Question
A gas expands with temperature according to the relation $V = k{T^{2/3}}.$ What is the work done when the temperature changes by ${30^o}C$
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Answer
(b) $W = \int_{}^{} {PdV = \int_{}^{} {\frac{{RT}}{V}dV} } $
Since $V = k{T^{2/3}}$ ==> $dV = \frac{2}{3}K{T^{ - 1/3}}dT$
Eliminating $K,$ we find $\frac{{dV}}{V} = \frac{2}{3}\frac{{dT}}{T}$
Hence$W = \int_{{T_1}}^{{T_2}} {\,\frac{2}{3}\frac{{RT}}{T}dT} = \frac{2}{3}R({T_2} - {T_1}) = \frac{2}{3}R(30) = 20\,R$
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