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 :

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-20 kJ

20 J

-12 kJ

20 kJ

answer is A.

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Detailed Solution

∆U=nCV∆T and T=PVnR So ∆T=T2-T1=P2V2-P1V1nR So ∆U=nRγ-1P2V2-P1V1nR=P2V2-P1V1γ-1 ⇒ ∆U=-8×1032/5=-20 kJ

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