Thermodynamics- 1 Practice Questions - DPP for NEET

 Page 1


1. (b) Differentiate 
PV
= constant w.r .t. V
0
PV
P V VP
PV
DD
ÞD+D=Þ=
2. (a) In isothermal compression, there is always an increase
of heat which must flow out the gas.
( 0) QU W QW U D =D +DÞD =D\D =
4
4
1.5 10
1.5 10
4.18
QJ
´
ÞD=- ´=
cal 
3
3.6 10 =-´
cal
3. (c) D = D +D Q UW
Þ
 D = D -D U QW 2240 168 2072 . = -= J
4. (b)
2
1
20
log 1 8.31 300log 1728
10
isoee
V
WRTJ
V
=m =´´=
5. (d) For adiabatic process 
1
T
P
g
g
-
 = constant
1
2 12
12
300
T PT
TP
-g
g æö
Þ=Þ
ç÷
èø
(1 1.4)
1.4 4
1
-
æö
=
ç÷
èø
0.4
1.4
2
300(4) T Þ=
6. (d) PV
g
 = constant
21
12
PV
PV
æö
Þ=
ç÷
èø
g
 
5/3
2 11
(8) 32 P PP Þ==
7. (b)
1
21
12
TV
TV
-
æö
=
ç÷
èø
g
52
1
33
2
27 27
300 300
88
T
-
æö æö
Þ==
ç÷ ç÷
èø èø
         
2
1/32
273
300 800 675
82
K
ìü
ïï æ ö æö
= ==
íý ç ÷ ç÷
è ø èø
ïï
îþ
675 300 375 TK ÞD= -=
8. (b) In adiabatic change Q = constant 0 Q ÞD=
So () W U Q UW D = -D \ D = D +D
9. (d)
1
TV
g-
= constant
1
1
21
2
927
V
TTC
V
o
g-
æö
Þ==
ç÷
èø
10. (d)
( ) ()
6 ( 4)
5 1
1
3
iff
f
RT T RT T
W R T TK
--
= Þ = Þ =-
g- æö
-
ç÷
èø
11. (c) PV
g
constant : Differentiating both sides
1
0
dP dV
P V dV V dP
PV
g-g
g + = Þ = -g
12. (c) V µ T at constant pressure
1 1 12
2
221
300 280
280ml
300
V T VT
V
VTT
´
Þ=Þ= ==
13. (b) In thermodynamic process, work done is equal to the
area covered by the PV curve with volume axis.
Hence, according to graph shown
W
adiabatic 
< W
isothermal
 < W
isobaric
Isobaric
Isothermal
Adiabatic
P
V
1
V
2
V
14. (d) W = P DV = 2.4 × 10
4
 × 1 × 10
5
 = 24J
15. (c) For isobaric process 
22
2
11
274
273
VT
VV
VT
= Þ =´
Increase = 
274
273 273
VV
V -=
16. (c) W = P DV = nR DT = 0.1 × 2 × 300 = 60 cal
17. (d) Fraction of supplied energy which increases the
internal energy is given by
() 1
( ) ()
VV
P PP
Q CT U
f
Q Q CT
D mD D
= = ==
D D m Dg
For diatomic gas, 
75
57
f g= Þ=
18. (a) In isothermal change, temperature remains constant,
Hence DU = 0.
Also from QU W QW D=D +D ÞD=D
Page 2


1. (b) Differentiate 
PV
= constant w.r .t. V
0
PV
P V VP
PV
DD
ÞD+D=Þ=
2. (a) In isothermal compression, there is always an increase
of heat which must flow out the gas.
( 0) QU W QW U D =D +DÞD =D\D =
4
4
1.5 10
1.5 10
4.18
QJ
´
ÞD=- ´=
cal 
3
3.6 10 =-´
cal
3. (c) D = D +D Q UW
Þ
 D = D -D U QW 2240 168 2072 . = -= J
4. (b)
2
1
20
log 1 8.31 300log 1728
10
isoee
V
WRTJ
V
=m =´´=
5. (d) For adiabatic process 
1
T
P
g
g
-
 = constant
1
2 12
12
300
T PT
TP
-g
g æö
Þ=Þ
ç÷
èø
(1 1.4)
1.4 4
1
-
æö
=
ç÷
èø
0.4
1.4
2
300(4) T Þ=
6. (d) PV
g
 = constant
21
12
PV
PV
æö
Þ=
ç÷
èø
g
 
5/3
2 11
(8) 32 P PP Þ==
7. (b)
1
21
12
TV
TV
-
æö
=
ç÷
èø
g
52
1
33
2
27 27
300 300
88
T
-
æö æö
Þ==
ç÷ ç÷
èø èø
         
2
1/32
273
300 800 675
82
K
ìü
ïï æ ö æö
= ==
íý ç ÷ ç÷
è ø èø
ïï
îþ
675 300 375 TK ÞD= -=
8. (b) In adiabatic change Q = constant 0 Q ÞD=
So () W U Q UW D = -D \ D = D +D
9. (d)
1
TV
g-
= constant
1
1
21
2
927
V
TTC
V
o
g-
æö
Þ==
ç÷
èø
10. (d)
( ) ()
6 ( 4)
5 1
1
3
iff
f
RT T RT T
W R T TK
--
= Þ = Þ =-
g- æö
-
ç÷
èø
11. (c) PV
g
constant : Differentiating both sides
1
0
dP dV
P V dV V dP
PV
g-g
g + = Þ = -g
12. (c) V µ T at constant pressure
1 1 12
2
221
300 280
280ml
300
V T VT
V
VTT
´
Þ=Þ= ==
13. (b) In thermodynamic process, work done is equal to the
area covered by the PV curve with volume axis.
Hence, according to graph shown
W
adiabatic 
< W
isothermal
 < W
isobaric
Isobaric
Isothermal
Adiabatic
P
V
1
V
2
V
14. (d) W = P DV = 2.4 × 10
4
 × 1 × 10
5
 = 24J
15. (c) For isobaric process 
22
2
11
274
273
VT
VV
VT
= Þ =´
Increase = 
274
273 273
VV
V -=
16. (c) W = P DV = nR DT = 0.1 × 2 × 300 = 60 cal
17. (d) Fraction of supplied energy which increases the
internal energy is given by
() 1
( ) ()
VV
P PP
Q CT U
f
Q Q CT
D mD D
= = ==
D D m Dg
For diatomic gas, 
75
57
f g= Þ=
18. (a) In isothermal change, temperature remains constant,
Hence DU = 0.
Also from QU W QW D=D +D ÞD=D
DPP/ P 24
71
19. (c) From graph it is clear that 
31
PP > .
C
A
B
V
3
P
1
P
2
P
1
() EV
2
( ) D V
Since area under adiabatic process (BCED) is greater
than that of isothermal process  (ABDE).Therefore net
work done
()0
i A Ai
W W W W WW = +- \ > Þ<
20. (b)
2
PV = constant represents adiabatic equation. So
during the expansion of ideal gas internal energy of
gas decreases and temperature falls.
21. (a) For adiabatic process
1
1
b
TV
g-
= Constant
For bc curve
1 1
12c
b
TV TV
g- g-
=
or
1
2
1
b
c
V T
TV
g-
æö
=
ç÷
èø
....(i)
For ad curve
1 1
12 a d
TV TV
g- g-
=
or
1
2
1
a
d
V T
TV
g-
æö
=
ç÷
èø
.....(ii)
From equation (i) and (ii)
ba
cd
VV
VV
=
22. (d) There is a decrease in volume during melting on an ice
slab at 273K. Therefore, negative work is done by
ice-water system on the atmosphere or positive work
is done on the ice-water system by the atmosphere.
Hence option (b) is correct. Secondly heat is absorbed
during melting (i.e. DQ is positive) and as we have
seen, work done by ice-water system is negative ( DW
is negative). Therefore, from first law of
thermodynamics ?U ?Q ?W =-
Change in internal energy of ice-water system, DU will
be positive or internal energy will increase.
23. (a) From graph it is clear that 
31
P P. >
Since area under adiabatic process (BCED) is greater
than that of isothermal process (ABDE). Therefore net
work done
( )
i A Ai
W W W W W W0 = +- > Þ< Q
P
3
P
1
P
2
E(V)
1
D(V)
2
V
B
A
C
24. (d) W ork done = Area of ABC with V -axis
           = P
0
(2V
0
 –V
0
) + 0 = P
0
V
0
 = nRT
0
 = RT
0
Change in internal energy = nC
V
DT
= 1 × 
3
R
2
× (4T
0
 – T T
0
) = 
9
2
RTT
0
Heat absorbed = 
9
2
RTT
0
  + RT
0
  = 
11
2
RTT
0
25. (b) AB is an isothermal process then
P × 2V = P
B
 × 6V  Þ 
3
B
P
P =
P
P
A C
B
2V 6V
V
Now BC is an isochoric process then
C B
BC
P P
TT
=
;   
0
3
C
PP
TT
=
;  T
C
 = 3T
0
26. (a) Heat absorbed during BC is given by
Q = nC
v 
DT  = 
3
()
2
CB
R
n TT ´-
   = 
0
3
(2)
2
R
nT ´ = 3nRT
0
.
27. (b) Heat capacity is given by
C = 
1 dQ
n dT
;  C = 
0
1
2
Q
nT
28. (c) As isothermal processes are very slow and so the
different isothermal curves have different slopes so
they cannot intersect each other.
29. (d) Adiabatic compression is a rapid action and both the
internal energy and the temperature increases.
30. (a)
.
=
Dq
Q
c
m
; a gas may be heated by putting pressure,
so it can have values for 0 to ¥ .
C
P
 and C
V
 are it’s two principle specific heats, out of
infinite possible values.
In adiabatic process C = 0 and in isothermal process
C = ¥ .