Work done by gas
Question

# A gas expands with temperature according to the relation $\mathrm{V}={\mathrm{KT}}^{1/3}.$  What is the work done when the temperature changes by  ${30}^{\mathrm{o}}\mathrm{C}$

Moderate
Solution

## $\mathrm{V}={\mathrm{KT}}^{1/3}$$⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{dV}=\frac{1}{3}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{KT}}^{-\frac{2}{3}}\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{dT}$$⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{1}{3}\text{\hspace{0.17em}\hspace{0.17em}}\frac{\mathrm{V}}{{\mathrm{T}}^{1/3}}\text{\hspace{0.17em}}{\mathrm{T}}^{-\frac{2}{3}}\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{dT}$$⇒\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{dV}\text{\hspace{0.17em}}=\frac{1}{3}\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{V}.\text{\hspace{0.17em}\hspace{0.17em}}{\mathrm{T}}^{-1}\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{dT}$$⇒\text{\hspace{0.17em}\hspace{0.17em}}\frac{\mathrm{dV}}{\mathrm{V}}\text{\hspace{0.17em}}=\frac{1}{3}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{\mathrm{dT}}{\mathrm{T}}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{W}=\int \begin{array}{c}{\mathrm{T}}_{2}\\ {\mathrm{T}}_{1}\end{array}\text{\hspace{0.17em}\hspace{0.17em}}\frac{\mathrm{RT}}{\mathrm{T}}\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{dT}$$=\frac{1}{3}\mathrm{ }\mathrm{R}\mathrm{ }\left({\mathrm{T}}_{2}-{\mathrm{T}}_{1}\right)$$⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{W}=\frac{1}{3}\text{\hspace{0.17em}\hspace{0.17em}}×\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{R}\text{\hspace{0.17em}\hspace{0.17em}}×\text{\hspace{0.17em}\hspace{0.17em}}30=10\mathrm{R}$

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