Thermodynamics- 2 Practice Questions - DPP for NEET

 Page 1


1. (c) Processes A to B and C to D are parts of straight line
graphs of the form y mx =
P
T
A
D
B
C
A
V
B
V
C
V
D
V
1
P
2
P
2
P
600 =
A
T K
800 =
B
T K
2200 =
C
T K
1200 =
D
T K
Also ( 6)
R
PT
V
m
= m=
PT Þµ
. So volume remains constant for the graphs
AB and CD.
So no work is done during processes for A to B and C
to D i.e., 0
AB CD
WW == and
2
( ) ()
BC C B CB
W PV V RTT = - =m-
          6 (2200 800) 6 1400 R RJ = - =´
Also 
1
( ) ()
DA A D AB
W PV V RTT = - =m-
                   6 (600 1200) 6 600 R RJ = - =-´
Hence work done in complete cycle
AB BC CD DA
WW WWW = +++
     
0 6 1400 0 6 600 RR = + ´ +-´
     6 900 6 8.3 800 40 R kJ = ´ =´ ´=
2. ( d) W = Area bonded by the indicator diagram with V-axis)
    
1
( )()
2
A B BA
P P VV = +-
3. ( d) For path ab :   ( ) 7000
ab
UJ D=
By using 
V
U CT D =mD
5
7000 700 0.48
2
R =m´ ´ Þm=
For path ca :
( ) ( ) ()
ca ca ca
Q UW D = D +D .......(i)
() () () 0
ab bc ca
U UU \ D +D +D =
7000 0 ( ) 0 ( ) 7000
ca ca
U UJ \ ++D=ÞD=- .... (ii)
Also 
11 2 12
( ) ( ) ()
ca
W PV V RTT D = - =m-
0.48 8.31 (300 1000) 2792.16J = ´ ´ - =- ......(iii)
on solving equations (i), (ii) and (iii)
( ) 7000 2792.16 9792.16 9800
ca
Q JJ D =- - = - =-
4. ( d) In all given cases, process is cyclic and in cyclic process
0 U D=
.
5. ( d)
1
11 12
12 12
21
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g-
g- g-
æö
= Þ=
ç÷
èø
 
52
1
33
22
11
LAL
LAL
-
æ ö æö
==
ç ÷ ç÷
è ø èø
6. ( d) Oxygen is diatomic gas, hence its energy of two moles
5
25
2
RT RT =´=
Argon is a monoatomic gas, hence its internal energy
of 4 moles 
3
46
2
RT RT =´=
Total internal energy (6 5) 11 RT RT =+=
7. (c)
23 12
1 1 22
AB
AB
TT TTWW
T Q TQ
- -
h= = Þh= =
23 111
2 2 1 22
AB
TT QTT
WW
Q T T TT
-
\ = ´ = \=
-
13
2
800 300
550
22
TT
TK
+ +
\= ==
8. (d) Initially 
121
11
(273 7)
0.5
TTT
TT
- -+
h= Þ=
1
1
1
280 1
560
2
T
TK
T
-
Þ = Þ=
Finally 
, 12
1
1
TT
T
¢
¢
-
= h
1
1
1
(273 7)
0.7 933
T
TK
T
¢
¢
¢
-+
Þ = Þ=
\
 increase in temperature 
933 560 373 380 KK = - =»
9. ( d) In both cylinders A and B the gases are diatomic
( 1.4) g= . Piston A is free to move i.e. it is isobaric
process. Piston B is fixed i.e. it is isochoric process. If
same amount of heat Q D is given to both then
() () D =D
isobaric isochoric
QQ
() () Þ D =D
PAB
C T CT
u
mm
() () () 1.430 42
P
B AA
C
T TTK
C
ÞD = D=gD= ´ =
u
10. (b) In first case, 
12
1
1
TT
T
-
h=
In second case, 
1 2 12
2
11
22
2
T T TT
TT
--
h = = =h
Page 2


1. (c) Processes A to B and C to D are parts of straight line
graphs of the form y mx =
P
T
A
D
B
C
A
V
B
V
C
V
D
V
1
P
2
P
2
P
600 =
A
T K
800 =
B
T K
2200 =
C
T K
1200 =
D
T K
Also ( 6)
R
PT
V
m
= m=
PT Þµ
. So volume remains constant for the graphs
AB and CD.
So no work is done during processes for A to B and C
to D i.e., 0
AB CD
WW == and
2
( ) ()
BC C B CB
W PV V RTT = - =m-
          6 (2200 800) 6 1400 R RJ = - =´
Also 
1
( ) ()
DA A D AB
W PV V RTT = - =m-
                   6 (600 1200) 6 600 R RJ = - =-´
Hence work done in complete cycle
AB BC CD DA
WW WWW = +++
     
0 6 1400 0 6 600 RR = + ´ +-´
     6 900 6 8.3 800 40 R kJ = ´ =´ ´=
2. ( d) W = Area bonded by the indicator diagram with V-axis)
    
1
( )()
2
A B BA
P P VV = +-
3. ( d) For path ab :   ( ) 7000
ab
UJ D=
By using 
V
U CT D =mD
5
7000 700 0.48
2
R =m´ ´ Þm=
For path ca :
( ) ( ) ()
ca ca ca
Q UW D = D +D .......(i)
() () () 0
ab bc ca
U UU \ D +D +D =
7000 0 ( ) 0 ( ) 7000
ca ca
U UJ \ ++D=ÞD=- .... (ii)
Also 
11 2 12
( ) ( ) ()
ca
W PV V RTT D = - =m-
0.48 8.31 (300 1000) 2792.16J = ´ ´ - =- ......(iii)
on solving equations (i), (ii) and (iii)
( ) 7000 2792.16 9792.16 9800
ca
Q JJ D =- - = - =-
4. ( d) In all given cases, process is cyclic and in cyclic process
0 U D=
.
5. ( d)
1
11 12
12 12
21
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TV
g-
g- g-
æö
= Þ=
ç÷
èø
 
52
1
33
22
11
LAL
LAL
-
æ ö æö
==
ç ÷ ç÷
è ø èø
6. ( d) Oxygen is diatomic gas, hence its energy of two moles
5
25
2
RT RT =´=
Argon is a monoatomic gas, hence its internal energy
of 4 moles 
3
46
2
RT RT =´=
Total internal energy (6 5) 11 RT RT =+=
7. (c)
23 12
1 1 22
AB
AB
TT TTWW
T Q TQ
- -
h= = Þh= =
23 111
2 2 1 22
AB
TT QTT
WW
Q T T TT
-
\ = ´ = \=
-
13
2
800 300
550
22
TT
TK
+ +
\= ==
8. (d) Initially 
121
11
(273 7)
0.5
TTT
TT
- -+
h= Þ=
1
1
1
280 1
560
2
T
TK
T
-
Þ = Þ=
Finally 
, 12
1
1
TT
T
¢
¢
-
= h
1
1
1
(273 7)
0.7 933
T
TK
T
¢
¢
¢
-+
Þ = Þ=
\
 increase in temperature 
933 560 373 380 KK = - =»
9. ( d) In both cylinders A and B the gases are diatomic
( 1.4) g= . Piston A is free to move i.e. it is isobaric
process. Piston B is fixed i.e. it is isochoric process. If
same amount of heat Q D is given to both then
() () D =D
isobaric isochoric
QQ
() () Þ D =D
PAB
C T CT
u
mm
() () () 1.430 42
P
B AA
C
T TTK
C
ÞD = D=gD= ´ =
u
10. (b) In first case, 
12
1
1
TT
T
-
h=
In second case, 
1 2 12
2
11
22
2
T T TT
TT
--
h = = =h
DPP/ P 25
73
11. (b)
2
1 11
1 500 500 1
11
22
T
T TT
h=- Þ =- Þ = ...(i)
''
22
11
602
1
100 5
=- Þ=
TT
TT
...(ii)
Dividing equation (i) by (ii),
,
2
,
2
500 5
400
4
TK
T
= Þ=
12. (a)
2
max
1
3001
1 1 25%
4004
T
T
h =- =- = =
So 26% efficiency is impossible.
13. (a)
2
1
(273 123) 150 1
1 1 1 50%
(273 27) 300 2
T
T
+
h=- =- =- = =
+
14. (c)
2
1 11
25 300 1 300
1 11
1004
T
T TT
h= - Þ = - Þ =-
1
400 127 T KC ==
o
15. ( b) QUW D = D +D
ÞD =D -D =- U Q W QW (using proper sign)
16. (b)
17. (a) , JQ U W U JQW D = D +D D = D -D
4.18 300 600 654 U D = ´ -=
Joule
18. (b) D = D +D Q UW
110 40 70 WQUJ ÞD =D -D = -=
19. (a)
20. (a) Fro m FLOT
( 0) ( 0) dU dQ dW dU dQ dW Þ= -Þ =< \ =
dU Þ< 0  So, temperature will decrease.
21. ( b) From FLOT QUW D = D +D
Work done at constant pressure
( ) ()
PP
W QU D = D -D
( ) ()
PV
QQ D -D (As we know ())
V
QU D =D
Also ()
PP
Q mcT D=D and ()
VV
Q mcT D=D
( ) ()
p PV
W m c cT Þ D = -D
3 34
( ) 1 (3.4 10 2.4 10)10 10 Cal.
P
W ÞD =´ ´ - ´=
22. (a) Slow isothermal expansion or compression of an ideal
gas is reversible process, while the other given pro-
cess are irreversible in nature.
23. (a) For a reversible process 
dQ
0
T
=
ò
24. (d)
2T
0
T
0
B
A
D
C
P
V
2V
0
V
0
Here W < 0 Þ Q < 0 and  | W | = R T
0
 ln 
0
0
V
2V
æö
ç÷
èø
 + 2RT T
0
ln 
0
0
2V
V
æö
ç÷
èø
 = R TT
0
 ln 2
(25) (c) (26)  (b)     (27)  (b)
(i)
b
a
d
c
P
V
Anticlockwise cycle Þ W < 0
(ii) Process ab : W
ab
 = 0, DU
ab
 > 0 Þ Q
ab
 > 0
Process bc : W
bc
 < 0, DU
bc
 = 0 Þ Q
bc
 < 0
Process cd : W
cd
 = 0, DU
cd
 < 0 Þ Q
cd
 < 0
Process da : W
da
 > 0, DU
da
 = 0 Þ Q
da
 > 0
(iii) da and bc are isothermal process.
28. (a) Second law of thermodynamics can be explained with
the help of example of refrigerator, as we know that
refrigerator, the working substance extracts heat from
colder body and rejcts a large amount of heat to a hotter
body with the help of an external agency i.e., the electric
supply of the refrigerator. No refrigerator can ever work
without external supply of electric energy to it.
29. (d) When the door of refrigerator is kept open, heat rejected
by the refrigerator to the room will be more than the
heat taken by the refrigerator from the room (by an
amount equal to work done by the compressor).
Therefore, temperature of room will increase and so it
will be warmed gradually . As according to second law
of thermodynamics, heat cannot be transferred on its
own, from a body at lower temperature to another at
higher temperature.
30. (c) As there is no change in internal energy of the system
during an isothermal change. Hence, the energy taken
by the gas is utilised by doing work against external
pressure. According to FLOT
            Q U pV D =D +D
Hence, Q U pV D =D =D
Therefore, statement-2 is true and statement-1 is false.