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A gas is governed by an equation $V = \frac{a T^{3}}{P}$ where P, V and T are pressure, volume and temperature of the gas respectively and ‘a’ is a constant. If the temperature of the gas is doubled at constant pressure then the work done by the gas is

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$8 a T^{3}$

$7 a T^{3}$

$6 a T^{3}$

$9 a T^{3}$

answer is D.

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Detailed Solution

$W = P (v_{2} - v_{1})$

$v_{1} = \frac{a {T_{1}}^{3}}{P}$

$V_{2} = \frac{a {(2 T_{1})}^{3}}{P} = \frac{8 {aT}_{1}^{3}}{P}$

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