Sol. The work for the constant-volume from state 1 to state 2 is zero since dV = 0. For the constant-pressure process the work is

W 2-3 = ∫ PdV = P (V 3 – V 2 ) = (100) (10 – 2) = 800 kJ

The work needed for the isothermal process is

W 3–1 = dV = mRT = mRT ln

To find W 3–1 we need the temperature. It is found from state 3 to be

T 3 = = = 1742 ºR

Thus, the work for the constant-temperature process is

W 3–1 = (2) (0.287) (1742) In = –1609 kJ

Finally, the net work is

W net = + W 2–3 + W 3–1 = 800 – 1609 = – 809 kJ

The negative sign means that there must be a net input of work to complete the cycle in the order shown above.