Question:
Two boxes are at the same temperature. The first box contains gas with molecular mass \(m_1\) and rms speed \(v_1\). The second box contains gas with molecular mass \(m_2\) and average speed \(v_2\). If \(v_1 = 1.5 v_2\), find \(\frac{m_1}{m_2}\).
Two boxes are at the same temperature. The first box contains gas with molecular mass \(m_1\) and rms speed \(v_1\). The second box contains gas with molecular mass \(m_2\) and average speed \(v_2\). If \(v_1 = 1.5 v_2\), find \(\frac{m_1}{m_2}\).
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For gases at same temperature, \(v_{\text{rms}} \propto 1/\sqrt{m}\). Use this to relate molecular masses when rms speeds are known.
Updated On: Jul 18, 2026
- 1.25
- 0.74
- 0.52
- 0.26
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The Correct Option is C
Solution and Explanation
Step 1: Recall rms speed relation.
For a gas at temperature \(T\), root mean square speed is:
\[ v_{\text{rms}} = \sqrt{\frac{3 k_B T}{m}} \]
Step 2: Express ratio of molecular masses.
Given \(v_1 = 1.5 v_2\), and both gases at same \(T\):
\[ v_1/v_2 = \sqrt{m_2/m_1} = 1.5 \]
Step 3: Solve for mass ratio.
\[ \sqrt{m_2/m_1} = 1.5 \implies m_2/m_1 = 2.25 \implies m_1/m_2 = 1/2.25 \approx 0.444 \]
Step 4: Consider rms vs average speed.
If \(v_2\) is average speed \(v_{\text{avg}} = \sqrt{8 k_B T/ \pi m}\), adjust factor:
\[ v_1/v_2 = 1.5 \implies \sqrt{3/m_1} / \sqrt{8/(\pi m_2)} \implies m_1/m_2 \approx 0.52 \]
Step 5: Check reasoning.
Ratio less than 1 since faster rms speed corresponds to lighter molecule. Calculation matches problem statement.
Step 6: Final conclusion.
Hence, the ratio of molecular masses is:
\[ \boxed{0.52} \]
For a gas at temperature \(T\), root mean square speed is:
\[ v_{\text{rms}} = \sqrt{\frac{3 k_B T}{m}} \]
Step 2: Express ratio of molecular masses.
Given \(v_1 = 1.5 v_2\), and both gases at same \(T\):
\[ v_1/v_2 = \sqrt{m_2/m_1} = 1.5 \]
Step 3: Solve for mass ratio.
\[ \sqrt{m_2/m_1} = 1.5 \implies m_2/m_1 = 2.25 \implies m_1/m_2 = 1/2.25 \approx 0.444 \]
Step 4: Consider rms vs average speed.
If \(v_2\) is average speed \(v_{\text{avg}} = \sqrt{8 k_B T/ \pi m}\), adjust factor:
\[ v_1/v_2 = 1.5 \implies \sqrt{3/m_1} / \sqrt{8/(\pi m_2)} \implies m_1/m_2 \approx 0.52 \]
Step 5: Check reasoning.
Ratio less than 1 since faster rms speed corresponds to lighter molecule. Calculation matches problem statement.
Step 6: Final conclusion.
Hence, the ratio of molecular masses is:
\[ \boxed{0.52} \]
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