Question

One mole of an ideal monoatomic gas is taken round the cyclic process ABCA as shown in the figure, calculate the work done by the gas.

Answer 1

Work done is area under the curve.

Work done under $W_{ABC}=W_{AB}+W_{BC}+W_{CA}$

Work done along AB is zero as there is no change in volume.

$W_{ABC}=0+W_{BC}+W_{AC}=\frac{1}{2}\times AB\times AC$ i.e., the area of triangle ABC.

$=\frac{1}{2}\times (2V_o-V_o)\times (3P_o-Po)$

$=P_o{V}_o$