DC Pandey Solutions: First Law of Thermodynamics- 1 | DC Pandey Solutions for JEE Physics PDF Download

 Page 1


18. First Law of Thermodynamics
Introductory Exercise 18.1
1. (a) D D W p V = = - ´ - 1.7 1.2 0.8 10
5
( ) J 
= - ´ 6.8 10
4
 J
(b)   DV = ´ 1.1 10
5
 J 
Þ D D D Q U W = + = - ´ 17.8 10
4
 J
i.e., 1.78 ´ 10
5
 J of heat has flown out of
the gas.
(c) No, it is independent of the type of
the gas.
2. (a) In p V - graph of cyclic process,
clockwise rotation gives positive
work and anticlockwise gives
negative work. And as  loop 1 has
greater area than loop 2, that is why 
total work done by the system is
positive.
(b)As in cyclic process change in
internal energy is zero, that’s why
for positive work done by the
system, heat flows into the system.
(c) In loop ‘1’ work done is positive so,
heat flows into the system and in
loop ‘2’ work done is negative so
heat flows out of the system.
3. As the box is insulated i.e., no heat
exchange takes place with surrounding
and as the gas expands against vacuum
i.e., zero pressure that’s why no work
has been done and there is no change in
internal energy. Thus, temperature do
not change, internal energy and gas does 
not do any work.
4. U
f
nRT nRT = =
2
3
2
 
Þ n
U
RT
= =
´
´ ´
2
3
2 100
3 300 8.314
    = 0.0267 mole.
5. D D Q ms = = ´ ´ q 1 387 30 J = 11610 J
   D D D V V
m
= = ´ g q
r
a q 3
     =
´
´ ´ ´ ´
-
1
10
3 7 10 30
3
6
8.92
    = ´
-
7.1 10
8
 m
- 3
 D D W p V = = ´ ´ ´
-
1.01 7.06 10 10
5 8
    = ´
-
7.13 10
3
 J
  D D D U Q W = - = 11609.99 J
Introductory Exercise 18.2
1. (a) At constant volume, 
DU = 0 Þ DW = 0
D D Q nC T
V
= 
Þ    D
D
= =
´ ´
=
Q
nC
V
200
1
3
2
8.314
16.04 K
\  T T T
f i
= + = + D 300 16.04  = 316.04
K
(b) At constant pressure, 
D
D
T
Q
nC
p
= =
´ ´
200
1
5
2
314 8.
 = 9.62  K
Þ  T
f
= + 300 9.62 K = 309.62 K
2. For adiabatic process,
pVg = constant = c (say)
\ pdV
c
V
dV c V dV
V
V
V
V
V
V
i
f
i
f
i
f
ò ò ò
= =
-
g
g
Page 2


18. First Law of Thermodynamics
Introductory Exercise 18.1
1. (a) D D W p V = = - ´ - 1.7 1.2 0.8 10
5
( ) J 
= - ´ 6.8 10
4
 J
(b)   DV = ´ 1.1 10
5
 J 
Þ D D D Q U W = + = - ´ 17.8 10
4
 J
i.e., 1.78 ´ 10
5
 J of heat has flown out of
the gas.
(c) No, it is independent of the type of
the gas.
2. (a) In p V - graph of cyclic process,
clockwise rotation gives positive
work and anticlockwise gives
negative work. And as  loop 1 has
greater area than loop 2, that is why 
total work done by the system is
positive.
(b)As in cyclic process change in
internal energy is zero, that’s why
for positive work done by the
system, heat flows into the system.
(c) In loop ‘1’ work done is positive so,
heat flows into the system and in
loop ‘2’ work done is negative so
heat flows out of the system.
3. As the box is insulated i.e., no heat
exchange takes place with surrounding
and as the gas expands against vacuum
i.e., zero pressure that’s why no work
has been done and there is no change in
internal energy. Thus, temperature do
not change, internal energy and gas does 
not do any work.
4. U
f
nRT nRT = =
2
3
2
 
Þ n
U
RT
= =
´
´ ´
2
3
2 100
3 300 8.314
    = 0.0267 mole.
5. D D Q ms = = ´ ´ q 1 387 30 J = 11610 J
   D D D V V
m
= = ´ g q
r
a q 3
     =
´
´ ´ ´ ´
-
1
10
3 7 10 30
3
6
8.92
    = ´
-
7.1 10
8
 m
- 3
 D D W p V = = ´ ´ ´
-
1.01 7.06 10 10
5 8
    = ´
-
7.13 10
3
 J
  D D D U Q W = - = 11609.99 J
Introductory Exercise 18.2
1. (a) At constant volume, 
DU = 0 Þ DW = 0
D D Q nC T
V
= 
Þ    D
D
= =
´ ´
=
Q
nC
V
200
1
3
2
8.314
16.04 K
\  T T T
f i
= + = + D 300 16.04  = 316.04
K
(b) At constant pressure, 
D
D
T
Q
nC
p
= =
´ ´
200
1
5
2
314 8.
 = 9.62  K
Þ  T
f
= + 300 9.62 K = 309.62 K
2. For adiabatic process,
pVg = constant = c (say)
\ pdV
c
V
dV c V dV
V
V
V
V
V
V
i
f
i
f
i
f
ò ò ò
= =
-
g
g
=
- +
- +
c
V
V
V
i
f g
g
1
1
=
-
-
- + - +
c
V V
f i
g g
g
1 1
1
=
× - ×
-
- + - +
p V V p V V
f f f
i i f
g g g
g
1 1
1
=
-
-
=
-
-
p V p V p V p V
f f i i i i f f
1 1 g g
 (Proved)
3. DW
AB
= + 500 J,DQ
AB
= + 250 J
Þ DU
AB
= - 250 J
DW
AC
= + 700 J, DQ
AC
= + 300 J
Þ DU
AC
= - 400 J
(a) Path BC is isochoric process, i.e., 
DW
BC
= 0
\               D D D D Q U U U
BC BC AC AB
= = -
= - - - = - 400 250 150 J J J ( ) 
(b) D D D W W W
CDA CD DA
= +
= - + = - 800 0 800 J J
(Work is negative as volume is
decreasing)
D D D U U U
CDA AC AC
= = - = 400 J
Þ D D D Q W U
CDA CDA CDA
= +
= - + = - 800 400 400 J J J
4. (a)     T
pV
nR
= =
´ ´ ´
´
-
1 10 2 10
1
2 5
8.314
= 240.6 K
(b)  D
D
W
p V
=
-
=
´ ´ ´
-
-
g 1
2 10 5 10
5
3
1
5 3
=
10
2 3
3
/
 J
5. (a) 
DK
p
m
p
m
p
m m
i f i f
= - = -
æ
è
ç
ç
ö
ø
÷
÷
2 2 2
2 2 2
1 1
    =
´ ´
´
-
é
ë
ê
ù
û
ú
-
-
( ) 10 10 200
2
1
10 10
1
3 2
3
2.01
= -
é
ë
ê
ù
û
ú
= 2 100
1
2
199 J
(b) D D Q nC T
V
= Þ D
D D
T
Q
nC
Q
m
M
C
V
V
= =
=
´
=
´
´ ´
M Q
m R
D
3
200 199
2010 3 8.314
= ° 0.8 C
6. D D D D D W W W W W
A B BC CD DA
= + + +
=
æ
è
ç
ç
ö
ø
÷
÷
+ - nRT m
p
p
p V V
C BC 1
1
2
2
( )
+
æ
è
ç
ç
ö
ø
÷
÷
+ - nRTm
p
p
p V V
2
1
1 1 2
( )
= -
æ
è
ç
ç
ö
ø
÷
÷
+ - nR T T
p
p
p V p V ( ) ln
2 1
2
1
1 2 1 1
+ - p V p V
1 1 1 2
= -
æ
è
ç
ç
ö
ø
÷
÷
( ) ln p V p V
p
p
2 2 1 1
2
1
7. D W
ABCA
= (+)ve Þ DW
AB
= (+)ve,
DW
BC
= 0, DW
CA
= - ( )ve
For BC, DQ = - ( )ve Þ DU
BC
= - ( )ve and 
     DW
BC
= 0
For CA, DU = - ( )ve Þ DQ
CA
= - ( )ve as 
DW
CA
= - ( )ve.
DU DW DQ
AB + + +
BC - 0 -
CA - - -
Total 0 + +
For AB, as DU
ABCA
= 0 and
DU
BC
= - ( )ve,
DU
CA
= - ( ) ve
Þ DU
AB
= - ( ) ve
As      D D Q W
ABCA ABCA
= = + ( )ve and
DQ
BC
= - ( )ve
DQ
CA
= - ( )ve Þ DQ
AB
= + ( )ve
In isobaric process, D D D W p V nR T = =
= ´ ´ - = 0.2 8.314 166.3 ( ) 300 200 J
9. DW pdV V dV V = = =
ò ò
a a
2 3
1
3
= ´ ´ ´ ´ -
1
1.01
3
5 10 2 1
5 3 3
( )
= ´ 1.18 10
6
 J
49  |  First Law of Thermodynamics
Page 3


18. First Law of Thermodynamics
Introductory Exercise 18.1
1. (a) D D W p V = = - ´ - 1.7 1.2 0.8 10
5
( ) J 
= - ´ 6.8 10
4
 J
(b)   DV = ´ 1.1 10
5
 J 
Þ D D D Q U W = + = - ´ 17.8 10
4
 J
i.e., 1.78 ´ 10
5
 J of heat has flown out of
the gas.
(c) No, it is independent of the type of
the gas.
2. (a) In p V - graph of cyclic process,
clockwise rotation gives positive
work and anticlockwise gives
negative work. And as  loop 1 has
greater area than loop 2, that is why 
total work done by the system is
positive.
(b)As in cyclic process change in
internal energy is zero, that’s why
for positive work done by the
system, heat flows into the system.
(c) In loop ‘1’ work done is positive so,
heat flows into the system and in
loop ‘2’ work done is negative so
heat flows out of the system.
3. As the box is insulated i.e., no heat
exchange takes place with surrounding
and as the gas expands against vacuum
i.e., zero pressure that’s why no work
has been done and there is no change in
internal energy. Thus, temperature do
not change, internal energy and gas does 
not do any work.
4. U
f
nRT nRT = =
2
3
2
 
Þ n
U
RT
= =
´
´ ´
2
3
2 100
3 300 8.314
    = 0.0267 mole.
5. D D Q ms = = ´ ´ q 1 387 30 J = 11610 J
   D D D V V
m
= = ´ g q
r
a q 3
     =
´
´ ´ ´ ´
-
1
10
3 7 10 30
3
6
8.92
    = ´
-
7.1 10
8
 m
- 3
 D D W p V = = ´ ´ ´
-
1.01 7.06 10 10
5 8
    = ´
-
7.13 10
3
 J
  D D D U Q W = - = 11609.99 J
Introductory Exercise 18.2
1. (a) At constant volume, 
DU = 0 Þ DW = 0
D D Q nC T
V
= 
Þ    D
D
= =
´ ´
=
Q
nC
V
200
1
3
2
8.314
16.04 K
\  T T T
f i
= + = + D 300 16.04  = 316.04
K
(b) At constant pressure, 
D
D
T
Q
nC
p
= =
´ ´
200
1
5
2
314 8.
 = 9.62  K
Þ  T
f
= + 300 9.62 K = 309.62 K
2. For adiabatic process,
pVg = constant = c (say)
\ pdV
c
V
dV c V dV
V
V
V
V
V
V
i
f
i
f
i
f
ò ò ò
= =
-
g
g
=
- +
- +
c
V
V
V
i
f g
g
1
1
=
-
-
- + - +
c
V V
f i
g g
g
1 1
1
=
× - ×
-
- + - +
p V V p V V
f f f
i i f
g g g
g
1 1
1
=
-
-
=
-
-
p V p V p V p V
f f i i i i f f
1 1 g g
 (Proved)
3. DW
AB
= + 500 J,DQ
AB
= + 250 J
Þ DU
AB
= - 250 J
DW
AC
= + 700 J, DQ
AC
= + 300 J
Þ DU
AC
= - 400 J
(a) Path BC is isochoric process, i.e., 
DW
BC
= 0
\               D D D D Q U U U
BC BC AC AB
= = -
= - - - = - 400 250 150 J J J ( ) 
(b) D D D W W W
CDA CD DA
= +
= - + = - 800 0 800 J J
(Work is negative as volume is
decreasing)
D D D U U U
CDA AC AC
= = - = 400 J
Þ D D D Q W U
CDA CDA CDA
= +
= - + = - 800 400 400 J J J
4. (a)     T
pV
nR
= =
´ ´ ´
´
-
1 10 2 10
1
2 5
8.314
= 240.6 K
(b)  D
D
W
p V
=
-
=
´ ´ ´
-
-
g 1
2 10 5 10
5
3
1
5 3
=
10
2 3
3
/
 J
5. (a) 
DK
p
m
p
m
p
m m
i f i f
= - = -
æ
è
ç
ç
ö
ø
÷
÷
2 2 2
2 2 2
1 1
    =
´ ´
´
-
é
ë
ê
ù
û
ú
-
-
( ) 10 10 200
2
1
10 10
1
3 2
3
2.01
= -
é
ë
ê
ù
û
ú
= 2 100
1
2
199 J
(b) D D Q nC T
V
= Þ D
D D
T
Q
nC
Q
m
M
C
V
V
= =
=
´
=
´
´ ´
M Q
m R
D
3
200 199
2010 3 8.314
= ° 0.8 C
6. D D D D D W W W W W
A B BC CD DA
= + + +
=
æ
è
ç
ç
ö
ø
÷
÷
+ - nRT m
p
p
p V V
C BC 1
1
2
2
( )
+
æ
è
ç
ç
ö
ø
÷
÷
+ - nRTm
p
p
p V V
2
1
1 1 2
( )
= -
æ
è
ç
ç
ö
ø
÷
÷
+ - nR T T
p
p
p V p V ( ) ln
2 1
2
1
1 2 1 1
+ - p V p V
1 1 1 2
= -
æ
è
ç
ç
ö
ø
÷
÷
( ) ln p V p V
p
p
2 2 1 1
2
1
7. D W
ABCA
= (+)ve Þ DW
AB
= (+)ve,
DW
BC
= 0, DW
CA
= - ( )ve
For BC, DQ = - ( )ve Þ DU
BC
= - ( )ve and 
     DW
BC
= 0
For CA, DU = - ( )ve Þ DQ
CA
= - ( )ve as 
DW
CA
= - ( )ve.
DU DW DQ
AB + + +
BC - 0 -
CA - - -
Total 0 + +
For AB, as DU
ABCA
= 0 and
DU
BC
= - ( )ve,
DU
CA
= - ( ) ve
Þ DU
AB
= - ( ) ve
As      D D Q W
ABCA ABCA
= = + ( )ve and
DQ
BC
= - ( )ve
DQ
CA
= - ( )ve Þ DQ
AB
= + ( )ve
In isobaric process, D D D W p V nR T = =
= ´ ´ - = 0.2 8.314 166.3 ( ) 300 200 J
9. DW pdV V dV V = = =
ò ò
a a
2 3
1
3
= ´ ´ ´ ´ -
1
1.01
3
5 10 2 1
5 3 3
( )
= ´ 1.18 10
6
 J
49  |  First Law of Thermodynamics
Introductory Exercise 18.3
1.
DW nRT
BB
=  ln ln
V
V
R
B
A
æ
è
ç
ç
ö
ø
÷
÷
= ´ 3 273 5
= 10959 J
DW
BC
= 0
D D D Q U W = +
D D D U Q W = -
       = - 80000 10959
       = 69041
T T
f i
= = ´ 5 5 273 K = 1365 K
D D D Q Q Q
ABC AB BC
= +
= + + + D D W U
BC BC
0 0
    D D Q nC T
BC V
=  Þ C
Q
n T
V
BC
=
D
D
=
´ ´
=
69041
3 4 273
21.07
   C C R
p V
= + = 29.39
Þ g = = =
C
C
p
V
29.39
21.07
1.4
2. D D D Q U W = + ; D D Q nC T
p
=
Þ 1600 1 72 = × × C
p
 
Þ     C
p
= 22.22
C C R
V p
= - = 13.9 Þ g = =
C
C
p
V
1.6
       D D D W Q U = - = - 1600 nC T
V
D
= - ´ ´ 1600 1 72 13.9
= - 1600 1000.8 J
= 599.2 J
and D D U nC T
V
= = ´ ´ 1 72 13.9 
= 1001
= 1 kJ
3. D D D W p V =
1
2
= ´ ´ ´ ´ ´
-
1
2
20 10 1 10
5 3
1.01
= ´ = 10 101 1010 J
\  p
n W
t
= =
´ D
D
100 1010
60
J
s
  = 1.68 kW
AIEEE Corner
¢ Subjectve Questions (Level 1)
1. D D D U Q W = - = + 254 73 J J
  = 327 J
2. (a) D
D D
T
Q
nC
Q
nR
V
= = =
´
´ ´
2
3
2 200
2 1 8.314
= 16 K 
Þ    T T T
f i
= + = D 316 K
(b) D
D D
T
Q
nC
Q
nR
p
¢ = = =
´
´ ´
2
5
2 200
5 1 8.314
= 9.6 K
Þ T T T
f i
¢ = + ¢ = D 309.6 K
3. D D U nC T
V
= , in adiabatic process,
DQ = 0 and D D U W = -
 First Law of Thermodynamics   | 50
p
0/5
p
0
p
C 
  A
V
0
5V
0
V
B
Page 4


18. First Law of Thermodynamics
Introductory Exercise 18.1
1. (a) D D W p V = = - ´ - 1.7 1.2 0.8 10
5
( ) J 
= - ´ 6.8 10
4
 J
(b)   DV = ´ 1.1 10
5
 J 
Þ D D D Q U W = + = - ´ 17.8 10
4
 J
i.e., 1.78 ´ 10
5
 J of heat has flown out of
the gas.
(c) No, it is independent of the type of
the gas.
2. (a) In p V - graph of cyclic process,
clockwise rotation gives positive
work and anticlockwise gives
negative work. And as  loop 1 has
greater area than loop 2, that is why 
total work done by the system is
positive.
(b)As in cyclic process change in
internal energy is zero, that’s why
for positive work done by the
system, heat flows into the system.
(c) In loop ‘1’ work done is positive so,
heat flows into the system and in
loop ‘2’ work done is negative so
heat flows out of the system.
3. As the box is insulated i.e., no heat
exchange takes place with surrounding
and as the gas expands against vacuum
i.e., zero pressure that’s why no work
has been done and there is no change in
internal energy. Thus, temperature do
not change, internal energy and gas does 
not do any work.
4. U
f
nRT nRT = =
2
3
2
 
Þ n
U
RT
= =
´
´ ´
2
3
2 100
3 300 8.314
    = 0.0267 mole.
5. D D Q ms = = ´ ´ q 1 387 30 J = 11610 J
   D D D V V
m
= = ´ g q
r
a q 3
     =
´
´ ´ ´ ´
-
1
10
3 7 10 30
3
6
8.92
    = ´
-
7.1 10
8
 m
- 3
 D D W p V = = ´ ´ ´
-
1.01 7.06 10 10
5 8
    = ´
-
7.13 10
3
 J
  D D D U Q W = - = 11609.99 J
Introductory Exercise 18.2
1. (a) At constant volume, 
DU = 0 Þ DW = 0
D D Q nC T
V
= 
Þ    D
D
= =
´ ´
=
Q
nC
V
200
1
3
2
8.314
16.04 K
\  T T T
f i
= + = + D 300 16.04  = 316.04
K
(b) At constant pressure, 
D
D
T
Q
nC
p
= =
´ ´
200
1
5
2
314 8.
 = 9.62  K
Þ  T
f
= + 300 9.62 K = 309.62 K
2. For adiabatic process,
pVg = constant = c (say)
\ pdV
c
V
dV c V dV
V
V
V
V
V
V
i
f
i
f
i
f
ò ò ò
= =
-
g
g
=
- +
- +
c
V
V
V
i
f g
g
1
1
=
-
-
- + - +
c
V V
f i
g g
g
1 1
1
=
× - ×
-
- + - +
p V V p V V
f f f
i i f
g g g
g
1 1
1
=
-
-
=
-
-
p V p V p V p V
f f i i i i f f
1 1 g g
 (Proved)
3. DW
AB
= + 500 J,DQ
AB
= + 250 J
Þ DU
AB
= - 250 J
DW
AC
= + 700 J, DQ
AC
= + 300 J
Þ DU
AC
= - 400 J
(a) Path BC is isochoric process, i.e., 
DW
BC
= 0
\               D D D D Q U U U
BC BC AC AB
= = -
= - - - = - 400 250 150 J J J ( ) 
(b) D D D W W W
CDA CD DA
= +
= - + = - 800 0 800 J J
(Work is negative as volume is
decreasing)
D D D U U U
CDA AC AC
= = - = 400 J
Þ D D D Q W U
CDA CDA CDA
= +
= - + = - 800 400 400 J J J
4. (a)     T
pV
nR
= =
´ ´ ´
´
-
1 10 2 10
1
2 5
8.314
= 240.6 K
(b)  D
D
W
p V
=
-
=
´ ´ ´
-
-
g 1
2 10 5 10
5
3
1
5 3
=
10
2 3
3
/
 J
5. (a) 
DK
p
m
p
m
p
m m
i f i f
= - = -
æ
è
ç
ç
ö
ø
÷
÷
2 2 2
2 2 2
1 1
    =
´ ´
´
-
é
ë
ê
ù
û
ú
-
-
( ) 10 10 200
2
1
10 10
1
3 2
3
2.01
= -
é
ë
ê
ù
û
ú
= 2 100
1
2
199 J
(b) D D Q nC T
V
= Þ D
D D
T
Q
nC
Q
m
M
C
V
V
= =
=
´
=
´
´ ´
M Q
m R
D
3
200 199
2010 3 8.314
= ° 0.8 C
6. D D D D D W W W W W
A B BC CD DA
= + + +
=
æ
è
ç
ç
ö
ø
÷
÷
+ - nRT m
p
p
p V V
C BC 1
1
2
2
( )
+
æ
è
ç
ç
ö
ø
÷
÷
+ - nRTm
p
p
p V V
2
1
1 1 2
( )
= -
æ
è
ç
ç
ö
ø
÷
÷
+ - nR T T
p
p
p V p V ( ) ln
2 1
2
1
1 2 1 1
+ - p V p V
1 1 1 2
= -
æ
è
ç
ç
ö
ø
÷
÷
( ) ln p V p V
p
p
2 2 1 1
2
1
7. D W
ABCA
= (+)ve Þ DW
AB
= (+)ve,
DW
BC
= 0, DW
CA
= - ( )ve
For BC, DQ = - ( )ve Þ DU
BC
= - ( )ve and 
     DW
BC
= 0
For CA, DU = - ( )ve Þ DQ
CA
= - ( )ve as 
DW
CA
= - ( )ve.
DU DW DQ
AB + + +
BC - 0 -
CA - - -
Total 0 + +
For AB, as DU
ABCA
= 0 and
DU
BC
= - ( )ve,
DU
CA
= - ( ) ve
Þ DU
AB
= - ( ) ve
As      D D Q W
ABCA ABCA
= = + ( )ve and
DQ
BC
= - ( )ve
DQ
CA
= - ( )ve Þ DQ
AB
= + ( )ve
In isobaric process, D D D W p V nR T = =
= ´ ´ - = 0.2 8.314 166.3 ( ) 300 200 J
9. DW pdV V dV V = = =
ò ò
a a
2 3
1
3
= ´ ´ ´ ´ -
1
1.01
3
5 10 2 1
5 3 3
( )
= ´ 1.18 10
6
 J
49  |  First Law of Thermodynamics
Introductory Exercise 18.3
1.
DW nRT
BB
=  ln ln
V
V
R
B
A
æ
è
ç
ç
ö
ø
÷
÷
= ´ 3 273 5
= 10959 J
DW
BC
= 0
D D D Q U W = +
D D D U Q W = -
       = - 80000 10959
       = 69041
T T
f i
= = ´ 5 5 273 K = 1365 K
D D D Q Q Q
ABC AB BC
= +
= + + + D D W U
BC BC
0 0
    D D Q nC T
BC V
=  Þ C
Q
n T
V
BC
=
D
D
=
´ ´
=
69041
3 4 273
21.07
   C C R
p V
= + = 29.39
Þ g = = =
C
C
p
V
29.39
21.07
1.4
2. D D D Q U W = + ; D D Q nC T
p
=
Þ 1600 1 72 = × × C
p
 
Þ     C
p
= 22.22
C C R
V p
= - = 13.9 Þ g = =
C
C
p
V
1.6
       D D D W Q U = - = - 1600 nC T
V
D
= - ´ ´ 1600 1 72 13.9
= - 1600 1000.8 J
= 599.2 J
and D D U nC T
V
= = ´ ´ 1 72 13.9 
= 1001
= 1 kJ
3. D D D W p V =
1
2
= ´ ´ ´ ´ ´
-
1
2
20 10 1 10
5 3
1.01
= ´ = 10 101 1010 J
\  p
n W
t
= =
´ D
D
100 1010
60
J
s
  = 1.68 kW
AIEEE Corner
¢ Subjectve Questions (Level 1)
1. D D D U Q W = - = + 254 73 J J
  = 327 J
2. (a) D
D D
T
Q
nC
Q
nR
V
= = =
´
´ ´
2
3
2 200
2 1 8.314
= 16 K 
Þ    T T T
f i
= + = D 316 K
(b) D
D D
T
Q
nC
Q
nR
p
¢ = = =
´
´ ´
2
5
2 200
5 1 8.314
= 9.6 K
Þ T T T
f i
¢ = + ¢ = D 309.6 K
3. D D U nC T
V
= , in adiabatic process,
DQ = 0 and D D U W = -
 First Law of Thermodynamics   | 50
p
0/5
p
0
p
C 
  A
V
0
5V
0
V
B
where, D
D
W
nR T
=
- 1 g
Þ D
D
U
nR T
=
- g 1
 for all process.
4.   DV = 0  Þ DW = 0
\  D D D Q U nC T
V
= = = × n R T
5
2
D
= -
5
2
( ) p V p V
f f i i
 = -
5
2
( ) p V V
f i
= ´ - ´ ´
-
5
2
5 10 10 10 10
5 5 3
( )
= ´ ´ ´ =
-
5
2
4 10 10 10
5 2 4
J
5. D D D Q U nC T n R T T
V i i 1 1
5
2
3 = = = × - ( )
   = 5 nRT
i
D D Q nC T n R T T
p i i 2
5
2
6 3 = = × - ( )
   = 7.5nRT
i
\  c
Q
n T
nRT
n T T
R
i
i i
= =
-
=
D
D
12.5 12.5
( ) 6 5
 = 2.5 R
6. DW
AB
= 0 , DW
p
V p V
BC
= ´ =
0
0 0 0
2
1
2
        = × =
1
2
300
0
nRT R
D D D D D Q U W U W
AB BC
= + + + ( ) ( )
 = + + D D D U U W
AB BC BC
 = + 0 300 R 
(As T T
A C
= )
 = ´ 2.49 10
3
 J = 2.49 kJ
7. D D D U Q W = - = - 1200 2100 J J
= 900 J
D
D
T
U
nC
V
= =
´ ´
=
900
5
3
2
8.314
14.43
Þ T T T
f i
= - = ° - ° D 127 C 14.43 C
      = ° 112.6 C
8. When gas expands it does positive work
on the surrounding and for this purpose
heat has to be supplied into the system.
9. D D W V V V
f i
= = - r r( )
= -
æ
è
ç
ç
ö
ø
÷
÷
= -
æ
è
ç
ö
ø
÷
r
r r
r m m
f i
1 1 1
1000
1
999.9
= -
´ ´
´
= -
10 2
1000
5
0.1
999.9
0.02 J
(work done is negative as volume
decreases)
D D Q ms = = ´ ´ q 2 4200 4
 = 33600 J
D D D U Q W = - = 33600.02 J
10. D D W p V pV p
m
f
= » =
r
=
´ ´
=
-
10 10 10
5 3
0.6
1666.67 J
D D Q ms mL = + q
= ´
-
10 4200
2
 ´ + ´ ´
-
100 10 25 10
2 6
.
        = + = 4200 25000 29200 J J J
D D D U Q W = - = 29200 J - 1666.67 J
  = 27533.33 J = ´ 2.75 10
4
 J
11. D D W p V =
  = ´ ´ ´
-
1.013 10 1670 10
5 6
  = ´ 1.013 167 J = 169.2 J
 DQ mL = = ´ ´
-
10 10
3 6
2.256 J
  = 2256 J
\   D D D U Q W = - = - ( ) 2256 169.2 J
      = 2086.8 J » 2087 J
12.   D D W p V = = - ´ ´ 2.3 0.5 10
5
  
      = - ´ 1.15 10
5
 J
    DU = - ´ 1.4 10
5
 J
Þ D D D Q U W = + 
51  |  First Law of Thermodynamics
p /5
0
p
0
p
C 
  A
V
0
2V
0
V
B
Page 5


18. First Law of Thermodynamics
Introductory Exercise 18.1
1. (a) D D W p V = = - ´ - 1.7 1.2 0.8 10
5
( ) J 
= - ´ 6.8 10
4
 J
(b)   DV = ´ 1.1 10
5
 J 
Þ D D D Q U W = + = - ´ 17.8 10
4
 J
i.e., 1.78 ´ 10
5
 J of heat has flown out of
the gas.
(c) No, it is independent of the type of
the gas.
2. (a) In p V - graph of cyclic process,
clockwise rotation gives positive
work and anticlockwise gives
negative work. And as  loop 1 has
greater area than loop 2, that is why 
total work done by the system is
positive.
(b)As in cyclic process change in
internal energy is zero, that’s why
for positive work done by the
system, heat flows into the system.
(c) In loop ‘1’ work done is positive so,
heat flows into the system and in
loop ‘2’ work done is negative so
heat flows out of the system.
3. As the box is insulated i.e., no heat
exchange takes place with surrounding
and as the gas expands against vacuum
i.e., zero pressure that’s why no work
has been done and there is no change in
internal energy. Thus, temperature do
not change, internal energy and gas does 
not do any work.
4. U
f
nRT nRT = =
2
3
2
 
Þ n
U
RT
= =
´
´ ´
2
3
2 100
3 300 8.314
    = 0.0267 mole.
5. D D Q ms = = ´ ´ q 1 387 30 J = 11610 J
   D D D V V
m
= = ´ g q
r
a q 3
     =
´
´ ´ ´ ´
-
1
10
3 7 10 30
3
6
8.92
    = ´
-
7.1 10
8
 m
- 3
 D D W p V = = ´ ´ ´
-
1.01 7.06 10 10
5 8
    = ´
-
7.13 10
3
 J
  D D D U Q W = - = 11609.99 J
Introductory Exercise 18.2
1. (a) At constant volume, 
DU = 0 Þ DW = 0
D D Q nC T
V
= 
Þ    D
D
= =
´ ´
=
Q
nC
V
200
1
3
2
8.314
16.04 K
\  T T T
f i
= + = + D 300 16.04  = 316.04
K
(b) At constant pressure, 
D
D
T
Q
nC
p
= =
´ ´
200
1
5
2
314 8.
 = 9.62  K
Þ  T
f
= + 300 9.62 K = 309.62 K
2. For adiabatic process,
pVg = constant = c (say)
\ pdV
c
V
dV c V dV
V
V
V
V
V
V
i
f
i
f
i
f
ò ò ò
= =
-
g
g
=
- +
- +
c
V
V
V
i
f g
g
1
1
=
-
-
- + - +
c
V V
f i
g g
g
1 1
1
=
× - ×
-
- + - +
p V V p V V
f f f
i i f
g g g
g
1 1
1
=
-
-
=
-
-
p V p V p V p V
f f i i i i f f
1 1 g g
 (Proved)
3. DW
AB
= + 500 J,DQ
AB
= + 250 J
Þ DU
AB
= - 250 J
DW
AC
= + 700 J, DQ
AC
= + 300 J
Þ DU
AC
= - 400 J
(a) Path BC is isochoric process, i.e., 
DW
BC
= 0
\               D D D D Q U U U
BC BC AC AB
= = -
= - - - = - 400 250 150 J J J ( ) 
(b) D D D W W W
CDA CD DA
= +
= - + = - 800 0 800 J J
(Work is negative as volume is
decreasing)
D D D U U U
CDA AC AC
= = - = 400 J
Þ D D D Q W U
CDA CDA CDA
= +
= - + = - 800 400 400 J J J
4. (a)     T
pV
nR
= =
´ ´ ´
´
-
1 10 2 10
1
2 5
8.314
= 240.6 K
(b)  D
D
W
p V
=
-
=
´ ´ ´
-
-
g 1
2 10 5 10
5
3
1
5 3
=
10
2 3
3
/
 J
5. (a) 
DK
p
m
p
m
p
m m
i f i f
= - = -
æ
è
ç
ç
ö
ø
÷
÷
2 2 2
2 2 2
1 1
    =
´ ´
´
-
é
ë
ê
ù
û
ú
-
-
( ) 10 10 200
2
1
10 10
1
3 2
3
2.01
= -
é
ë
ê
ù
û
ú
= 2 100
1
2
199 J
(b) D D Q nC T
V
= Þ D
D D
T
Q
nC
Q
m
M
C
V
V
= =
=
´
=
´
´ ´
M Q
m R
D
3
200 199
2010 3 8.314
= ° 0.8 C
6. D D D D D W W W W W
A B BC CD DA
= + + +
=
æ
è
ç
ç
ö
ø
÷
÷
+ - nRT m
p
p
p V V
C BC 1
1
2
2
( )
+
æ
è
ç
ç
ö
ø
÷
÷
+ - nRTm
p
p
p V V
2
1
1 1 2
( )
= -
æ
è
ç
ç
ö
ø
÷
÷
+ - nR T T
p
p
p V p V ( ) ln
2 1
2
1
1 2 1 1
+ - p V p V
1 1 1 2
= -
æ
è
ç
ç
ö
ø
÷
÷
( ) ln p V p V
p
p
2 2 1 1
2
1
7. D W
ABCA
= (+)ve Þ DW
AB
= (+)ve,
DW
BC
= 0, DW
CA
= - ( )ve
For BC, DQ = - ( )ve Þ DU
BC
= - ( )ve and 
     DW
BC
= 0
For CA, DU = - ( )ve Þ DQ
CA
= - ( )ve as 
DW
CA
= - ( )ve.
DU DW DQ
AB + + +
BC - 0 -
CA - - -
Total 0 + +
For AB, as DU
ABCA
= 0 and
DU
BC
= - ( )ve,
DU
CA
= - ( ) ve
Þ DU
AB
= - ( ) ve
As      D D Q W
ABCA ABCA
= = + ( )ve and
DQ
BC
= - ( )ve
DQ
CA
= - ( )ve Þ DQ
AB
= + ( )ve
In isobaric process, D D D W p V nR T = =
= ´ ´ - = 0.2 8.314 166.3 ( ) 300 200 J
9. DW pdV V dV V = = =
ò ò
a a
2 3
1
3
= ´ ´ ´ ´ -
1
1.01
3
5 10 2 1
5 3 3
( )
= ´ 1.18 10
6
 J
49  |  First Law of Thermodynamics
Introductory Exercise 18.3
1.
DW nRT
BB
=  ln ln
V
V
R
B
A
æ
è
ç
ç
ö
ø
÷
÷
= ´ 3 273 5
= 10959 J
DW
BC
= 0
D D D Q U W = +
D D D U Q W = -
       = - 80000 10959
       = 69041
T T
f i
= = ´ 5 5 273 K = 1365 K
D D D Q Q Q
ABC AB BC
= +
= + + + D D W U
BC BC
0 0
    D D Q nC T
BC V
=  Þ C
Q
n T
V
BC
=
D
D
=
´ ´
=
69041
3 4 273
21.07
   C C R
p V
= + = 29.39
Þ g = = =
C
C
p
V
29.39
21.07
1.4
2. D D D Q U W = + ; D D Q nC T
p
=
Þ 1600 1 72 = × × C
p
 
Þ     C
p
= 22.22
C C R
V p
= - = 13.9 Þ g = =
C
C
p
V
1.6
       D D D W Q U = - = - 1600 nC T
V
D
= - ´ ´ 1600 1 72 13.9
= - 1600 1000.8 J
= 599.2 J
and D D U nC T
V
= = ´ ´ 1 72 13.9 
= 1001
= 1 kJ
3. D D D W p V =
1
2
= ´ ´ ´ ´ ´
-
1
2
20 10 1 10
5 3
1.01
= ´ = 10 101 1010 J
\  p
n W
t
= =
´ D
D
100 1010
60
J
s
  = 1.68 kW
AIEEE Corner
¢ Subjectve Questions (Level 1)
1. D D D U Q W = - = + 254 73 J J
  = 327 J
2. (a) D
D D
T
Q
nC
Q
nR
V
= = =
´
´ ´
2
3
2 200
2 1 8.314
= 16 K 
Þ    T T T
f i
= + = D 316 K
(b) D
D D
T
Q
nC
Q
nR
p
¢ = = =
´
´ ´
2
5
2 200
5 1 8.314
= 9.6 K
Þ T T T
f i
¢ = + ¢ = D 309.6 K
3. D D U nC T
V
= , in adiabatic process,
DQ = 0 and D D U W = -
 First Law of Thermodynamics   | 50
p
0/5
p
0
p
C 
  A
V
0
5V
0
V
B
where, D
D
W
nR T
=
- 1 g
Þ D
D
U
nR T
=
- g 1
 for all process.
4.   DV = 0  Þ DW = 0
\  D D D Q U nC T
V
= = = × n R T
5
2
D
= -
5
2
( ) p V p V
f f i i
 = -
5
2
( ) p V V
f i
= ´ - ´ ´
-
5
2
5 10 10 10 10
5 5 3
( )
= ´ ´ ´ =
-
5
2
4 10 10 10
5 2 4
J
5. D D D Q U nC T n R T T
V i i 1 1
5
2
3 = = = × - ( )
   = 5 nRT
i
D D Q nC T n R T T
p i i 2
5
2
6 3 = = × - ( )
   = 7.5nRT
i
\  c
Q
n T
nRT
n T T
R
i
i i
= =
-
=
D
D
12.5 12.5
( ) 6 5
 = 2.5 R
6. DW
AB
= 0 , DW
p
V p V
BC
= ´ =
0
0 0 0
2
1
2
        = × =
1
2
300
0
nRT R
D D D D D Q U W U W
AB BC
= + + + ( ) ( )
 = + + D D D U U W
AB BC BC
 = + 0 300 R 
(As T T
A C
= )
 = ´ 2.49 10
3
 J = 2.49 kJ
7. D D D U Q W = - = - 1200 2100 J J
= 900 J
D
D
T
U
nC
V
= =
´ ´
=
900
5
3
2
8.314
14.43
Þ T T T
f i
= - = ° - ° D 127 C 14.43 C
      = ° 112.6 C
8. When gas expands it does positive work
on the surrounding and for this purpose
heat has to be supplied into the system.
9. D D W V V V
f i
= = - r r( )
= -
æ
è
ç
ç
ö
ø
÷
÷
= -
æ
è
ç
ö
ø
÷
r
r r
r m m
f i
1 1 1
1000
1
999.9
= -
´ ´
´
= -
10 2
1000
5
0.1
999.9
0.02 J
(work done is negative as volume
decreases)
D D Q ms = = ´ ´ q 2 4200 4
 = 33600 J
D D D U Q W = - = 33600.02 J
10. D D W p V pV p
m
f
= » =
r
=
´ ´
=
-
10 10 10
5 3
0.6
1666.67 J
D D Q ms mL = + q
= ´
-
10 4200
2
 ´ + ´ ´
-
100 10 25 10
2 6
.
        = + = 4200 25000 29200 J J J
D D D U Q W = - = 29200 J - 1666.67 J
  = 27533.33 J = ´ 2.75 10
4
 J
11. D D W p V =
  = ´ ´ ´
-
1.013 10 1670 10
5 6
  = ´ 1.013 167 J = 169.2 J
 DQ mL = = ´ ´
-
10 10
3 6
2.256 J
  = 2256 J
\   D D D U Q W = - = - ( ) 2256 169.2 J
      = 2086.8 J » 2087 J
12.   D D W p V = = - ´ ´ 2.3 0.5 10
5
  
      = - ´ 1.15 10
5
 J
    DU = - ´ 1.4 10
5
 J
Þ D D D Q U W = + 
51  |  First Law of Thermodynamics
p /5
0
p
0
p
C 
  A
V
0
2V
0
V
B
      = - ´ (1.4 + 1.15) 10
5
J
      = - ´ 2.55 10
5
 J
Thus, 2.55 ´ 10
5
 J of heat flows out of the 
system and it is independent of the type
of the gas.
13. In a cyclic process, U = 0 Þ D D Q W =
(a) \ W Q Q Q Q
h
= + + + ( )
1 2 3 4
- + + ( ) W W W
1 2 3
= - - + ( ) 5960 5585 2980 3645
- - - ( ) 2200 825 1100
= - = 1040 275 765 J
(b) h = = =
work done
heat supplied
10.83%
1040
9605
14. (a) DW = ´
¾® ¾®
1
2
AB AC = ´ ´
1
2
2
0 0
p V
= p V
0 0
(b) T
p V
nR
C
=
2
0 0
 and T
p V
nR
A
=
0 0
Þ D D Q nC T nC
p V
nR
CA p p
= = - ×
0 0
= - × = -
5
2
5
2
0 0
0 0
R
p V
R
p V
   T
p V
nR
B
=
3
0 0
, D D Q nC T
AB V
=
= ´ -
æ
è
ç
ö
ø
÷
n R
p V
nR
p V
nR
3
2
3
0 0 0 0
= ´
3
2
2
0 0
p V = 3
0 0
p V
(c) D D D D Q Q Q W
AB BC CA
+ + =
= + - = 3
5
2
0 0 0 0 0 0
p V Q p V p V
BC
D
Þ DQ
p V
BC
=
0 0
2
(d) Temperature is maximum at a point 
D lying somewhere between B and C
where the product pV is maximum.
  p
p
V
p = - +
2
5
0
0
0
Þ pV
p
V
V p V =
-
+
æ
è
ç
ç
ö
ø
÷
÷
2
5
0
0
0
       = - +
2
5
0
0
2
0
p
V
V p V
For pV = maximum 
d
dV
pV ( ) = 0 
Þ        - × + = 2
2
5 0
0
0
0
V
p
V
p
Þ V
V
=
5
4
0
         p
p
V
V
p p = - × + =
2 5
4
5
5
2
0
0
0
0 0
\    T
pV
nR
max
max
( )
=
\ p
p V
R
p V
R
=
×
×
=
5
2
5
4
1
25
8
0 0
0 0
15. V
nRT
p
R
R
A
A
A
= =
´
´
= ´
-
2 300
2 10
3 10
5
3
V
R
R
B
=
´
´
= ´
-
2 400
2 10
4 10
5
3
,
V
R
R
C
=
´
= ´
-
2 400
10
8 10
5
3
V
R
R
0
3
2 300
10
6 10 =
´
= ´
-
DW R = ´ - ´
-
2 10 4 3 10
5 3
( )
+ ´
æ
è
ç
ö
ø
÷
+ 2 400
8
4
1 R ln ´ - ´
-
10 6 8 10
5 3
( )
K
+ ´
æ
è
ç
ö
ø
÷
2 300
3
6
R ln
\   DW R R R = + - 200 800 2 200 ln
- 600 2 R ln
= = 2000 2 1153 R ln J
As D D Q W = = 1153 J and DU = 0  cyclic
process.
 First Law of Thermodynamics   | 52
1 atm
2 atm
p
C 
  A
V
  B
D