One mole of an ideal diatomic gas undergoes a transition from A to B along path AB shown in the P–V graph. A is at (V = 4 m³, P = 5 kPa), and B is at (V = 6 m³, P = 2 kPa). The change in internal energy of the gas is:
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Question

One mole of an ideal diatomic gas undergoes a transition from A to B along path AB shown in the P–V graph. A is at (V = 4 m³, P = 5 kPa), and B is at (V = 6 m³, P = 2 kPa). The change in internal energy of the gas is:

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G1 Physics · Accepted Answer

Step 1: For one mole of a diatomic ideal gas with vibrational modes ignored, molar heat capacity Cᵥ = 5R/2. Step 2: Internal energy depends only on temperature: ΔU = nCᵥΔT = (5/2)nR(T at B − T at A). Step 3: From PV = nRT, we can use nRΔT = (PV at B) − (PV at A). Step 4: At state A: PV = (5 kPa)(4 m³) = 20 kJ. Step 5: At state B: PV = (2 kPa)(6 m³) = 12 kJ. Step 6: ΔU = (5/2)(12 − 20) kJ = (5/2)(−8) kJ = −20 kJ. Final answer: (D) −20 kJ. The path shape does not affect ΔU.