For monoatomic gas work done at constant pressure is \(W\).For the same rise in temperature of the gas, the heat supplied at constant volume is

At constant pressure, work done is:
\(\begin{aligned} & \int_0^W d W=p \int_{V_1}^{V_2} d V \\ & \Rightarrow W=p\left(V_2-V_1\right)\end{aligned}\)
Using, \(p V=n R T\)
\(W=n R\left(T_2-T_1\right)=n R \Delta T\)
Now, at constant volume, heat supplied is change in internal energy:
\(\begin{aligned} & d Q=n C_v \Delta T \\ & \Rightarrow \int_0^Q d Q=\int_{T_1}^{T_2} n\left(\frac{R}{\gamma-1}\right) d T \\ & \Rightarrow Q=\frac{n R\left(T_2-T_1\right)}{\left(\frac{5}{3}-1\right)}=\frac{3}{2}(n R \Delta T)=\frac{3 W}{2} \\ & Q=\frac{n R T}{\gamma-1}, \gamma=\frac{5}{3} \\ & \therefore Q=\left(\frac{3}{2}\right) n R T\end{aligned}\)