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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a
-20 kJ
b
20 J
c
-12 kJ
d
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