A) 10 R
B) 30 R
C) 40 R
D) 20 R
Correct Answer: C
Solution :
[c]\[dW=\,pdV=\,\frac{RT}{V}dV\] . (i) |
As, \[V=\,k{{T}^{2/3}},\,dV=\,k\frac{2}{3}\,\,{{T}^{-2/3}}\,\,dT\] |
\[\frac{dV}{V}=\,\frac{k\frac{2}{3}\,\,{{T}^{-2/3}}dT}{k{{T}^{2/3}}}\,=\,\frac{2}{3}\,\frac{dT}{T}\] |
From Eq. (i), \[W=\int_{{{T}_{1}}}^{{{T}_{2}}}{RT}\,\frac{dV}{V}=\int_{{{T}_{1}}}^{{{T}_{2}}}{RT\,\frac{2}{3}\,\,\frac{dT}{T}}\] |
\[W\,=\,\frac{2}{3}\,R\,\,({{T}_{2}}-\,{{T}_{1}})\] |
\[=\frac{2}{3}\,R\times \,60\,=\,40R\] |
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