One mole of an ideal diatomic gas undergoes a transition from A to B along a path AB as shown in the figure below. The change in internal energy of the gas during the transition is :

- 12 kJ

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20 kJ

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- 20 kJ

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20 J

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Solution

The correct option is C
- 20 kJ

$Δ U = n C_{v} Δ T = n \frac{5 R}{2} Δ T = \frac{5}{2} n R Δ T = \frac{5}{2} (P_{f} V_{f} - P_{i} V_{i}) = \frac{5}{2} (2 \times 6 - 5 \times 4) = \frac{5}{2} (- 8) = - 20 k J$

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One mole of an ideal diatomic gas undergoes a transition from A to B along a path AB as shown in the figure below. The change in internal energy of the gas during the transition is :

The correct option is C - 20 kJ ΔU=nCvΔT=n5R2ΔT=52nRΔT=52(PfVf−PiVi)=52(2×6−5×4)=52(−8)=−20kJ

The correct option is C
- 20 kJ

$Δ U = n C_{v} Δ T = n \frac{5 R}{2} Δ T = \frac{5}{2} n R Δ T = \frac{5}{2} (P_{f} V_{f} - P_{i} V_{i}) = \frac{5}{2} (2 \times 6 - 5 \times 4) = \frac{5}{2} (- 8) = - 20 k J$