Q149 / 180 · COMEDK 2023 (Evening Shift) · Physics
The molecules of a given mass of a gas have root mean square speed of $120 \mathrm{~m} / \mathrm{s}$ at $88^{\circ} \mathrm{C}$ and 1 atmospheric pressure. The root mean square speed of the molecules at $127^{\circ} \mathrm{C}$ and 2 atmospheric pressure is
- A1.443 m/s
- Correct answerB126.3 m/s
- C88.12 m/s
- D105.2 m/s
Solution
The root mean square speed $v_{rms}$ of gas molecules is given by the formula $v_{rms} = \sqrt{\dfrac{3RT}{M}}$.
From this formula, it is evident that $v_{rms}$ depends only on the temperature $T$ and the molar mass $M$ of the gas. It is independent of the pressure $P$ of the gas.
Given $v_{rms,1} = 120 \text{ m/s}$ at $T_1 = 88^{\circ}\text{C} = 88 + 273 = 361 \text{ K}$.
We need to find $v_{rms,2}$ at $T_2 = 127^{\circ}\text{C} = 127 + 273 = 400 \text{ K}$.
Taking the ratio of the two speeds:
$\dfrac{v_{rms,2}}{v_{rms,1}} = \sqrt{\dfrac{T_2}{T_1}}$
Substituting the values:
$\dfrac{v_{rms,2}}{120} = \sqrt{\dfrac{400}{361}}$
$\dfrac{v_{rms,2}}{120} = \dfrac{20}{19}$
$v_{rms,2} = 120 \times \dfrac{20}{19} = \dfrac{2400}{19} \approx 126.315 \text{ m/s}$.
Answer: 126.3 m/s
From this formula, it is evident that $v_{rms}$ depends only on the temperature $T$ and the molar mass $M$ of the gas. It is independent of the pressure $P$ of the gas.
Given $v_{rms,1} = 120 \text{ m/s}$ at $T_1 = 88^{\circ}\text{C} = 88 + 273 = 361 \text{ K}$.
We need to find $v_{rms,2}$ at $T_2 = 127^{\circ}\text{C} = 127 + 273 = 400 \text{ K}$.
Taking the ratio of the two speeds:
$\dfrac{v_{rms,2}}{v_{rms,1}} = \sqrt{\dfrac{T_2}{T_1}}$
Substituting the values:
$\dfrac{v_{rms,2}}{120} = \sqrt{\dfrac{400}{361}}$
$\dfrac{v_{rms,2}}{120} = \dfrac{20}{19}$
$v_{rms,2} = 120 \times \dfrac{20}{19} = \dfrac{2400}{19} \approx 126.315 \text{ m/s}$.
Answer: 126.3 m/s
Hence, the correct answer is option: 'B'.