Question

If the rms speed of oxygen molecules at 0\(^{\circ}\)C is 160 m/s, find the rms speed of hydrogen molecules at 0\(^{\circ}\)C.

• A. 332 m/s
• B. 80 m/s
• C. 640 m/s
• D. 40 m/s

Hint:

Lighter gases move faster. Hydrogen is the lightest gas. Always expect its speed to be significantly higher than heavier gases like oxygen at the same temperature. The ratio of molar masses of O\(_2\) to H\(_2\) is 16, so the ratio of speeds will be \( \sqrt{16} = 4 \). This is a common comparison in exams.

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
We are given the root-mean-square (rms) speed of oxygen molecules at a certain temperature and asked to find the rms speed of hydrogen molecules at the same temperature.
Step 2: Key Formula or Approach:
The rms speed of gas molecules is given by the formula from the kinetic theory of gases:
\[ v_{\text{rms}} = \sqrt{\frac{3RT}{M}} \] where:
- \( R \) is the universal gas constant.
- \( T \) is the absolute temperature in Kelvin.
- \( M \) is the molar mass of the gas.
From this formula, we can see that at a constant temperature \( T \), the rms speed is inversely proportional to the square root of the molar mass:
\[ v_{\text{rms}} \propto \frac{1}{\sqrt{M}} \] Step 3: Detailed Explanation:
Let \( v_{\text{O}_2} \) and \( v_{\text{H}_2} \) be the rms speeds of oxygen and hydrogen, respectively.
Let \( M_{\text{O}_2} \) and \( M_{\text{H}_2} \) be their molar masses.
Using the proportionality, we can write the ratio:
\[ \frac{v_{\text{H}_2}}{v_{\text{O}_2}} = \sqrt{\frac{M_{\text{O}_2}}{M_{\text{H}_2}}} \] We need the molar masses of oxygen (\(O_2\)) and hydrogen (\(H_2\)).
Molar mass of Oxygen (\(O_2\)): \( M_{\text{O}_2} = 32 \, \text{g/mol} \).
Molar mass of Hydrogen (\(H_2\)): \( M_{\text{H}_2} = 2 \, \text{g/mol} \).
Now substitute the values into the ratio equation:
\[ \frac{v_{\text{H}_2}}{v_{\text{O}_2}} = \sqrt{\frac{32}{2}} = \sqrt{16} = 4 \] This means the rms speed of hydrogen is 4 times the rms speed of oxygen at the same temperature.
\[ v_{\text{H}_2} = 4 \times v_{\text{O}_2} \] We are given \( v_{\text{O}_2} = 160 \, \text{m/s} \).
\[ v_{\text{H}_2} = 4 \times 160 \, \text{m/s} = 640 \, \text{m/s} \] (Note: The temperature of 0\(^{\circ}\)C is the same for both gases, so it doesn't need to be converted to Kelvin for this ratio calculation).
Step 4: Final Answer:
The rms speed of hydrogen molecules at 0\(^{\circ}\)C is 640 m/s.