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(D) Given the relation $V = K T^2$.Using the ideal gas equation $PV = RT$ for one mole of gas,we have $P = \frac{RT}{V}$.Substituting $V = K T^2$ into

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$120 R$

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(D) Given the relation $V = K T^2$.Using the ideal gas equation $PV = RT$ for one mole of gas,we have $P = \frac{RT}{V}$.Substituting $V = K T^2$ into the expression for $P$,we get $P = \frac{RT}{K T^2} = \frac{R}{KT}$.The work done $W$ is given by $W = \int P dV$.Since $V = K T^2$,differentiating with respect to $T$ gives $dV = 2KT dT$.Substituting $P$ and $dV$ into the work integral:$W = \int_{T_1}^{T_2} \left( \frac{R}{KT} ight) (2KT dT) = \int_{T_1}^{T_2} 2R dT$.Given the change in temperature $\Delta T = T_2 - T_1 = 60 	ext{ K}$.Therefore,$W = 2R (T_2 - T_1) = 2R(60) = 120R$.

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 $60 \text{ K}$,then the work done by the gas is ($R$ is the universal gas constant).

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