Question:
The temperature of an ideal gas is increased from 100 K to 400 K. If the rms speed of the gas molecule is $v$ at 100 K then at 400 K it becomes
The temperature of an ideal gas is increased from 100 K to 400 K. If the rms speed of the gas molecule is $v$ at 100 K then at 400 K it becomes
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To double the average speed of gas molecules, you must increase the absolute temperature by a factor of four.
Updated On: Apr 29, 2026
- $2v$
- $4v$
- $0.5v$
- $0.25v$
- $v$
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The Correct Option is A
Solution and Explanation
Concept:
The root mean square (rms) speed of an ideal gas molecule is given by the formula: \[ v_{rms} = \sqrt{\frac{3RT}{M}} \] where $R$ is the gas constant, $T$ is the absolute temperature, and $M$ is the molar mass. This indicates that $v_{rms} \propto \sqrt{T}$.
Step 1: Set up the ratio for the two temperatures.
Let $v_1 = v$ at $T_1 = 100$ K, and $v_2$ be the speed at $T_2 = 400$ K. \[ \frac{v_2}{v_1} = \sqrt{\frac{T_2}{T_1}} \]
Step 2: Calculate the final speed.
\[ \frac{v_2}{v} = \sqrt{\frac{400}{100}} = \sqrt{4} = 2 \] \[ v_2 = 2v \]
The root mean square (rms) speed of an ideal gas molecule is given by the formula: \[ v_{rms} = \sqrt{\frac{3RT}{M}} \] where $R$ is the gas constant, $T$ is the absolute temperature, and $M$ is the molar mass. This indicates that $v_{rms} \propto \sqrt{T}$.
Step 1: Set up the ratio for the two temperatures.
Let $v_1 = v$ at $T_1 = 100$ K, and $v_2$ be the speed at $T_2 = 400$ K. \[ \frac{v_2}{v_1} = \sqrt{\frac{T_2}{T_1}} \]
Step 2: Calculate the final speed.
\[ \frac{v_2}{v} = \sqrt{\frac{400}{100}} = \sqrt{4} = 2 \] \[ v_2 = 2v \]
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