Question

Find work done by the gas in the process shown in the figure above in atm$\times$L

• A. $\frac{5}{2}\pi$
• B. $\frac{5}{2}$
• C. $-\frac{3}{2}\pi$
• D. $-\frac{5}{4}\pi$

Answer 1

The correct option is D $-\frac{5}{4}\pi$

The work done is Area under the curve $ABCD$

Work done =area of semicircle $ABC$ +area of semicircle $CAD$

ACB= radius is $\frac{AC}{2}=\frac{4-2}{2}=1$

CDA= radius is $\frac{3.5-2.5}{2}=\frac{1}{2}$

Therefore work done is $W=-PV==(\frac{\pi \ast 1^2}{2}+$$\frac{\pi \ast 1^2}{2\ast 2^2})=-\frac{5\pi }{4}$