A) \[20.1\,R\]
B) \[30.2\,R\]
C) \[26.6\,R\]
D) \[35.6\,R\]
Correct Answer: C
Solution :
\[W=\int{pdV}\] \[=\int{\frac{\mu RT}{V}}dV\] \[=\int{\frac{\mu RT}{K{{T}^{\frac{2}{3}}}}}dV\] ...(i) \[\because \] \[V=K{{T}^{\frac{2}{3}}}\] \[dV=K\frac{2}{3}{{T}^{-1/3}}dT\] ...(ii) From Eqs. (i) and (ii) \[W=\int{\frac{\mu RT}{K{{T}^{\frac{2}{3}}}}\times k\frac{2}{3}{{T}^{-1/3}}}dT\] \[=\frac{2}{3}R\int{{{T}^{-1-\frac{1}{3}-\frac{2}{3}}}}dT\] \[=\frac{2}{3}R\int_{0}^{40}{{{T}^{0}}d}T=\frac{2}{3}R(T)_{0}^{40}\] \[=\frac{2}{3}R(40-0)=\frac{80R}{3}=26.6\,R\]
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