Figure shows two paths by which a gas goes from A to C. A = (2 × 10⁻³ m³, 2 × 10⁴ Pa), B = (2 × 10⁻³ m³, 6 × 10⁴ Pa), and C = (4 × 10⁻³ m³, 6 × 10⁴ Pa). In process AB, 400 J of heat is added, and in process BC, 100 J of heat is added. Heat absorbed in process AC (straight line) is:
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Question

Figure shows two paths by which a gas goes from A to C. A = (2 × 10⁻³ m³, 2 × 10⁴ Pa), B = (2 × 10⁻³ m³, 6 × 10⁴ Pa), and C = (4 × 10⁻³ m³, 6 × 10⁴ Pa). In process AB, 400 J of heat is added, and in process BC, 100 J of heat is added. Heat absorbed in process AC (straight line) is:

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G1 Physics · Accepted Answer

Step 1: Apply the first law separately along A→B and B→C: ΔU = Q − W, with W positive for expansion. Step 2: Along A→B, volume is fixed. Hence W = 0 and ΔU = Q = +400 J. Step 3: Along B→C, pressure is constant at 6 × 10⁴ Pa and ΔV = (4 − 2) × 10⁻³ = 2 × 10⁻³ m³. Step 4: Work along B→C: W = PΔV = (6 × 10⁴)(2 × 10⁻³) = 120 J. Step 5: Internal-energy change along B→C: ΔU = Q − W = 100 − 120 = −20 J. Step 6: Since ΔU is a state function, total change from A to C = 400 + (−20) = 380 J. Step 7: For t…