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Tardigrade
Question
Physics
The figure shows the P-V plot of an ideal gas taken through a cycle ABCDA. The part ABC is a semi-circle and CDA is half of an ellipse. Then,
Q. The figure shows the P-V plot of an ideal gas taken through a cycle
A
BC
D
A
. The part
A
BC
is a semi-circle and
C
D
A
is half of an ellipse. Then,
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A
the process during the path
A
→
B
is isothermal
B
heat flows out of the gas during the path
B
→
C
→
D
C
work done during the path
A
→
B
→
C
is zero
D
positive work is done by the gas in the cycle
A
BC
D
A
Solution:
Δ
Q
=
Δ
U
+
W
For process
B
→
C
→
D
Δ
U
is negative as well as
W
is also negative