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

Step-by-step solution

Δ W AB = p Δ V = 1 0 2 - 1 = 10 J Δ W BC = 0                 as V = constant From first law of thermodynamics,            Δ Q = Δ W + Δ U Δ U = 0         Process ABCA is cyclic, ∴   Δ Q = Δ W AB + Δ W BC + Δ W CA ∴   Δ W CA = Δ Q - Δ W AB - Δ W BC = 5 - 1 0 - 0 = - 5 J