For monoatomic gas work done at constant pressure is $W$.For the same rise in temperature of the gas, the…
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
$\frac{W}{2}$
$2 W$
$\frac{3}{2} w$
$\frac{5}{2} W$
Solution
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:
$\begin{aligned} & d Q=n C_v \Delta T \\ & \Rightarrow \int_0^Q d Q=\int_{T_1}^{T_2} n\left(\frac{R}{\gamma-1}\right) d T \\ & \Rightarrow Q=\frac{n R\left(T_2-T_1\right)}{\left(\frac{5}{3}-1\right)}=\frac{3}{2}(n R \Delta T)=\frac{3 W}{2} \\ & Q=\frac{n R T}{\gamma-1}, \gamma=\frac{5}{3} \\ & \therefore Q=\left(\frac{3}{2}\right) n R T\end{aligned}$