Q112 / 150 · MHT CET 2024 (03 May Shift 1) · Physics
The r.m.s. velocity of hydrogen at S.T.P. is ' u ' $\mathrm{m} / \mathrm{s}$. If the gas is heated at constant pressure till its volume becomes three times, then the final temperature of the gas and the r.m.s. speed are respectively
- Correct answerA$819 \mathrm{~K},(\sqrt{3}) \mathrm{u} \mathrm{m} / \mathrm{s}$
- B$\quad 1092 \mathrm{~K}, 3 \mathrm{u} \mathrm{m} / \mathrm{s}$
- C$819 \mathrm{~K}, \frac{\mathrm{u}}{\sqrt{3}} \mathrm{~m} / \mathrm{s}$
- D$\quad 1092 \mathrm{~K}, \frac{\mathrm{u}}{3} \mathrm{~m} / \mathrm{s}$
Solution
As the process is adiabatic,
$\begin{array}{rlrl}
& & \frac{V_2}{V_1} & =\frac{T_2}{T_1} \Rightarrow \frac{3 V_1}{V_1}=\frac{T_2}{T_1} \\
\therefore & 3 T_1 & =T_2 \\
\therefore & T_2 & =3 \times 273=819 \mathrm{~K}
\end{array}$
Also, $\begin{aligned} & \mathrm{V}_{\mathrm{rms}}=\sqrt{\frac{3 \mathrm{RT}}{\mathrm{M}_0}} \\ & \Rightarrow \mathrm{~V}_{\mathrm{rms}} \propto \sqrt{\mathrm{~T}} \\ & \frac{\mathrm{~V}_{\mathrm{rms}}^{\prime}}{\mathrm{V}_{\mathrm{rms}}}=\sqrt{\frac{\mathrm{T}_2}{\mathrm{~T}_1}} \end{aligned}$ $\therefore \quad \mathrm{V}_{\mathrm{rms}}^{\prime}=\mathrm{V}_{\mathrm{rms}} \sqrt{\frac{819}{273}}=\sqrt{3} \mathrm{um} / \mathrm{s} \ldots\left(\text { given } \mathrm{V}_{\mathrm{rms}}=\mathrm{u}\right)$
Also, $\begin{aligned} & \mathrm{V}_{\mathrm{rms}}=\sqrt{\frac{3 \mathrm{RT}}{\mathrm{M}_0}} \\ & \Rightarrow \mathrm{~V}_{\mathrm{rms}} \propto \sqrt{\mathrm{~T}} \\ & \frac{\mathrm{~V}_{\mathrm{rms}}^{\prime}}{\mathrm{V}_{\mathrm{rms}}}=\sqrt{\frac{\mathrm{T}_2}{\mathrm{~T}_1}} \end{aligned}$ $\therefore \quad \mathrm{V}_{\mathrm{rms}}^{\prime}=\mathrm{V}_{\mathrm{rms}} \sqrt{\frac{819}{273}}=\sqrt{3} \mathrm{um} / \mathrm{s} \ldots\left(\text { given } \mathrm{V}_{\mathrm{rms}}=\mathrm{u}\right)$
Hence, the correct answer is option: 'A'.