Question

A gas has molar heat capacity $C=37.35\text{J}{\text{mol}}^{-1}{\text{K}}^{-1}$ in the process $PT=\text{constant}$. Find the number of degrees of freedom of molecules in the gas.

• A. $4$
• B. $5$
• C. $2$
• D. $3$

Answer 1

The correct option is B $5$

Given,
$PT=K....\left(1\right)$
where, $K$ is constant.

For one mole of an ideal gas, we can write,
$PV=RT....\left(2\right)$

From equation $\left(1\right)$ and $\left(2\right)$, we have

$V=\frac{RT}{P}$

$\Rightarrow V=\frac{R{T}^2}{K}$

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

$\therefore$ Work done, when temperature changes from $T$ to $(T+dT)$,

$dW=PdV$

Substituting values,
$dW=\frac{K}{T}\left(\frac{2RT}{K}\right)dT$

$\Rightarrow dW=2RdT$

Now from first law of thermodynamics,

$dQ=dU+dW$

Substituting the values for ideal equation,

$CdT=C_{v}dT+2RdT$

$\Rightarrow C_v=C-2R...\left(3\right)$

For an idea gas,
$C_v=\frac{n}{2}R$
where, $n$ is degree of freedom of the molecules of the gas.
Substituting it in the above equation,

$\Rightarrow \frac{n}{2}R=C-2R$

$\Rightarrow n=\frac{2(C-2R)}{R}$

Given, $C=37.35\text{J}{\text{mol}}^{-1}{\text{K}}^{-1}$ and we know $R=8.3$.
Substituting the values in the above equation, we get

$n=\frac{2(37.35-2\times 8.3)}{8.3}$

$\therefore n=5$
Hence, degree of freedom is $5$ .