When 5 moles of He gas expand isothermally and reversibly at 300 K from 10 litre to 20 litre, the ma — Thermodynamics and Thermochemistry Chemistry Question
Question
When 5 moles of He gas expand isothermally and reversibly at 300 K from 10 litre to 20 litre, the magnitude of the maximum work obtained is ______ J. [nearest integer] (Given: R = 8.3 J K mol and log 2 = 0.3010) –1 –1
💡 Solution & Explanation
**Step 1: Identify the process and formula** For an isothermal reversible expansion, the maximum work obtained is: $$W = nRT \ln\left(\frac{V_f}{V_i}\right)$$ **Step 2: List given values** - n = 5 moles - R = 8.3 J K⁻¹ mol⁻¹ - T = 300 K - V₁ = 10 L, V₂ = 20 L - log 2 = 0.3010 **Step 3: Calculate the volume ratio** $$\frac{V_f}{V_i} = \frac{20}{10} = 2$$ **Step 4: Convert natural logarithm using log relationship** $$\ln(2) = 2.303 \times \log_{10}(2)$$ $$\ln(2) = 2.303 \times 0.3010 = 0.6933$$ **Step 5: Calculate work** $$W = nRT \ln\left(\frac{V_f}{V_i}\right)$$ $$W = 5 \times 8.3 \times 300 \times 0.6933$$ $$W = 5 \times 8.3 \times 300 \times 0.6933$$ $$W = 5 \times 2479.9 \times 0.6933$$ $$W = 8629.5 \text{ J}$$ **Step 6: Round to nearest integer** W ≈ 8630 J Therefore, the answer is 8630.