Thank you for reporting, we will resolve it shortly
Q.
A gas has molar heat capacity $C = 37.55\, J\, mole^{-1} \,K^{-1}$, in the process $PT = $ constant. The number of degrees of freedom of the molecules of the gas.
Kinetic Theory
Solution:
Here, $C = 37.55 \,J \,mole^{-1}\, K^{-1}$; and
$PT= K$ (constant) $\quad...\left(i\right)$
According to standard gas equation
$PV = RT$ or $P = RT/V$
From $\left(i\right)$, $\frac{RT}{V} \times T = K$ or $V = \frac{RT^{2}}{K}$
$\therefore \quad \frac{dV}{dt} = \frac{2RT}{K}$
But $\frac{T}{K} = \frac{1}{P} \quad$ rom eqn. $\left(i\right)$, therefore, $\frac{dV}{dT} = \frac{2R}{P}\quad...\left(ii \right)$
As, $C = C_{v} + P \frac{dV}{dT}$, therefore, using $\left(ii\right)$
$C = C_{v} + P \times \frac{2R}{P} = C_{v} + 2R$ or $C_{v}=C - 2R$
As, $C_{v} = \frac{n}{2} R$
$\therefore \quad \frac{n}{2}R = C - 2R$ or $n = \frac{2\left(C - 2R\right)}{R} = \frac{2\left(37.55-2\times8.3\right)}{8.3} = 5$