Question

A sample contains a mixture of helium and oxygen gas. The ratio of root mean square speed of helium and oxygen in the sample is:

• A. $\frac{1}{\sqrt{32}}$
• B. $2\sqrt{2}$
• C. $\frac{1}{4}$
• D. $\frac{1}{2\sqrt{2}}$

Answer 1

The Correct Option is B

The root mean square speed (\(V_{\text{rms}}\)) is given by:

\[ V_{\text{rms}} = \sqrt{\frac{3RT}{M_w}} \]

where \(M_w\) is the molar mass of the gas.

The ratio of root mean square speeds of helium (\(V_{\text{He}}\)) and oxygen (\(V_{\text{O}_2}\)) is:

\[ \frac{V_{\text{O}_2}}{V_{\text{He}}} = \sqrt{\frac{M_{w,\text{He}}}{M_{w,\text{O}_2}}} \]

Substituting the values:

• \(M_{w,\text{He}} = 4 \, \text{g/mol}\)
• \(M_{w,\text{O}_2} = 32 \, \text{g/mol}\)

\[ \frac{V_{\text{O}_2}}{V_{\text{He}}} = \sqrt{\frac{4}{32}} = \frac{1}{2\sqrt{2}} \]

The ratio \(V_{\text{He}} / V_{\text{O}_2}\) is:

\[ \frac{V_{\text{He}}}{V_{\text{O}_2}} = \frac{2\sqrt{2}}{1} \]

Answer 2

To determine the ratio of the root mean square (rms) speed of helium and oxygen in the sample, we utilize the formula for the root mean square speed in gases:

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

where:

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

The ratio of the rms speeds of two gases \(1\) and \(2\) is given by the formula:

\(\frac{v_{\text{rms, 1}}}{v_{\text{rms, 2}}} = \sqrt{\frac{M_2}{M_1}}\)

For helium (He) and oxygen (O₂):

• The molar mass of helium, \(M_{\text{He}} = 4 \,\text{g/mol}\)
• The molar mass of oxygen, \(M_{\text{O}_2} = 32 \,\text{g/mol}\)

Thus, the ratio of rms speeds is:

\(\frac{v_{\text{rms, He}}}{v_{\text{rms, O}_2}} = \sqrt{\frac{M_{\text{O}_2}}{M_{\text{He}}}} = \sqrt{\frac{32}{4}} = \sqrt{8} = 2\sqrt{2}\)

Therefore, the ratio of the root mean square speed of helium to that of oxygen in the sample is \(2\sqrt{2}\). The correct answer is:

• \(2\sqrt{2}\)