Q68 / 90 · KVPY 2016 (SB/SX) · Physics
The heat capacity of one mole an ideal is found to be $\mathrm{CV}=3 \mathrm{R}(1+\mathrm{aRT}) / 2$ where a is a constant. The equation obeyed by this gas during a reversible adiabatic expansion is -
- Correct answerA$\mathrm{TV}^{3 / 2} \mathrm{e}^{\mathrm{aRT}}=$ constant
- B$\mathrm{TV}^{3 / 2} \mathrm{e}^{3 \mathrm{aRT} / 2}=$ constant
- C$\mathrm{TV}^{3 / 2}=$ constant
- D$\mathrm{TV}^{3 / 2} \mathrm{e}^{2 \mathrm{aRT} / 3}=\mathrm{constant}$
Solution
Adiabatic process
$$
\begin{array}{l}
\mathrm{TV}^{\gamma-1}=\mathrm{C} \\
\gamma=1+\frac{2}{\mathrm{f}} \\
\mathrm{TV}^{\frac{2}{\mathrm{f}}}=\mathrm{C} \\
\mathrm{C}_{\mathrm{V}}=\frac{\mathrm{f} \mathrm{R}}{2}=\frac{3 \mathrm{R}(1+\mathrm{aRT})}{2} \\
\frac{\mathrm{f} \mathrm{R}}{2}=\frac{3 \mathrm{Re}^{\mathrm{aRT}}}{2} \\
\frac{2}{\mathrm{f}}=\frac{2}{3 \mathrm{e}^{\mathrm{aRT}}}
\end{array}
$$
$$
\begin{array}{l}
\mathrm{TV}^{\frac{2}{3 \mathrm{e}^{\mathrm{aRT}}}}=\mathrm{C} \\
\mathrm{TV}^{\frac{3 \mathrm{e}^{\mathrm{aRT}}}{2}}=\mathrm{C}
\end{array}
$$
Ans. given is $\mathrm{TV}^{\frac{3}{2}} \mathrm{e}^{\mathrm{aRT}}$
So no option is matching may be due to printing mistake.
Hence, the correct answer is option: 'A'.