Question

In the cyclic process shown on the P - V diagram the magnitude of the work done is

• A. $\pi {\left(\frac{P_2-P_1}{2}\right)}^2$
• B. $\pi {\left(\frac{V_2-V_1}{2}\right)}^2$
• C. $\frac{\pi }{4}(P_2-P_1)(V_2-V_1)$
• D. $\pi [P_2{V}_2-P_1{V}_1]$

Answer 1

The correct option is C $\frac{\pi }{4}(P_2-P_1)(V_2-V_1)$

Work done in a p-v diagram is $W={\int }_{V_1}^{V_2}Pdv$
In the given diagram, the enclosed area is an ellipse.

Area of the ellipse, $A=\pi$ $\times$ semi-major axis(a) $\times$ semi-minor axis(b)

In the given diagram,
Major axis: $2b=V_2-V_1$
Major axis: $2a=P_2-P_1$
Therefore, $A=\pi \times a\times b=\pi \times \frac{P_2-P_1}{2}\times \frac{V_2-V_1}{2}=\frac{\pi }{4}(P_2-P_1)(V_2-V_1)$
Here, area is positive i.e is work is done by the system is positive. When work is done on the system, the area will be negative.