Given, $C=37.55 \mathrm{J} {\mathrm{mol}}^{-1} {\mathrm{K}}^{-1}$

Also, $PT=$ constant $K$

According to ideal gas equation, $PV=RT$

$\Rightarrow p=\frac{RT}{V} .....\left(\mathrm{i}\right)$

Putting the value of $p$ in Eq. (i), we get

$\frac{RT}{V}\times T=K \Rightarrow V=\frac{R{T}^2}{K} .....\left(\mathrm{ii}\right)$

On differentiating above equation both sides, we get

$\frac{dV}{dT}=\frac{2RT}{K}....\left(\mathrm{iii}\right)$
But $\frac{T}{K}=\frac{1}{p} \left[\mathrm{from}\mathrm{Eq}. \left(\mathrm{i}\right)\right]$
Hence, Eq. $\left(\mathrm{iii}\right)$becomes

$\frac{dV}{dT}=\frac{2 \mathrm{R}}{p}$
So, $C=C_V+\frac{pdV}{dT}$
or $C=C_V+\frac{p\times 2 \mathrm{R}}{p}=C_V+2 \mathrm{R}$
or $C_V=C-2 \mathrm{R} ...\left(\mathrm{iv}\right)$

As $C_V=\frac{nR}{2},$ where $n=$ number of degrees of freedom.
Putting the value of $C_V$ in Eq. (iv), we get

$\frac{nR}{2}=C-2 \mathrm{R}$

$n=\frac{2 \left(C-2 \mathrm{R}\right)}{R}=\frac{2 \left(37.55-2\times 8.3\right)}{8.3}=5.048\approx 5$