A gas expands with temperature according to t | vedclass

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(D) Given the relation $V = k T^{2/3}$.From the ideal gas law,$PV = RT$ (for $1 \ mole$),so $P = \frac{RT}{V}$.The work done is given by $W = \int P \

Vedclass · Accepted Answer

$40$

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(D) Given the relation $V = k T^{2/3}$.From the ideal gas law,$PV = RT$ (for $1 \ mole$),so $P = \frac{RT}{V}$.The work done is given by $W = \int P \ dV = \int \frac{RT}{V} \ dV$.From $V = k T^{2/3}$,we differentiate with respect to $T$:$dV = k \cdot \frac{2}{3} T^{-1/3} \ dT$.Substituting $V$ and $dV$ into the work integral:$W = \int \frac{RT}{k T^{2/3}} \cdot (k \cdot \frac{2}{3} T^{-1/3} \ dT) = \int R \cdot \frac{2}{3} \cdot \frac{T \cdot T^{-1/3}}{T^{2/3}} \ dT = \int \frac{2}{3} R \ dT$.Given the change in temperature $\Delta T = 60 \ K$,the work done is:$W = \frac{2}{3} R \int_{T_1}^{T_2} dT = \frac{2}{3} R \Delta T = \frac{2}{3} R (60) = 40 R$.

$A$ gas expands with temperature according to the relation,$V=k T^{2/3}$,where $k$ is a constant. Work done when the temperature changes by $60 \ K$ is ($R=$ universal gas constant.) (in $R$)

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