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 \[V=K{{T}^{2/3}}\therefore dV=K\frac{2}{3}{{T}^{-1/3}}dT\] \[\therefore \]\[\frac{dV}{V}=\frac{K\frac{2}{3}{{T}^{-1/3}}dT}{K{{T}^{2/3}}}=\frac{2}{3}\frac{dT}{T}\] From (i),\[W=\int\limits_{{{T}_{1}}}^{{{T}_{2}}}{RT}\frac{dV}{V}=\int\limits_{{{T}_{1}}}^{{{T}_{2}}}{RT}\frac{2}{3}\frac{dT}{T}\] \[W=\frac{2}{3}R({{T}_{2}}-{{T}_{1}})=\frac{2}{3}R\times 60=40R\]
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