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 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?
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$vᵣms \propto 1/\sqrtM$ at constant temperature.
Updated On: May 24, 2026
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The Correct Option is A
Solution and Explanation
To find the ratio of the root mean square (RMS) speeds of the molecules of two gases, we can use the formula for the root mean square speed of gas molecules:
\(v_{\text{rms}} = \sqrt{\frac{3kT}{m}}\)
where:
- \(v_{\text{rms}}\) is the root mean square speed,
- \(k\) is the Boltzmann constant,
- \(T\) is the absolute temperature, and
- \(m\) is the mass of the gas molecule.
The above equation can also be expressed in terms of molar mass (\(M\)) as:
\(v_{\text{rms}} = \sqrt{\frac{3RT}{M}}\)
where:
- \(R\) is the ideal gas constant,
- \(M\) is the molar mass of the gas.
Given:
- Gas 1 is monoatomic, and Gas 2 is diatomic.
- The ratio of their molar masses \(\frac{M_1}{M_2} = \frac{1}{4}\).
We need to find the ratio of the RMS speeds \(\frac{v_{\text{rms1}}}{v_{\text{rms2}}}\) for gases 1 and 2:
\(\frac{v_{\text{rms1}}}{v_{\text{rms2}}} = \sqrt{\frac{M_2}{M_1}}\)
Substitute the given mass ratio:
\(\frac{v_{\text{rms1}}}{v_{\text{rms2}}} = \sqrt{\frac{4}{1}} = 2\)
Thus, the ratio of the root mean square speeds of the molecules of the two gases is 2, which matches with the given correct answer.
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