Question

The figure shows the p-V plot an ideal gas taken through a cycle ABCDA. The part ABC is a semi-circle and CDA is half of an ellipse. Then,

• A. the process during the path A $\rightarrow$ B is isothermal
• B. heat flows out of the gas during the path B $\rightarrow$ C $\rightarrow$ D
• C. work done during the path A$\rightarrow$ B $\rightarrow$ C is zero
• D. positive work is done by the gas in the cycle ABCDA

Answer 1

Answer: (B), (D)

(A) P-V graph is not rectangular hyperbola. Therefore, process A - B is not isothermal.
(B) In process BCD, product of pV (therefore temperature and internal energy) is decreasing. Further, volume is decreasing. Hence, work done is also negative. Hence, Q will be negative or heat will flow out of the gas.
(C) W$_{ABC}$ = positive
(D) For clockwise cycle on p-V diagram with P on y-axis, net work done is positive.