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JEE Main Chemistry Class 11 Chemical Thermodynamics Part 2 Questions
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© examsnet.com
Question : 6
Total: 88
An ideal gas,
C
V
=
5
2
R
, is expanded adiabatically against a constant pressure of
1
atm
untill it doubles in volume. If the initial temperature and pressure is
298
K
and
5
atm
, respectively then the final temperature is _______
K
(nearest integer). [
C
V
is the molar heat capacity at constant volume]
[6 Apr 2024 Shift 1]
Your Answer:
Validate
Solution:
∆
U
=
q
+
w
(
q
=
0
)
nC
V
∆
T
=
−
P
ext
(
V
2
−
V
1
)
V
2
=
2
V
1
nRT
2
P
2
=
2
nRT
1
P
1
P
1
=
5
,
T
1
=
298
P
2
=
5
T
2
2
×
298
n
5
2
R
(
T
2
−
T
1
)
=
−
1
(
nRT
2
P
1
−
nRT
1
P
1
)
Put
T
1
=
298
and
P
2
=
5
T
2
2
×
298
Solve and we get
T
2
=
274.16
K
T
2
≈
274
K
© examsnet.com
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