2
A cyclic process on an ideal monatomic gas is
shown in figure. The correct statement is
Р.
5
B
V
(1) Work done by gas in process AB is more than
that in the process BC
(2) Net heat energy has been supplied to the system
(3) Temperature of the gas is maximum at state B
(4) In process CA, heat energy is absorbed by sysstem
Answers
Answer:
Clockwise cyclic process
a) Work done by gas
W = area(△ABC)
=
2
1
(2V
o
−V
o
)(3P
o
−P
o
)
=P
o
V
o
b) n = 1 monoatomic gas
C
V
=
2
3
R⇒
R
C
V
=
2
3
C
P
=
2
5
R⇒
R
C
P
=
2
5
Heat rejected by gas in the path CA
CA → It is isochoric
dθ
CA
=
2
5
(P
o
V
o
−2P
o
V
o
)=
2
−5
(P
o
V
o
−2P
o
V
o
)
Heat absorbed by the gas in path AB
AB → isochoric
∴dθ
AB
=C
V
dT
=C
V
(T
final
−T
initial
)
=C
V
(
R
P
f
V
f
−
R
P
i
V
i
)
=
R
C
V
(P
f
V
f
−P
i
V
i
)
=
2
3
(P
f
V
f
−P
i
V
i
)
=
2
3
(3P
o
V
o
−P
o
V
o
)=3P
o
V
o
c) Total heat absorbed during the process
dθ=dθ
AB
+dθ
BC
+dθ
CA
=3P
o
V
o
+dθ
BC
+(
2
−5
P
o
V
o
)
dθ=
2
P
o
V
o
+dθ
BC
Change in internal energy
dU=0
⇒dθ=dW(dθ=dU+dW)
2
P
o
V
o
+dθ
BC
=P
o
V
o
dθ
BC
=
2
P
o
V
o
d) PV equation for the process BC is
P=−mV+C (max temperature will be between B and C)
here
m=
V
o
2P
o
,C=5P
o
∴P=−(
V
o
2P
o
)V+5P
o
PV=−(
V
o
2P
o
)V
2
+5P
o
V
RT=−(
V
o
2P
o
)V
2
+5P
o
V[PV=RT;forn=1]
T=
R
1
[5P
o
V−(
V
o
2P
o
)V
2
]
dV
dT
=T
⇒5P
o
−
V
o
4P
o
V=0⇒V=
4
5V
o
at,V=
4
5V
o
;T=maximum
T
max
=
R
1
⎣
⎢
⎡
5P
o
(
4
5V
o
)−(
V
o
2P
o
)(
4
5V
o
)
2
⎦
⎥
⎤
T
max
=
8
25
R
P
o
V
o