A thermodynamic process is shown in the P–V diagram. The pressure at A is 3 × 10⁴ Pa, the volume at A is 2 × 10⁻³ m³, the pressure at B is 8 × 10⁴ Pa, and the volume at D is 5 × 10⁻³ m³. In process AB, 600 J of heat is added to the system. The change in internal energy of the system in process AC would be:
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Question

A thermodynamic process is shown in the P–V diagram. The pressure at A is 3 × 10⁴ Pa, the volume at A is 2 × 10⁻³ m³, the pressure at B is 8 × 10⁴ Pa, and the volume at D is 5 × 10⁻³ m³. In process AB, 600 J of heat is added to the system. The change in internal energy of the system in process AC would be:

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G1 Physics · Accepted Answer

Step 1: The provided path A→B is constant-volume, so work along A→B is W = 0. Step 2: Heat added along A→B is 600 J. The first law ΔU = Q − W therefore gives ΔU = 600 J for this segment. Step 3: Along B→C the pressure remains at 8 × 10⁴ Pa, while the volume increases from 2 × 10⁻³ to 5 × 10⁻³ m³. Step 4: Expansion work along B→C: W = PΔV = (8 × 10⁴)(3 × 10⁻³) = 240 J. Step 5: Heat along B→C is NOT specified in the printed question. Let it be q J. Step 6: For B→C, ΔU = q − 240 J. Therefore total…