One mole of an ideal gas undergoes a process in which $T = T_{0} + a V^{3}$ , where $T_{0}$ and ' $a$ ' are positive constants and $V$ is the volume. The volume for which the pressure of the gas will be minimum is

{(\frac{T_{0}}{2 a})}^{1 / 3}

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{(\frac{T_{0}}{3 a})}^{1 / 3}

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{(\frac{a}{2 T_{0}})}^{2 / 3}

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{(\frac{a}{2 T_{0}})}^{2 / 3}

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Solution

The correct option is A ${(\frac{T_{0}}{2 a})}^{1 / 3}$
Given that, $T = T_{0} + a V^{3} . . . . . . . (1)$
We know from the ideal gas equation,
$P V = n R T$
From this, we can write that,
$T = \frac{P V}{n R} . . . . . . . (2)$
From $(1)$ and $(2)$ we get,
$P V = n R T_{0} + a V^{3} (n R)$
$\Rightarrow P = \frac{n R T_{0}}{V} + a V^{2} (n R) . . . . . . . (3)$
To find the volume for which pressure of the gas will be minimum, we have to show that, $\frac{d P}{d V} = 0$ and $\frac{d^{2} P}{d V^{2}} > 0$
Thus, differentiating $(3)$ on both sides we get,
$\frac{d P}{d V} = \frac{- n R T_{0}}{V^{2}} + 2 a V (n R) . . . . . . (4)$
$\Rightarrow \frac{- n R T_{0}}{V^{2}} + 2 a V (n R) = 0$
$\Rightarrow V = {(\frac{T_{0}}{2 a})}^{1 / 3} . . . . . (5)$
Differentiating $(4)$ again we get,
$\frac{d^{2} P}{d V^{2}} = \frac{2 n R T_{0}}{V^{3}} + 2 a (n R)$
By substituting $(5)$ in the above equation we get,
$\frac{d^{2} P}{d V^{2}} = 6 a n R > 0$
$∴$ At Volume $V = {(\frac{T_{0}}{2 a})}^{1 / 3}$ , the pressure of the gas is minimum
Thus, option (a) is the correct answer.

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