Question

The molecular mass of a gas having r.m.s. speed four times as that of another gas having molecular mass $32$ is

• A. 2
• B. 4
• C. 16
• D. 32

Hint:

Higher r.m.s. speed always implies a lower molecular mass at a constant temperature.

Answer 1

The Correct Option is A

Step 1: Recall the formula for the root mean square (r.m.s.) speed of gas molecules. \[ v_{rms} = \sqrt{\frac{3RT}{M \]
Step 2: Establish the relationship between r.m.s. speed and molecular mass at the same temperature. \[ v_{rms} \propto \frac{1}{\sqrt{M \implies \frac{v_1}{v_2} = \sqrt{\frac{M_2}{M_1 \]
Step 3: Substitute the given values ($v_1 = 4v_2$ and $M_2 = 32$) into the ratio formula. \[ \frac{4v_2}{v_2} = \sqrt{\frac{32}{M_1 \] \[ 4 = \sqrt{\frac{32}{M_1 \]
Step 4: Square both sides and solve for $M_1$. \[ 16 = \frac{32}{M_1} \implies M_1 = \frac{32}{16} = 2 \]