In the given elliptical `P-V` diagram

Options

A. The work done is positve
B. The change in internal energy is non-zero
C. The work done `=-((pi)/(4))(P_(2)-P_(1))(V_(2)-V_(1))`
D. The work done `=(pi)(V_(2)-V_(1))^(2)=pi(P_(2)P_(1))^(2)`

Answer

C

Explanation

The work done =area of P-V diagram
`a=(V_(2)-V_(1))/(2),b=(P_(2)-P_(1))/(2)`
`W=-pi((V_(2)-V_(1))/(2))((P_(2)-P_(1))/(2))`
But the cyclic process is anticlockwise.
Hence, the work done is negative..