DC Pandey Solutions: First Law of Thermodynamics- 2 - DC Pandey Solutions for JEE Physics

 Page 1


27. D D W W
12 13
< can be seen from area
under the curve, while D D V V
1 2
= 
Þ D D Q Q
12 13
<
Þ      Q Q
2 1
< or  Q Q
1 2
>
28. DW p V V p V
CA
= - = -
0 0 0 0 0
2 ( )
and DU p V
CA
= -
3
2
0 0
Þ DQ p V
CA
= -
5
2
0 0
29. DQ
AB
= 200 kJ = nC T
V
D ;
DU
BC
= - 100 kJ and DW
BC
= - 50 kJ
DW
AB
= 0 Þ DU
AB
= 200 kJ, DQ
CA
= 0
D D D D U U U U
ABC AB BC CA
= + + = 0
or 200 100 0 kJ kJ - + = DU
CA
 DU
CA
= - 100 kJ
D D D Q Q Q
AB BC CA
+ +
= + - - + 200 100 50 0 kJ kJ kJ ( )
= 50 kJ
D D D W W W
A B BC CA
+ +
= + + 0 200 kJ DW
CA
= = DQ
A BC
50 kJ
\  DW
CA
= - 150 kJ
30. D D Q W ab = = = ´
´
´
´
-
p p
20 10
2
20 10
2
3 3
= 10
2
p J
31. DW nRT
V
Q
AB
B
AB
=
æ
è
ç
ö
ø
÷ = = ´ ln
1
9 10
4
J
Þ 800 9 10
4
T V
B
ln = ´ J
Þ T V
D
ln =
225
2
  D D D W W W
ABCD AB BC
= +
+ + D D W W
CD DA
=
æ
è
ç
ç
ö
ø
÷
÷
+
-
-
nRT
V
V
p p p p
B
A
C C B B
ln
1 g
+ - + p V V
C D C
( ) 0
= ´ +
-
-
+ - 9 10
10
1
5
3
10 2 1
4
5
5
nRT
B
( )
= ´ - - 19 10
3
2
10 800
4 5
( ) T
B
= ´ + 4 10 1200
4
T
B
= ´ + ´
´ ´
´
4 10 1200
10 1
100 8
4
5
2.4
= ´ 4 10
5
J
31. D D W W
p V p V
A B
C C B B
= +
-
- 1 g
 + DW
C D
= ´ +
´ - ´
-
- ´ 9 10
2 10 9 10
1 5 3
1 10
4
5 4
5
/
= ´ + ´ ´ - ´ 9 10
3
2
11 10 10 10
4 4 4
= -
æ
è
ç
ö
ø
÷
´ = ´
33
2
1 10 10
4 4
15.5
32. D W p d V kV d V k V = = =
ò ò
1
2
2
 
 First Law of Thermodynamics   | 58
p
V
B
A
D
1
C
1 2
2.4
p
V
B
A
C
3
m
B
A
D
1
C
1 2
2.4
32
10N/m
Page 2


27. D D W W
12 13
< can be seen from area
under the curve, while D D V V
1 2
= 
Þ D D Q Q
12 13
<
Þ      Q Q
2 1
< or  Q Q
1 2
>
28. DW p V V p V
CA
= - = -
0 0 0 0 0
2 ( )
and DU p V
CA
= -
3
2
0 0
Þ DQ p V
CA
= -
5
2
0 0
29. DQ
AB
= 200 kJ = nC T
V
D ;
DU
BC
= - 100 kJ and DW
BC
= - 50 kJ
DW
AB
= 0 Þ DU
AB
= 200 kJ, DQ
CA
= 0
D D D D U U U U
ABC AB BC CA
= + + = 0
or 200 100 0 kJ kJ - + = DU
CA
 DU
CA
= - 100 kJ
D D D Q Q Q
AB BC CA
+ +
= + - - + 200 100 50 0 kJ kJ kJ ( )
= 50 kJ
D D D W W W
A B BC CA
+ +
= + + 0 200 kJ DW
CA
= = DQ
A BC
50 kJ
\  DW
CA
= - 150 kJ
30. D D Q W ab = = = ´
´
´
´
-
p p
20 10
2
20 10
2
3 3
= 10
2
p J
31. DW nRT
V
Q
AB
B
AB
=
æ
è
ç
ö
ø
÷ = = ´ ln
1
9 10
4
J
Þ 800 9 10
4
T V
B
ln = ´ J
Þ T V
D
ln =
225
2
  D D D W W W
ABCD AB BC
= +
+ + D D W W
CD DA
=
æ
è
ç
ç
ö
ø
÷
÷
+
-
-
nRT
V
V
p p p p
B
A
C C B B
ln
1 g
+ - + p V V
C D C
( ) 0
= ´ +
-
-
+ - 9 10
10
1
5
3
10 2 1
4
5
5
nRT
B
( )
= ´ - - 19 10
3
2
10 800
4 5
( ) T
B
= ´ + 4 10 1200
4
T
B
= ´ + ´
´ ´
´
4 10 1200
10 1
100 8
4
5
2.4
= ´ 4 10
5
J
31. D D W W
p V p V
A B
C C B B
= +
-
- 1 g
 + DW
C D
= ´ +
´ - ´
-
- ´ 9 10
2 10 9 10
1 5 3
1 10
4
5 4
5
/
= ´ + ´ ´ - ´ 9 10
3
2
11 10 10 10
4 4 4
= -
æ
è
ç
ö
ø
÷
´ = ´
33
2
1 10 10
4 4
15.5
32. D W p d V kV d V k V = = =
ò ò
1
2
2
 
 First Law of Thermodynamics   | 58
p
V
B
A
D
1
C
1 2
2.4
p
V
B
A
C
3
m
B
A
D
1
C
1 2
2.4
32
10N/m
= = =
1
2
1
2
1
2
0 0
pV nRT RT
D D U nC T RT
V
= = × 1
3
2
0
Þ DQ RT RT = +
æ
è
ç
ö
ø
÷
=
3
2
1
2
2
0 0
33. pT = constant = = p
pV
nR
p V
nR
2
Þ p V
2
= constant 
\            p V
p
V
0
2
0
0
2
2
=
æ
è
ç
ö
ø
÷
 Þ V V = 4
0
Þ     T
p
V
nR
p V
nR
T =
×
= =
0
0
0 0
0
2
4
2 2
\ D D U nC T R T T
V
= = ´ - 2
3
2
2
0 0
( )
       = × = 3
2
3
2
0 0
0 0
R
p V
R
p V
35. DW nRT
V
V
BC
C
B
=
æ
è
ç
ç
ö
ø
÷
÷
0
ln
=
æ
è
ç
ç
ö
ø
÷
÷
nRT
p
p
B
C
0
ln = ×
æ
è
ç
ç
ö
ø
÷
÷
2
0
nRT
V
V
B
A
ln
=
æ
è
ç
ç
ö
ø
÷
÷
2
0
nRT
p
p
A
B
ln
\    ln ln
/
ln
p
p
p
p
B
C
æ
è
ç
ç
ö
ø
÷
÷
=
æ
è
ç
ç
ö
ø
÷
÷
=
0
0
2
2
4
Þ p p
B C
= 4
Þ p
p p
C
B
= =
4 8
0
36. As, D D W W
a b
> Þ D D W W
1 2
> 
while, D D U U
1 2
= Þ D D Q Q
1 2
>
37. h = - = - 1 1
300
600
T
T
sin k
sou r c e
 
= - = = 1
1
2
50 0.5 %
38. As the volume is adiabatically
decreased, temperature of the gas
increases and as the time elapsed,
temperature normalizes i.e., decreases
and so pressure also decreases.
39. As the compression is quick , the process
is adiabatic while leads to heating of the
gas.
40. pV
g
= constant 
= =
-
nRT
V
V nRTV
g g 1
Þ TV
g -
=
1
constant
T
T
V
V
L
L
L
L
1
2
2
1
1
2
1
5
3
1
2
1
=
æ
è
ç
ç
ö
ø
÷
÷
=
æ
è
ç
ç
ö
ø
÷
÷
=
æ
è
ç
ç
ö
ø
÷
÷
- - g
2
3
41. pV
g
= constant =
æ
è
ç
ö
ø
÷ p
n T
p
g
g
Þ p T
1 -
=
g g
constant
Þ p T
g g -
µ
1
 
Þ p T µ
-
g
g 1
As 
g
g -
=
-
=
1
7 5
7
5
1
7
2
/
  for diatom gases.
\ p T µ
3.5
 Þ a = 3.5
42. pV
x
= constant , D
D
W
nR T
x
=
- 1
,
D D U n R T = ×
5
2
C
Q
n T
nR T
x
nR T
n T
= =
-
+
D
D
D
D
D
1
5
2
    = +
-
<
5
2 1
0 R
R
x
5
2 1
R
R
x
<
-
 Þ x - < 1
2
5
59  |  First Law of Thermodynamics
p
B 
  C
V
A
Page 3


27. D D W W
12 13
< can be seen from area
under the curve, while D D V V
1 2
= 
Þ D D Q Q
12 13
<
Þ      Q Q
2 1
< or  Q Q
1 2
>
28. DW p V V p V
CA
= - = -
0 0 0 0 0
2 ( )
and DU p V
CA
= -
3
2
0 0
Þ DQ p V
CA
= -
5
2
0 0
29. DQ
AB
= 200 kJ = nC T
V
D ;
DU
BC
= - 100 kJ and DW
BC
= - 50 kJ
DW
AB
= 0 Þ DU
AB
= 200 kJ, DQ
CA
= 0
D D D D U U U U
ABC AB BC CA
= + + = 0
or 200 100 0 kJ kJ - + = DU
CA
 DU
CA
= - 100 kJ
D D D Q Q Q
AB BC CA
+ +
= + - - + 200 100 50 0 kJ kJ kJ ( )
= 50 kJ
D D D W W W
A B BC CA
+ +
= + + 0 200 kJ DW
CA
= = DQ
A BC
50 kJ
\  DW
CA
= - 150 kJ
30. D D Q W ab = = = ´
´
´
´
-
p p
20 10
2
20 10
2
3 3
= 10
2
p J
31. DW nRT
V
Q
AB
B
AB
=
æ
è
ç
ö
ø
÷ = = ´ ln
1
9 10
4
J
Þ 800 9 10
4
T V
B
ln = ´ J
Þ T V
D
ln =
225
2
  D D D W W W
ABCD AB BC
= +
+ + D D W W
CD DA
=
æ
è
ç
ç
ö
ø
÷
÷
+
-
-
nRT
V
V
p p p p
B
A
C C B B
ln
1 g
+ - + p V V
C D C
( ) 0
= ´ +
-
-
+ - 9 10
10
1
5
3
10 2 1
4
5
5
nRT
B
( )
= ´ - - 19 10
3
2
10 800
4 5
( ) T
B
= ´ + 4 10 1200
4
T
B
= ´ + ´
´ ´
´
4 10 1200
10 1
100 8
4
5
2.4
= ´ 4 10
5
J
31. D D W W
p V p V
A B
C C B B
= +
-
- 1 g
 + DW
C D
= ´ +
´ - ´
-
- ´ 9 10
2 10 9 10
1 5 3
1 10
4
5 4
5
/
= ´ + ´ ´ - ´ 9 10
3
2
11 10 10 10
4 4 4
= -
æ
è
ç
ö
ø
÷
´ = ´
33
2
1 10 10
4 4
15.5
32. D W p d V kV d V k V = = =
ò ò
1
2
2
 
 First Law of Thermodynamics   | 58
p
V
B
A
D
1
C
1 2
2.4
p
V
B
A
C
3
m
B
A
D
1
C
1 2
2.4
32
10N/m
= = =
1
2
1
2
1
2
0 0
pV nRT RT
D D U nC T RT
V
= = × 1
3
2
0
Þ DQ RT RT = +
æ
è
ç
ö
ø
÷
=
3
2
1
2
2
0 0
33. pT = constant = = p
pV
nR
p V
nR
2
Þ p V
2
= constant 
\            p V
p
V
0
2
0
0
2
2
=
æ
è
ç
ö
ø
÷
 Þ V V = 4
0
Þ     T
p
V
nR
p V
nR
T =
×
= =
0
0
0 0
0
2
4
2 2
\ D D U nC T R T T
V
= = ´ - 2
3
2
2
0 0
( )
       = × = 3
2
3
2
0 0
0 0
R
p V
R
p V
35. DW nRT
V
V
BC
C
B
=
æ
è
ç
ç
ö
ø
÷
÷
0
ln
=
æ
è
ç
ç
ö
ø
÷
÷
nRT
p
p
B
C
0
ln = ×
æ
è
ç
ç
ö
ø
÷
÷
2
0
nRT
V
V
B
A
ln
=
æ
è
ç
ç
ö
ø
÷
÷
2
0
nRT
p
p
A
B
ln
\    ln ln
/
ln
p
p
p
p
B
C
æ
è
ç
ç
ö
ø
÷
÷
=
æ
è
ç
ç
ö
ø
÷
÷
=
0
0
2
2
4
Þ p p
B C
= 4
Þ p
p p
C
B
= =
4 8
0
36. As, D D W W
a b
> Þ D D W W
1 2
> 
while, D D U U
1 2
= Þ D D Q Q
1 2
>
37. h = - = - 1 1
300
600
T
T
sin k
sou r c e
 
= - = = 1
1
2
50 0.5 %
38. As the volume is adiabatically
decreased, temperature of the gas
increases and as the time elapsed,
temperature normalizes i.e., decreases
and so pressure also decreases.
39. As the compression is quick , the process
is adiabatic while leads to heating of the
gas.
40. pV
g
= constant 
= =
-
nRT
V
V nRTV
g g 1
Þ TV
g -
=
1
constant
T
T
V
V
L
L
L
L
1
2
2
1
1
2
1
5
3
1
2
1
=
æ
è
ç
ç
ö
ø
÷
÷
=
æ
è
ç
ç
ö
ø
÷
÷
=
æ
è
ç
ç
ö
ø
÷
÷
- - g
2
3
41. pV
g
= constant =
æ
è
ç
ö
ø
÷ p
n T
p
g
g
Þ p T
1 -
=
g g
constant
Þ p T
g g -
µ
1
 
Þ p T µ
-
g
g 1
As 
g
g -
=
-
=
1
7 5
7
5
1
7
2
/
  for diatom gases.
\ p T µ
3.5
 Þ a = 3.5
42. pV
x
= constant , D
D
W
nR T
x
=
- 1
,
D D U n R T = ×
5
2
C
Q
n T
nR T
x
nR T
n T
= =
-
+
D
D
D
D
D
1
5
2
    = +
-
<
5
2 1
0 R
R
x
5
2 1
R
R
x
<
-
 Þ x - < 1
2
5
59  |  First Law of Thermodynamics
p
B 
  C
V
A
x <
7
5
 Þ x < 1.4 but x > 1  as for x < 1, 
C will become positive.
\      1 < < x 1.4
43. C
n C n C
n n
R
V
V V
=
+
+
=
1 2
1 2
1 2
13
6
(a) 
2
5
2
4
5
2
2 4
15
6
´ + ´
+
=
R R
R
(b) 
2
5
2
4
3
2
2 4
11
6
´ + ´
+
=
R R
R
(c) 
2
3
2
4
5
2
2 4
13
6
´ + ´
+
=
R R
R and
(d) 
2
6
2
4
3
2
2 4
12
6
´ + ´
+
=
R R
R
Passage 44 & 45
44. D D W p V
pV
Q
ABCA
= ´ ´ = =
1
2 2
net
45. CA ® isobaric and BC ® isochoric,
\ 
C
C
p
v
= = g
5
3
46. pV
g
= constant  =
æ
è
ç
ö
ø
÷ p
nRT
p
g
Þ p T
1 -
=
g g
constant
Þ T p µ
- g
g
1
\ T p µ
- 5 3 1
5 3
/
/
Þ T p µ
2 5 /
\ 
T
T
p
p
p
p
B
A
B
A
c
c
=
æ
è
ç
ç
ö
ø
÷
÷
=
æ
è
ç
ç
ö
ø
÷
÷
=
2 5 2 5
2
3
/ /
0.85
\ T T
B A
= = 0.85 K 850
47. DW
nRT
AB
=
-
=
´ ´
-
1
1
25
3
150
5
3
1
g
= ´ = 75 25 1875 J J
48. DW
BC
= 0 , D D Q U
BC BC
=
= ´ - n R T T
C B
3
2
( )
= ´ -
æ
è
ç
ö
ø
÷ n R
p V
nR
p V
nR
C B
3
2
= -
æ
è
ç
ö
ø
÷
3
2
1
3
2
3
p p V
A A
= - = - ´ = - ×
1
2
1
2
3
2
3
4
p V p V nRT
A B B
= - ´ ´ ´ = -
3
4
1
25
3
850 5312.5 J
49. DW
AB
= + ( ) ve, T T
A B
= 
p
p
V
V p = - +
0
0
0
2
3
2
Þ     
nRT
V
p
V
V p = - +
0
0
0
2
3
2
or           T
p
nRV
V
p
nR
V = - +
0
0
2 0
0
2
3
2
Þ   y ax bx = +
2
 is parabola .
Again, p
p
V
nRT
p
p = - × +
2
3
2
0
0
Þ is also equation of parabola.
While going from A to B temperature
first increases ad than decreases.
50. pV
2
= constant
DW pdV
k
V
dV k
V
= = = -
æ
è
ç
ö
ø
÷
ò ò 2
1
= - = - pV p V p V
i
f
i i f f
= - nR T T
i f
( ) = - - = - nR T T
f i
( ) ( ) ve
as T T
f i
>
as T T
i f
< Þ U U
i f
< 
Þ DU = + ( ) ve
 First Law of Thermodynamics   | 60
p
B 
  A
V
p
0/2
p
0
2V
0
V
0
Page 4


27. D D W W
12 13
< can be seen from area
under the curve, while D D V V
1 2
= 
Þ D D Q Q
12 13
<
Þ      Q Q
2 1
< or  Q Q
1 2
>
28. DW p V V p V
CA
= - = -
0 0 0 0 0
2 ( )
and DU p V
CA
= -
3
2
0 0
Þ DQ p V
CA
= -
5
2
0 0
29. DQ
AB
= 200 kJ = nC T
V
D ;
DU
BC
= - 100 kJ and DW
BC
= - 50 kJ
DW
AB
= 0 Þ DU
AB
= 200 kJ, DQ
CA
= 0
D D D D U U U U
ABC AB BC CA
= + + = 0
or 200 100 0 kJ kJ - + = DU
CA
 DU
CA
= - 100 kJ
D D D Q Q Q
AB BC CA
+ +
= + - - + 200 100 50 0 kJ kJ kJ ( )
= 50 kJ
D D D W W W
A B BC CA
+ +
= + + 0 200 kJ DW
CA
= = DQ
A BC
50 kJ
\  DW
CA
= - 150 kJ
30. D D Q W ab = = = ´
´
´
´
-
p p
20 10
2
20 10
2
3 3
= 10
2
p J
31. DW nRT
V
Q
AB
B
AB
=
æ
è
ç
ö
ø
÷ = = ´ ln
1
9 10
4
J
Þ 800 9 10
4
T V
B
ln = ´ J
Þ T V
D
ln =
225
2
  D D D W W W
ABCD AB BC
= +
+ + D D W W
CD DA
=
æ
è
ç
ç
ö
ø
÷
÷
+
-
-
nRT
V
V
p p p p
B
A
C C B B
ln
1 g
+ - + p V V
C D C
( ) 0
= ´ +
-
-
+ - 9 10
10
1
5
3
10 2 1
4
5
5
nRT
B
( )
= ´ - - 19 10
3
2
10 800
4 5
( ) T
B
= ´ + 4 10 1200
4
T
B
= ´ + ´
´ ´
´
4 10 1200
10 1
100 8
4
5
2.4
= ´ 4 10
5
J
31. D D W W
p V p V
A B
C C B B
= +
-
- 1 g
 + DW
C D
= ´ +
´ - ´
-
- ´ 9 10
2 10 9 10
1 5 3
1 10
4
5 4
5
/
= ´ + ´ ´ - ´ 9 10
3
2
11 10 10 10
4 4 4
= -
æ
è
ç
ö
ø
÷
´ = ´
33
2
1 10 10
4 4
15.5
32. D W p d V kV d V k V = = =
ò ò
1
2
2
 
 First Law of Thermodynamics   | 58
p
V
B
A
D
1
C
1 2
2.4
p
V
B
A
C
3
m
B
A
D
1
C
1 2
2.4
32
10N/m
= = =
1
2
1
2
1
2
0 0
pV nRT RT
D D U nC T RT
V
= = × 1
3
2
0
Þ DQ RT RT = +
æ
è
ç
ö
ø
÷
=
3
2
1
2
2
0 0
33. pT = constant = = p
pV
nR
p V
nR
2
Þ p V
2
= constant 
\            p V
p
V
0
2
0
0
2
2
=
æ
è
ç
ö
ø
÷
 Þ V V = 4
0
Þ     T
p
V
nR
p V
nR
T =
×
= =
0
0
0 0
0
2
4
2 2
\ D D U nC T R T T
V
= = ´ - 2
3
2
2
0 0
( )
       = × = 3
2
3
2
0 0
0 0
R
p V
R
p V
35. DW nRT
V
V
BC
C
B
=
æ
è
ç
ç
ö
ø
÷
÷
0
ln
=
æ
è
ç
ç
ö
ø
÷
÷
nRT
p
p
B
C
0
ln = ×
æ
è
ç
ç
ö
ø
÷
÷
2
0
nRT
V
V
B
A
ln
=
æ
è
ç
ç
ö
ø
÷
÷
2
0
nRT
p
p
A
B
ln
\    ln ln
/
ln
p
p
p
p
B
C
æ
è
ç
ç
ö
ø
÷
÷
=
æ
è
ç
ç
ö
ø
÷
÷
=
0
0
2
2
4
Þ p p
B C
= 4
Þ p
p p
C
B
= =
4 8
0
36. As, D D W W
a b
> Þ D D W W
1 2
> 
while, D D U U
1 2
= Þ D D Q Q
1 2
>
37. h = - = - 1 1
300
600
T
T
sin k
sou r c e
 
= - = = 1
1
2
50 0.5 %
38. As the volume is adiabatically
decreased, temperature of the gas
increases and as the time elapsed,
temperature normalizes i.e., decreases
and so pressure also decreases.
39. As the compression is quick , the process
is adiabatic while leads to heating of the
gas.
40. pV
g
= constant 
= =
-
nRT
V
V nRTV
g g 1
Þ TV
g -
=
1
constant
T
T
V
V
L
L
L
L
1
2
2
1
1
2
1
5
3
1
2
1
=
æ
è
ç
ç
ö
ø
÷
÷
=
æ
è
ç
ç
ö
ø
÷
÷
=
æ
è
ç
ç
ö
ø
÷
÷
- - g
2
3
41. pV
g
= constant =
æ
è
ç
ö
ø
÷ p
n T
p
g
g
Þ p T
1 -
=
g g
constant
Þ p T
g g -
µ
1
 
Þ p T µ
-
g
g 1
As 
g
g -
=
-
=
1
7 5
7
5
1
7
2
/
  for diatom gases.
\ p T µ
3.5
 Þ a = 3.5
42. pV
x
= constant , D
D
W
nR T
x
=
- 1
,
D D U n R T = ×
5
2
C
Q
n T
nR T
x
nR T
n T
= =
-
+
D
D
D
D
D
1
5
2
    = +
-
<
5
2 1
0 R
R
x
5
2 1
R
R
x
<
-
 Þ x - < 1
2
5
59  |  First Law of Thermodynamics
p
B 
  C
V
A
x <
7
5
 Þ x < 1.4 but x > 1  as for x < 1, 
C will become positive.
\      1 < < x 1.4
43. C
n C n C
n n
R
V
V V
=
+
+
=
1 2
1 2
1 2
13
6
(a) 
2
5
2
4
5
2
2 4
15
6
´ + ´
+
=
R R
R
(b) 
2
5
2
4
3
2
2 4
11
6
´ + ´
+
=
R R
R
(c) 
2
3
2
4
5
2
2 4
13
6
´ + ´
+
=
R R
R and
(d) 
2
6
2
4
3
2
2 4
12
6
´ + ´
+
=
R R
R
Passage 44 & 45
44. D D W p V
pV
Q
ABCA
= ´ ´ = =
1
2 2
net
45. CA ® isobaric and BC ® isochoric,
\ 
C
C
p
v
= = g
5
3
46. pV
g
= constant  =
æ
è
ç
ö
ø
÷ p
nRT
p
g
Þ p T
1 -
=
g g
constant
Þ T p µ
- g
g
1
\ T p µ
- 5 3 1
5 3
/
/
Þ T p µ
2 5 /
\ 
T
T
p
p
p
p
B
A
B
A
c
c
=
æ
è
ç
ç
ö
ø
÷
÷
=
æ
è
ç
ç
ö
ø
÷
÷
=
2 5 2 5
2
3
/ /
0.85
\ T T
B A
= = 0.85 K 850
47. DW
nRT
AB
=
-
=
´ ´
-
1
1
25
3
150
5
3
1
g
= ´ = 75 25 1875 J J
48. DW
BC
= 0 , D D Q U
BC BC
=
= ´ - n R T T
C B
3
2
( )
= ´ -
æ
è
ç
ö
ø
÷ n R
p V
nR
p V
nR
C B
3
2
= -
æ
è
ç
ö
ø
÷
3
2
1
3
2
3
p p V
A A
= - = - ´ = - ×
1
2
1
2
3
2
3
4
p V p V nRT
A B B
= - ´ ´ ´ = -
3
4
1
25
3
850 5312.5 J
49. DW
AB
= + ( ) ve, T T
A B
= 
p
p
V
V p = - +
0
0
0
2
3
2
Þ     
nRT
V
p
V
V p = - +
0
0
0
2
3
2
or           T
p
nRV
V
p
nR
V = - +
0
0
2 0
0
2
3
2
Þ   y ax bx = +
2
 is parabola .
Again, p
p
V
nRT
p
p = - × +
2
3
2
0
0
Þ is also equation of parabola.
While going from A to B temperature
first increases ad than decreases.
50. pV
2
= constant
DW pdV
k
V
dV k
V
= = = -
æ
è
ç
ö
ø
÷
ò ò 2
1
= - = - pV p V p V
i
f
i i f f
= - nR T T
i f
( ) = - - = - nR T T
f i
( ) ( ) ve
as T T
f i
>
as T T
i f
< Þ U U
i f
< 
Þ DU = + ( ) ve
 First Law of Thermodynamics   | 60
p
B 
  A
V
p
0/2
p
0
2V
0
V
0
D D D D Q nC T nR T n C R T
V V
= - = - ( )
= + ( ) ve as C R
V
>
i.e., heat is given to the system.
51. In cyclic process, DU = 0
DW nR T
V
V
= +
æ
è
ç
ç
ö
ø
÷
÷
0 2
2
0
0
0
ln
+ +
æ
è
ç
ç
ö
ø
÷
÷
0
2
0
0
0
nRT
V
V
ln
   = - 2 2 2
0 0
nRT nRT ln ln
    = = + nRT
0
2 ln ( ) ve
i.e., DW > 0
D D D Q U W
ab bc supplied
= +
= - +
æ
è
ç
ç
ö
ø
÷
÷
nC T T nR T
V
V
V
( ) ln 2 2
2
0 0 0
0
0
= ´ + 2
3
2
4 2
0 0
RT RT ln
= + 3 4 2
0 0
RT RT ln
52. ab ® isochoric, bc ® isobaric and
ca ® isothermal.
DW
ab
= 0, DU
ca
= 0 
as in ca  density is increasing, so
volume is decreasing i.e., 
DW
ca
= - ( ) ve, i.e., DW
ca
< 0
in isochoric process DQ
ab
 is positive for
increase in temperature.
53. In isochoric process DW = 0.
and in adiabatic process
DQ = 0   Þ Q
3
 to be minimum
Þ     Q Q Q
2 1 3
> >
JEE Corner
¢ Assertion & Reasons
1. In adiabatic expression, DW= + ( )ve
while DQ=0 and as according to first law 
of thermodynamics,
D D D Q U W = + Þ D D U W = -
i.e., DU = - ( ) ve this implies decrease in
temperature. So, Assertion and reason
are both true but not correct
explanation.
2. Assertion is false, as work done is a path
function and not a state function i.e., it
depends on the path through which the
gas was taken from initial to find state.
3. Assertion is false, as first law can be
applied for both real and ideal gases.
4. During melting of ice its volume
decreases, so work done by it is negative
and that by atmosphere is positive. So,
reason is true explanation of assertion.
5. As D D D Q U W = + Þ D D D U Q W = - ,
where DU is state function while DQ and 
DW are path function as for definite
61  |  First Law of Thermodynamics
p
V
b
a
c
p
d 
b
V
a
T
0
2T
0
2V
0
V
0
c
Page 5


27. D D W W
12 13
< can be seen from area
under the curve, while D D V V
1 2
= 
Þ D D Q Q
12 13
<
Þ      Q Q
2 1
< or  Q Q
1 2
>
28. DW p V V p V
CA
= - = -
0 0 0 0 0
2 ( )
and DU p V
CA
= -
3
2
0 0
Þ DQ p V
CA
= -
5
2
0 0
29. DQ
AB
= 200 kJ = nC T
V
D ;
DU
BC
= - 100 kJ and DW
BC
= - 50 kJ
DW
AB
= 0 Þ DU
AB
= 200 kJ, DQ
CA
= 0
D D D D U U U U
ABC AB BC CA
= + + = 0
or 200 100 0 kJ kJ - + = DU
CA
 DU
CA
= - 100 kJ
D D D Q Q Q
AB BC CA
+ +
= + - - + 200 100 50 0 kJ kJ kJ ( )
= 50 kJ
D D D W W W
A B BC CA
+ +
= + + 0 200 kJ DW
CA
= = DQ
A BC
50 kJ
\  DW
CA
= - 150 kJ
30. D D Q W ab = = = ´
´
´
´
-
p p
20 10
2
20 10
2
3 3
= 10
2
p J
31. DW nRT
V
Q
AB
B
AB
=
æ
è
ç
ö
ø
÷ = = ´ ln
1
9 10
4
J
Þ 800 9 10
4
T V
B
ln = ´ J
Þ T V
D
ln =
225
2
  D D D W W W
ABCD AB BC
= +
+ + D D W W
CD DA
=
æ
è
ç
ç
ö
ø
÷
÷
+
-
-
nRT
V
V
p p p p
B
A
C C B B
ln
1 g
+ - + p V V
C D C
( ) 0
= ´ +
-
-
+ - 9 10
10
1
5
3
10 2 1
4
5
5
nRT
B
( )
= ´ - - 19 10
3
2
10 800
4 5
( ) T
B
= ´ + 4 10 1200
4
T
B
= ´ + ´
´ ´
´
4 10 1200
10 1
100 8
4
5
2.4
= ´ 4 10
5
J
31. D D W W
p V p V
A B
C C B B
= +
-
- 1 g
 + DW
C D
= ´ +
´ - ´
-
- ´ 9 10
2 10 9 10
1 5 3
1 10
4
5 4
5
/
= ´ + ´ ´ - ´ 9 10
3
2
11 10 10 10
4 4 4
= -
æ
è
ç
ö
ø
÷
´ = ´
33
2
1 10 10
4 4
15.5
32. D W p d V kV d V k V = = =
ò ò
1
2
2
 
 First Law of Thermodynamics   | 58
p
V
B
A
D
1
C
1 2
2.4
p
V
B
A
C
3
m
B
A
D
1
C
1 2
2.4
32
10N/m
= = =
1
2
1
2
1
2
0 0
pV nRT RT
D D U nC T RT
V
= = × 1
3
2
0
Þ DQ RT RT = +
æ
è
ç
ö
ø
÷
=
3
2
1
2
2
0 0
33. pT = constant = = p
pV
nR
p V
nR
2
Þ p V
2
= constant 
\            p V
p
V
0
2
0
0
2
2
=
æ
è
ç
ö
ø
÷
 Þ V V = 4
0
Þ     T
p
V
nR
p V
nR
T =
×
= =
0
0
0 0
0
2
4
2 2
\ D D U nC T R T T
V
= = ´ - 2
3
2
2
0 0
( )
       = × = 3
2
3
2
0 0
0 0
R
p V
R
p V
35. DW nRT
V
V
BC
C
B
=
æ
è
ç
ç
ö
ø
÷
÷
0
ln
=
æ
è
ç
ç
ö
ø
÷
÷
nRT
p
p
B
C
0
ln = ×
æ
è
ç
ç
ö
ø
÷
÷
2
0
nRT
V
V
B
A
ln
=
æ
è
ç
ç
ö
ø
÷
÷
2
0
nRT
p
p
A
B
ln
\    ln ln
/
ln
p
p
p
p
B
C
æ
è
ç
ç
ö
ø
÷
÷
=
æ
è
ç
ç
ö
ø
÷
÷
=
0
0
2
2
4
Þ p p
B C
= 4
Þ p
p p
C
B
= =
4 8
0
36. As, D D W W
a b
> Þ D D W W
1 2
> 
while, D D U U
1 2
= Þ D D Q Q
1 2
>
37. h = - = - 1 1
300
600
T
T
sin k
sou r c e
 
= - = = 1
1
2
50 0.5 %
38. As the volume is adiabatically
decreased, temperature of the gas
increases and as the time elapsed,
temperature normalizes i.e., decreases
and so pressure also decreases.
39. As the compression is quick , the process
is adiabatic while leads to heating of the
gas.
40. pV
g
= constant 
= =
-
nRT
V
V nRTV
g g 1
Þ TV
g -
=
1
constant
T
T
V
V
L
L
L
L
1
2
2
1
1
2
1
5
3
1
2
1
=
æ
è
ç
ç
ö
ø
÷
÷
=
æ
è
ç
ç
ö
ø
÷
÷
=
æ
è
ç
ç
ö
ø
÷
÷
- - g
2
3
41. pV
g
= constant =
æ
è
ç
ö
ø
÷ p
n T
p
g
g
Þ p T
1 -
=
g g
constant
Þ p T
g g -
µ
1
 
Þ p T µ
-
g
g 1
As 
g
g -
=
-
=
1
7 5
7
5
1
7
2
/
  for diatom gases.
\ p T µ
3.5
 Þ a = 3.5
42. pV
x
= constant , D
D
W
nR T
x
=
- 1
,
D D U n R T = ×
5
2
C
Q
n T
nR T
x
nR T
n T
= =
-
+
D
D
D
D
D
1
5
2
    = +
-
<
5
2 1
0 R
R
x
5
2 1
R
R
x
<
-
 Þ x - < 1
2
5
59  |  First Law of Thermodynamics
p
B 
  C
V
A
x <
7
5
 Þ x < 1.4 but x > 1  as for x < 1, 
C will become positive.
\      1 < < x 1.4
43. C
n C n C
n n
R
V
V V
=
+
+
=
1 2
1 2
1 2
13
6
(a) 
2
5
2
4
5
2
2 4
15
6
´ + ´
+
=
R R
R
(b) 
2
5
2
4
3
2
2 4
11
6
´ + ´
+
=
R R
R
(c) 
2
3
2
4
5
2
2 4
13
6
´ + ´
+
=
R R
R and
(d) 
2
6
2
4
3
2
2 4
12
6
´ + ´
+
=
R R
R
Passage 44 & 45
44. D D W p V
pV
Q
ABCA
= ´ ´ = =
1
2 2
net
45. CA ® isobaric and BC ® isochoric,
\ 
C
C
p
v
= = g
5
3
46. pV
g
= constant  =
æ
è
ç
ö
ø
÷ p
nRT
p
g
Þ p T
1 -
=
g g
constant
Þ T p µ
- g
g
1
\ T p µ
- 5 3 1
5 3
/
/
Þ T p µ
2 5 /
\ 
T
T
p
p
p
p
B
A
B
A
c
c
=
æ
è
ç
ç
ö
ø
÷
÷
=
æ
è
ç
ç
ö
ø
÷
÷
=
2 5 2 5
2
3
/ /
0.85
\ T T
B A
= = 0.85 K 850
47. DW
nRT
AB
=
-
=
´ ´
-
1
1
25
3
150
5
3
1
g
= ´ = 75 25 1875 J J
48. DW
BC
= 0 , D D Q U
BC BC
=
= ´ - n R T T
C B
3
2
( )
= ´ -
æ
è
ç
ö
ø
÷ n R
p V
nR
p V
nR
C B
3
2
= -
æ
è
ç
ö
ø
÷
3
2
1
3
2
3
p p V
A A
= - = - ´ = - ×
1
2
1
2
3
2
3
4
p V p V nRT
A B B
= - ´ ´ ´ = -
3
4
1
25
3
850 5312.5 J
49. DW
AB
= + ( ) ve, T T
A B
= 
p
p
V
V p = - +
0
0
0
2
3
2
Þ     
nRT
V
p
V
V p = - +
0
0
0
2
3
2
or           T
p
nRV
V
p
nR
V = - +
0
0
2 0
0
2
3
2
Þ   y ax bx = +
2
 is parabola .
Again, p
p
V
nRT
p
p = - × +
2
3
2
0
0
Þ is also equation of parabola.
While going from A to B temperature
first increases ad than decreases.
50. pV
2
= constant
DW pdV
k
V
dV k
V
= = = -
æ
è
ç
ö
ø
÷
ò ò 2
1
= - = - pV p V p V
i
f
i i f f
= - nR T T
i f
( ) = - - = - nR T T
f i
( ) ( ) ve
as T T
f i
>
as T T
i f
< Þ U U
i f
< 
Þ DU = + ( ) ve
 First Law of Thermodynamics   | 60
p
B 
  A
V
p
0/2
p
0
2V
0
V
0
D D D D Q nC T nR T n C R T
V V
= - = - ( )
= + ( ) ve as C R
V
>
i.e., heat is given to the system.
51. In cyclic process, DU = 0
DW nR T
V
V
= +
æ
è
ç
ç
ö
ø
÷
÷
0 2
2
0
0
0
ln
+ +
æ
è
ç
ç
ö
ø
÷
÷
0
2
0
0
0
nRT
V
V
ln
   = - 2 2 2
0 0
nRT nRT ln ln
    = = + nRT
0
2 ln ( ) ve
i.e., DW > 0
D D D Q U W
ab bc supplied
= +
= - +
æ
è
ç
ç
ö
ø
÷
÷
nC T T nR T
V
V
V
( ) ln 2 2
2
0 0 0
0
0
= ´ + 2
3
2
4 2
0 0
RT RT ln
= + 3 4 2
0 0
RT RT ln
52. ab ® isochoric, bc ® isobaric and
ca ® isothermal.
DW
ab
= 0, DU
ca
= 0 
as in ca  density is increasing, so
volume is decreasing i.e., 
DW
ca
= - ( ) ve, i.e., DW
ca
< 0
in isochoric process DQ
ab
 is positive for
increase in temperature.
53. In isochoric process DW = 0.
and in adiabatic process
DQ = 0   Þ Q
3
 to be minimum
Þ     Q Q Q
2 1 3
> >
JEE Corner
¢ Assertion & Reasons
1. In adiabatic expression, DW= + ( )ve
while DQ=0 and as according to first law 
of thermodynamics,
D D D Q U W = + Þ D D U W = -
i.e., DU = - ( ) ve this implies decrease in
temperature. So, Assertion and reason
are both true but not correct
explanation.
2. Assertion is false, as work done is a path
function and not a state function i.e., it
depends on the path through which the
gas was taken from initial to find state.
3. Assertion is false, as first law can be
applied for both real and ideal gases.
4. During melting of ice its volume
decreases, so work done by it is negative
and that by atmosphere is positive. So,
reason is true explanation of assertion.
5. As D D D Q U W = + Þ D D D U Q W = - ,
where DU is state function while DQ and 
DW are path function as for definite
61  |  First Law of Thermodynamics
p
V
b
a
c
p
d 
b
V
a
T
0
2T
0
2V
0
V
0
c
initial and final state DU is constant and
so is Q W - . Thus assertion and reason
are both true  but not correct
explanation.
6. Carnot’s engine is ideal heat engine with 
maximum efficiency but it is not also
100%. So assertion and reason are both
true but not correct explanation.
7. pT = constant = × = p
pV
R
p V
nR
2
Þ p V
2
= constant 
\  DW pdV k
dV
V
k
V
= = = ×
ò ò
1 2
1 2
/
/
      = 2 k V = = 2 2
2
kV p V /
      = = - = 2 2 2 pV nR T T nRT T
f i
( ) D
\  DW = + ( ) ve for DT = + ( ) ve
 and 
nRT
V
T = constant. 
Þ T V
2
µ
or, V T µ
2
Thus assertion is true but reason is
false.
8. In adiabatic changes for free expansion, 
Q = 0, W = 0 and DU = 0
as in free expansion no work is done
against any force.
For ideal gases pV = constant as DU = 0
Þ T = constant So, assertion and reason 
are both true but not correct
explanation.
9. Assertion and reason are both true and
correct explanation.
10. Assertion and reason are both true and
correct explanation.
¢ Match the Columns
1. (a) DW p dV pV nR T T
f i
= = = -
ò
( ) 
= = nRT RT 2 ¾® r
(b) DU nC T R T t
V
= = ´ - 2
3
2
2 ( )
= 3RT  ¾® p
(c) DW
nR T T
RT =
-
-
= - ´
( )
/
2
1 5 3
3
2
2
= - 3RT ¾® s
(d)D D U nC T RT
V
= = 3  ¾® p
2. (a) In ab slope is more so, pressure is less
as V
nR
p
T = × , but is constant and in
isobaric process. D D D W p V nR T = = and 
as DT is same in both process so, DW is
same for both ¾® r
(b) As D D U nC T
V
= - is same for both
process ¾® r
(c) As D D D Q U W = + , it is also same for 
both process ¾® s
(d) Nothing can be said about molar
heat capacity ® s
3. (a) D =
ò
W pdV
      = =
ò ò
k
V
dV k
dV
V
   
     = = = 2 2 2 kV pV nR T D  ¾® p
(b) D D D U nC T nR T
V
= =
3
2
 ¾® s
(c) D D D Q nR T nR T = + 2
3
2
       =
7
2
nR T D  ¾® s
(d) ¾® s
4. (a) D D D W p V nR T = = and D D U nC T
V
=  
Þ D D W U < ¾® q
(b) DW = 0 Þ D D Q U = ,DU = - ( ) ve ® p r ,
(c) DW = + ( ) ve, DU = - ( ) ve, DQ = 0 ® p
(d) DW = + ( ) ve, DU = 0, DQ = + ( ) ve  ® p
5. (a) DW p V p V
AB
= +
0 0 0 0
1
2
 First Law of Thermodynamics   | 62