Question

The root mean square speed of molecules of a given mass of a gas at 27°C and 1 atmosphere pressure is 200 m s⁻¹. The root mean square speed of molecules of the gas at 127°C and 2 atmosphere pressure is \( \dfrac{x}{\sqrt{3}} \) m s⁻¹. The value of \( x \) will be __________.

Hint:

Always convert temperatures to Kelvin for Gas Law problems. Pressure is a "distractor" here as $v_{rms}$ depends only on temperature for a given gas.

Answer 1

Correct Answer: 400

Step 1: $v_{rms} = \sqrt{\frac{3RT}{M}}$, so $v_{rms} \propto \sqrt{T}$. Pressure change does not affect $v_{rms}$ at a given temperature.
Step 2: $T_1 = 27 + 273 = 300 \text{ K}$, $T_2 = 127 + 273 = 400 \text{ K}$.
Step 3: $\frac{v_2}{v_1} = \sqrt{\frac{T_2}{T_1}} = \sqrt{\frac{400}{300}} = \frac{2}{\sqrt{3}}$.
Step 4: $v_2 = v_1 \times \frac{2}{\sqrt{3}} = 200 \times \frac{2}{\sqrt{3}} = \frac{400}{\sqrt{3}}$. Comparing to $x/\sqrt{3}$, $x = 400$.