Question
An ideal gas undergoes a process maintaining relation between pressure (P) and volume (V) as $P = P_{o}\left( 1 + \left( \frac{V_{o}}{V} \right)^{2} \right)^{- 1}$, where $P_{o}$ and $V_{o}$ are constants. If two samples A and B (two moles each) with initial volume $V_{o}$ and $3{\text{ }V}_{o}$ respectively undergo above mentioned process and attain same pressure, then the difference at the temperatures of these samples, $T_{B} - T_{A}$ is $\_\_\_\_$
( $R =$ gas constant)
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Solution
Sol. [Key Concept: Apply $PV = 2RT$ at $V_{0}$ and at $3V_{0}$ for the given $P(V)$, then subtract to get $T_{B} - T_{A}$.]
For $V = V_{0},P = \frac{P_{0}}{2}$
$$\begin{matrix} & \frac{P_{0}}{2} \times V_{0} = 2 \times RT_{A} \\ & T_{A} = \frac{P_{0}V_{0}}{4R} \end{matrix}$$
for $V = 3{\text{ }V}_{0}\ P = \frac{9P_{0}}{10}$
$$\begin{matrix} & \frac{9P_{0}}{10} \times 3{\text{ }V}_{0} = 2 \times {RT}_{B} \\ & {\text{ }T}_{B} = \frac{27P_{0}{\text{ }V}_{0}}{20R} \end{matrix}$$
So, $T_{B} - T_{A}$
$$= \frac{11P_{0}{\text{ }V}_{0}}{10R}$$
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