One mole of an ideal gas passes through a process where pressure and volume obey the relation $P = P_{0} [1 - \frac{1}{2} {(\frac{V_{0}}{V})}^{2}]$ Here $P_{0}, V_{0}$ are constants. Calculate the change in the temperature of the gas, if its volume charges from $V_{0}$ to $2 V_{0}$

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$\frac{1}{2} \frac{P_{o} V_{0}}{R}$

$\frac{1}{4} \frac{P_{o} V_{0}}{R}$

$\frac{3}{4} \frac{P_{o} V_{o}}{R}$

$\frac{5}{4} \frac{P_{o} V_{o}}{R}$

answer is D.

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Detailed Solution

Given process equation for 1 mole of an ideal gas is
$p = p_{o} (1 - \frac{1}{2} {(\frac{V_{o}}{V})}^{2})$ …(i)
Also, for 1 mole of ideal gas,
$p V = R T$
$p = \frac{R T}{V}$ …(ii)
So from equation (i) (ii) we have
$\frac{R T}{V} = p_{o} (1 - \frac{1}{2} {(\frac{V_{o}}{V})}^{2})$
$T = \frac{p_{o} V}{R} (1 - \frac{1}{2} {(\frac{V_{o}}{V})}^{2})$ …(iii)
When volume of gas is $V_{o}$ then by substituting $V = V_{o}$ in equation (iii), we get
Temperature of gas is
$T_{1} = \frac{ρ_{o} V_{0}}{R} (1 - \frac{1}{2} {(\frac{V_{o}}{V_{o}})}^{2}) = \frac{p_{o} V_{o}}{2 R}$
Similarly, at volume, $V = 2 V_{o}$
Temperature of gas is $T_{2} = \frac{p_{o} (2 V_{o})}{R} (1 - \frac{1}{2} {(\frac{V_{o}}{2 V_{o}})}^{2}) = \frac{7}{4} \frac{p_{o} V_{o}}{R}$
So change in temperature as volume changes from $V_{o} t o 2 V_{o}$ is
$Δ T = T_{2} - T_{1} = (\frac{7}{4} - \frac{1}{2}) \frac{p_{o} V_{o}}{R} = \frac{5}{4} \frac{p_{o} V_{o}}{R}$