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JEE Main Physics Class 11 Thermodynamics Part 2 Questions
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© examsnet.com
Question : 72
Total: 82
Thermodynamic process is shown below on a
p
-V diagram for one mole of an ideal gas.
If
V
2
=
2
V
1
, then the ratio of temperature
T
2
T
1
is
[25 Feb 2021 Shift 2]
1
√
2
√
2
1
2
2
Validate
Solution:
Here,
p
1
and
p
2
,
T
1
and
T
2
,
V
1
and
V
2
are initial and final pressures, temperatures and volumes, respectively.
Given,
V
2
=
2
V
1
p
V
1
∕
2
=
constant
From graph,
γ
=
adiabatic constant
=
1
∕
2
⇒
p
1
V
1
1
∕
2
=
p
2
V
2
1
∕
2
⇒
p
1
p
2
=
(
2
V
1
V
1
)
1
∕
2
=
2
1
∕
2
Also,
p
1
−
γ
T
γ
=
constant (for adiabatic process)
⇒
p
1
1
−
γ
T
1
γ
=
p
2
1
−
γ
T
2
γ
⇒
(
p
1
p
2
)
1
−
γ
=
(
T
2
T
1
)
γ
⇒
(
p
1
p
2
)
1
−
γ
γ
=
T
2
T
1
∴
T
2
∕
T
1
=
(
2
1
∕
2
)
1
−
0.5
0.5
=
√
2
© examsnet.com
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