Question

One mole of an ideal monoatomic gas is taken in cyclic process ABCA as shown in the figure

Calculate: (A) The work done by the gas (B) The heat rejected by the gas in the path CA

Answer 1

A) P0V0, B) 1/2P0V0

Answer 2

(A) The work done by the gas is the area enclosed by the cycle ABCA, which is a triangle with vertices A $(V_0, 3P_0)$, B $(V_0, P_0)$, C $(2V_0, P_0)$. The area is $P_0 V_0$. The cycle is clockwise, so the work done by the gas is positive. $W_{cycle} = P_0 V_0$.

(B) The heat rejected by the gas in the path CA. Path CA is from C $(2V_0, P_0)$ to A $(V_0, 3P_0)$. We calculated $Q_{CA} = -\frac{1}{2} P_0 V_0$. Since $Q_{CA}$ is negative, heat is rejected by the gas in path CA. The heat rejected is $|Q_{CA}| = |-\frac{1}{2} P_0 V_0| = \frac{1}{2} P_0 V_0$.