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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
$-20\, kJ$
$20\, J$
$-12\, kJ$
$20\, kJ$
Solution
$\Delta \mathrm{U}=\mathrm{n} \mathrm{C}_{\mathrm{V}} \Delta \mathrm{T} \quad $ and $ \quad \mathrm{T}=\frac{\mathrm{PV}}{\mathrm{nR}}$
so $\Delta \mathrm{T}=\mathrm{T}_{2}-\mathrm{T}_{1}=| \frac{\mathrm{P}_{2} \mathrm{V}_{2}-\mathrm{P}_{1} \mathrm{V}_{1}}{\mathrm{n} \mathrm{R}}$
so $\Delta \mathrm{U}=\frac{\mathrm{nR}}{\gamma-1}\left(\frac{\mathrm{P}_{2} \mathrm{V}_{2}-\mathrm{P}_{1} \mathrm{V}_{1}}{\mathrm{nR}}\right)=\frac{\mathrm{P}_{2} \mathrm{V}_{2}-\mathrm{P}_{1} \mathrm{V}_{1}}{\gamma-1}$
$\Rightarrow \Delta \mathrm{U}=\frac{-8 \times 10^{3}}{2 / 5}=-20 \mathrm{\,kJ}$
