Question

$2$ moles of a diatomic gas undergoes a thermodynamic process $\frac{P{T}^2}{V}=$ constant. The molar heat capacity of the gas is $\frac{nR}{2}.$ Find $n.$

Answer 1

Answer: 8

$\frac{P{T}^2}{V}=k\Rightarrow \frac{nRT}{V}\cdot \frac{T^2}{V}=k\Rightarrow T^3=\frac{k{V}^2}{nR}$
Now, Differentiating,
$3T^2\frac{dT}{dV}=\frac{2kV}{nR}\Rightarrow \frac{dV}{dT}=\frac{3nR{T}^2}{2kV}=\frac{3}{2}\frac{nR}{P}$
Molar Heat capacity,
$C=C_V+\frac{P}{n}\frac{dV}{dT}=\frac{5R}{2}+\frac{P}{n}\cdot \frac{3}{2}\frac{nR}{P}=\frac{8R}{2}$