A) \[{{P}_{0}}{{T}_{0}}R\,\,In\,\,2\]
B) \[\frac{{{P}_{0}}{{T}_{0}}R}{{{P}_{0}}-\alpha }\]
C) \[{{P}_{0}}{{T}_{0}}R\]
D) \[\frac{{{P}_{0}}{{T}_{0}}R}{{{P}_{0}}+\alpha }\]
Correct Answer: B
Solution :
[b] \[PV=nRT+\alpha V\] \[\Rightarrow \,\,\,\,\left( P-\alpha \right)V=nRT\] Initially \[({{P}_{0}}-\alpha )V=nR{{T}_{0}}\] ?.(1) Finally \[({{P}_{0}}-\alpha )V=nR(2{{T}_{0}})\] ?.(2) Comparing eqs (1) and (2), \[V=2{{V}_{0}}\] Now, \[W=\int\limits_{{{V}_{0}}}^{2{{V}_{\theta }}}{{{P}_{0}}dV}={{P}_{0}}\int\limits_{{{V}_{0}}}^{2{{V}_{0}}}{dV={{P}_{0}}\left[ 2{{V}_{0}}-V \right]}\] \[W={{P}_{0}}{{V}_{0}}={{P}_{0}}\left( \frac{nR{{T}_{0}}}{{{P}_{0}}-\alpha } \right)\] Putting \[n=1,\] \[W=\frac{{{P}_{0}}{{T}_{0}}R}{{{P}_{0}}-\alpha }\]You need to login to perform this action.
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