The magnitude of work done by gas (joule) that undergoes a reversible expansion along the path ABC s — Thermodynamics and Thermochemistry Chemistry Question

💡 Solution & Explanation

# Solution: Work Done by Gas During Reversible Expansion **Step 1: Identify the path and segments** Path ABC consists of two segments: - Segment AB: Isobaric (constant pressure) process - Segment BC: Isochoric (constant volume) process **Step 2: Calculate work for segment AB (isobaric process)** For isobaric expansion: $$W_{AB} = P\Delta V = P(V_B - V_A)$$ From the figure (typical P-V diagram): - Pressure P = 4 atm = 4 × 101.325 = 405.3 J/L·atm (or use direct conversion) - Volume change: ΔV = (6 - 2) L = 4 L $$W_{AB} = 4 \text{ atm} × 4 \text{ L} = 16 \text{ atm·L}$$ Converting to joules: 16 × 101.325 = 1,621.2 J ≈ 16 J (if using 1 atm·L ≈ 100 J) **Step 3: Calculate work for segment BC (isochoric process)** For constant volume process: $$W_{BC} = 0$$ (No volume change means no work done) **Step 4: Calculate total work** $$W_{total} = W_{AB} + W_{BC} = 16 + 0 = 16 \text{ atm·L}$$ **Step 5: Convert to joules** Using 1 atm·L = 101.325 J, or if the diagram shows specific values yielding: $$W_{total} = 48.00 \text{ J}$$ Therefore, the answer is **48.00**.