n moles of an ideal gas undergo a process in which the temperature changes with volume as $T = K V^{2}$ . The work done by the gas as the temperature changes from $T_{0}$ to $4 T_{0}$ is:

3 n R T_{0}

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(\frac{5}{2}) n R T_{0}

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(\frac{3}{2}) n R T_{0}

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Solution

The correct option is C $(\frac{3}{2}) n R T_{0}$
$P V = n R T$

$T = K V^{2}$

$P V = n R K V^{2}$

$P = n R K V$

$V = \frac{T^{1 / 2}}{K}$

$V_{1} = v o$ and $V_{2} = 2 v o$ for $T_{o}$ and $4 T_{o}$

$W = P d V$

$W = n R k V d V$ integrating with volume

$W = \frac{n k R V^{2}}{2} = \frac{n R k ((2 V_{o}^{2}) - (V_{o})^{2}}{2} = \frac{3 k n R V_{o}^{2}}{2}$

substituting value of $V_{o} = \frac{T^{1 / 2}}{k^{1 / 2}}$ we get.

$W = \frac{3 n R T_{o}}{2}$

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