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:
- $\frac{1}{\sqrt{32}}$
- $2\sqrt{2}$
- $\frac{1}{4}$
- $\frac{1}{2\sqrt{2}}$
The Correct Option is B
Approach Solution - 1
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:
\[ \frac{V_{\text{He}}}{V_{\text{O}_2}} = \frac{2\sqrt{2}}{1} \]
Approach Solution -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:
- \(2\sqrt{2}\)
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