Q140 / 160 · AP EAMCET 2024 (22 May Shift 2) · Physics
The total internal energy of 2 moles of a monotomic gas at a temperature $27^{\circ} \mathrm{C}$ is U . The total internal energy of 3 moles of a diatomic gas at a temperature $127^{\circ} \mathrm{C}$ is
- AU
- Correct answerB$\frac{10 \mathrm{U}}{3}$
- C2 U
- D$\frac{2 U}{3}$
Solution
$\mathrm{n}_1=2$ moles, $\mathrm{T}_1=27^{\circ} \mathrm{c}=300 \mathrm{~K}$
For monoatomic gas, $f_1=3$
$\therefore \quad \mathrm{U}=\frac{\mathrm{n}_1 \mathrm{f}_1 \mathrm{RT}_1}{2}=\frac{2 \times 3 \times \mathrm{R} \times 300}{2}=900 \mathrm{R}$
For diatomic gas, $\mathrm{f}_2=5$ $\begin{aligned} & \mathrm{n}_2=3 \text { moles, } \mathrm{T}_2=127^{\circ} \mathrm{c}=400 \mathrm{~K} \\ & \mathrm{U}^{\prime}=\frac{\mathrm{n}_2 \mathrm{f}_2 \mathrm{RT}_2}{2}=\frac{3 \times 5 \times \mathrm{R} \times 400}{2}=3000 \mathrm{R} \\ & \therefore \frac{\mathrm{U}^{\prime}}{\mathrm{U}}=\frac{3000 \mathrm{R}}{900 \mathrm{R}}=\frac{10}{3} \\ & \therefore \quad U^{\prime}=\frac{10 \mathrm{U}}{3} \end{aligned}$
For diatomic gas, $\mathrm{f}_2=5$ $\begin{aligned} & \mathrm{n}_2=3 \text { moles, } \mathrm{T}_2=127^{\circ} \mathrm{c}=400 \mathrm{~K} \\ & \mathrm{U}^{\prime}=\frac{\mathrm{n}_2 \mathrm{f}_2 \mathrm{RT}_2}{2}=\frac{3 \times 5 \times \mathrm{R} \times 400}{2}=3000 \mathrm{R} \\ & \therefore \frac{\mathrm{U}^{\prime}}{\mathrm{U}}=\frac{3000 \mathrm{R}}{900 \mathrm{R}}=\frac{10}{3} \\ & \therefore \quad U^{\prime}=\frac{10 \mathrm{U}}{3} \end{aligned}$
Hence, the correct answer is option: 'B'.