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JEE Main Chemistry Class 11 Chemical Thermodynamics Part 2 Questions
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© examsnet.com
Question : 42
Total: 88
One mole of an ideal monoatomic gas is subjected to changes as shown in the graph. The magnitude of the work done (by the system or on the system) is
J
(nearest integer).
Given :
log
2
=
0.3
,
ln
10
=
2.3
[24-Jan-2023 Shift 2]
Your Answer:
Validate
Solution:
1
→
2
⇒
Isobaric process
2
→
3
⇒
Isochoric process
3
→
1
⇒
Isothermal process
W
=
W
1
→
2
+
W
2
→
3
+
W
3
→
1
=
(
−
P
(
V
2
−
V
1
)
+
0
[
−
P
1
V
1
ln
(
V
2
V
1
)
]
)
=
[
−
1
×
(
40
−
20
)
+
0
+
[
−
1
×
20
ln
(
20
40
)
]
]
=
−
20
+
20
ln
2
=
−
20
+
20
×
2.3
×
0.3
=
−
6.2
bar
L
|
W
|
=
6.2
bar
1
=
620
J
© examsnet.com
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