First slide

Work done by gas

Question

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

Moderate

A

$- 1125 π kJ$
B

$- \frac{125 π}{2} kJ$
C

$- \frac{200}{3} kJ$
D

$- \frac{325 π}{3} kJ$

Solution

As cyclic process is in anticlockwise sense, hence work done by the gas should be negative and equal to area enclosed
$W = - \frac{π}{4} (P_{2} - P_{1}) (V_{2} - V_{1})$
$W = - \frac{π}{4} (4 - 1) \times 10^{5} \times (20 - 5)$
= $- \frac{π}{4} \times 3 \times 15 \times 10^{5} = - \frac{225 π}{2} \times 10^{3} J$
= $- 1125 π kJ$

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