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Q.

A gas expands with temperature according to the relation, $V = k T^{2 / 3}$ , where ‘k’ is a constant. Work done when the temperature change by 60 K is (R = universal gas constant)

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a

10 R

b

20 R

c

40 R

d

50 R

answer is D.

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Detailed Solution

$dW = PdV$

$dW = \frac{RT}{V} dV = \frac{RT}{{kT}^{2 / 3}} dV$

Now, $V = {kT}^{2 / 3}$

$\Rightarrow dV = \frac{2}{3} {kT}^{- 1 / 3} dT$

$\Rightarrow dW = \frac{2}{3} RdT$

$\Rightarrow W = \frac{2 R}{3} ∆ T = 40 R$

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