Q146 / 150 · MHT CET 2024 (09 May Shift 1) · Physics
The temperature of a gas is $-80^{\circ} \mathrm{C}$. To what temperature the gas should be heated so that the r.m.s. speed is increased by 2 times?
- Correct answerA$499^{\circ} \mathrm{C}$
- B$772^{\circ} \mathrm{C}$
- C$1464^{\circ} \mathrm{C}$
- D$1737^{\circ} \mathrm{C}$
Solution
$T_1=-80^{\circ} \mathrm{C}=-80+273 \mathrm{~K}=193 \mathrm{~K}$
R.M.S. velocity, $\mathrm{v}_{\mathrm{rms}} \propto \sqrt{\mathrm{T}}$
As the speed is increased by two times, final speed becomes $\left(2 \mathrm{~V}_{\mathrm{rms}}\right)_1$ $\begin{array}{ll} & \frac{\left(v_{\mathrm{mss}}\right)_2}{\left(v_{\mathrm{ms}}\right)_1}=\sqrt{\frac{T_2}{T_1}}=2 \\ \therefore \quad & \frac{T_2}{T_1}=4 \\ \therefore \quad & T_2=4 \times 193=772 \mathrm{~K}=499^{\circ} \mathrm{C} \end{array}$
As the speed is increased by two times, final speed becomes $\left(2 \mathrm{~V}_{\mathrm{rms}}\right)_1$ $\begin{array}{ll} & \frac{\left(v_{\mathrm{mss}}\right)_2}{\left(v_{\mathrm{ms}}\right)_1}=\sqrt{\frac{T_2}{T_1}}=2 \\ \therefore \quad & \frac{T_2}{T_1}=4 \\ \therefore \quad & T_2=4 \times 193=772 \mathrm{~K}=499^{\circ} \mathrm{C} \end{array}$
Hence, the correct answer is option: 'A'.