First slide

Work done by gas

Question

T-V curve of cyclic process is shown below, number of moles of the gas are n find the total work done during the cycle.

Moderate

A

$\frac{3}{2} {nRT}_{0} \ln 2$
B

${nRT}_{0} \ln 2$
C

$\frac{{nRT}_{0} \ln 2}{2}$
D

$2 {nRT}_{0} \ln 2$

Solution

Since path AB and CD arc isochoric therefore work done is zero during path AB and CD. Process BC and DA are isothermal, therefore
$W_{BC} = nR 2 T_{0} \ln \frac{V_{C}}{V_{B}} = 2 {nRT}_{0} \ln 2$
$W_{DA} = {nRT}_{0} \ln \frac{V_{A}}{V_{D}} = - {nRT}_{0} \ln 2$
Total work done
$W_{BC} + W_{DA} = 2 {nRT}_{0} \ln 2 - {nRT}_{0} \ln 2 = {nRT}_{0} \ln 2$

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