Question

A mixture of two gases is contained in a vessel. The gas 1 is monoatomic and gas 2 is diatomic and the ratio of their molecular masses $M₁/M₂ = 1/4$. What is the ratio of root mean square speeds of the molecules of two gases?

• A. 2
• B. 4
• C. 8
• D. 16

Hint:

$vᵣms \propto 1/\sqrtM$ at constant temperature.

Answer 1

The Correct Option is A

To determine the ratio of the root mean square (RMS) speeds of two gas molecules, we will start by recalling the formula for RMS speed:

\(v_{\text{rms}} = \sqrt{\frac{3kT}{M}}\)

where:

• \(v_{\text{rms}}\) is the root mean square speed,
• \(k\) is the Boltzmann constant,
• \(T\) is the temperature in Kelvin,
• \(M\) is the molar mass of the gas.

The problem provides that gas 1 is monoatomic, and gas 2 is diatomic, with a molecular mass ratio of \(\frac{M_1}{M_2} = \frac{1}{4}\).

Now, let's find the ratio of their RMS speeds:

\(\frac{v_{\text{rms1}}}{v_{\text{rms2}}} = \sqrt{\frac{3kT/M_1}{3kT/M_2}} = \sqrt{\frac{M_2}{M_1}}\)

Substituting the given ratio \(\frac{M_1}{M_2} = \frac{1}{4}\), we have:

\(\frac{v_{\text{rms1}}}{v_{\text{rms2}}} = \sqrt{\frac{M_2}{M_1}} = \sqrt{4} = 2\)

Thus, the ratio of the root mean square speeds of the two gases is 2.

Therefore, the correct answer is:

• 2