Question

The work done, V/during an isothermal process in which 1 mole of the gas expands from an initial volume $ {{V}_{1}} $ to a final volume $ {{V}_{2}} $ is given by: (R=gas constant, T= temperature)

• 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}_{2}}}{{{V}_{1}}} \right) $
• C. $ RT{{\log }_{e}}\left( \frac{{{V}_{2}}}{{{V}_{1}}} \right) $
• D. $ 2RT{{\log }_{e}}\left( \frac{{{V}_{1}}}{{{V}_{2}}} \right) $

Answer 1

Answer: (C)

Work done during isothermal process in expanding volume of gas from $ {{V}_{1}} $ to $ {{V}_{2}} $ is given by
$ W=\int_\limits{{{V}_{1}}}^{{{V}_{2}}}{PdV} $
$ =\int_\limits{{{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=cons\tan t $ )
$ =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.