Q138 / 160 · TS EAMCET 2024 (10 May Shift 2) · Physics
The temperature at which the rms speed of oxygen molecules is $75 \%$ of rms speed of nitrogen molecules at a temperature of $287^{\circ} \mathrm{C}$ is
- Correct answerA$87^{\circ} \mathrm{C}$
- B$127^{\circ} \mathrm{C}$
- C$227^{\circ} \mathrm{C}$
- D$360^{\circ} \mathrm{C}$
Solution
$R M S$ speed, $v_{\mathrm{rms}}=\sqrt{\frac{3 R T}{M}} \Rightarrow v_{\mathrm{rms}} \propto \sqrt{\frac{\mathrm{T}}{\mathrm{M}}}$
$\mathrm{v}_{\mathrm{rms}, \mathrm{o}}=75 \%$ of $\mathrm{v}_{\mathrm{rms}, \mathrm{N}}=\frac{3}{4} \mathrm{v}_{\mathrm{rms}, \mathrm{N}}$
$\therefore \frac{\mathrm{v}_{\mathrm{rms}, \mathrm{O}}}{\mathrm{v}_{\mathrm{rms}, \mathrm{N}}}=\sqrt{\frac{\mathrm{T}_{0}}{\mathrm{T}_{\mathrm{N}}} \times \frac{\mathrm{M}_{\mathrm{N}}}{\mathrm{M}_{\mathrm{o}}}}=\sqrt{\frac{\mathrm{T}_{\mathrm{c}}}{(273+287)} \times \frac{28}{32}}$
$\Rightarrow\left(\frac{3}{4}\right)^2=\frac{T_0}{20 \times 32} \Rightarrow T_0=360 \mathrm{~K}=(360-273)^{\circ} \mathrm{C}$
$=87^{\circ} \mathrm{C}$
Hence, the correct answer is option: 'A'.