A) \[3nR{{T}_{0}}\]
B) \[\left( \frac{5}{2}nR{{T}_{0}} \right)\]
C) \[\left( \frac{3}{2} \right)nR{{T}_{0}}\]
D) 0
Correct Answer: C
Solution :
\[T=K{{v}^{2}},dT=2kvdV\]or \[dV=\left( \frac{dT}{2kv} \right)\] Further, \[P=\frac{nRT}{V},\]Now \[W=\int_{{}}^{{}}{Pdv=\int_{{}}^{{}}{\frac{nRT}{v}}.\frac{dT}{2kV}}\] \[=\int\limits_{{{T}_{0}}}^{4{{T}_{0}}}{\frac{nR}{2}dT}\,\,\]\[(as\,k{{v}^{2}}=T)\] \[=\frac{3nR{{T}_{0}}}{2}\] Hence, the correction option is [c].
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