (c) work done by the gas during the process AB is negative (d) the change in internal energy from A to B is zero 13. A gas is found to obey the law P2V == constant. The initial temperature and volume are To and Vo. If the gas expands to a volume 3Vo' then: (a) final temperature becomes J3 To (b) internal energy of the gas will increase (c) final tempemture becomes
C
~
V3
(d) internal energy of the gas decreases 14. For an ideal gas : (a) The change in internal energy in a constant pressure process from temperature T1 and T2 is equal to nC ~ (T 2  'Ii), where Cv is the molar heat capacity at constant volume and 11 the number of moles of the gas (b) The change in internal energy of the gas and the work done by the gas are equal in magnitude in an adiabatic process (c) the internal energy does not change in an isothermal process (d) no heat is added or removed in an adiabatic process. 15. An ideal gas is taken from the state A(Pressure P, volume V) to the state B (pressure PI2, volume 2\1) along a straight line path in the P~Vdiagram. Select the correct statements from the following: (a) The work done by the gas in the process A to B exceed the work that would be done by it if the system were taken from A to B along an isothermal (b) In the TV diagram, the path AB becomes a part of a parabola
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16.
17.
18.
19.
THERMODYNAMICS ee) In the poT diagram, the path ABbecomes a part of a hyperbola Cd) In going from A to B, the temperature T of the gas first increases to a maximum value and then decreases An ideal gas can be expanded from an initial state to certain volume through two different processes (i) PV2 "" constant and (ii) P = KV2 where K is a positive constant. Then: (a) final temperature in (i) will be greater than in (ii) (b) final temperature in (ij) will be greater than in (i) (e) total heat given to the gas in (0 case is greater than in (ii) ,'I \'"1 Cd) total heat given to the gas in eii) case is greater than in (i) The total kinetic energy of translatory motion of all the molecules of 5 Htres of nitrogen exening a pressure P is 3000 J : (a) the total KE. of 10 litres of N2 at a pressure of 2P is 3000J (b) the toral KE. of 10 litres of He at a pressure of2P is 3000J (c) the total KE. of 10 litres of O2 at a pressure of 2P is 2000J (d) th~ total KE. of 10 litres of Neat a pressure of 2P is 1200J Two spheres A and B have same radius but the heat capacity of A is greater than that of B. The surfaces of both are painted black. They are heated to the same temperature and allowed to cool. Then: (a) A cools faster than B . (b) both A and B cool at the same rate (c) at any temperature the ratio of their rates of cooling is a constant (d) B cools faster than A A thermally insulated chamber of volume 2Vo B is divided by a frictionless piston of area 5 into two equal parts A and B. Pan A has an ideal gas at pressure Po and temperature To and in part B is vacuum. A massless spring of force constant k is connected with piston and the wall of the container as shown. Initially spring is unstretched. Gas in chamber A is allowed to expand. Let in equilibrium spring is compressed by xo' Then: (a) final pressure of the gas is
kxo
S 1 , (b) work done by the gas kxo 2
(c) change in internal energy of the gas is ..!. kx~ 2
(d) temperature of the gas is decreased 20. n moles of an ideal gas p undergo an isothermal process at temperature T. pv graph of the process is shown in the Fig. A point A(V"P,) is located on the PV curve. Tangent at point A, cuts the Vaxis at point D. 0 D AD is the line joining the point A to the origin D of PV diagram. Then : (a) coordinate of points D is ( 3~1
v
,oJ
(b) coordinate of points D is (2V,,0) (c) area of the triangle ADD is nRT (d) area of the triangle ADD is ~ nRT 4
2). A gas may expand either adiabatically or isothermally. A number of PVcurves are drawn for the two process over different ranges of pressure and volume. It will be found that: (a) two adiabatic curves do not intersect (b) two isothermal curves do not intersect (c) an adiabatic curve and an isothennal curve may intersect (d) the magnitude of the slope of an isothennal curve for the same value of pressure and volume is less than that of adiabatic curve 22. During the melting of a slab of ice at 273 K at atmospheric pressure : (a) positive work is done by the icewater system on the atmosphere (b) positive work is done on the icewater system by the atmosphere (c) the internal energy of the icewater increases (d) the internal energy of the icewater system decreases 23. A bimetallic strip is fanned out of two identical Strip one of copper and the other of brass. The coefficients of linear expansion of the (\NO metals are ar and UB' On heating, the temperature of the strip goes up by fiT and the strip bends to form an arc of radius of curvature R. Then R is : (a) proportional to l!.T (b) inversely proportional to fiT (c) proportional to lOB ucl (d) inversely proportional to laB ucl 24. The Fig. shows a (lIp) versus T graph for an ideal gas. Select the correct statement(s) :
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THERMOOYNAMICS (b) is infinite for an isothermal process
(lIp)
2
.,//
o'
(c) depends only on the nature of the gas for a process in which either volume or pressure is constant (d) is equal to the product of the molecular weight and specific heat capacity for any process 28. During an experiment, an ideal gas is found to ooey a
/ 1
condition p2 = constant [p = density of the gas]. The
T
(a) The graph represents an isobaric expansion (b) L..,rgethe slope of straight line higher the pressure ee) Internal energy of the gas increases Cd) Work done during the process is positive 25. The Fig. shmvs PV versus T graph of ideal gas. Choose the correct alternative(s) : PV
T
(a) The slope of the straight line is independent of the monoatomic or diatomic nature of the gas (b) The slope of the straight line depends on the molar mass of the gas ee) The straight line denotes a process in which the gas is heated up Cd) The slope of straight line represents the universal constant 26. A rectangular narrow Vtube has equal arm lengths and base length, each equal to I. The venical arms are filled with mercury up to l!2 and then one end is sealed. By heating the enclosed gas all the mercury is expelled. If atmospheric pressure is Po, the density of mercury is p and cross. sectional area is S, 1/2 then: • •• (a) Work done by the gas . agamst t h e atmosp henc. pressure IS. 51 PoS
1.
2
(b) Work done by the gas against the gravity is ~ Spg/2 (c) Work done by the gas against the atmospheric pressure is Post (d) Work done by the gas against the gravity is ~ Spg/2 27. The molar heat capacity for an ideal gas: (a) is lero for an adiabatic process
p
gas is initially at temperature T, pressure P and density p. The gas expands such that density changes to 1'/2 : (a) the pressure of the gas changes to .J2 P (b) the temperature of the gas changes to ..Pi T (c) the graph of the above process on the P.T diagram is parabola (d) the graph of the above process on the PT diagram is hyperbola . 29. Monoatomic, diatomic and triatomic gases whose initial volume and pressure are same, each is compressed till their pressure becomes tvvice the initial pressure. Then : (a) if the compression is isothermal, then their final volumes will be same (b) if the compression is adiabatic, then their final volumes will be different (c) if the compression is adiabatic, then triatomic gas will have maximum final volume (d) if the compression is adiabatic, then monoatomic gas will have maximum final volume 30. Two gases have the same initial pressure, volume and temperature. They expand to the same final volume, one adiabatically and the other isothermal. Then : (a) th~ final temperature is greater for the isothennal process (b) the final pressure is greater for the isothermal process (c) the work done by the gas is greater for the isothennal process (d) all the above options are incorrect 31. Some of the thermodynamic parameters are state variables while some are process variables. Some grouping of the parameters are given. Select the correct: (a) State variables : Temperature, No. of moles; Process variables: Internal energy, work done by the gas (b) State variables : Volume, Temperature; Process variables: Internal energy, work done by the gas (c) State ••.. ariables : work done by the gas, heat rejected by the gas; Process variables Temperature, volume (d) State variables: Internal energy, •... olume; Process variables: Work done by the gas, heat absorbed by the gas
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Study Anurag Mishra with www.puucho.com THERMODYNAMICS During an experiment, an ideal gas found to obey a condition vp2 = constant. The gas is initially at a temperature T, pressure P and volume V. The gas expands to volume 4V : (a) The pressure of gas changes to ~ 2 (b) The temperature of gas changes to 4T (c) The graph of above process on the P.T diagram is parabola (d) The graph of above process on the PoT diagram is hyperbola During an experiment, an ideal gas is found to obey a ".
p2
can d ltIon 
= constant [p = density of the gas]. The
p
33. An ideal gas expands in such a way that PV2 = constant throughout the process: (a) The graph of the process of TV diagram is a parabola (b) The graph of the process of TV diagram is a straight line (c) Such an expansion is possible only with heating (d) Such an expansion is possible only with cooling 34. A gas expands such that its initial and final temperature are equal. Also, the process followed by the gas traces a straight line on the PV diagram: (a) The temperature of the gas remains constant throughout (b) The temperature of the gas first increases and then deceases (c) The temperature of the gas first decreases and then increases (d) The straight line has a negative slope 35. Fig. shows the pressure P versus volume V graphs for two different gas sample at a given temperature. M A and M B are masses of two samples, nA and nB are numbers of moles. Which of the following must be incorrect. (a) MA >M8 (c)
nA >nB
(b)MA T
W
=.!:.:L Tyl
>0
(where i = degrees of freedom)
.6,W
= constt. l
51. [a b, c, d]
56. [a. c] C,
p( R:r
CoI'T'fd
+
1.06)( 105 N/m'l
RT
PV=RT:::>V=
P
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T
PV
4PV=(3VVI) V,
energy of
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Problem 3
Pevel
. ComRrehension Based Problems
Indicator diagram for 1 mole of a monoatomic ideal gas is shown in Fig. it consist of two isobar and two isotherm. The temperature of isotherms are 300 K and 600 K. The minimum volume of gas is given VI and maximum volume is 41), heat absorbed is process (1), (2), (3) and (4) are Q" Q2• Q3 and Q4 respectively and area of cycle is A. Now answer the following question.
=The Fig. shows an insulated cylinder of volume V containing monoatomic gas in both the compartments. The piston is diathermic. Initially the piston is kept fixed and the' system is allowed to acquire a state of thermal equilibrium. The initial pressures and temperatures are as shown in the Fig. Calculate:

,~
P....:Q,V' P,V, 4
2
.••••••••
1. The final temperature: (a) (Pt P2)TIT2
P DP.V43
v 1. Efficiency of cycle : A Ca) 
Q,
Ce)
A Q1 +Q2 +QJ +Q4
(b) (PI +P2)TIT2
~~~~ (c) (PI
~~~~ (d) (PI + P2)T1T2
+ P2)TIT2
~~+~~ (b)
A Q1 +Q2
Cd) Q, +Q2 +Q3 +Q4
Q,
2. The value of Q1 : Ca) 450R (b) 750R (e) 200R Cd) zero 3. Change in internal energy in process '3' : Ca) '450R Cb) 200R Ce) 200R Cd) 750R
~~~~
2. The final pressures of each companments (a) PI PI (PI +,P2)T2 'P _P_,_CP_,_+_P_,_lT_, 2 PIT2 +P2TI (PIT2 +P2TI)
,
Cb)P Pl(PI+P2)T2 PIT2 +P2TI
.,
P P2(PIP2)TI (P1T2 +P2TI)
(c) PI PI (PI +P2)T2 'P
PIT2 P2TI
d) P PI(PI +P2)T1
C
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,
P]T2 +P2T2
2
P2(P1 P2)TI (PIT2 +P2TI)
P P2(PI +P )T
, 'c PIT2 +P22TI)1
:
Study Anurag Mishra with www.puucho.com 261
THERMODYNAMiCS 3. The heat that flows from RHS to LHS, given '1"2> Ti• Ca) ~PIP2V (12 +T1) 4 PITz  PZT}
(b) ~PIP2V tTl TI} 4 PITz + P2Y}
(e) ~PIP2V {TlTll 3 PJTZ +PZT]
(d)~P1P2V tTzTll 4 PJTZ +P2TJ
1 V
1. The initial pressure of the gas is : (a) IN/m2 (c) 1.l0N/m
(b) l02N/m2. (d) L12N/m 2.
2.
2. Wo:k done by the gas in t = 5s is : Ca) 300 J Cb) 400 J (e) 114.3 J Cd) 153.6 J 3. Increase in temperature of gas in 5 s is : Ca) 6.9 x 1O3K
Two mole of a monoatomic ideal gas follows the process PV4/3 "" const. 1. which of the following statement is correct about this process? (a) It is an adiabatic process (b) It is a process in which t.Q =  ~ ~U 5
ee) It is a process in which dQ = dU (ahvays) Cd) It is a process in which dQ =  dU (always) 2. If this process is possible: (a) its molar heat capacity is negative (b) its molar heat capacity is positive (e) its molar heat capacity is zero Cd) molar heat capacity can't be defined until volume or pressure is constant 3. If initial temperature of system is To and it is compressed to '1 times its initiCll volume then work done by system is :
(c)
3RTo(l
(11)11,)
m'"
Cb)
6R1ol' 
Cd)
3RTo[I(~r3)
J
2 kmol of an ideal diatomic gas is enclosed in a vertical cylinder fitted with a piston and spring as shown in the Fig_ Initially, the spring is compressed by 5 em and then the electric heater starts supplying energy to the gas at constant rate of 100 J/s and due to conduction through walls of cylinder and radiation, 20 J/s has been lost to surroundings.
(c)
83
X
(d) 96
x
lO2.K
The Fig. shmvs tvo/Oidentical cylinders A and B which contain equal amounts of an ideal gas with adiabatic exponent g. The piston of cylinder A is frce and that of cylinder B is fixed. B
A
, Heat
, Heat
1. If the same amount of heat is absorbed by each cylinder, then in which cylinder the temperature rise is more? (a) Temperature of A will be more (bl Information is not sufficient to reach the conclusion (c) The temperature of both A and B will be same (d) Temperature of B will be more 2. If the temperature rise in cylinder A is To, then determine the temperature risc in cylinder B. T (a) .....Q. (b) To y (e) 11" Cd)Zfa 3. An ideal gas with adiabatic exponent y is heated at constant pressure. It absorbs Q amount of heat. Detennine the fractions of heat absorbed in raising the internal energy and performing the work : Ca) y + 2
Cbl'i
2y (c) y+1
Y
[Take k = 1000 N/m, g = 10m/s2, Atmospheric pressure, 2 Po = lOs N/m 2.. Crosssection area of piston Ao = 50cm • Mass of piston m = 1kg, R = 8.3kJ/molK) Based on above information, answer the following questions:
Cb) 6.9 K
104 K
Cd) "(2
Y
Fig. shows variation of internal energy U with the density p of 1 mole of ideal monoatomic gas for a thermodynamic cycle ABO\. Here process AB is a part of rectangular hyperbola.
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THERMODYNAMICS u
.:::KJ. ,
, ,, ,
1. When the lower piston is moved upwards a distance h!2, the compression : (a) leads to cooling (b) takes place isothermally (c) takes place adiabatically (d) leads to heating
p
1. The PV diagram of above cycle: p
p
(a)
(b)
v
v
p
(e)
Cd) none of these
c....c==;>
2. Which of the following statements is correct? (a) Heat given in compression is used to do work against elastic force only (b) Heat given in compression is used to do work against elastic force and atmospheric force (c) No work is done during compression (d) There is no change in internal energy 3. The value of h is : . , (a) 1 m
'V
(b) 1.4 m
(c) 1.6 m
(d) 2 m
2. The total amount of heat absorbed by the cycle is : (a)
C30!n2S2)Uo
(b)
Cs
O!nOA2)Uo
(e) sauo Cd) None of these 3. The work done in process AB is : (a) Va (b) 2Uo ee) SUo Cd) None of these 4. Efficiency of the cycle is : (a) 1Oln2.56 (b) 5In2.56 10102.5+9 5102.5+9 (e)
10102.53 5102.5+9
An horizontal insulated cylindrical vessel of length 21 is separated by a thin insulating piston into two equal pans as shown in Fig. Each part contains n moles of an ideal monoatomic gas at temperature T. The piston is connected to end faces of the cylinder with the help of two springs each of spring constant k. Initially both the springs are in relaxed state. The temperature of left part of cylinder is kept constant by some external means while heat is supplied to right pan with the help of filament. Based on above information, answer the following questions:
(dlNoneofthese
• M ideal gas at NTP is enclosed in an adiabatic vertical cylinder having an area of cross section A 27 em 2 between mo light movable pistons as shown in the Fig. Spring with force constant k = 3700N/m is in a relaxed state initially. Now the lower position is moved upwards a distance hl2, h being the initial length of gas column. It is observed that the upper piston moves up by a distance 'h / 16. Final temperature of gas = 413 x 273 K. Take y for: the gas to be 3/2. ._
k
=
k
o. T
o. T
1. If energy (heat) Q is supplied to right part so that the piston moves to the left by
i, then
the change in 4 internal energy of right part of cylinder is : (a) 15 kl2 + nRT 16 (b)
zr]
3nR[5kl' + 2 4nR 3
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Study Anurag Mishra with www.puucho.com THERMODYIIAMICS ee} ~k12 +2nRT 16 Cd) None of the above 2. The heat given by left part of cylinder to external arrangement is (for Q. 1) : (a) Q  kl2  nRT
kl2
(b) Q 

2nRT
(a) PV J
ee) Q_15 kl2 nRT
3. The temperature of the gas in right part in final equilibrium is (for information of Q. 1) : 2
+ ZT BnR 3
Skl
(b)
2
15k/
+ 51' 3
16nR
2
,
(e) 15kl + zr 16nR 3
3
A cylinder containing an ideal gas (see diagram) and closed by a movable piston is submerged in an icewater mixture. The piston is quickly pushed down from position (1) tD position (2) (process AB). The piston is held at position (2) until the gas is again at 0" C (process Be). Then the piston is slowly raised back to position (1) (process CA) :
000
o o o
00
o
0
:: 2 0 0 ~:.. 0 
0
o
0 0 00 0 0 0
,
:
lc f2>. ,,
P
B~ (a)
v, p
(e)
,, ,
Nc
V, p
A
:8
v,
v,
,, ,,
'
,, 'B
V, V
:
ANC
(b)
,, ,
(d)
' ' ',
,
V
i.I
1. The molar heat capacity of gas for this process is : (a) 2.5R (b) 2R (e) 4,5R (d) 3,SR 2. The amount of heat transferred between system and surrounding so that the internal energy of system increases by 300 J is : (a) 300 J (b) 240 J (e) 420 J (d) 540 J 3. If 2 moles of gas is taken in mentioned process, then the change in temperature of gas for Q. 12 is : [R o8.314J/molk] (a) 7.2 K
(b) 10K
(c) 72K
(d) 100 K
0
O'oooooe.o
1. Which of the following PV diagrams will correctly represent the processes AB, BC and CA and the cycle ABCA ? A'
!P" " " " "I ~i(j1 __ Vo.i_lJ_~ An ideal diatomic gas undergoes a process in which its internal energy relates to its volume as, U = kJV, where k is a positive constant. Based on given information, answer the following questions:
+ 51'
Cd) Ski 8nR
~o o o
3 (d) None of these
(c) PV J
Cd) None of the above
o
(b) 2PV J
2
16
(a)
2. If 100 gm of ice is melted during the cycle ABQ\, how much work is done on the gas? (a) S kcal (b) 5 kcal (e) 2.1 kJ (d) 4.2 kJ 3. If the change in the volume is (VI  Vl) = Vml, the work done (in N/m 2) during the cycle is, with P being the atmospheric pressure acting on the piston.
A container of volume 4Vo made of a perfectly nonconducting material is divided into two equal parts by a fixed rigid wall whose lower half is nonconducting and upper half is purely conducting. The right side of the wall is divided into equal parts (initially) by means of a massless nonconducting piston free to move as shown. Section A contains 2 mol of a gas while the section Band C contain 1 mol each of the same gas (y = 1.5) at pressure Po. The heater in left part is switched on till the final pressure in section C becomes.125/27 Po, Calculate:
V, V
s'
:
C~A
'
,
v,
v, v
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Study Anurag Mishra with www.puucho.com
~'j A
~
•
B
," \ "',.'1 .•.. {JI ~ •.P." Vol_tJ ~
V,
P,
,
P,
I
I'
An ideal gas undergoes a change in its state from 1 to 2 through three difference processes, A, Band C. as shown in Fig. P
P,
1. Final temperature
in part A :
(a) 12PoVo
v
(b) PaVo R d) 20SPoVo
13R ee) IOSPoVo 13R (27R 2. Final temperature in pan C : (a) PoVo
1. If the work done during the three processes are WA, WB and We, respectively, then: (a) WA =WB =We (b)WA >WB >Wc (e) WA =WC>WB
tfP" ,~,~" 1'1 ~ r.TI@ __ vv_;::;v~ 5 mole of helium are mixed with 2 mole of hydrogen to form a mixture. Take molar mass of helium M I 4g and that of hydrogen M:2 = 2g
=
9
"f) v~ V ~ __ vv_,.
4{>
ti'
,~ \~
1. The equivalent molar mass of the mixture is:
An ideal diatomic gas is expanded so that the amount of. heat transferred to the gas is equal to the decrease in its internal energy. 1. The molar specific heat of the gas in this process is given by C whose value is : (a) 5R (b) 3R
2
2
Ce) 2R
Cd) 5R 2 2. The process can be represented by the equation 711" = constant where the value of n is :
Wn=! (e)
5
n =~
(b)n=! 5
(d)
n =~
2 5 3. If in the above process, the initial temperature of the gas be To and the final volume be 32 times the initial volume, the work done by the gas during the process will be, in joules : (b) 5RT, (a) RTo 2 (d) RT, (e) 2RTo 2
,
>WB
2. If the heat supplied to the gas are QA' QB and Qe respectively, then : (a) QA =QB =Qc (b) QA < Q8 < Qe (e) QA > QB > Qe (d)QA =Qn >Qc
(b) PaVo
R 3R (e) SPaVo Cd) SPaVo 3R R 3. The heat supplied by the heater: Ca) 405 p ~ (b) 113 p ~ 800 500 315 368 (e) PaVo Cd) PaVo
9
(d)We>WA
(a) 6 g
Cb) 13g
(e) lBg
7 (d) none
7 2. The equivalent degree of freedom f of the mixture is: (a) 3.57 Cb)1.14 (c) 4.4 (d) none 3. The equivalent value of y is: (a) 1.59 Cb)1.53 (c) 1.56 (d) none 4. If the internal energy of He sample of 100J and that of the hydrogen sample is 200 J, then the internal energy of the mixture is: (a) 900 J (b) 128.5 J (e) 171.4 J (d) 300 J
'A monoatomic ideal gas of two moles is taken through a cyclic process starring from A as shown in Fig. The volume ratios are VB = 2 and VD = 4. If the temperature TA at A is VA VA 27:.J:" then _
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Study Anurag Mishra with www.puucho.com 265
THERMODYNAMICS tJ:!.. "" vv \~ " .•.
Cb) 600 K Cd) 3001n2K
,
2. Hear absorbed by the gas in the process A to B is : Ca) 600 R ee) 1500 R
./'f ., ~
A massless piston, which can move without friction, doses a sample of Helium in a vertical, thermally insulated cylinder, which is dosed ac its baHam, and the crosssection of which is A = 2 dm 2. Above the piston there is a fixed stand to which an unstretched spring of spring constant 2000N/rn is attached, whose bottom end is at a distance of 2 drn from the piston when the volume of the gas is Vo = 8dm 2. The external pressure is Po = lOs Pa, g = 10m/ S2. The gas confined in the cylinder is heated with an electric heating element. (1 dm = 101 m)
1. The temperature of the gas at point B is : (a) 1200 K Ce) 400 K
f'/
•.•.O~
4JF:t
Cb) 900 R Cd) 1200Rln2
3. The total work done by the gas during the complete cycle is :
d
VO
Cb)ISOOR (d) 900Rln2
Ca) 600 R Ce) 1200 R
The dot in Fig. represents the initial state of a gas. An adiabat divides the P.V diagram into regions 1 and 2 as shown. p
1. How much heat is supplied by the elemcnl till it reaches the spring: (a) zero (b) 600 J (e) 8000 J Cd) 1000 J 2. How will process look like on a PV diagmm :
(a)
P)'~B
,
B
v
(b)
Pl'~B
Cd)
v 1. For
which
corresponding positive:
of
the
following
processes,
the
heat supplied to the system Q is (e)
(a) the gas moves up along the adiabat (b) it moves down along the adiabat ee} it moves to anywhere in region 1 Cd) it moves to anywhere in region 2 2. As the gas moves down temperature:
v
.v
3. What is the ratio of work done by gas in moving the piston from initial position to A, and from there to B:
along the adiabatic,
(a) increases (b) decreases ee} remains constant Cd) variation depends on type of gas
P ./_A_ L
the
Cal 10
11 (e)~
3
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(b) ~
10 (d)~
5
Study Anurag Mishra with www.puucho.com 266
THEIlJIIOD'/NAMI 4. Choose the correct PT diagram for the given cyclic process:
t!"_ •.••.• ,","":.':{:, ~ •. t,;  ~ The Fig. shows p~vdiagram of a thermodynamic cycle. P 2P.
..~
(a)
c
PIl?". a
p. 0
v.
3Vo
temperature
a
during
(a)
the cycle
(d) aU the above 3. Identify the diagram which correctly represents the heat inflow and outflow of the system:
' t:] a,
A
0
a. 0',,,;';;;,. v P
t:l
a, (b)
a,
A
0
P
04
AOa,
a, (e)
D
04
P
0
,
Cd) none of these
~c aL~=.T
~"
," ," tl ",:~/;' ~ ..... _•.••.• •... ~ Two cylinder A and B having piston connected by massless rod (as shown in Fig.). The crosssectional area of two cylinders are same and equal to •S' . The cylinder A contains m gm of an ideal gas at pressure Po and temperature To. The cylinder B contain identical gas at same temperature To but has different mass. The piston is held at the state in the position so that volume of gas in cylinder A and cylinder B are same and is equal to Vo'
'H:'::.1.....i:;~
i( :::: :.'.
B
ifhe walls and piston of cylinder A are thermally insulated, whereas cylinder B is maintained at temperature To reservoir. The whole system is in vacuum. Now the piston is slowly released and it moves towards left and mechanical equilibrium is reached at the state when the volume of gas
1. The mass of gas in cylinder B :
a,
04
A\\
in cylinder A becomes Vo . Then (here y for gas:::: 1.5). 2
a, o
(e) v
A
C
W
(b)
J=_r
B
V
Cd)
VI,~lt
V
a, o
T
5. Choose the correct V.T diagram for the given cyclic process:
(b) To = 3To ee) Ts = 2To
a, c
Cd) none of these
(e) P ~c
(a) 2PoVo (b) 3PoVo (e) PoVo Cd) 6PoVo 2. IfTA,TB,Tc and To are the respective temperature at A, B,C and D. Then, choose the correct statement if TA = To :
(a)
T
V
1. The work done by the cycle is :
P
JC_
T
o
"A:
(a) The maximum occurs at C.
PI;lt'"
(b)
(a)
C
(c) V
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2./2 m J2 m
(b) 3./2 m (d) none of these
•
Study Anurag Mishra with www.puucho.com
_. 267
THERMODYNAMICS 2. The change in internal energy of gas in cylinder A : (a) (e)
(12 l)PoVo PoVo
(b) 2(../2 l)PoVo
Cd) none of these
(a)
(,/21)
a
3. If work done by gas in cylinder B is WB and work done by gas in cylinder A is WA' then : (a) WA =WE
(b) IW,i>iW,1 (e) IW,I< IWBI
(e)
Cd) we can't say anything
4. What will be the compressive force in connecting rod at equilibrium: (b) ,/ZPS (a) PS 32 Cd) none of these ee) 2 pS
The Fig. represents a cyclic process in which the process CA is adiabatic
1>7". T
VIp"
aLT
A sample of ideal gas is taken through the cyclic process shown in the Fig. The temperature of the gas at state A is T A = 200 K. At states Band C the temperature of the gas is the same.
P
P
aL_'v 3V,
1. Select the correct statement(s) : (a) Positive work is done during the isobaric process (b) Negative work is done during the adiabatic process
ee) The system cools dovl'D. during the process Be Cd) The system reject heat during the process CA 2. If TA,TB and Tc are the temperatures at A,B and C respectively, then identify the correct statement : (a) TA ;;;:r:=:
VA
VrJ2 Vo 3V"''''' ••••••••
nRT = 3PA/VA
'
V2 + 12PA V
nRT dT = _ 3P (2V) + 12P A dV VA
Passage.20 1. [a, b, e] (a) Since volume is increasing (b) Since volume is decreasing (c) Since PV is deceasing
dTjdV=O
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,
... (1)
Study Anurag Mishra with www.puucho.com !HERM DYNAMICS (a) Isothennal process, 6,V = 0, W = +ve ~ 8Q = +ve Since volume is increasing in both processes (b) A is adiabatic with ve work done => tlU = +ve B is isothennal (c) Same as a (d) Same as b 15. (a) A to C along circle net work done is +ve and tlV is ve But (tlW  tlV) > 0 SO, 6Q = tlW  tlU
14.
By equation (1) 3PA nRT = (2VA)
2
VA
= l2PA VA
+ 12PA(2VA)
+ 24PA VA
nRT=l2PAVA nRT = 12nRTA
.'
>
T = 12TA
=12x200
tlQ>O (b) B to C along circle there is compression so work done is(ve) and tlU is (ve) So tlQ is ve
= 2400K
(c) Same as B to C along circle Matching
Type Problem Assertion and Reason
8.
dQ = dU + dW
(a)
Here dQ = + SOOx 4184 J = 2092 1000 dW = +400J, since the work is done by the system Hence, dU = dQ  dW = 20924000= 1700J (b)
~U=~Q~W = (300)(4.184)  (420) = 1680J
(e)
~U= (~Q ~W) = (1200)(4.184)  0 = 5000J
[Note that dQ is positive when heat is added to the
system and .:i,W is positive when the system does the work} Wittl + Wif = Riaf
12.
20+uij = 50 uif = 30 and
Wibj +Uif
Wibj
Wibf = Wibf = 6
(a)
13.
= Hib/ + 30 = 36
6
(b)
Hji=Wfi+Ufi =13+30=43
(e)
Vif =30
Cd)
Hib"",Wib+Uib
PV1
U;=10
Vf =40
= 6+(2210) = 18 = constant, V(yI)T = constant, P(lY)T"f
p~
7. It is true that for real gases, and for the real procedures in the laboratory, an adiabatic process is a sudden and large change of the system obtained in a small time. However, for an ideal gas, and for an ideal procedure we assume for such a gas, an ideal adiabatic process is considered as quasi.static. in which the system passes through a continuous succession of equilibrium states, which we can plot on a PV diagram. Such a process is also assumed to be reversible. though of course. a quasistatic process may not be reversible. Such ideal conditions as we assume may not be obtained in particle but almost all real systems deviate little from the results of our ideal procedures, which may therefore considered as approximations to real processes. 8. Work is done only expansions or compressions of the gas where volume changes occur. In a quasistatic condition, a small change of volume dV under pressure P involves the work dW = p.dV. In a succession of quasistatic equilibrium states, total work done can be obtained by integration. 9. Any given thennodynamic state has a definite temperature T. The internal energy of a system in an ideal gas is a function of only the absolute temperature and is independent of its pressure and volume. The internal energy per mole of a gas is directly proportional to its absolute temperature.
= constant
_1_.
P
oc
oc
(L E. )yi{yI)
T1'/(yll.
V
WI > W3 (c) WI > Wz > W3
(b) W2 > W3 > WI
Cd) WI > W3 > W2 20. An ideal gas is initially at temperatureT and volume V.
Its volume is increased by.1.V due to an increase in temperature ~T, pressure remaining constant. The , u, = ~v vanes , Wit' h temperature as : quannry V~T
(liT 1999) (a) 4RT
(b) 15RT
(e) 9RT
(d) llRT
15. The ratio of the speed of sound in nitrogen gas to that in helium gas, at 300 Kis: (lIT 1999) (a) AlI7) (b) J(l/7J (e) (,[3)/5 [d) (.J6)/5
16. Two monoawmic ideal gases 1 and 2 of molecular masses M1 and M z respectively are enclosed in separate containers kept at the same temperature. The ratio of the speed of sound in gas 1 to that in gas 2 is given by: (lIT 2000) IMj (b) 1M2 (e) Mj (d) M2 iM, \Mj M2 M[ 17. A monoatomic ideal gas, initially at temperature T1, is enclosed in a cylinder fitted with a frictionless pistion. The gas is allowed to expand adiabatically to a temperature Tz by releasing the piston suddenly. If L} and Lz are the lengths of the gas column before and after expansion respectively, then T11Tz is given by: (a)
21. The plots of intensity versus wavelength for three black bodies at temperature TI,Tz and T3 respectively are as shown. Their temperature are such that:
y
T,
) fT2 >T3
(b) T1 >T3 >T2
ee) T2 >T3 >T}
(d)T3
>T2 >1i.
22. Three rods made of same ~900C material and having the same crosssection have been joined DOC as shown in the figure. Each rod is of the same length. The left 90°C and right ends are kept at DOC and 90"C respectively. The temperature of the junction aCthe three rods will be: (liT 2001) (a) 45' (b) 60'C Ce)30'C Cd)20'C 23. In a given process on an ideal gas dW = 0 and dQ < 0. Thenfor the gas: (lIT 2001) (a) the temperature will decrease (b) the volume will increase (e) the pressure will remain constam Cd) the temperature will increase 24. PV plots for two gases during adiabatic processes are shown in the figure. Plots 1 and 2 should correspond respectively to :
(liT 2001)
p
v (a) Heand02 (e) He and Ar
(b) 02 andHe Cd)02 and N2
25. Which of the following graphs correctly represents the "" fA dV;dP Wit "h p lor , vanatlOn 0 I'= an I"dea I gas at V
constant temperature?
ca)P~
(lIT 2002)
(a)
(b)P~ p
27. An ideal blackbody at room temperature is thrown into a furnace. It is observed that: (lIT 2002) (a) initially it is the darkest body and at later times the brightest (b) it is the darkest body at all times (c) it cannot be distinguished at all times Cd) initially it is the darkest body and at later times it cannot be distinguished 28. The graph, shown in the adjacent diagram, represents the x ~ variation of temperature (T) of two bodies, x and y having same T surface area, with time (t) due to the emission of radiation. Find the correct relation between the emissivity and absorptivity powers of the two bodies: (lIT 2003J (a) Ex >Ey and ax Ey and ax >ay(d)Ex
4
=
or
Only One Alternative Is COrrect
=KA("T)
'"
I
A(T)
y for all gases
T
is same.
(b) The average translational molecule
kinetic energy per
is ~ kT. 2
(c) When pressure increases, mean free path of molecule
90"
decreases.
B
;f2n
Cd) Average K.E. does not depend on the gas. Hence each component in a gaseous mixture has same average translational kinetic energy. So (d) is correct. 4 5. (d) According to Stefan's law, 6Q = eaAT M Also, 6Q
= ms 6T
.'. ms 6T
c (Td
Since B is at higher temperature than A, heat flows from B to A, A to C and then C to B, for steady state.
= eaAT4 M
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Study Anurag Mishra with www.puucho.com ~Q =KA(rrc)
FarAe,
at
ForCB,
.J'2n.
or
rzr)
dQ = KA (Tc M a
l
or
Equate the two equations for steady state.
... KA(Th:C)=KA(TC/Zf) or TTc =.J2TcZf or
. .,
Tc = T
'.'
3 (.J2+I)
or
/3kT, ~ \ m
V2 =
=
vI
{T; XVI
vIi
T, T1
1 I', •
= 1 _ (127 + 273) = 1_ 400 = ~
(227 + 273)
fG
o..m)STs
(e)
z
2
P
T
P,
Zf
=
and does not depend
or
= SRT+ 6RT = llRT (e)
Velocity of sound in a gas
for v =
Jr
RT
= ~YRT M
dV
flt
v=
(':densityxvolume=
VN2 =
YN2 MHe x __
VHe
"tHe
M)
MN2
YN2 = 7/5, "tHe = 5/3
where
7/5 4 J3 = x=_. 5/3 28 5
or Pz =2P
Energy radiated •.
..
VA = 4x3xRT =6RT 2 Total internal energy = U 0 + UA
average translational viz 0.048 eV.
C",'V, nand R are same)
 SRT
Again, for 4 mole of argon,
15.
SlO
11. (d) According to Stefan's law,
:. W = 2000J.
2
=350=0.69
(A'm h 3 Average K. E. (K) =  kT 2 It depends on temperature on molar mass. o'. For both the gases, kinetic energy will be same
=P T
104 = 2000J
=
= (Am)NSTNS
10. (e) pv= nRT IJ. T1
X
where n = number of mole of the gas f number of degree of freedom .. For 2 mole of oxygen, U0 = 2x5xRT
= ~600 x484 300
TNS
9.
=.!5
5
2
vT;
Ts =o.'m)NS
..
W = llClJ.
500
nfRT Internal energy = __
14. (d)
= J2 x 484 = lA14x 484 = 684mjs 8. (bl According to Wien's displacement law, AmY = constant (b). Let 5 = Sun,NS = North star, ..
=450X(~)x(\6)=450X4
~=I
..
=
Ii
r}
= ~ x (8.3) x (l00) = 1.25 x 103 J 2
](2 =12.42xlOZ1J Vrrru
l
= laDOW .
P2
K1 _ T2 K1 _ x 600 ('21 = 1i or K2 _  x ~ x 10'1) K2 T2 1i 300
or
=('1/2)' x(Z!j)4
dU=CvdT=(3:JdT
13. (al
7. (d) Average K. E. (K) = ~ kT 2
P2
or
12. (d)
3T=TdJ2+1)or
P, 450
16. (b) Velocity of sound in a gas v :,.
flt
E = eAoy4(.6.t)
or
.!.=ex(41tr2)xor4
v = ~YRT
dv
at
Power (P) =
r
(4necr)r2 4
r r
i {~ x[~~
or
..
v = jrRT
'M
.:'L = tl2
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whered =densityofgas,
JMM}2 as y, R, T are constant
(","dV .e M)
factors.
Study Anurag Mishra with www.puucho.com THE 17. (d) For an adiabatic
process, IVy1 = constant
=AL2 where A denotes area of crosssection cylinder.
VI =~,V2
T1 (V,)3
.•
_
or
~1 .
,as y = 5 for monoatomtC. gas. 3
= _
Tz
VI
.!L = (L, Tz
L}
of the gas
O'C
O"C
B
2Q
A
)2'3
Q AB=BD=BE=/
18. (a) The change of ice at _lOoe into steam at lOOC occurs in four stages: It is represented by curve (a).
I stage
The temperature lOoe roO°C.
of ice changes
from
Heat will flow by conduction along the rods from 90° e ends to junction and from junction to 0° C end.
IV
lOO"C
Steam
Along rod DB, Q =
KA(90e)
... (1)
l
AlongEB,Q = KA(9~e) :.2Q=KA(80)
AlongBA,heat=2Q or
Ice
II stage Ice at
aoc changes in water at Doe.
The state changes as heat is supplied. III stage Water at DOGchange into water at
100°C. IV stage Water at 100°C changes
19.
(a)
into steam at
100°C. The work done is equal to area under the curve and volume axis
00
the PV diagram.
Obviously W2 > WI > W3. 20. (e) For a given mass of gas at constant V  = constant T or VT+V6T=VT+T6V 1 6V 1 _= __ or =0 T V6T T The equation represents or
ofthefonnxy
pressure,
V+.6.V V =T +.1.T T or V6T:T6V .. 
I
from eq. (1) and (3) eliminate 2KA(90e) KA(e)
I
II
... (2)
Q.
1
or 2(908) = e or3e = 180 or e = 60°C According to first law of thermodynamics, 23. (a) dQ=dU+dW or dQ=dU+O (':dW=o, given) or
dQ=dU
since dQ < 0 as per the question, :. dQ is negative :. dU is also negative. Thus internal energy decreases. :. Internal energy decreases when the temperature decreases Hence the temperature will decrease. 24. (b) For adiabatic
process, pv"Y = constant
p
or oT=l a rectangular
hyperbola
= ,2.
Hence the &T variation is represented (e). 21. (b) According to Wien's displacement law, AT = constant. From!'A graph, Al < A3 < A2' Hence T1 > T3 > T2. 22. (b) Let the temperature of junction be 8°e.
...(3)
by graph V V
For He gas, '{ = 1.67 Helium is a monoatomic For oxygen gas, '{ = 1.4
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Study Anurag Mishra with www.puucho.com PROBIIMS
296,
Oxygen is diatomic gas For a given value of v, 11 > P2' as is clear from graph. Since PV'Y = constant.
:. 11 V'YI
or
= P2VY2
1.. = V'Y2 P2
25.
(a)
= (V)Y2Yl
V'YI
since 11 > P2 :. 12 > 11 since YHe > 10 2 ,', Curve 2 represents helium Cmonoatomic) and Curve 1 represents oxygen (diatomic) ,", Plot 1 is for oxygen, Plot 2 is for helium. At a constant temperaNre, for a given mass of gas, PV constant, according to Boyle's law. ,', PV = constant :. Pdv + VdP = 0 dV V dV 1 or = or  __ =+_ dPP VdPP 1 or ~= or /3P=l.
=
P
26.
(a)
The equation between /3and P is of the form xy = constant which represents a rectangular hyperbola. ,', Graph between p and P will be a rectangular hyperbola represented by graph (a). For cyclic process ABCA, Q=WAB+WBC+WD\. WAH = P x.1.V lOx 1) 5 = 10+ 0+ W~
(,;
=
27.
(a)
28.
(e)
or Wo. = 510 or WQ\ = SJ. According [0 Kirchhoffs law, good absorbers are good emitters and bad reflectors. While at lower temperature, a blackbody absorbs all the incident radiations. Ie does not reflect any radiation incident upon it when it is thrown in the surface. Initially it is therefore the darkest body. At later times, the black body attains the temperature of the hot furnace and so it radiates maximum energy. Ie becomes the brightest of all. Option (a) represents the answer. The two bodies x and y have same surface area. Emissivity (E)oc rate of cooling T
From graph, consider cooling of x and y from temperature T2 to TI. Time taken by x = (t2  (1) Time taken by y = (£3 tl) x takes less time and soEx > Ey ...(1) According to Kirchhoff s law, a good emitter is also a good absorber. ax > ay
..
° =~, L x liT ,,, ,, ,:' TI .. :, , ,,
or
Eoc(~~)"
y
= change
in length.
..
M =aLliT Ma =Oa1]dT and M$ =0$12liT Ma = M$ or 0a1IliT = 0sl2liT
°
12 =~ 'I
or
"
whereM
..
or
"
...(2)
29. (b) Heat lost by water at 20"C = msw liT H = 5 x 1 x 20 = lOOkcal At first, ice at (20"C)will take heat to change into ice at DOC. H = msiceliT = (2kg) x O.5 x 20 = 20 kcal :. After this, heat available = (100  20) = SOkcal This heat will now be gained by ice at O"C to melt into water at 0" C. Let m kg of ice melt. .'. m x SO = SO :. m = lkg. Out of 2 kg of ice, 1 kg of ice melts into water and 1 kg of ice remains unmelted in container. :. Amount of water in container = 5 + 1 = 6kg. 30. (b) An inspection of the P.T graph reveals that (1) AC is an adiabatic process (given) (2) AB is an isothermal process at T constant. (3) BC is an isobaric process at P constant. (i) In all the four options, BC is an isobaric process. (ii) AB is an isothermal process. :. PV = constant :. Graph between P and V is a rectangular hyperbola of the form .ry = constant. Graphs (b) and (d) satisfy this condition. Options (a) and (c) are therefore incorrect. (iii) AC is an adiabatic process (given). In PV diagram, slope of AC > slope of AB because slope of adiabatic curve > slope of isothermal curve. This condition is certainly not satisfied in option (d) where AC is a straight line at constant V. Here AC is an isochoric process. Ie is not an adiabatic process. (d) is incorrect. Hence the ultimate correct process is (b). 31. (e) By coefficient of linear expansion,
05
°
12 =>+1=~+1 11
12+1] 0a+os _=___
0$
11 .. __
II 05 11+12 32. (b) According to Wien's displacement law, AT = b = Wien's constant.
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0$_
ll.a+o$
.•
Study Anurag Mishra with www.puucho.com
'
THERMODYNAMICS
297,' ..
Further I WHel >1 WAn I Net work done = WAn + (WBC) W = A negative value
b
AATA = b or TA = 
3 X 107
b =
TA
X
7
10 3
wherez=(bx107)
Z
=_ 7
=
4
_bXI07
T c
4
Z
S
5
Again, According
QA = (7tR~) xa x (TA)4 QA = 1[(2x 102)2 xcr x
QA = (rrO"x10
4
K
xz
4
(~r
)x
..
q2 = KA x 100 x I , ~ = .!. q} 21 2KA x 100 ql 4
81
35. (el (i) Oxygen supplied is in liquid state. When heat is supplied at constant rate, at first liquid oxygen at 50 K will be raised to its boiling temperature, say OK, .. Q = ms(8  50) = msTl, where TI = (8  50) K
= 81
= l5.. ::::O.062k
(4)4
(5)4
16
or
= 36k =0.057k 625
~time
r
(t)
,, 1 .• "
'" '
•j
,
• 1
..I
1 j
• •
•
as heat is supplied at constant rate.
r
(constant)xt Hence (T. r) graph will be a straight line in this region. It is inclined to time axis. (ii) At boiling poim, the heat supplied is used in converting liquid oxygen into gaseous state, keeping temperature at e K. •
33. (bl The slope of adiabatic process, at a given state (P, V, T) is greater than the slope of isothermal process. Hence AB is isothermal and Be is adiabatic. Work done in a process is given by area under the curve and Vaxis. p
msTI =
rr t)
c
Ql=mL
graph vvill be a straight line parallel to taxis as heat is supplied at constant rate. (iii) When the vaporisation is complete, the gas at e will now be raised to 300 K.
.
1
B
,, ,, ,,
VI
msT} =
Q
g = constant,
Hence QB is maximum.
p~•.••.
=
xK K 2K 2 (2K)A(lOO  0) 2KA x 100 q} "" ~~ = I I _(KJAOOOO) _ KAxlOO q2 2 I 21
4
Put nO"x 104 x z4 = k = constant 4k or QA = o.049k
Similarly, QB = k x (4)2
=K
s
2' 34
QA =1(OX104xz4x
Qc = kx(6)'
WAB)
[n series combination, Ks
where A = area of disc = 1tR2.
or
WBC >
Kp =K+K=2K to Stefan's law,
Q ::::Power radiated by black body = AcrT4
or
"
(.:
From the graph itself, P3 > 11 :. W < Oand P3 > 11 Hence option (b) represents correct ans\ver. 34. (d) In parallel combination of rods, resultant conductivity K p = K} + K 2
an d
Z
..\• ..
W A2 :. W1 > W2 Option (a) is correct. (b) To study V.T diagram. In the given process, AB is a straight line. If has a negative slope and a positive intercept. ... (1) The equation ofline is P = aV + p Where a and P are positive constants . ... (2)
PV=nRT PnRTp" = . V
..
..
ut
Equation, ~T
(C P )mixture = (C v )mixture + R
=2R+R=3R _(Cp)mixture_3R_15
.
Ymixture 
 
(C V )miXlure
2R

.
or
liQ=liU+6W 0= t.U +.dW tJ.T =
(:.lJ.U=nCv!1T) process, tlQ = 0
.1U = nCvdT tlQ = nCp!l.T :. 6U nCvdT Gv 6Q= nCpdT=Cp
1
5
='1=7
8. (b) According to Maxwell's distribution
gas, the average speed of molecules
.. v
of Ted
v=
9. (a, b, d) (a)
P
p A
P
........ A B
v Givcn prucess WI (i)
dT = 2nV dV nR
When
2
or P P~
lsothennal process W~ (ii)
= Q.nRT ... (4)
dT =0, dV
d2r
nR
+1
... (5)
oR
V=1
... (6)
2n
2a
2u
=+0=dV2 oR oR 2 . . d T..
Th e secon d d envatlve 
dV2
IS
h negatlve. It means T as
some maximum value.
!.is the value of maxima
of temperature. 2n Also PAVA = PBVB ,RTA = RTn orTA = TB In going from A to B, the temperature of the gas first increases to maximum at V '" 13/2a Then the temperantre decreases and resorted to original value. Hence option (d) is correct. V =
P/2 ........
P/2
... (3)
oR
SRT rrM
ft, for a definite
p
nR
anR allR This represents a parabola in terms ofT and P. :. The path AB becomes a part of parabola. :. The option (c) is not correct as AB does not a part of hyperbola. (d) Variation ofT along AB 2 From (3). T = aV + pV
of
gas. Since temperature of A andC vessels are same, the average speed of oxygen (°2) molecules will be equal in A and C. Average speed of O2 molecules in vessel C=l'l' 00:
+'W
,JR
_p2 pp T=+
0
Hence all the four options are correct. 7. (d)
_aV
P
6U=O
Cd) In an adiabatic
2
aV +13
=
,JRT P = a+~
or IM~=I~WI (e) In an isothermal process, ..
=
In eq. ()1
This represents a parabola in terms ofT and V. :. The path AB becomes a part of parabola. Option (b) is correct. (c) To study P.T diagram For P.T diagram. eliminate V between (1) and (2)
6. (a. b, c, d) (a) !lU = nCvDT = nCv(Tz TI) (b) In an adiabatic process t1Q = a ..
T
or
It
10. (a, b)
(a) To calculate temperature of B
Work done = Area under curve and V.axis.
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3011i
Power of bpdy A = eAaT~x (area) Power of body B = eBaT 84 x (area)
Bulk modulus = (P.6V) = P ~V :. Isothermal Bulk modulus.= P. 14. (d) The piswn of A is free to move. Volume of gas in cylinder A will change hut pressure •••• ;11 remainconstant. Heat will be gained by the gas at constant pressure.
The cwo powers are equal.
..
eBaT:
x (area) =
eAaTJ
ri = (:: )rJ
or
x (area)
..
or
ri = (0.01) x (5802)4 0.B1
ri =
or
:. (~Q)A = nCp(~TlA ...(1) The piston of Bis held fixed. Heat will be gained by the gas at constant volume.
(~r
x (5802)4
or
Ta = 58 °2 3
:. (~Q). = nCy ( (Cp,C1')monoatomic
22. (b, d) (a) process is not isothermal (b) volume decreases and temperature decreases ilU = negative, 6W = negative 6Q = negative (c) work done in process A 7 B 7 C is positive (d) cycle is clockwise, so work done by the gas is positive.
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302
=:I True/False
M/2
/3RT M
:,vrms
=~3RT M
v'= 2v.
1. False According to kinetic theory of gases, vrrm '= \
V,=~3RXzr
For atomic oxygen O.
7. False
depends on molar mass.
It will be different for different ideal gases. 2. False
v
v,
v,
T 1. C
v
PV
For a gas, 
T
= constant
For a given temperature T1,
11. VI
= Pz V2 1\ V = 2 Pz \.'t
From graph, Vi. > V2 3. False Vrms
=
\/3~T. It
means that
Vrms
depends on M.
Hence different gases have different Vrms' 4. True The slope of PV graph is more for adiabatic process than for isothermal process. From the graph it is clear that slope for curve B is greater than the slope for curve A. Hence the isothermal process is represented by the curve A. 5. True Cp > Cv.
When volume is kept constant, the heat supplied is used only for one purpose of increasing temperature of gas. When pressure is kept constant, the heat is required for two purpose viz for increasing temperature and for adding work against constant pressure. Hence Cp > Cv.
6. False vnm
=
J3~T
= nlCV1 +n2CV2
=
{W)+I(~)
=2R.
nl +n2
1+1
2. A to B represents a change which occurs at constant rem perature when heat is added to the system. From 0 to A = Temperature increases at solid state. A = It represents solid state. AB = It represents change of state from solid to liquid when heat is added. Since P lies on AB, the material is partly solid and partly liquid. At B, the state is liquid. 3. GiVen: VP2 = constant ...(1) According to gas equation, PV = RT ...(2) Eliminate P from eq. (1) and (2) .. m
RT)2
V( V
(T2)2
l1i
= constant => =
(V2 )
=>
l \.l
T2 = T ~
=
T2 V = constant T2 = T,
rv;
vV;
Fzr :. T2 = Fzr.
4. Ice at 0° C gains heat first melt into water at DOC and then this water gains heat for rise of its temperature if heat energy is available. Heat required ice for melting = mL ..
mL=100x80x4.2J=3360DJ
...(1)
Heat lost by water at 25°C to fall to DOC = m.s6.t ..
m.s6.T=3DDx4.2x2S=315DDJ
...(2)
Water does not provide adequate energy and so the whole of ice does not melt, as (2) is less than (1). :. Final temperature will be DOC. 5. According to Stefan's law, sun at temperature T radiates energy.
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303
7. Let inner radius of shell is a and outer radius is b. If thickness of shell isc,a
..
.. ..
Equate (1) and (2) or
r4
= 1400xr
_:.,
b
2
.. , . '
, ,
.._.
+:.
2
,
b
'
=R
...
I'
'.
Energy radiated per unit time per unit area = cry4 .. Energy = (crT4)x (area 47tR2) ... (1) time .. Energy received on earth = 1400 x 41'1:r2 ... (2)
=R
.i
.'
,
oR' or
r4
=
1400 x (1.5 x lOll )2
CS.67 X 108)(7 x 108)2 24 or . T4 = 14x2.25xl0 or T=5803K. 5.67x49xl08 6. The solid copper sphere is losing heat by radiation when its temperature falls from 200 K to 100 K. According
... (1)
3 aAT4 = _ 4Jtr ps dT 3 dt
... (2)
3 or dt == _ 41tr ps dT 3Acr T4
R2_~",R2 4
dt = _ 4n:r ps
dT 3(41t£2)0" T4
I r dt
1
= _ prs 100 3cr 200
o
t =
.E! = prs I zoo
dT 3cr 100 T4
r4
3T 3 100
120" 7prs x 106
= 72 t
= prs x
9cr c = 7ps x 106 =
7'21J c
x (5.67 x 108)
(_I100 )'[1_.1]8 6
.7prs x 10
72x(S.67xlO8)
H = dQ = L\ T [in steady state] dt RTh 2 ~ t = 4rtKR T P
=
8. Let L latent heat of fusion of substance Heat energy lost in solidification = ML .. (Power P) x (time t) = ML L = Pc M
..
prs [ __ 1_] 200 = 7ps x 106 3cr
t 2; 4nKR 2 p = 4nKR T
RTh =
9. The whole system is thermally isolated from the surroundings. Hence the expansion is adiabatic: obviously dQ = O.The other part of the container is vacuum. Work done (dW) the gas is zero.
3
or
R»c,
t
From (1) and (2),
or
, If
to Stefan's law,
energy radiated per second = aAy4 Heat lost = msdr. The negative sign means loss, or Heat lost =: (volume x density)sdT Heat lost 4 2 dT OC =TCT pstime 3 dt
OC
1
=4,K
dQ=dW+dU
dU=zero
Temperature remains constant. The temperature will now be 300 K. 10, When expansion is isothermal P PV=~ x2V or ~ =
... (1)
2
When expansion is adiabatic; PVY = Pa
= l.71prs
X
(2V)Y
or
Pa
p =
2'
P P 2 2 2 a =x = = = 0.628 ~ 2Y P ZY 21.67
= 1.7lprssec.
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... (2)
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," .
PREVIOUS YEAR PROBLEMS
•
t
..
=1400xO.2xt Q =280,
0.628.
, crosssectIOns.
•
t"
.
""
r t ~
~ => => =>
(lOO°CT) 1
=
(T  O°C)
H=K2A
forradB, ,
For rod A, H = KIA I
KIA (lOO"CT) = K A.!:. 2 I I 300(lOO"C  T) = 200T 300°C3T=2T => ST=300"C T = 60"C
12. The movable stopper S moves irside the ring till the pressure on both sides are equal.
:.
or t = 5.Smin .
remains
constant
When pressure decreases, T also decreases. .ection .. Force required to maintain the second rod = .. YxA x 2x 105 x100= 2.21 x 108 A
P2
SOO P2 We get = _~_
0
Increase in length of copper rod 0.051 ern L12S = L25 (l + P x 100) 70(1 + 100/3) where J3 is the coefficient of linear expansion of the second rod. Increase in length of second rod "" 7000~ Total increase in length = 0.051 + 7000p = 0.191 em 0.191 = 0.051+ 7000~ 7000~ = 0140
K be 100 .
Applying the formula Ii = P2 ,
=
=
(T ~)
V,
B
,C ':
V
V,
B + C. BC is parallel to Paxis. Volume remains constant at V2• temperature decreases from T2 to 1i and pressure decreases from Ii .to P2. C + A. CA represents an" isothermal process. At constant temperature 1i.. volume decreases from V2 to Vi while pressure increases from P2 to 11.
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307 7. The kinetic energy of bullet, when stopped by an obstacle, is converted into heat. 75% of the heat generated is retained by the bullet so that it is heated till it melts down.
.. 75 [..!:.mpZ]=
ms~e+mL
100 2
or
v2 =~(S6.8+L)
or
vZ =~[(O.03x4.2x300)+(6x4.2)]
or
v2:::; ~ (37.8+2S.2]
3
3 3
= 8x63 3 1
= 168
v = 12 .96 m sec8. The compression of one mole of a perfect gas is an adiabatic process. or
...(1)
..
Cp Cv =R Cp :::;R +Cv whereCv 3R
SR
2
2
Cp =R+=
or
5
... (2)
ory= =3R,'2 3
From (2) and (1), Pz =
2 ...
105(~r'3
= 6.24xl0sNm
PVI = constant K
(3)
:. P = 1(1,,1
Ii;'
=: 5.3xlO23g
..
Mass of one molecule of oxygen m =: 5.3 x 1026kg
..
1)'
2
or
j(\1,y \!,
11 \!,1
W =
or
1)' 1)' 3 W = (6.26xlOS)x(2xlO3)(lOS)(6xIO )
2
.
1
1
or
W = P2 V2
or
2
5.3 x 1023g ...(1)
mu  (mu) =: 2mu
.. Change in momentum in n collisions ~ 2mnu .. Change in momentum per sec per m 2 ~ 2mnu Change .. __ ~ in momentum ~ 2m nu or pressure Time x area
or
1
K,rY V 
=:
Change in momentum per collision ~
u ~
[Vr+l _ \1.y+l]
W =~
ms(de) dt
where _de '" Rare of fall of temperature. dr de IdQ de I :. _"' __ or _
2x SOD = lOOOm/s 1
= 
or or (b) ..
.'. \13RT M
or
or
0.149
To  400 = 20 or To = 420 K.
42. When pressure is kept constant, T1 = T2 VI
arK.E. =~x(1.38xl023)(l60)J
2
K.E.= 3.312 x 1021 J Total mass of helium in the box.
PV = nRT
m= PVM
Volume = Area x height = Ah =~ or lIZ = T2h1 Ah1 Ah2 T1 or h _400xl_~m 2300 3 The process is adiabatic. When the gas is compressed without exchange of heat,
(100)(1)(4)
3
Vpg = V'p"g
where V,V' = volume of cubical body immersed p,p' = densities of liquid at tvo/Otemperatures
or m=o.3g
25 x160 10 3 40. Decrease in kinetic energy
(Ah)p = IA(I + 2n,"T)h](
= Increase in internal energy
or
~mv6=nCv~T
or
liT = Muo
or
~mv6=(:)(3:)~T
2
3R
41. (a) The rate of heat loss per unit area due to radiation from the lid is given by Stefan's law. .. E = eo(Tl4  T04) = (Q6)e:
x 108jr(400l4  (300)4]
8 x 108[2563 (b) Temperarure of oil: To:
= O.6x 17 x 10
Rate ofheatconduetion=
81] = 595 w/m2
KA (To  TI)
or
p I+YI!!T
)
(l + y,!!T) = 1+ 'liJ..sfiT
or y,fiT = 'liJ..s!!T or Yl = 'liJ..s. 44. The lateral surface of the cylindrical rod does not lose heat either by conduction or by radiation, Energy is radiated by the circular cross section at temperature T2. Heat flows by conduction along the rod. .'. Heat conducted Q = KA(T1  T2 )tJ.r I Heat radiated E = eoA(T24  Ts4)M For equilibrium, E = Q ~4
T4). _ KA(1} T2)tJ.r s o.t  ~~
or
ea:A\'
or
(Ii _T2)=eO(Tz4Ts4)l K
or
T)  (Ts + fiT) = eol [(Ts K
or
TtTs!!T=KTs
... (1)
2 
t
Rate of heatradiation = 595 x area of lidtop = 595A
K
43 . When the temperature is increased by !!T, the volume of the cube increases but the density of liquid decreases, The depth upto which the cube is submerged in the liquid remains the same. Obviously the upthrust remains the same. Initial upthrust = Final upthrust
RT
m==
V2
:.
Average kinetic energy per atom of helium:
or
2 400 = 595 x (0.5 x 10 )
1
3 x 25
PV= m R'I' M
_
:. ;~=(~:r T'=400(~r'S
x3
T=160K
2 or (e)
= 1000 ....
3xR T= lOOOxlOOOx4xlO~3
K.E.=~kT
T
K
,. 2L
BY k.meuc. t heory, vrms = ~3RT M T = 1000 x 1000 x M
or
T _TI=595xl o
a
500
:.
KA(To Til  595A thickness (t)
..
... (2)
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I
+ fiT)4  Ts4]
eol 4[( l+r:; "T)4
1
1
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ealT' [ 1+1 4~T ='
or
1iTs
or
4ealTs3 tJ.T T} Ts =+~ K
or ..
1j T,
K
=[
T,
4e~T,3
.. 'li + Q2
= Heat supplied 400(T  20) + O.OOl(T  20) = 20000 or 400.00l(T  20) = 20000
] +t1T
T
or or
+1]~T
3
dt
= (viscous force F) x (velocity v) , = 6m"'1J
= (61tl1rv)(v)
= 61tll{~ (0  p)r2g]'
9 ::: (constant)
46.
=
(fir[ x 4g2 x (0  p)2 ) ,,5
11 x rS
81xll or
Rate of heating oc: ,5
Hence the rate of heating is proportional to fifth power of radius of body when it attains terminal velocity. (a) Heat supplied to the cylinder is used up in (0 raising temperature of cylinder = msliT (ii) doing work = Pann (V2 \1) (i)