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One mole of a 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 $60^{\circ} \mathrm{C}$, then work done by the gas is ( $R$ is universal gas constant).
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Verified Answer
The correct answer is:
$120 R$
$\mathrm{V}=K T^2 \Rightarrow d V=2 K T d T$
So, $\quad W=\int P d V=R \int \frac{T}{v} d V$
$$
=R \int \frac{T}{K T^2} 2 K T=2 R \int_0^{60} d T=120 R
$$
So, $\quad W=\int P d V=R \int \frac{T}{v} d V$
$$
=R \int \frac{T}{K T^2} 2 K T=2 R \int_0^{60} d T=120 R
$$
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