An ideal gas is taken through the cycle A → B → C → A, as shown in the figure. If the net heat supplied to the gas in the cycle is 5 J, the work done by the gas in the process C → A is

For cyclic process. Total work done =WAB+WBC+WCAWAB=P∆V = 10(2 – 1) = 10J  and WBC =0(as V = constant)From FLOT, Q = ∆U+∆W=U + W∆U = 0 (Process ABCA is cyclic)Q =WAB+WBC+WCA 5 = 10 + 0 + WCAWCA = – 5 J