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  4. An ideal gas with heat capacity at constant volume Cv undergoes a quasistatic process described by pVα in a p-V diagram, where a is a constant. The heat capacity of the gas during this process is given by

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Q. An ideal gas with heat capacity at constant volume $C_v$ undergoes a quasistatic process described by $pV^{\alpha} $ in a $p-V$ diagram, where a is a constant. The heat capacity of the gas during this process is given by

KVPYKVPY 2010

Solution:

Process equation is
$pV^{\alpha}=$constant $\left(k\right)$
$\Rightarrow p=\frac{k}{V^{\alpha}}$
Work done by the gas in given process is
$\Delta W\int_{_{V_i}}^{^{V_{f}}}pdV $
$=\int_{V_i}^{V_{f}} \frac{kdV}{V^{\alpha}}=\left[\frac{kV^{1}-\alpha}{1-\alpha}\right] ^{V_{f}}$
$=\left[\frac{pV}{1-\alpha}\right]_{_{_{V_i}}}^{^{V_{f}}}=\frac{p\left(V_{f}-V_{i}\right)}{1-\alpha} $
$=\frac{p\Delta V}{1-\alpha}=\frac{nR\Delta T}{1-\alpha}$
The change of internal energy of gas inthis process will be
$\Delta U=C_{V}\Delta T$
And if $\Delta Q$ is heat supplied to the gasthen,
$\Delta Q=c\Delta T $
Now, by first law of thermodynamics, wehave
$\Delta Q=\Delta U+\Delta W $
$\Rightarrow C\Delta T=C_{V}\Delta T+\frac{nR\Delta T}{1-\alpha}$
Heat capacity of the gas is
$\Rightarrow C=C_{V}+\frac{nR}{1-\alpha}$