ThermodynamicsHard
Question
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,
Options
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 ABCDA
Solution
ᐃQ = ᐃU + w
For process B → C → D
ᐃU is negative as well as W is also negative
For process B → C → D
ᐃU is negative as well as W is also negative
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