The figure shows the cycle for a heat engine that uses a gas having gamma =1.25. The initial temperature is T1=300K, and this engine operates at 20 cycles per second. a.) What is the power output of the engine? b.) What is the engine's thermal efficiency? The link to this image is here: http://session.masteringphysics.com/problemAsset/1074111/4/knight_Figure_19_54.jpg This is how I attempted to solve this problem: Given: T1 = T2 = 300K P2 = P3 P1 = 1 atm = 1.013 * 10^(5) Pa gamma = 1.25 V1 = V3 = 600 cm^(3) V2 = 200 cm^(3) R = 8.31 J/mol*K or 0.08206 atm *L/mol*K Relevant equations: PV = nRT η = Wout/QH Power = Wnet/time Q_12 = W_12 Q_23 = nc_pΔT Q_31 = nc_vΔT Finding number of moles n = PV/RT n = [1.013 * 10^(5) Pa(0.0006 m^(3))]/[(8.31J/(mol*K))(300K)]= 0.024 mol Finding P2 and P3 P1V1 = P2V2 P2 = [P1V1]/V2 = [1atm(0.6L)]/0.2L P2 = 3atm P2 = P3 Finding T3 P3V3 = nRT3 T3 = P3V3/nR = 900K Finding W_12 W_12 = nRT(ln(vf/vi)) = 0.024mol(8.31J/mol*K) (300K)(ln(0.0006m^(3)/0.0002m^(3))) = 65.73 J Finding W_23 W_23 = P(vf-vi) = 3atm[(0.0006 - 0.0002)m^(3)] * 101325J = 121.59 J W_31 = 0 because it is an isochoric process Wnet = 187.32J Power P = Wnet/time = 187.32J(20(s^(-1))) = 3746.4 W = 3.746 kW I got it wrong. Why was gamma given? I thought it was only used for adiabatic processes. I also don't know which c_p and c_v to use since the type of gas is not given.