The magnitude of heat exchanged by a system for the given cyclic process ABCA (as shown in figure) is (in SI unit) :

For a cyclic process,

$$ΔU=0$$

So net heat exchanged equals net work done:

$$Q=W$$

And work done in a cycle is area enclosed in P-V diagram.

The path ABCA encloses the upper semicircle.

From figure:

• Center at

$$(300\text{ cc},300\text{ kPa})$$

• Radius:

$$r=100\text{ cc}$$

Area of semicircle:

$$W=\frac{1}{2}\pi r^2$$

$$=\frac{1}{2}\pi(100)^2$$

$$=5000\pi$$

Units are

$$(\text{kPa})(\text{cc})$$

Convert to SI:

$$1(\text{kPa})(\text{cc})=10^3\times10^{-6}=10^{-3}\text{ J}$$

So

$$Q=5000\pi\times10^{-3}$$

$$=5\pi J$$