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

Question

Aicharya · Accepted Answer

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.

2 moles of an ideal gas are expanded isothermally and reversibly from 20 L to 40 L at 300 K . Calculate work…

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