An ideal gas on isothermal reversible compression from \(10 \mathrm{~L}\) to \(5 \mathrm{~L}\) performs \(1730 \mathrm{~J}\) of work at \(300 \mathrm{~K}\). Calculate number of moles of gas involved in compression? \(\left(\mathrm{R}=8.314 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}\right)\)

\(\mathrm{V}_1=10 \mathrm{~L}, \mathrm{~V}_2=5 \mathrm{~L}, \mathrm{~W}=1730 \mathrm{~J}, \mathrm{~T}\) \(=300 \mathrm{~K}, \mathrm{R}=8.314 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}, \mathrm{n} \)
\( =? \)
\( \mathrm{~W}_{\max }=-2.303 \mathrm{nRT} \log _{10} \frac{\mathrm{V}_2}{\mathrm{~V}_1} \)
\( \therefore 1730 \mathrm{~J}=-2.303 \mathrm{n} \times 8.314 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}\) \(\times 300 \mathrm{~K} \times \log _{10}\left(\frac{5}{10}\right) \)
\( \therefore \mathrm{n}=\frac{1730 \mathrm{~J}}{-2.303 \times 8.314 \mathrm{JK}^{-1} \mathrm{mo}^{-1} \times 300 \mathrm{~K} \times \log _{10}(0.5)} \)
\( \therefore \mathrm{n}=\frac{1730}{-5744.14 \times(-0.3010)} \mathrm{mol} \)
\( \therefore \mathrm{n}=\frac{1730}{1729} \approx 1 \mathrm{~mol}\)