For an ideal gas the instantaneous change in pressure ‘p’ with volume ‘v’ is given by the equation $\frac{d p}{d v} = - a p .$ If $p = p_{0}$ at $v = 0$ is the given boundary condition, then the maximum temperature one mole of gas can attain is (Here R is the gas constant).

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$\frac{a p_{0}}{e R}$

$\frac{p_{0}}{a e R}$

infinity

answer is A.

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Detailed Solution

Given $\frac{d p}{d v} = - a p$

Intergrating both sides we get $\Rightarrow \int_{P_{0}}^{P} \frac{d p}{P} = - a \int_{0}^{v} d v \Rightarrow l n (\frac{P}{P_{0}}) = - a v \Rightarrow p = P_{0} e^{- a v}$

Using ideal gas equation $p v = n R T$

$T = \frac{p v}{n R} = \frac{P_{0} v e^{- a v}}{R} (∵ n = 1)$

$\frac{d T}{d v} = 0 \Rightarrow \frac{P_{0}}{R} (e^{- a v} + v e^{- a v} (- a)) \Rightarrow \frac{P_{0} v e^{- a v}}{R} {1 - a v} = 0 \Rightarrow v = \frac{1}{a}$