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(B) The work done by a gas is given by $W = \int P dV$.Using the ideal gas law $PV = RT$ (for $1$ mole),we have $P = \frac{RT}{V}$.Thus,$W = \int \fra

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$20$

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(B) The work done by a gas is given by $W = \int P dV$.Using the ideal gas law $PV = RT$ (for $1$ mole),we have $P = \frac{RT}{V}$.Thus,$W = \int \frac{RT}{V} dV$.Given the relation $V = k T^{2/3}$,we differentiate with respect to $T$:$dV = k \cdot \frac{2}{3} T^{-1/3} dT$.Substituting $k = \frac{V}{T^{2/3}}$ into the expression for $dV$:$dV = \left( \frac{V}{T^{2/3}} \right) \cdot \frac{2}{3} T^{-1/3} dT = \frac{2}{3} \frac{V}{T} dT$.Therefore,$\frac{dV}{V} = \frac{2}{3} \frac{dT}{T}$.Substituting this into the work integral:$W = \int_{T_1}^{T_2} R T \left( \frac{2}{3} \frac{dT}{T} \right) = \int_{T_1}^{T_2} \frac{2}{3} R dT$.$W = \frac{2}{3} R (T_2 - T_1)$.Given the change in temperature $\Delta T = T_2 - T_1 = 30^{\circ}C = 30 \ K$:$W = \frac{2}{3} R (30) = 20 R$.

$A$ gas expands with temperature according to the relation $V = k T^{2/3}$. What is the work done when the temperature changes by $30^{\circ}C$ (in $R$)?

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