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One mole of gas expands such that its volume ‘v’ changes with absolute temperature ‘T’ in accordance with the relation $V = K T^{2}$ where K is a constant. If the temperature of the gas changes by $60^{0} C$ , then work done by the gas is (R is universal gas constant)

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$K R \log_{e} 60$

$R \log_{e} 60$

120 R

40 R

answer is C.

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Detailed Solution

$P V = n R T$

Given $V = k T^{2}$

$P V = (R) {(\frac{V}{K})}^{1 / 2}$

$P V^{1 / 2} = c o n s t$ , comparing with $P V^{γ} = c o n s \tan t$

then

$γ = \frac{1}{2}$ for this process.

$∴$ work done in this process =

$W = \frac{n R (T_{i} - T_{f})}{γ - 1} = \frac{R 60}{1 / 2} = 120 R$

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