Question

In the cyclic process shown in 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. $\pi (V_2-V_1{)}^2$

Answer 1

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

Work done in the process will be equal to the area under the curve.
From the figure we can see that it is an ellipse,
So,
work done = Area under the curve
$w=\frac{\pi }{4}\left(d_1\right)\left(d_2\right)$
where,
$d_1$ = diameter of major axis of ellipse (which is along x-axis)
$d_2$ = diameter of minor axis of ellipse (which is along y-axis)

where,
$\left(d_1\right)=(P_2-P_1)and\left(d_2\right)=(V_2-V_1)$

putting these values we get.
$w=\frac{\pi }{4}(P_2-P_1)(V_2-V_1)$