A real gas within a closed chamber at $$27°C$$ undergoes the cyclic process as shown in figure. The gas obeys $$PV^3 = RT$$ equation for the path $$A$$ to $$B$$. The net work done in the complete cycle is (assuming $$R = 8 \text{ J/molK}$$):

or a cyclic process, net work done is the sum of work in each path.

Cycle is

$$C→A→B→C$$

For C→A, volume is constant, so

$$W_{CA}=0$$

For A→B, given

$$PV^3=RT$$

Temperature is

$$27^{\circ}C=300K$$

and

R=8

So

$$PV^3=8(300)=2400$$

Thus

$$P=\frac{2400}{V^3}$$

Work done from A to B is

$$W_{AB}=\int PdV$$

$$=\int_2^4\frac{2400}{V^3}dV$$

$$=2400\int_2^4V^{-3}dV$$

$$=2400\left(\frac{-1}{2V^2}\right)_2^4$$

$$2400\left(\frac{1}{8}-\frac{1}{32}\right)$$

$$=2400\cdot\frac{3}{32}$$

=225J

For B→C, pressure is constant at

P=10

So

=10(2−4)

=−20J

Hence net work done in the complete cycle is

W=0+225−20W

=205J