Question

A quantity of $2mol$ of helium gas undergoes a thermodynamic process, in which molar specific heat capacity $\left(C\right)$ of the gas depends on absolute temperature $T$, according to relation:
$C=\frac{3RT}{4T_0}$
Where $T_0$ is initial temperature of gas. It is observed that when temperature is increased, volume of gas first decreases and then increases. The total work done on the gas until it reaches minimum
volume is $\frac{\gamma }{\delta }R{T}_0$ Find $(\gamma +\delta )$.

Answer 1

According to given condition,
$C=\frac{3RT}{4T_0}$
We know
$C_v=\frac{fR}{2}$
For helium degree of freedom, $f=3$
$\Rightarrow C_v=\frac{3R}{2}$
When it reaches minimum volume, the process can be treated as isochoric.
$\therefore C=C_v$
$C=\frac{3RT}{4T_0}=\frac{3R}{2}$
$T=2T_0$
From the first law of thermodynamics
$\delta Q=dU+\delta W$
${\int }_{T_0}^{2T_0}\left(2\right)\frac{3RT}{4T_0}=\left(2\right)\frac{3RT}{2}(2T_0-T_0)+W$
$\frac{6R}{4T_0}{\left[\frac{T^2}{2}\right]}_{T_0}^{2T_0}=3R{T}_0+W$
$W=\frac{9R{T}_0}{2}-3R{T}_0$
$W=-\frac{3R{T}_0}{4}$
So, work done on the gas is $\frac{3R{T}_0}{4}$
Final answer: $\left(7\right)$