If one mole of an ideal monoatomic gas undergoes
a reversible cyclic process as given in volume
versus temperature graph. Volume is taken in litre
and temperature is taken in kelvin.
V (litre)
С
16
V. T = constant
8
B
А
1000K
→ T(K)
500
It is found that the magnitude of work done by the
gas in complete process i.e. during entire cycle is
XR (where Rishiwersal gas constant The value
X
of
to nearest integer is (use In2 = 0.7)
5
Answers
Answer:
Correct option is A)
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