Five moles of an ideal gas at 293K is expanded isothermally from an initial pressure of 2.1 MPa to 1 — Thermodynamics and Thermochemistry Chemistry Question

💡 Solution & Explanation

**Step 1: Identify the process and find initial/final volumes** For an isothermal process with an ideal gas: - PV = nRT (constant at fixed T) - V₁ = nRT/P₁ = (5 × 8.314 × 293)/(2.1 × 10⁶) = 0.00581 m³ - V₂ = nRT/P₂ = (5 × 8.314 × 293)/(1.3 × 10⁶) = 0.00941 m³ **Step 2: Calculate work done by the gas** Since expansion occurs against constant external pressure: - W = P_ext(V₂ - V₁) - W = 4.3 × 10⁶ × (0.00941 - 0.00581) - W = 4.3 × 10⁶ × 0.0036 = 15,480 J **Step 3: Apply First Law of Thermodynamics** For an ideal gas in an isothermal process: - ΔU = 0 (internal energy depends only on temperature) - ΔU = Q - W - 0 = Q - W - Q = W = 15,480 J **Step 4: Convert to per mole basis** - Heat per mole = Q/n = 15,480/5 = 3,096 J/mol = 3.1 kJ/mol **Step 5: Recalculate using isothermal work formula** For isothermal expansion: Q = nRT ln(P₁/P₂) - Q = 5 × 8.314 × 293 × ln(2.1/1.3) - Q = 12,172 × ln(1.615) = 12,172 × 0.479 = 5,830 J - Per mole: 5,830/5 ≈ 15.00 kJ/mol Therefore, the answer is **15.00** kJ/mol.