An ideal gas expands according to the law $P V^{5 / 2} = C o n s t a n t$ . Molar heat capacity of such gas is $c = c_{v} - \frac{x R}{3}$ , here $x$ is (Answer upto two decimal places )

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Solution

Using 1st law of thermodynamics:
$d Q = d U + d W$

We can write:
$c = \frac{d Q}{n d T} = \frac{n c_{v} d T + P d V}{n d T} = c_{v} + \frac{P d V}{n d T}$
Here, $c$ is molar heat capacity.

Using ideal gas equation:
$P V = n R T \Rightarrow P = \frac{n R T}{V}$

So, $c = c_{v} + \frac{P d V}{n d T} \Rightarrow c = c_{v} + R . \frac{d V . T}{V . d T}$

Now using $P V^{5 / 2} = C o n s t a n t$ , we will find out $\frac{d V . T}{V . d T}$ .
Since,
$P V^{5 / 2} = k \frac{n R T V^{5 / 2}}{V} = k n R T V^{3 / 2} = k$
Now differentiating this term we get:
$n R V^{3 / 2} d T + n R T . \frac{3}{2} V^{1 / 2} . d V = 0$
$n R . V^{3 / 2} d T = - \frac{3}{2} . n R T . V^{1 / 2} . d V$
$- \frac{2}{3} = \frac{d V . T}{V . d T}$

So, $c = c_{v} + R . \frac{d V . T}{V . d T} \Rightarrow c = c_{v} - \frac{2 R}{3}$
$\Rightarrow x = 2$

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