Question

A heat engine composed of 1.6 moles of an ideal, monotonic gas is initially at 350 K and 1x10^5 Pa. The first step is an isothermal expansion to a pressure of 5x10^4 Pa. Second, the gas is compressed at constant pressure back to the inital volume. Finally the gas returns, at constant volume to the initial state. What is the total work done by the gas during this cycle? What is the efficiency of this cycle?

Answer 1

Using ideal gas equation P1*V1 = n*R*T1

1*10^5*V1 = 1.6*8.314*350

V1 = 0.04655 m^3

During the cycle ,if we drawn the figure ,Pressure Versus Volume we will get a right angle triangle of height (P2-P1) and base (V2-V1)

then Work done is W = area of that triangle = 0.5*b*h = 0.5*(P2-P1)*(V2-V1)

During First step

P1*V1 = P2*V2

1*10^5*0.04655 = 5*10^4*V2

V2 = 0.0931 m^3

then required work done is W = 0.5*(P2-P1)*(V2-V1) = 0.5*(0.5-1)*10^5*(0.0931-0.04655) = -1163.75 J

Efficiency is Eta = W/Q1 = 1163.75 / (Q1)

Q1 = Work doen in isothermal process = 2.3026*n*R*T*log(V2/V1) = 2.3026*1.6*8.314*350*log(0.0931/0.04655) = 3227.2 J

eta = 1163.75/3227.2 = 0.3606 = 36.06 %