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11.Thermodynamics
normal
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
A
$-\,20\,kJ$
B
$20\,J$
C
$-\,12\,kJ$
D
$20\,kJ$
Solution
$\Delta \mathrm{U}=\mathrm{n} \mathrm{C}_{\mathrm{V}} \Delta \mathrm{T} \quad \& \quad \mathrm{T}=\frac{\mathrm{PV}}{\mathrm{nR}}$
so $\quad \Delta \mathrm{T}=\mathrm{T}_{2}-\mathrm{T}_{1}=\frac{\mathrm{P}_{2} \mathrm{V}_{2}-\mathrm{P}_{1} \mathrm{V}_{1}}{\mathrm{n} \mathrm{R}}$
so $\Delta U=\frac{n R}{\gamma-1}\left(\frac{P_{2} V_{2}-P_{1} V_{1}}{n R}\right)=\frac{P_{2} V_{2}-P_{1} V_{1}}{\gamma-1}$
$\Rightarrow \Delta U=\frac{-8 \times 10^{3}}{2 / 5}=-20 \mathrm{kJ}$
Standard 11
Physics
