Question:
The rms speed of oxygen molecule at some temperature is $150 \text{ ms}^{-1}$. Then the rms speed of hydrogen molecule at the same temperature is
The rms speed of oxygen molecule at some temperature is $150 \text{ ms}^{-1}$. Then the rms speed of hydrogen molecule at the same temperature is
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Lighter gas molecules move much faster! Since hydrogen is 16 times lighter than oxygen, its molecules move $\sqrt{16} = 4$ times faster at any given temperature.
Updated On: Jun 3, 2026
- $400 \text{ ms}^{-1}$
- $600 \text{ ms}^{-1}$
- $200 \text{ ms}^{-1}$
- $800 \text{ ms}^{-1}$
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The Correct Option is B
Solution and Explanation
Step 1: Concept
The root-mean-square speed ($v_{\text{rms}}$) of gas molecules is given by the formula $v_{\text{rms}} = \sqrt{\frac{3RT}{M}}$, where $R$ is the universal gas constant, $T$ is the absolute temperature, and $M$ is the molar mass of the gas.
Step 2: Meaning
At a fixed uniform temperature $T$, the root-mean-square speed of a gas molecule is inversely proportional to the square root of its molecular weight: $v_{\text{rms}} \propto \frac{1}{\sqrt{M}}$.
Step 3: Analysis
Let $v_1$ and $M_1$ represent oxygen ($O_2$), and $v_2$ and $M_2$ represent hydrogen ($H_2$). The molecular weights are $M_1 = 32 \text{ g mol}^{-1}$ and $M_2 = 2 \text{ g mol}^{-1}$ respectively. Setting up the ratio equation: $\frac{v_2}{v_1} = \sqrt{\frac{M_1}{M_2}} = \sqrt{\frac{32}{2}} = \sqrt{16} = 4$. Given that $v_1 = 150 \text{ ms}^{-1}$, we find $v_2 = 4 \times 150 = 600 \text{ ms}^{-1}$.
Step 4: Conclusion
Hence, the root-mean-square speed of the hydrogen molecule under the same thermal condition is $600 \text{ ms}^{-1}$.
Final Answer: (B)
The root-mean-square speed ($v_{\text{rms}}$) of gas molecules is given by the formula $v_{\text{rms}} = \sqrt{\frac{3RT}{M}}$, where $R$ is the universal gas constant, $T$ is the absolute temperature, and $M$ is the molar mass of the gas.
Step 2: Meaning
At a fixed uniform temperature $T$, the root-mean-square speed of a gas molecule is inversely proportional to the square root of its molecular weight: $v_{\text{rms}} \propto \frac{1}{\sqrt{M}}$.
Step 3: Analysis
Let $v_1$ and $M_1$ represent oxygen ($O_2$), and $v_2$ and $M_2$ represent hydrogen ($H_2$). The molecular weights are $M_1 = 32 \text{ g mol}^{-1}$ and $M_2 = 2 \text{ g mol}^{-1}$ respectively. Setting up the ratio equation: $\frac{v_2}{v_1} = \sqrt{\frac{M_1}{M_2}} = \sqrt{\frac{32}{2}} = \sqrt{16} = 4$. Given that $v_1 = 150 \text{ ms}^{-1}$, we find $v_2 = 4 \times 150 = 600 \text{ ms}^{-1}$.
Step 4: Conclusion
Hence, the root-mean-square speed of the hydrogen molecule under the same thermal condition is $600 \text{ ms}^{-1}$.
Final Answer: (B)
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