Question

Find the work done by gas going through a cyclic process shown in figure?

• A. $-\frac{2250\pi }{2}kJ$
• B. $-\frac{125\pi }{2}kJ$
• C. $-\frac{200\pi }{3}kJ$
• D. $-\frac{325\pi }{3}kJ$

Answer 1

Answer: (A)

$-\frac{2250\pi }{2}kJ$

$\text{Solution}:$

As cyclic process is in anticlockwise sense, hence work done by the gas should be negative and equal to area enclosed

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

$W=-\frac{\pi }{4}(4-1)\times {10}^5\times (20-5)=-\frac{\pi }{4}\times 3\times 15\times {10}^5=-\frac{225\pi }{2}\times {10}^{4}J=-\frac{2250\pi }{2}kJ$

$\text{Hence A is the correct option}$