1 , text{mole} of a gas expands with temperat | vedclass

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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 $1 \, 	ext{mole}$),we have $P = \frac{RT}{V}$.Given

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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 $1 \, 	ext{mole}$),we have $P = \frac{RT}{V}$.Given the relation $V = KT^{2/3} \dots (i)$.Differentiating with respect to $T$,we get $dV = K \cdot \frac{2}{3} T^{-1/3} \, dT \dots (ii)$.Dividing equation $(ii)$ by $(i)$,we get $\frac{dV}{V} = \frac{2}{3} \frac{dT}{T}$.Substituting $P = \frac{RT}{V}$ into the work integral: $W = \int \frac{RT}{V} dV$.Substituting $\frac{dV}{V} = \frac{2}{3} \frac{dT}{T}$,we get $W = \int R T \left( \frac{2}{3} \frac{dT}{T} ight) = \int \frac{2}{3} R \, dT$.Integrating over the temperature change $\Delta T = 30^oC$,we get $W = \frac{2}{3} R \Delta T$.$W = \frac{2}{3} 	imes 30R = 20R$.

$1 \, \text{mole}$ of a gas expands with temperature as $V = KT^{2/3}$. What is the work done when temperature changes by $30^oC$ (in $R$)?

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