A gas has molar heat capacity C = 37 . 55J mol ^ - 1 K^ - 1 , in...

Given, $C = 37 . 55J{mol}^{- 1}K^{- 1}$

Also, $PT =$ constant $K$

According to ideal gas equation, $PV = RT$

$⇒ p = \dfrac{RT}{V} . . . . . \left(i\right)$

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

\dfrac{RT}{V} \timesT =$K ⇒ V =$\dfrac{RT^{2}}{K} . . . . . \left(ii\right)$</p><p>On differentiating above equation both sides, we get</p><p>\dfrac{dV}{dT} = \dfrac{2RT}{K} . . . .$\left(iii\right)$<br />But~$\dfrac{T}{K} = \dfrac{1}{p} \left[fromEq . \left(i\right)\right]$<br />Hence, Eq. \left(iii\right)$becomes

\dfrac{dV}{dT} = \dfrac{2R}{p}
So,~$C = {C}_{V} +$\dfrac{pdV}{dT}
or~$C = {C}_{V} +$\dfrac{p \times 2R}{p} = {C}_{V} +$2R$
or~${C}_{V} = C - 2R . . . \left(iv\right)$

As ${C}_{V} = \dfrac{nR}{2},$~where $n =$ number of degrees of freedom.
Putting the value of ${C}_{V}$ in Eq. (iv), we get

$\dfrac{nR}{2} =C - 2R$

$n = \dfrac{2\left(C - 2R\right)}{R} = \dfrac{2\left(37 . 55 - 2 \times 8 . 3\right)}{8 . 3} =5 . 048 \approx5$