A gas has molar heat capacity $C=37.55 J mo{l}^{-1} K^{-1}$, in the process $PT =$constant. The number of degrees of freedom of the molecules of the gas.

Here, $C=37.55 J mo{l}^{-1} K^{-1}$; and

$PT = K$(constant)  ……(i)

According to standard gas equation
$PV= RT$ or $P=RT/V$

From (i) , $\frac{RT}{V}\times T=K$ or $V=\frac{R{T}^2}{K}$

Therefore , $\frac{dV}{dT}=\frac{2RT}{K}$

But $\frac{T}{K}=\frac{1}{P}$ from eqn. (i), therefore, $\frac{dV}{dT}=\frac{2R}{P}$…….(ii)

As , $C=C_v+P\frac{dV}{dT},$therefore, using (ii)

$C=C_V+P\times \frac{2R}{P}=C_V+2R$ or $C_V=C-2R$

As,   $C_V=\frac{n}{2}R$

Therefore , $\frac{n}{2}R=C-2R$ or $n=\frac{2\left(C-2R\right)}{R}=\frac{2\left(37.55-2\times 8.3\right)}{8.3}=5$