1. (c) Here,  = , P1 = P, 1 = ;P2 = P’, 2 = ’,

P1V1 = p2V2,


= ( ) = ( ) = ( )7/5 = (32)7/5,

PV = constant
For an adiabatic process,
= (25)7/5 = 27 = 128
P  V or PV-1 = constant
2. (b) As P – V diagram is a straight line passing through origin, therefore,
In the process, PVx = constant, molar heat capacity is given by,
where x = - 1 here and  =1.4 for diatomic gas.
C=
max T = b,
3. (a) According to Wien’s displacement law,
T=
C=
+

=
=
Radiation intensity, R = T4
R = (5.67 X 10-8 W m-2 K-4) (104 K)4,
= 5.67 X 108 W m-2
WBC = 0,
WCD = 4P0V0,
C = 3R
= 104 K
=
W = W1 + W2 + W3 + W4,
U = Uf – Ui = 0,
For a cyclic process, Uf = Ui,
Q = U + W,
For first law of thermodynamics,
5. (c) WAB = - P0V0,
,
= 5960 J – 5585 J – 2980 J + 3645 J = 1040 J,
= 2200 J – 825 J – 1100 J + W4 = 275 J + W4,
+
R+ ,
According to Stefan Boltzmann law,
4. (c) Q = Q1 + Q2 + Q3 + Q4,

1040 J = 0 + 275 J + W4 or W4 = 765 J
WABCD = WAB + WBC + WCD,
0
6. (a) Heat required by ice to convert totally into water at 100 C,
= - P0V0 + 0 + 4P0V0 = 3P0V0
Q1 = 1 X 80 + 1 X 1 X 100 = 180 cal
Heat supplied by steam if it was to condense totally and convert into water at 100 0C,
As Q2 > Q1, entire steam will not condense and final temperature = 100 0C.
Q2 = 1 X 540 = 540 cal,
Both water and steam will exist together in equilibrium at 100 0C.
7. (c) In cylinder A, heat is supplied at constant pressure while in cylinder B, heat is supplied at constant volume.
(Q)A = nCp (T)A,
or
(T)B =
and
(Q)B = nCV (T)B,
(T)a =  (T)A
Given:
( =
(Q)A = (Q)B
),
= ( ) (42 K) = 70 K
8. (a) Let T1 be the temperature of diatomic gas at volume V and T2 is the temperature at volume 32V.
TV - 1 = constant,
For an adiabatic process,
= ( ) - 1 = (
)- 1 =

For diatomic gas,  = ,
,
Efficiency of engine,  = 1 -
T1V1 - 1 = T2V2 - 1,
,
=1-
=
(Using (i)),
=
=
=
=
….(i)
= 0.75
9. (a) Let L and A be length and area of cross-section of each rod. At steady state,
H 1 = H 2 + H 3,
=
300 – 3T = 2T – 100 + T,
10. (b) Here,  = ,

T,
=
12. (a) t =

tT,
=
=
0
=
C
= 41500 J = 41.5 X 103 J = 41.5 kJ, Work done = change
Helium is a monatomic gas.
T1V1 - 1 = T2V2 - 1 or T2 = T1 ( ) - 1,
No. of moles of helium, n =
W=
T=
(Q = 0 for adiabatic process), Change or increase in internal energy, = 41.5 kJ
11. (a) Here, V1 = 5.6 litre, V2 = 0.7 litre, W = ?,
TV - 1 = constant,
3 (100 – T) = 2 (T – 50) + T,
,
6T = 400,
Work done =
in internal energy
+
= ,
=-
RT1,
T2 = T1 (
For helium process,
–
= T1 (8)2/3 = 4T1
Work done during an adiabatic process is
Here – ve sign shows that work is done on the gas.
X 9 X 10-7 0C-1 X 0.5 s X (300C – 200C)
13. (b) Work done by the gas during the isothermal process
14. (b) Thermal capacity, S = Specific heat (s) X Mass (m),
= 22.5 X 10-7 s = 2.25 X 10-6 s
W = nRT ln
=
=


,
= (1) RT ln (
, = ( )3


) = RTln2
= ( )3 X ( ) X ( ) =
,
15. (c) As nitrogen is diatomic, its molar specific heat at constant pressure is
No. of moles, n =
= ,
or
16. (b) According to Newton’s law of cooling,
=K[
surroundings.
Divide (ii) by (i), we get,
X
=[
17. (c) Heat supplied to a gas,
=1-=1-
respectively.
T2 =
where TS is the temperature of the
- TS],
,
3
,
X 300 = 750 K,
19. (a) 20. (c) Let m1 = m2 = m3 = m,
X
R X 40 = 70 R
=K[
- 20],
100 - 2 = 10 + ,
=K[
3 = 90 or
]
…(ii)
= 300C
Work done by the gas,
2
= 5.25 X 10 N m = 525 J,
U = Q - W,
According to first law
= 1500 J – 525 J = 975 J
where T1 and T2 are the temperatures of source and sink
= ,
T1 =
T2 =
X 300 K = 500 K, Increase in efficiency = 50%
New efficiency, ’ = 40% + 20% = 60%
of 40% = 20%,
T’1 =
-3
Q = W + U,
18. (d) Efficiency of Carnot engine,  = 1 -
=K[
Q = 1500 J,
-2
of thermodynamics,
=
]X
W = PV = 2.1 X 10 X N m X 2.5 X 10 m ,
5
R
Q = nCpT,
4 = K [40] ….(i) and
- 20],
CP =
= 1 - ’ = 1 -
Increase in temperature of source = T’1 – T1,
= ,
= 750 – 500 = 250 K
Let s1, s2, s3 be the respectively specific heats of the three liquids.
(1) When A and B are mixed, temperature of mixture = 160C, As heat gained by A = heat lost by B,
ms1 (16 – 12) = ms2 (19 – 16),
4s1 = 3s2 …(i) (2) When B and C are mixed, temperature of mixture = 23 0C.
ms2 (23 – 19) = ms3 (28 – 23), 4s2 = 5s3
As heat gained by B = heat lost by C,
From eqs. (i) and (ii), we get,
s1 =
temperature of mixture = T,
heat gained by A = heat lost by C,
ms1 (T – 12) = ms3 (29 – T)
15T – 180 = 448 – 16T,
= 20.20C
s3 (T – 12) = s3 (28 – T),
s2 =
s3,
…(ii)
When A and C are mixed, suppose
T=
21. (a) The relation of P and V when temperature is constant is given by Boyle’s law as PV = a constant. The
variation of P and V will be a curve as shown in figure (a).
22. (b) Given: VP3 = constant = k or P =
V2/3 =
V = nRT,
, rms 
23. (d) rms = √
√
,
,
According to ideal gas equation,
Hence, ( )2/3 =
,
[


(At a given temperature T),
24. (b) Molar specific heat of the mixture at constant volume is
25. (b) Here, P = 1 atm,
As
PV =
Rt,
]2/3 =
(CV)mixture =
or T2 = 9T1 = 9T
=√
=√
,
+
27. (d)The value of
28. (a)
,
T = 270C = 300 K,
R = 0.0821 L atm mol-1 K-1
m=
=
= 14.56 g = 0.01456 kg
,
,
P X 1000 = 1 X 100 + 0.5 X 50,
or
= 2.1R
n = n1 + n2
P=
atm = 0.125 atm
for one mole of an ideal gas = gas constant = 2 cal mol -1 K-1
29. (b)The thermal energy or internal energy is U =
freedom as the gas is diatomic),
But PV = nRT,
P = 8 X104 N m-2 (Given),
or
30. (a) Mean free path,  =
√
,
d = Diameter of the molecule,
where,
=
=√
V = 11.2 litres,
26. (d) No. of moles in mixture is the sum of separate no. of moles of gases.
=
PV = nRT
U=
nRT for diatomic gases. (5 is the degrees of
V=
X 8 X 104 X
=
=
m3
= 5 X 104 J
where, n = Number of molecules per unit volume
As PV = kBNT,
kB = Boltzmann constant,
n=
P = Pressure of gas,
=
or  =
√
T = Absolute temperature of the gas