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Question

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 :

Difficult

A

-20 kJ
B

20 J
C

-12 kJ
D

20 kJ

Solution

$∆ U = {nC}_{V} ∆ T and T = \frac{PV}{nR} So ∆ T = T_{2} - T_{1} = \frac{P_{2} V_{2} - P_{1} V_{1}}{nR} So ∆ U = \frac{nR}{γ - 1} (\frac{P_{2} V_{2} - P_{1} V_{1}}{nR}) = \frac{P_{2} V_{2} - P_{1} V_{1}}{γ - 1} \Rightarrow ∆ U = \frac{- 8 \times 10^{3}}{2 / 5} = - 20 kJ$

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