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JEE Main 8 Apr 2019 Paper 2 Solved Paper
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© examsnet.com
Question : 56
Total: 90
5 moles of an ideal gas at 100 K are allowed to undergo reversible compression till its temperature becomes 200 K. If
C
V
=
28
J
K
–
1
m
o
l
–
1
, calcuIate ΔU and ΔPV for this process.
(
R
=
8.0
J
K
−
1
m
o
l
−
1
)
Δ
U
=
14
k
J
;
Δ
(
p
V
)
=
18
k
J
Δ
U
=
14
J
;
Δ
(
p
V
)
=
0.8
J
Δ
U
=
14
k
J
;
Δ
(
p
V
)
=
4
k
J
Δ
U
=
2.8
k
J
;
Δ
(
p
V
)
=
0.8
k
J
Validate
Solution:
Δ
U
=
28
×
5
×
100
=
14
k
J
Δ
(
p
V
)
=
p
2
v
2
−
p
1
v
1
=
n
R
(
T
2
−
T
1
)
=
5
×
8.314
×
100
≈
4
k
J
© examsnet.com
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