Chapter 27 : Specific Heat Capacities of Gases - HC Verma Solution, Physics Class 11 Notes | EduRev

HC Verma and Irodov Solutions

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JEE : Chapter 27 : Specific Heat Capacities of Gases - HC Verma Solution, Physics Class 11 Notes | EduRev

 Page 1


27.1
CHAPTER – 27
SPECIFIC HEAT CAPACITIES OF GASES
1. N = 1 mole, W = 20 g/mol, V = 50 m/s
K.E. of the vessel = Internal energy of the gas
= (1/2) mv
2
= (1/2) × 20 × 10
–3
× 50 × 50 = 25 J
25 = n
2
3
r( ?T) ? 25 = 1 × 
2
3
× 8.31 × ?T ? ?T= 
3 . 8 3
50
?
˜ 2 k. ?
2. m = 5 g, ?t = 25 – 15 = 10°C
C
V
= 0.172 cal/g-°CJ = 4.2 J/Cal.
dQ = du + dw
Now, V = 0 (for a rigid body)
So, dw = 0.
So dQ = du.
Q = msdt = 5 × 0.172 × 10 = 8.6 cal = 8.6 × 4.2 = 36.12 Joule.
3. ? = 1.4, w or piston = 50 kg., A of piston = 100 cm
2
Po = 100 kpa, g = 10 m/s
2
,  x = 20 cm.
dw = pdv = Adx Po
A
mg
?
?
?
?
?
?
? = 
2 4 5
4
10 20 10 100 10
10 100
10 50
? ?
?
? ? ? ?
?
?
?
?
?
?
?
?
= 1.5 × 10
5
× 20 × 10
–4
= 300 J.
nRdt = 300 ? dT = 
nR
300
dQ = nCpdT = nCp × 
nR
300
= 
nR ) 1 (
300 R n
? ?
?
= 
4 . 0
4 . 1 300 ?
= 1050 J.
4. C
V
H
2
= 2.4 Cal/g°C, C
P
H
2
= 3.4 Cal/g°C
M = 2 g/ Mol, R = 8.3 × 10
7
erg/mol-°C
We know, C
P
– C
V
= 1 Cal/g°C
So, difference of molar specific heats 
= C
P
× M – C
V
× M = 1 × 2 = 2 Cal/g°C
Now, 2 × J = R ? 2 × J = 8.3 × 10
7
erg/mol-°C ? J = 4.15 × 10
7
erg/cal.
5.
V
P
C
C
= 7.6, n = 1 mole, ?T = 50K
(a) Keeping the pressure constant, dQ = du + dw,
?T = 50 K, ? = 7/6, m = 1 mole, 
dQ = du + dw ? nC
V
dT = du + RdT ? du = nCpdT – RdT
= RdT dT
1
R
1 ? ?
? ?
?
? = RdT dT
1
6
7
6
7
R
?
?
?
= DT – RdT = 7RdT – RdT = 6 RdT = 6 × 8.3 × 50 = 2490 J. ?
(b) Kipping Volume constant, dv = nC
V
dT
= dt
1
R
1 ?
? ?
? = 50
1
6
7
3 . 8 1
?
?
?
= 8.3 × 50 × 6 = 2490 J
(c) Adiabetically dQ = 0, du = – dw
= ? ?
?
?
?
?
?
?
?
? ?
?
2 1
T T
1
R n
= ? ?
1 2
T T
1
6
7
3 . 8 1
?
?
?
= 8.3 × 50 × 6 = 2490 J
Page 2


27.1
CHAPTER – 27
SPECIFIC HEAT CAPACITIES OF GASES
1. N = 1 mole, W = 20 g/mol, V = 50 m/s
K.E. of the vessel = Internal energy of the gas
= (1/2) mv
2
= (1/2) × 20 × 10
–3
× 50 × 50 = 25 J
25 = n
2
3
r( ?T) ? 25 = 1 × 
2
3
× 8.31 × ?T ? ?T= 
3 . 8 3
50
?
˜ 2 k. ?
2. m = 5 g, ?t = 25 – 15 = 10°C
C
V
= 0.172 cal/g-°CJ = 4.2 J/Cal.
dQ = du + dw
Now, V = 0 (for a rigid body)
So, dw = 0.
So dQ = du.
Q = msdt = 5 × 0.172 × 10 = 8.6 cal = 8.6 × 4.2 = 36.12 Joule.
3. ? = 1.4, w or piston = 50 kg., A of piston = 100 cm
2
Po = 100 kpa, g = 10 m/s
2
,  x = 20 cm.
dw = pdv = Adx Po
A
mg
?
?
?
?
?
?
? = 
2 4 5
4
10 20 10 100 10
10 100
10 50
? ?
?
? ? ? ?
?
?
?
?
?
?
?
?
= 1.5 × 10
5
× 20 × 10
–4
= 300 J.
nRdt = 300 ? dT = 
nR
300
dQ = nCpdT = nCp × 
nR
300
= 
nR ) 1 (
300 R n
? ?
?
= 
4 . 0
4 . 1 300 ?
= 1050 J.
4. C
V
H
2
= 2.4 Cal/g°C, C
P
H
2
= 3.4 Cal/g°C
M = 2 g/ Mol, R = 8.3 × 10
7
erg/mol-°C
We know, C
P
– C
V
= 1 Cal/g°C
So, difference of molar specific heats 
= C
P
× M – C
V
× M = 1 × 2 = 2 Cal/g°C
Now, 2 × J = R ? 2 × J = 8.3 × 10
7
erg/mol-°C ? J = 4.15 × 10
7
erg/cal.
5.
V
P
C
C
= 7.6, n = 1 mole, ?T = 50K
(a) Keeping the pressure constant, dQ = du + dw,
?T = 50 K, ? = 7/6, m = 1 mole, 
dQ = du + dw ? nC
V
dT = du + RdT ? du = nCpdT – RdT
= RdT dT
1
R
1 ? ?
? ?
?
? = RdT dT
1
6
7
6
7
R
?
?
?
= DT – RdT = 7RdT – RdT = 6 RdT = 6 × 8.3 × 50 = 2490 J. ?
(b) Kipping Volume constant, dv = nC
V
dT
= dt
1
R
1 ?
? ?
? = 50
1
6
7
3 . 8 1
?
?
?
= 8.3 × 50 × 6 = 2490 J
(c) Adiabetically dQ = 0, du = – dw
= ? ?
?
?
?
?
?
?
?
? ?
?
2 1
T T
1
R n
= ? ?
1 2
T T
1
6
7
3 . 8 1
?
?
?
= 8.3 × 50 × 6 = 2490 J
Specific Heat Capacities of Gases
27.2
6. m = 1.18 g, V = 1 × 10
3 
cm
3
= 1 L T = 300 k, P = 10
5
Pa
PV = nRT or n = 
RT
PV
= 10
5
= atm.
N = 
RT
PV
= 
2 2
10 3 10 2 . 8
1
? ? ?
?
= 
3 2 . 8
1
?
= 
6 . 24
1
Now, C
v
=
dt
Q
n
1
? = 24.6 × 2 = 49.2
C
p
= R + C
v
= 1.987 + 49.2 = 51.187
Q = nC
p
dT = 1 187 . 51
6 . 24
1
? ? = 2.08 Cal.
7. V
1
= 100 cm
3
, V
2
= 200 cm
3
P = 2 × 10
5
Pa, ?Q = 50J
(a) ?Q = du + dw ? 50 = du + 20× 10
5
(200 – 100 × 10
–6
) ? 50 = du + 20 ? du = 30 J
(b) 30 = n × 
2
3
× 8.3 × 300 [ U = 
2
3
nRT for monoatomic]
? n = 
83 3
2
?
= 
249
2
= 0.008
(c) du = nC
v
dT ? C
v
= 
dndTu
= 
300 008 . 0
30
?
= 12.5
C
p
= C
v
+ R = 12.5 + 8.3 = 20.3
(d) C
v
= 12.5 (Proved above)
8. Q = Amt of heat given
Work done = 
2
Q
, ?Q = W + ? U
for monoatomic gas ? ?U = Q –
2
Q
= 
2
Q
V = n
2
3
RT = 
2
Q
= nT× 
2
3
R = 3R × nT
Again Q = n CpdT Where C
P
> Molar heat capacity at const. pressure.
3RnT = ndTC
P
? C
P
= 3R ?
9. P = KV ?
V
nRT
= KV  ? RT = KV
2
  ? R ?T = 2KV ?U ?
KV 2
T R ?
= dv
dQ = du + dw ? mcdT = C
V
dT + pdv ? msdT = C
V
dT+ 
KV 2
PRdF
? ms = C
V
+ 
KV 2
RKV
? C
P
+ 
2
R
10.
V
P
C
C
= ?, C
P
– C
V 
= R, C
V
= 
1
r
? ?
, C
P
= 
1
R
? ?
?
Pdv = ? ? Rdt
1 b
1
?
? 0 = C
V
dT + ? ? Rdt
1 b
1
?
?
1 b
1
?
= 
R
C
V
?
? b + 1 = 
V
C
R ?
= 
? ?
V
V P
C
C C ? ?
= – ? +1 ? b = – ?
11. Considering two gases, in Gas(1) we have,
?, Cp
1
(Sp. Heat at const. ‘P’), Cv
1
(Sp. Heat at const. ‘V’), n
1
(No. of moles)
1
1
Cv
Cp
??? & Cp
1
– Cv
1
= R 
Page 3


27.1
CHAPTER – 27
SPECIFIC HEAT CAPACITIES OF GASES
1. N = 1 mole, W = 20 g/mol, V = 50 m/s
K.E. of the vessel = Internal energy of the gas
= (1/2) mv
2
= (1/2) × 20 × 10
–3
× 50 × 50 = 25 J
25 = n
2
3
r( ?T) ? 25 = 1 × 
2
3
× 8.31 × ?T ? ?T= 
3 . 8 3
50
?
˜ 2 k. ?
2. m = 5 g, ?t = 25 – 15 = 10°C
C
V
= 0.172 cal/g-°CJ = 4.2 J/Cal.
dQ = du + dw
Now, V = 0 (for a rigid body)
So, dw = 0.
So dQ = du.
Q = msdt = 5 × 0.172 × 10 = 8.6 cal = 8.6 × 4.2 = 36.12 Joule.
3. ? = 1.4, w or piston = 50 kg., A of piston = 100 cm
2
Po = 100 kpa, g = 10 m/s
2
,  x = 20 cm.
dw = pdv = Adx Po
A
mg
?
?
?
?
?
?
? = 
2 4 5
4
10 20 10 100 10
10 100
10 50
? ?
?
? ? ? ?
?
?
?
?
?
?
?
?
= 1.5 × 10
5
× 20 × 10
–4
= 300 J.
nRdt = 300 ? dT = 
nR
300
dQ = nCpdT = nCp × 
nR
300
= 
nR ) 1 (
300 R n
? ?
?
= 
4 . 0
4 . 1 300 ?
= 1050 J.
4. C
V
H
2
= 2.4 Cal/g°C, C
P
H
2
= 3.4 Cal/g°C
M = 2 g/ Mol, R = 8.3 × 10
7
erg/mol-°C
We know, C
P
– C
V
= 1 Cal/g°C
So, difference of molar specific heats 
= C
P
× M – C
V
× M = 1 × 2 = 2 Cal/g°C
Now, 2 × J = R ? 2 × J = 8.3 × 10
7
erg/mol-°C ? J = 4.15 × 10
7
erg/cal.
5.
V
P
C
C
= 7.6, n = 1 mole, ?T = 50K
(a) Keeping the pressure constant, dQ = du + dw,
?T = 50 K, ? = 7/6, m = 1 mole, 
dQ = du + dw ? nC
V
dT = du + RdT ? du = nCpdT – RdT
= RdT dT
1
R
1 ? ?
? ?
?
? = RdT dT
1
6
7
6
7
R
?
?
?
= DT – RdT = 7RdT – RdT = 6 RdT = 6 × 8.3 × 50 = 2490 J. ?
(b) Kipping Volume constant, dv = nC
V
dT
= dt
1
R
1 ?
? ?
? = 50
1
6
7
3 . 8 1
?
?
?
= 8.3 × 50 × 6 = 2490 J
(c) Adiabetically dQ = 0, du = – dw
= ? ?
?
?
?
?
?
?
?
? ?
?
2 1
T T
1
R n
= ? ?
1 2
T T
1
6
7
3 . 8 1
?
?
?
= 8.3 × 50 × 6 = 2490 J
Specific Heat Capacities of Gases
27.2
6. m = 1.18 g, V = 1 × 10
3 
cm
3
= 1 L T = 300 k, P = 10
5
Pa
PV = nRT or n = 
RT
PV
= 10
5
= atm.
N = 
RT
PV
= 
2 2
10 3 10 2 . 8
1
? ? ?
?
= 
3 2 . 8
1
?
= 
6 . 24
1
Now, C
v
=
dt
Q
n
1
? = 24.6 × 2 = 49.2
C
p
= R + C
v
= 1.987 + 49.2 = 51.187
Q = nC
p
dT = 1 187 . 51
6 . 24
1
? ? = 2.08 Cal.
7. V
1
= 100 cm
3
, V
2
= 200 cm
3
P = 2 × 10
5
Pa, ?Q = 50J
(a) ?Q = du + dw ? 50 = du + 20× 10
5
(200 – 100 × 10
–6
) ? 50 = du + 20 ? du = 30 J
(b) 30 = n × 
2
3
× 8.3 × 300 [ U = 
2
3
nRT for monoatomic]
? n = 
83 3
2
?
= 
249
2
= 0.008
(c) du = nC
v
dT ? C
v
= 
dndTu
= 
300 008 . 0
30
?
= 12.5
C
p
= C
v
+ R = 12.5 + 8.3 = 20.3
(d) C
v
= 12.5 (Proved above)
8. Q = Amt of heat given
Work done = 
2
Q
, ?Q = W + ? U
for monoatomic gas ? ?U = Q –
2
Q
= 
2
Q
V = n
2
3
RT = 
2
Q
= nT× 
2
3
R = 3R × nT
Again Q = n CpdT Where C
P
> Molar heat capacity at const. pressure.
3RnT = ndTC
P
? C
P
= 3R ?
9. P = KV ?
V
nRT
= KV  ? RT = KV
2
  ? R ?T = 2KV ?U ?
KV 2
T R ?
= dv
dQ = du + dw ? mcdT = C
V
dT + pdv ? msdT = C
V
dT+ 
KV 2
PRdF
? ms = C
V
+ 
KV 2
RKV
? C
P
+ 
2
R
10.
V
P
C
C
= ?, C
P
– C
V 
= R, C
V
= 
1
r
? ?
, C
P
= 
1
R
? ?
?
Pdv = ? ? Rdt
1 b
1
?
? 0 = C
V
dT + ? ? Rdt
1 b
1
?
?
1 b
1
?
= 
R
C
V
?
? b + 1 = 
V
C
R ?
= 
? ?
V
V P
C
C C ? ?
= – ? +1 ? b = – ?
11. Considering two gases, in Gas(1) we have,
?, Cp
1
(Sp. Heat at const. ‘P’), Cv
1
(Sp. Heat at const. ‘V’), n
1
(No. of moles)
1
1
Cv
Cp
??? & Cp
1
– Cv
1
= R 
Specific Heat Capacities of Gases
27.3
? ?Cv
1
– Cv
1
= R ? Cv
1
( ? – 1) = R 
? Cv
1
= 
1
R
? ?
& Cp
1
=
1
R
? ?
?
In Gas(2) we have, ?, Cp
2
(Sp. Heat at const. ‘P’), Cv
2
(Sp. Heat at const. ‘V’), n
2
(No. of moles)
2
2
Cv
Cp
??? & Cp
2
– Cv
2
= R ? ?Cv
2
– Cv
2
= R ? Cv
2
( ? – 1) = R ? Cv
2
= 
1
R
? ?
& Cp
2
=
1
R
? ?
?
Given n
1
: n
2
= 1 :2
dU
1
= nCv
1
dT & dU
2
= 2nCv
2
dT = 3nCvdT
? C
V
= 
3
Cv 2 Cv
2 1
?
= 
3
1
R 2
1
R
? ?
?
? ?
= 
) 1 ( 3
R 3
? ?
= 
1
R
? ?
…(1)
&Cp = ?Cv = 
1
r
? ?
?
…(2)
So, 
Cv
Cp
= ? [from (1) & (2)] ?
12. Cp ? = 2.5 RCp ? = 3.5 R
Cv ? = 1.5 R Cv ? = 2.5 R
n
1
= n
2
= 1 mol (n
1
+ n
2
)C
V
dT = n
1
C
v
dT + n
2
C
v
dT
? C
V
= 
2 1
2 1
n n
v C n v C n
?
? ? ? ?
= 
2
R 5 . 2 R 5 . 1 ?
2R
C
P
= C
V
+ R = 2R + R = 3R
? = 
V
p
C
C
= 
R 2
R 3
= 1.5
13. n = 
2
1
  mole, R =
3
25
J/mol-k, ? = 
3
5
(a) Temp at A = T
a
, P
a
V
a
= nRT
a
? T
a
= 
nR
V P
a a
= 
3
25
2
1
10 100 10 5000
3 6
?
? ? ?
?
= 120 k.
Similarly temperatures at point b = 240 k at C it is 480 k and at D it is 240 k.
(b) For ab process,
dQ = nCpdT [since ab is isobaric]
= ? ?
a b
T T
1
R
2
1
?
? ?
?
? = ) 120 240 (
1
3
5
3
5
3
35
2
1
? ?
?
?
? = 120
2
3
9
125
2
1
? ? ? = 1250 J
For bc, dQ = du + dw [dq = 0, Isochorie process]
? dQ = du = nC
v
dT = ? ?
a c
T T
1
nR
?
? ?
= ) 240 (
1
3
5
3
25
2
1
?
?
?
?
?
?
?
? = 240
2
3
3
25
2
1
? ? ? = 1500 J
(c) Heat liberated in cd = – nC
p
dT
= ? ?
c d
T T
1
nR
2
1
?
? ?
?
?
= 240
2
3
3
125
2
1
? ? ?
?
= 2500 J
Heat liberated in da = – nC
v
dT
= ? ?
d a
T T
1
R
2
1
?
? ?
?
?
= ) 240 120 (
2
25
2
1
? ? ?
?
= 750 J 
100 KPa
5000 cm
3
a b
c
d
Ta Tb
Tc
Td
10000 cm
3
200 KPa
Page 4


27.1
CHAPTER – 27
SPECIFIC HEAT CAPACITIES OF GASES
1. N = 1 mole, W = 20 g/mol, V = 50 m/s
K.E. of the vessel = Internal energy of the gas
= (1/2) mv
2
= (1/2) × 20 × 10
–3
× 50 × 50 = 25 J
25 = n
2
3
r( ?T) ? 25 = 1 × 
2
3
× 8.31 × ?T ? ?T= 
3 . 8 3
50
?
˜ 2 k. ?
2. m = 5 g, ?t = 25 – 15 = 10°C
C
V
= 0.172 cal/g-°CJ = 4.2 J/Cal.
dQ = du + dw
Now, V = 0 (for a rigid body)
So, dw = 0.
So dQ = du.
Q = msdt = 5 × 0.172 × 10 = 8.6 cal = 8.6 × 4.2 = 36.12 Joule.
3. ? = 1.4, w or piston = 50 kg., A of piston = 100 cm
2
Po = 100 kpa, g = 10 m/s
2
,  x = 20 cm.
dw = pdv = Adx Po
A
mg
?
?
?
?
?
?
? = 
2 4 5
4
10 20 10 100 10
10 100
10 50
? ?
?
? ? ? ?
?
?
?
?
?
?
?
?
= 1.5 × 10
5
× 20 × 10
–4
= 300 J.
nRdt = 300 ? dT = 
nR
300
dQ = nCpdT = nCp × 
nR
300
= 
nR ) 1 (
300 R n
? ?
?
= 
4 . 0
4 . 1 300 ?
= 1050 J.
4. C
V
H
2
= 2.4 Cal/g°C, C
P
H
2
= 3.4 Cal/g°C
M = 2 g/ Mol, R = 8.3 × 10
7
erg/mol-°C
We know, C
P
– C
V
= 1 Cal/g°C
So, difference of molar specific heats 
= C
P
× M – C
V
× M = 1 × 2 = 2 Cal/g°C
Now, 2 × J = R ? 2 × J = 8.3 × 10
7
erg/mol-°C ? J = 4.15 × 10
7
erg/cal.
5.
V
P
C
C
= 7.6, n = 1 mole, ?T = 50K
(a) Keeping the pressure constant, dQ = du + dw,
?T = 50 K, ? = 7/6, m = 1 mole, 
dQ = du + dw ? nC
V
dT = du + RdT ? du = nCpdT – RdT
= RdT dT
1
R
1 ? ?
? ?
?
? = RdT dT
1
6
7
6
7
R
?
?
?
= DT – RdT = 7RdT – RdT = 6 RdT = 6 × 8.3 × 50 = 2490 J. ?
(b) Kipping Volume constant, dv = nC
V
dT
= dt
1
R
1 ?
? ?
? = 50
1
6
7
3 . 8 1
?
?
?
= 8.3 × 50 × 6 = 2490 J
(c) Adiabetically dQ = 0, du = – dw
= ? ?
?
?
?
?
?
?
?
? ?
?
2 1
T T
1
R n
= ? ?
1 2
T T
1
6
7
3 . 8 1
?
?
?
= 8.3 × 50 × 6 = 2490 J
Specific Heat Capacities of Gases
27.2
6. m = 1.18 g, V = 1 × 10
3 
cm
3
= 1 L T = 300 k, P = 10
5
Pa
PV = nRT or n = 
RT
PV
= 10
5
= atm.
N = 
RT
PV
= 
2 2
10 3 10 2 . 8
1
? ? ?
?
= 
3 2 . 8
1
?
= 
6 . 24
1
Now, C
v
=
dt
Q
n
1
? = 24.6 × 2 = 49.2
C
p
= R + C
v
= 1.987 + 49.2 = 51.187
Q = nC
p
dT = 1 187 . 51
6 . 24
1
? ? = 2.08 Cal.
7. V
1
= 100 cm
3
, V
2
= 200 cm
3
P = 2 × 10
5
Pa, ?Q = 50J
(a) ?Q = du + dw ? 50 = du + 20× 10
5
(200 – 100 × 10
–6
) ? 50 = du + 20 ? du = 30 J
(b) 30 = n × 
2
3
× 8.3 × 300 [ U = 
2
3
nRT for monoatomic]
? n = 
83 3
2
?
= 
249
2
= 0.008
(c) du = nC
v
dT ? C
v
= 
dndTu
= 
300 008 . 0
30
?
= 12.5
C
p
= C
v
+ R = 12.5 + 8.3 = 20.3
(d) C
v
= 12.5 (Proved above)
8. Q = Amt of heat given
Work done = 
2
Q
, ?Q = W + ? U
for monoatomic gas ? ?U = Q –
2
Q
= 
2
Q
V = n
2
3
RT = 
2
Q
= nT× 
2
3
R = 3R × nT
Again Q = n CpdT Where C
P
> Molar heat capacity at const. pressure.
3RnT = ndTC
P
? C
P
= 3R ?
9. P = KV ?
V
nRT
= KV  ? RT = KV
2
  ? R ?T = 2KV ?U ?
KV 2
T R ?
= dv
dQ = du + dw ? mcdT = C
V
dT + pdv ? msdT = C
V
dT+ 
KV 2
PRdF
? ms = C
V
+ 
KV 2
RKV
? C
P
+ 
2
R
10.
V
P
C
C
= ?, C
P
– C
V 
= R, C
V
= 
1
r
? ?
, C
P
= 
1
R
? ?
?
Pdv = ? ? Rdt
1 b
1
?
? 0 = C
V
dT + ? ? Rdt
1 b
1
?
?
1 b
1
?
= 
R
C
V
?
? b + 1 = 
V
C
R ?
= 
? ?
V
V P
C
C C ? ?
= – ? +1 ? b = – ?
11. Considering two gases, in Gas(1) we have,
?, Cp
1
(Sp. Heat at const. ‘P’), Cv
1
(Sp. Heat at const. ‘V’), n
1
(No. of moles)
1
1
Cv
Cp
??? & Cp
1
– Cv
1
= R 
Specific Heat Capacities of Gases
27.3
? ?Cv
1
– Cv
1
= R ? Cv
1
( ? – 1) = R 
? Cv
1
= 
1
R
? ?
& Cp
1
=
1
R
? ?
?
In Gas(2) we have, ?, Cp
2
(Sp. Heat at const. ‘P’), Cv
2
(Sp. Heat at const. ‘V’), n
2
(No. of moles)
2
2
Cv
Cp
??? & Cp
2
– Cv
2
= R ? ?Cv
2
– Cv
2
= R ? Cv
2
( ? – 1) = R ? Cv
2
= 
1
R
? ?
& Cp
2
=
1
R
? ?
?
Given n
1
: n
2
= 1 :2
dU
1
= nCv
1
dT & dU
2
= 2nCv
2
dT = 3nCvdT
? C
V
= 
3
Cv 2 Cv
2 1
?
= 
3
1
R 2
1
R
? ?
?
? ?
= 
) 1 ( 3
R 3
? ?
= 
1
R
? ?
…(1)
&Cp = ?Cv = 
1
r
? ?
?
…(2)
So, 
Cv
Cp
= ? [from (1) & (2)] ?
12. Cp ? = 2.5 RCp ? = 3.5 R
Cv ? = 1.5 R Cv ? = 2.5 R
n
1
= n
2
= 1 mol (n
1
+ n
2
)C
V
dT = n
1
C
v
dT + n
2
C
v
dT
? C
V
= 
2 1
2 1
n n
v C n v C n
?
? ? ? ?
= 
2
R 5 . 2 R 5 . 1 ?
2R
C
P
= C
V
+ R = 2R + R = 3R
? = 
V
p
C
C
= 
R 2
R 3
= 1.5
13. n = 
2
1
  mole, R =
3
25
J/mol-k, ? = 
3
5
(a) Temp at A = T
a
, P
a
V
a
= nRT
a
? T
a
= 
nR
V P
a a
= 
3
25
2
1
10 100 10 5000
3 6
?
? ? ?
?
= 120 k.
Similarly temperatures at point b = 240 k at C it is 480 k and at D it is 240 k.
(b) For ab process,
dQ = nCpdT [since ab is isobaric]
= ? ?
a b
T T
1
R
2
1
?
? ?
?
? = ) 120 240 (
1
3
5
3
5
3
35
2
1
? ?
?
?
? = 120
2
3
9
125
2
1
? ? ? = 1250 J
For bc, dQ = du + dw [dq = 0, Isochorie process]
? dQ = du = nC
v
dT = ? ?
a c
T T
1
nR
?
? ?
= ) 240 (
1
3
5
3
25
2
1
?
?
?
?
?
?
?
? = 240
2
3
3
25
2
1
? ? ? = 1500 J
(c) Heat liberated in cd = – nC
p
dT
= ? ?
c d
T T
1
nR
2
1
?
? ?
?
?
= 240
2
3
3
125
2
1
? ? ?
?
= 2500 J
Heat liberated in da = – nC
v
dT
= ? ?
d a
T T
1
R
2
1
?
? ?
?
?
= ) 240 120 (
2
25
2
1
? ? ?
?
= 750 J 
100 KPa
5000 cm
3
a b
c
d
Ta Tb
Tc
Td
10000 cm
3
200 KPa
Specific Heat Capacities of Gases
27.4
14. (a) For a, b ’V’ is constant
So, 
2
2
1
1
T
P
T
P
? ?
300
100
= 
2
T
200
? T
2
= 
100
300 200 ?
= 600 k
For b,c ‘P’ is constant
So, 
2
2
1
1
T
V
T
V
? ?
600
100
= 
2
T
150
? T
2
= 
100
150 600 ?
= 900 k
(b) Work done = Area enclosed under the graph 50 cc × 200 kpa = 50 × 10
–6
× 200 × 10
3
J = 10 J
(c) ‘Q’ Supplied = nC
v
dT
Now, n = 
RT
PV
considering at pt. ‘b’
C
v
= dT
1
R
? ?
= 300 a, b.
Q
bc
= dT
1
R
RT
PV
? ?
? = 300
67 . 0 600
10 100 10 200
6 3
?
?
? ? ?
?
= 14.925 ( ?? = 1.67)
Q supplied to be nC
p
dT [ ?C
p
= 
1
R
? ?
?
]
= dT
1
R
RT
PV
? ?
?
? = 300
67 . 0
3 . 8 67 . 1
900 3 . 8
10 150 10 200
6 3
?
?
?
?
? ? ?
?
= 24.925
(d) Q = ?U + w
Now, ?U = Q – w = Heat supplied – Work done = (24.925 + 14.925) – 1 = 29.850 ?
15. In Joly’s differential steam calorimeter
C
v
= 
) ( m
L m
1 2 1
2
? ? ?
m
2
= Mass of steam condensed = 0.095 g, L = 540 Cal/g = 540 × 4.2 J/g
m
1
= Mass of gas present = 3 g, ?
1
= 20°C, ?
2
= 100°C
? C
v
= 
) 20 100 ( 3
2 . 4 540 095 . 0
?
? ?
= 0.89 ˜ 0.9 J/g-K ?
16. ? = 1.5
Since it is an adiabatic process, So PV
?
= const.
(a) P
1
V
1
?
= P
2
V
2
?
Given V
1
= 4 L, V
2
= 3 L, 
1
2
P
P
= ?
??
1
2
P
P
= 
?
?
?
?
?
?
?
?
?
2
1
V
V
= 
5 . 1
3
4
?
?
?
?
?
?
= 1.5396 ˜ 1.54
(b) TV
?–1
= Const.
T
1
V
1
?–1
= T
2
V
2
?–1
?
1
2
T
T
= 
1
2
1
V
V
? ?
?
?
?
?
?
?
?
?
= 
5 . 0
3
4
?
?
?
?
?
?
= 1.154 
17. P
1
= 2.5 × 10
5
Pa, V
1
= 100 cc, T
1
= 300 k
(a) P
1
V
1
?
= P
2
V
2
?
? 2.5 × 10
5
× V
1.5
= 
2
5 . 1
P
2
V
? ?
?
?
?
?
?
? P
2
= 2
1.5
× 2.5 × 10
5
= 7.07 × 10
5
˜ 7.1 × 10
5
(b) T
1
V
1
?–1
= T
2
V
2
?–1
? 300 × (100)
1.5 – 1
= T
2
× (50)
1.5 – 1
? T
2
= 
07 . 7
3000
= 424.32 k ˜ 424 k 
a
100 KPa 200 KPa
150 cm
3
100 cm
3
b
c
Page 5


27.1
CHAPTER – 27
SPECIFIC HEAT CAPACITIES OF GASES
1. N = 1 mole, W = 20 g/mol, V = 50 m/s
K.E. of the vessel = Internal energy of the gas
= (1/2) mv
2
= (1/2) × 20 × 10
–3
× 50 × 50 = 25 J
25 = n
2
3
r( ?T) ? 25 = 1 × 
2
3
× 8.31 × ?T ? ?T= 
3 . 8 3
50
?
˜ 2 k. ?
2. m = 5 g, ?t = 25 – 15 = 10°C
C
V
= 0.172 cal/g-°CJ = 4.2 J/Cal.
dQ = du + dw
Now, V = 0 (for a rigid body)
So, dw = 0.
So dQ = du.
Q = msdt = 5 × 0.172 × 10 = 8.6 cal = 8.6 × 4.2 = 36.12 Joule.
3. ? = 1.4, w or piston = 50 kg., A of piston = 100 cm
2
Po = 100 kpa, g = 10 m/s
2
,  x = 20 cm.
dw = pdv = Adx Po
A
mg
?
?
?
?
?
?
? = 
2 4 5
4
10 20 10 100 10
10 100
10 50
? ?
?
? ? ? ?
?
?
?
?
?
?
?
?
= 1.5 × 10
5
× 20 × 10
–4
= 300 J.
nRdt = 300 ? dT = 
nR
300
dQ = nCpdT = nCp × 
nR
300
= 
nR ) 1 (
300 R n
? ?
?
= 
4 . 0
4 . 1 300 ?
= 1050 J.
4. C
V
H
2
= 2.4 Cal/g°C, C
P
H
2
= 3.4 Cal/g°C
M = 2 g/ Mol, R = 8.3 × 10
7
erg/mol-°C
We know, C
P
– C
V
= 1 Cal/g°C
So, difference of molar specific heats 
= C
P
× M – C
V
× M = 1 × 2 = 2 Cal/g°C
Now, 2 × J = R ? 2 × J = 8.3 × 10
7
erg/mol-°C ? J = 4.15 × 10
7
erg/cal.
5.
V
P
C
C
= 7.6, n = 1 mole, ?T = 50K
(a) Keeping the pressure constant, dQ = du + dw,
?T = 50 K, ? = 7/6, m = 1 mole, 
dQ = du + dw ? nC
V
dT = du + RdT ? du = nCpdT – RdT
= RdT dT
1
R
1 ? ?
? ?
?
? = RdT dT
1
6
7
6
7
R
?
?
?
= DT – RdT = 7RdT – RdT = 6 RdT = 6 × 8.3 × 50 = 2490 J. ?
(b) Kipping Volume constant, dv = nC
V
dT
= dt
1
R
1 ?
? ?
? = 50
1
6
7
3 . 8 1
?
?
?
= 8.3 × 50 × 6 = 2490 J
(c) Adiabetically dQ = 0, du = – dw
= ? ?
?
?
?
?
?
?
?
? ?
?
2 1
T T
1
R n
= ? ?
1 2
T T
1
6
7
3 . 8 1
?
?
?
= 8.3 × 50 × 6 = 2490 J
Specific Heat Capacities of Gases
27.2
6. m = 1.18 g, V = 1 × 10
3 
cm
3
= 1 L T = 300 k, P = 10
5
Pa
PV = nRT or n = 
RT
PV
= 10
5
= atm.
N = 
RT
PV
= 
2 2
10 3 10 2 . 8
1
? ? ?
?
= 
3 2 . 8
1
?
= 
6 . 24
1
Now, C
v
=
dt
Q
n
1
? = 24.6 × 2 = 49.2
C
p
= R + C
v
= 1.987 + 49.2 = 51.187
Q = nC
p
dT = 1 187 . 51
6 . 24
1
? ? = 2.08 Cal.
7. V
1
= 100 cm
3
, V
2
= 200 cm
3
P = 2 × 10
5
Pa, ?Q = 50J
(a) ?Q = du + dw ? 50 = du + 20× 10
5
(200 – 100 × 10
–6
) ? 50 = du + 20 ? du = 30 J
(b) 30 = n × 
2
3
× 8.3 × 300 [ U = 
2
3
nRT for monoatomic]
? n = 
83 3
2
?
= 
249
2
= 0.008
(c) du = nC
v
dT ? C
v
= 
dndTu
= 
300 008 . 0
30
?
= 12.5
C
p
= C
v
+ R = 12.5 + 8.3 = 20.3
(d) C
v
= 12.5 (Proved above)
8. Q = Amt of heat given
Work done = 
2
Q
, ?Q = W + ? U
for monoatomic gas ? ?U = Q –
2
Q
= 
2
Q
V = n
2
3
RT = 
2
Q
= nT× 
2
3
R = 3R × nT
Again Q = n CpdT Where C
P
> Molar heat capacity at const. pressure.
3RnT = ndTC
P
? C
P
= 3R ?
9. P = KV ?
V
nRT
= KV  ? RT = KV
2
  ? R ?T = 2KV ?U ?
KV 2
T R ?
= dv
dQ = du + dw ? mcdT = C
V
dT + pdv ? msdT = C
V
dT+ 
KV 2
PRdF
? ms = C
V
+ 
KV 2
RKV
? C
P
+ 
2
R
10.
V
P
C
C
= ?, C
P
– C
V 
= R, C
V
= 
1
r
? ?
, C
P
= 
1
R
? ?
?
Pdv = ? ? Rdt
1 b
1
?
? 0 = C
V
dT + ? ? Rdt
1 b
1
?
?
1 b
1
?
= 
R
C
V
?
? b + 1 = 
V
C
R ?
= 
? ?
V
V P
C
C C ? ?
= – ? +1 ? b = – ?
11. Considering two gases, in Gas(1) we have,
?, Cp
1
(Sp. Heat at const. ‘P’), Cv
1
(Sp. Heat at const. ‘V’), n
1
(No. of moles)
1
1
Cv
Cp
??? & Cp
1
– Cv
1
= R 
Specific Heat Capacities of Gases
27.3
? ?Cv
1
– Cv
1
= R ? Cv
1
( ? – 1) = R 
? Cv
1
= 
1
R
? ?
& Cp
1
=
1
R
? ?
?
In Gas(2) we have, ?, Cp
2
(Sp. Heat at const. ‘P’), Cv
2
(Sp. Heat at const. ‘V’), n
2
(No. of moles)
2
2
Cv
Cp
??? & Cp
2
– Cv
2
= R ? ?Cv
2
– Cv
2
= R ? Cv
2
( ? – 1) = R ? Cv
2
= 
1
R
? ?
& Cp
2
=
1
R
? ?
?
Given n
1
: n
2
= 1 :2
dU
1
= nCv
1
dT & dU
2
= 2nCv
2
dT = 3nCvdT
? C
V
= 
3
Cv 2 Cv
2 1
?
= 
3
1
R 2
1
R
? ?
?
? ?
= 
) 1 ( 3
R 3
? ?
= 
1
R
? ?
…(1)
&Cp = ?Cv = 
1
r
? ?
?
…(2)
So, 
Cv
Cp
= ? [from (1) & (2)] ?
12. Cp ? = 2.5 RCp ? = 3.5 R
Cv ? = 1.5 R Cv ? = 2.5 R
n
1
= n
2
= 1 mol (n
1
+ n
2
)C
V
dT = n
1
C
v
dT + n
2
C
v
dT
? C
V
= 
2 1
2 1
n n
v C n v C n
?
? ? ? ?
= 
2
R 5 . 2 R 5 . 1 ?
2R
C
P
= C
V
+ R = 2R + R = 3R
? = 
V
p
C
C
= 
R 2
R 3
= 1.5
13. n = 
2
1
  mole, R =
3
25
J/mol-k, ? = 
3
5
(a) Temp at A = T
a
, P
a
V
a
= nRT
a
? T
a
= 
nR
V P
a a
= 
3
25
2
1
10 100 10 5000
3 6
?
? ? ?
?
= 120 k.
Similarly temperatures at point b = 240 k at C it is 480 k and at D it is 240 k.
(b) For ab process,
dQ = nCpdT [since ab is isobaric]
= ? ?
a b
T T
1
R
2
1
?
? ?
?
? = ) 120 240 (
1
3
5
3
5
3
35
2
1
? ?
?
?
? = 120
2
3
9
125
2
1
? ? ? = 1250 J
For bc, dQ = du + dw [dq = 0, Isochorie process]
? dQ = du = nC
v
dT = ? ?
a c
T T
1
nR
?
? ?
= ) 240 (
1
3
5
3
25
2
1
?
?
?
?
?
?
?
? = 240
2
3
3
25
2
1
? ? ? = 1500 J
(c) Heat liberated in cd = – nC
p
dT
= ? ?
c d
T T
1
nR
2
1
?
? ?
?
?
= 240
2
3
3
125
2
1
? ? ?
?
= 2500 J
Heat liberated in da = – nC
v
dT
= ? ?
d a
T T
1
R
2
1
?
? ?
?
?
= ) 240 120 (
2
25
2
1
? ? ?
?
= 750 J 
100 KPa
5000 cm
3
a b
c
d
Ta Tb
Tc
Td
10000 cm
3
200 KPa
Specific Heat Capacities of Gases
27.4
14. (a) For a, b ’V’ is constant
So, 
2
2
1
1
T
P
T
P
? ?
300
100
= 
2
T
200
? T
2
= 
100
300 200 ?
= 600 k
For b,c ‘P’ is constant
So, 
2
2
1
1
T
V
T
V
? ?
600
100
= 
2
T
150
? T
2
= 
100
150 600 ?
= 900 k
(b) Work done = Area enclosed under the graph 50 cc × 200 kpa = 50 × 10
–6
× 200 × 10
3
J = 10 J
(c) ‘Q’ Supplied = nC
v
dT
Now, n = 
RT
PV
considering at pt. ‘b’
C
v
= dT
1
R
? ?
= 300 a, b.
Q
bc
= dT
1
R
RT
PV
? ?
? = 300
67 . 0 600
10 100 10 200
6 3
?
?
? ? ?
?
= 14.925 ( ?? = 1.67)
Q supplied to be nC
p
dT [ ?C
p
= 
1
R
? ?
?
]
= dT
1
R
RT
PV
? ?
?
? = 300
67 . 0
3 . 8 67 . 1
900 3 . 8
10 150 10 200
6 3
?
?
?
?
? ? ?
?
= 24.925
(d) Q = ?U + w
Now, ?U = Q – w = Heat supplied – Work done = (24.925 + 14.925) – 1 = 29.850 ?
15. In Joly’s differential steam calorimeter
C
v
= 
) ( m
L m
1 2 1
2
? ? ?
m
2
= Mass of steam condensed = 0.095 g, L = 540 Cal/g = 540 × 4.2 J/g
m
1
= Mass of gas present = 3 g, ?
1
= 20°C, ?
2
= 100°C
? C
v
= 
) 20 100 ( 3
2 . 4 540 095 . 0
?
? ?
= 0.89 ˜ 0.9 J/g-K ?
16. ? = 1.5
Since it is an adiabatic process, So PV
?
= const.
(a) P
1
V
1
?
= P
2
V
2
?
Given V
1
= 4 L, V
2
= 3 L, 
1
2
P
P
= ?
??
1
2
P
P
= 
?
?
?
?
?
?
?
?
?
2
1
V
V
= 
5 . 1
3
4
?
?
?
?
?
?
= 1.5396 ˜ 1.54
(b) TV
?–1
= Const.
T
1
V
1
?–1
= T
2
V
2
?–1
?
1
2
T
T
= 
1
2
1
V
V
? ?
?
?
?
?
?
?
?
?
= 
5 . 0
3
4
?
?
?
?
?
?
= 1.154 
17. P
1
= 2.5 × 10
5
Pa, V
1
= 100 cc, T
1
= 300 k
(a) P
1
V
1
?
= P
2
V
2
?
? 2.5 × 10
5
× V
1.5
= 
2
5 . 1
P
2
V
? ?
?
?
?
?
?
? P
2
= 2
1.5
× 2.5 × 10
5
= 7.07 × 10
5
˜ 7.1 × 10
5
(b) T
1
V
1
?–1
= T
2
V
2
?–1
? 300 × (100)
1.5 – 1
= T
2
× (50)
1.5 – 1
? T
2
= 
07 . 7
3000
= 424.32 k ˜ 424 k 
a
100 KPa 200 KPa
150 cm
3
100 cm
3
b
c
Specific Heat Capacities of Gases
27.5
(c) Work done by the gas in the process
W = ? ?
1 2
T T
1
mR
?
? ?
= ? ?
1 2
1 1
T T
) 1 ( T
V P
?
? ?
= ] 300 424 [
) 1 5 , 1 ( 300
10 100 10 5 . 2
6 5
?
?
? ? ?
?
= 124
5 . 0 300
10 5 . 2
?
?
?
= 20.72 ˜ 21 J 
18. ? = 1.4, T
1
= 20°C = 293 k, P
1
  = 2 atm, p
2
  = 1 atm
We know for adiabatic process,
P
1
1– ?
× T
1
?
= P
2
1– ?
× T
2
?
?or (2)
1–1.4
× (293)
1.4
= (1)
1–1.4
× T
2
1.4
? (2)
0.4
× (293)
1.4
= T
2
1.4
? 2153.78 = T
2
1.4
? T
2
  = (2153.78)
1/1.4
= 240.3 K
19. P
1
= 100 KPa = 10
5
Pa, V
1
= 400 cm
3
= 400 × 10
–6
m
3
, T
1
  = 300 k, 
? = 
V
P
C
C
= 1.5
(a) Suddenly compressed to V
2
= 100 cm
3
P
1
V
1
?
= P
2
V
2
?
? 10
5
(400)
1.5
= P
2
× (100)
1.5
? P
2
= 10
5 
× (4)
1.5
= 800 KPa
T
1
V
1
?–1
= T
2
V
2
?–1
?  300 × (400)
1.5–1
= T
2
× (100)
1.5-1
? T
2
= 
10
20 300 ?
= 600 K
(b) Even if the container is slowly compressed the walls are adiabatic so heat transferred is 0.
Thus the values remain, P
2
= 800 KPa, T
2
= 600 K.
20. Given 
V
P
C
C
= ? P
0
(Initial Pressure), V
0
(Initial Volume)
(a) (i) Isothermal compression, P
1
V
1
= P
2
V
2
  or, P
0
V
0
= 
2
V P
0 2
? P
2
  = 2P
0
(ii) Adiabatic Compression P
1
V
1
?
= P
2
V
2
?
or 2P
0
?
?
?
?
?
?
?
2
V
0
= P
1
?
?
?
?
?
?
?
4
V
0
? P ? = 
?
?
?
?
? ?
0
0
o
V
4
P 2
2
V
= 2
?
× 2 P
0
? P
0
2
?+1
(b) (i) Adiabatic compression P
1
V
1
?
= P
2
V
2
?
or P
0
V
0
?
= 
?
?
?
?
?
?
?
?
2
V
P
0
? P ? = P
0
2
?
(ii) Isothermal compression P
1
V
1
= P
2
V
2
   or 2
?
P
0
× 
2
V
0
= P
2
× 
4
V
0
? P
2
= P
0
2
?+1
  
21. Initial pressure = P
0
Initial Volume = V
0
? =
V
P
C
C
(a) Isothermally to pressure 
2
P
0
P
0
V
0
= 
1
0
V
2
P
? V
1
= 2 V
0
Adiabetically to pressure = 
4
P
0
? ?
?
1
0
V
2
P
= ? ?
?
2
0
V
4
P
?
?
) V 2 (
2
P
0
0
= 
?
) V (
4
P
2
0
? 2
?+1
V
0
?
= V
2
?
? V
2
= 2
( ?+1)/ ?
V
0
? Final Volume = 2
( ?+1)/ ?
V
0
Read More
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