A) \[R\,({{V}_{2}}-{{V}_{1}}){{\log }_{e}}\left( \frac{{{T}_{1}}}{{{T}_{2}}} \right)\]
B) \[R\,({{T}_{2}}-{{T}_{1}}){{\log }_{e}}\left( \frac{{{V}_{1}}}{{{V}_{2}}} \right)\]
C) \[RT\,{{\log }_{e}}\left( \frac{{{V}_{2}}}{{{V}_{1}}} \right)\]
D) \[2RT\,{{\log }_{e}}\left( \frac{{{V}_{1}}}{{{V}_{2}}} \right)\]
Correct Answer: C
Solution :
Work done during isothermal process in expanding volume of gas from \[{{V}_{1}}\] to \[{{V}_{2}}\] is given by \[W=\int_{{{V}_{1}}}^{{{V}_{2}}}{PdV}\] \[=\int_{{{V}_{1}}}^{{{V}_{2}}}{\left( \frac{nRT}{V} \right)dV}\] \[\left( as\,\,P=\frac{nRT}{{}}V \right)\] \[=nRT\,\,\int_{{{V}_{1}}}^{{{V}_{2}}}{\frac{dV}{V}}\] (as T = constant) \[=nRT\,{{\log }_{e}}\left( \frac{{{V}_{2}}}{{{V}_{1}}} \right)\] For expansion of 1 mole of gas, i.e., \[n=1\] \[W=RT{{\log }_{e}}\left( \frac{{{V}_{2}}}{{{V}_{1}}} \right)\] Note: For a process to be isothermal, any heat flow into or out of the system must occur slowly enough, so that thermal equilibrium is maintained.
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