One mole of an ideal diatomic gas undergoes a transition from A to B along a path AB as shown in the figure,

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 $- 20 kJ$
As we know,
$Δ U = n C_{v} Δ T$

$= n \frac{5 R}{2} Δ T = \frac{5}{2} n R Δ T$

$\frac{5}{2} (P_{f} V_{f} - P_{i} V_{i}) = \frac{5}{2} (2 \times 6 - 5 \times 4) = \frac{5}{2} (- 8) = - 20 kJ$

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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, The change in internal energy of the gas during the transition is:

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

One mole of an ideal diatomic gas undergoes a transition from A to B along a path AB as shown in the figure,

The change in internal energy of the gas during the transition is:

The correct option is C $- 20 kJ$
As we know,
$Δ U = n C_{v} Δ T$

$= n \frac{5 R}{2} Δ T = \frac{5}{2} n R Δ T$

$\frac{5}{2} (P_{f} V_{f} - P_{i} V_{i}) = \frac{5}{2} (2 \times 6 - 5 \times 4) = \frac{5}{2} (- 8) = - 20 kJ$