For a monoatomic gas, work done at constant pressure is $\mathrm{W}$. The heat supplied at constant volume…
For a monoatomic gas, work done at constant pressure is $\mathrm{W}$. The heat supplied at constant volume for the same rise in temperature of the gas is
$\mathrm{W}$
$\frac{5 \mathrm{~W}}{2}$
$\frac{\mathrm{W}}{2}$
$\frac{3 \mathrm{~W}}{2}$
Solution
Heat supplied at constant pressure, $\mathrm{Q}_1=\mathrm{nC}_{\mathrm{p}} \Delta \mathrm{T}$
Heat supplied at constant volume, $\mathrm{Q}_2=\mathrm{nC}_{\mathrm{v}} \Delta \mathrm{T}$
Work done, $\mathrm{W}=\mathrm{Q}_1-\mathrm{Q}_2=\mathrm{n}\left(\mathrm{C}_{\mathrm{p}}-\mathrm{C}_{\mathrm{v}}\right) \mathrm{dT}$
$\begin{aligned}
& \frac{\mathrm{W}}{\mathrm{Q}_2}=\frac{\mathrm{C}_{\mathrm{p}}-\mathrm{C}_{\mathrm{v}}}{\mathrm{C}_{\mathrm{v}}}=\frac{\mathrm{C}_{\mathrm{p}}}{\mathrm{C}_{\mathrm{v}}}-1=\frac{5}{3}-1=\frac{2}{3} \\
& \mathrm{Q}_2=\frac{3 \mathrm{~W}}{2}
\end{aligned}$