2 moles of an ideal gas are expanded isothermally and reversibly from 20 L to 40 L at 300 K . Calculate work done. ( \(\mathrm{R}=8.314 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}\) )

Given:
- \(n=2 \mathrm{~mol}\),
- \(R=8.314 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}\),
- \(T=300 \mathrm{~K}\),
- \(V_1=20 \mathrm{~L}\),
- \(V_2=40 \mathrm{~L}\).
Formula for Work Done:
\(W=-2.303 n R T \log _{10}\left(\frac{V_2}{V_1}\right)\)
Step 1: Substitute the values:
\(W=-2.303 \cdot 2 \cdot 8.314 \cdot 300 \cdot \log _{10}\left(\frac{40}{20}\right)\)
Step 2: Simplify the logarithmic term:
\(\log _{10}\left(\frac{40}{20}\right)=\log _{10}(2) \approx 0.3010\)
Step 3: Calculate:
\(\begin{gathered}
W=-2.303 \cdot 2 \cdot 8.314 \cdot 300 \cdot 0.3010 \\
W=-2.303 \cdot 4988.4 \cdot 0.3010 \approx-3457.97 \mathrm{~J}
\end{gathered}\)
Answer: \(W=-3457.97 \mathrm{~J}\), closest to Option 4: -3,457.97 J.