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How much the change in internal energy of the gas during the transition of one mole of an ideal diatomic gas undergoes a transition from A to B along a path AB as shown in the figure,
is :

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$- 20 k J$

20 J

$- 12 k J$

20 kJ

answer is A.

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Detailed Solution

$ΔU = {nC}_{V} ΔT & T = \frac{PV}{nR}$

So, $ΔT = T_{2} - T_{1} = \frac{P_{2} V_{2} - P_{1} V_{1}}{nR}$

So,

$\begin{array}{l} Δ U = \frac{n R}{γ - 1} (\frac{P_{2} V_{2} - P_{1} V_{1}}{n R}) = \frac{P_{2} V_{2} - P_{1} V_{1}}{γ - 1} \\ \Rightarrow Δ U = \frac{- 8 \times 10^{3}}{2 / 5} = - 20 kJ \end{array}$

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