Question:
At \(T(K)\), the rms velocity of an ideal gas is \(x\,\text{m s}^{-1}\). At what temperature (in K), the rms velocity becomes \(3x\,\text{m s}^{-1}\)?
At \(T(K)\), the rms velocity of an ideal gas is \(x\,\text{m s}^{-1}\). At what temperature (in K), the rms velocity becomes \(3x\,\text{m s}^{-1}\)?
Show Hint
For an ideal gas:
\[
v_{rms} \propto \sqrt{T}
\]
If speed becomes \(n\) times, temperature becomes \(n^2\) times.
Updated On: Oct 6, 2026
- \(3T\)
- \(4T\)
- \(9T\)
- \(18T\)
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The Correct Option is C
Solution and Explanation
Concept:
The rms speed of an ideal gas is given by:
\[
v_{rms} = \sqrt{\frac{3RT}{M}}
\]
Thus:
\[
v_{rms} \propto \sqrt{T}
\]
This means:
\[
\frac{v_2}{v_1} = \sqrt{\frac{T_2}{T_1}}
\]
Therefore, when rms speed changes, temperature changes as the square of speed ratio.
Step 1: Write the proportional relation. Initially: \[ v_1 = x \] at temperature: \[ T_1 = T \] Finally: \[ v_2 = 3x \] Required temperature: \[ T_2 = ? \] Using: \[ \frac{v_2}{v_1} = \sqrt{\frac{T_2}{T_1}} \]
Step 2: Substitute the values. \[ \frac{3x}{x} = \sqrt{\frac{T_2}{T}} \] \[ 3 = \sqrt{\frac{T_2}{T}} \] Squaring both sides: \[ 9 = \frac{T_2}{T} \] Thus: \[ T_2 = 9T \]
Step 3: Write the final answer. Hence, required temperature is: \[ \boxed{9T} \] Therefore, the correct option is: \[ \boxed{(C)} \]
Step 1: Write the proportional relation. Initially: \[ v_1 = x \] at temperature: \[ T_1 = T \] Finally: \[ v_2 = 3x \] Required temperature: \[ T_2 = ? \] Using: \[ \frac{v_2}{v_1} = \sqrt{\frac{T_2}{T_1}} \]
Step 2: Substitute the values. \[ \frac{3x}{x} = \sqrt{\frac{T_2}{T}} \] \[ 3 = \sqrt{\frac{T_2}{T}} \] Squaring both sides: \[ 9 = \frac{T_2}{T} \] Thus: \[ T_2 = 9T \]
Step 3: Write the final answer. Hence, required temperature is: \[ \boxed{9T} \] Therefore, the correct option is: \[ \boxed{(C)} \]
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