A thermodynamic system is taken from an original state $$A$$ to an intermediate state $$B$$ by a linear process as shown in the figure. Its volume is then reduced to the original value from $$B$$ to $$C$$ by an isobaric process. The total work done by the gas from $$A$$ to $$B$$ and $$B$$ to $$C$$ would be :

Work done is the area under the P-V curve.

From A→BA\to BA→B, pressure changes linearly, so work done is area of trapezium:

$$W_{AB}=\frac{P_A+P_B}{2}(V_B-V_A)$$

Given

$$P_A=8000\ \text{dyne/cm}^2,\quad P_B=4000\ \text{dyne/cm}^2$$

$$V_A=3m^3\ and\ V_B=7m^3$$

So

$$W_{AB}=\frac{8000+4000}{2}(7-3)$$

$$=6000\times4=24000$$

From B→C, process is isobaric compression, so

$$W_{BC}=P(V_C-V_B)$$

$$=4000(3−7)$$

=−16000

Therefore total work done,

$$W=W_{AB}+W_{BC}$$

$$=24000−16000=8000$$

Now converting units:

$$1\ \text{dyne/cm}^2=0.1\ \text{Pa}$$

So

$$8000\ \text{dyne/cm}^2=800\ \text{Pa}$$

Thus,

$$W=800(m^3)=800J$$