Explanation:

\(\mathrm{V}_{1}=10 \mathrm{~L}, \mathrm{~V}_{2}=5 \mathrm{~L}, \mathrm{~W}=1730 \mathrm{~J}\),
\(\mathrm{T}=300 \mathrm{~K}, \mathrm{R}=8.314 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}, \mathrm{n}=\text { ? }\)
Workdone in isothermal reversible compression, \(\mathrm{W}_{\text {max }}=-2.303 \mathrm{nRT} \log _{10} \dfrac{\mathrm{V}_{2}}{\mathrm{~V}_{1}}\)
\(\therefore 1730\;{\rm{J}} = - 2.303 \times {\rm{n}} \times 8.314{\rm{J}}{{\rm{K}}^{ - 1}}\;{\rm{mo}}{{\rm{l}}^{ - 1}} \times \)
\(300\;{\rm{K}} \times {\log _{{\rm{I0}}}}\left( {\frac{5}{{10}}} \right)\)
\[\therefore {\text{n}} = \frac{{1730\;{\text{J}}}}{{\left( \begin{gathered}
- 2.303 \times 8.314\;{\text{J}}\;{{\text{K}}^{ - 1}}\;{\text{mo}}{{\text{l}}^{ - 1}} \times \hfill \\
\,\,\,\,\,\,\,\,300\;{\text{K}} \times {\log _{10}}(0.5) \hfill \\
\end{gathered} \right)}}\]
\(\therefore {\text{n}} = \frac{{1730}}{{ - 5744.14 \times ( - 0.3010)}}{\text{mol}}\)
\(\therefore {\text{n}} = \frac{{1730}}{{1729}} \approx 1\;{\text{mol}}\)