Question

The $PV$ diagram shown in the figure is in a circle for a cyclic process. The work done during the process cannot be given by:

• A. $\frac{\pi }{4}(P_2-P_1{)}^2$
• B. $\frac{\pi }{4}(V_2-V_1{)}^2$
• C. $\pi (P_2-P_1)(V_2-V_1)$
• D. $\frac{\pi }{4}(P_2-P_1)(V_2-V_1)$

Answer 1

The correct option is C $\pi (P_2-P_1)(V_2-V_1)$

Work done is the area under the $PV$ diagram.

$\therefore$ work done $=\frac{\pi }{4}\times d^2$

Here diameter d is equal to $(V_2-V_1)$ or $(P_2-P_1)$

$d=(V_2-V_1)=(P_2-P_1)$

$\therefore$ work done $=\frac{\pi }{4}{(V_2-V_1)}^2$

$=\frac{\pi }{4}{(P_2-P_1)}^2$

$=\frac{\pi }{4}(P_2-P_1)(V_2-V_1)$

Therefore, (C) is the correct answer.