Question

If the temperature of a gas is changed to \( 9 \) times the initial value, then the rms velocity of the gaseous molecule increases by

• A. \( 9 \) times
• B. \( 3 \) times
• C. \( \sqrt{3} \) times
• D. \( 18 \) times
• E. \( 12 \) times

Hint:

For gases, \[ v_{\text{rms}} \propto \sqrt{T} \] So if temperature becomes \( n \) times, rms speed becomes \( \sqrt{n} \) times.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
The root-mean-square (rms) velocity is a measure of the speed of particles in a gas. It is directly related to the absolute temperature of the gas. This question tests the understanding of this relationship.
Step 2: Key Formula or Approach:
The formula for the rms velocity (\(v_{rms}\)) of gas molecules is:
\[ v_{rms} = \sqrt{\frac{3RT}{M}} \] where \(R\) is the universal gas constant, \(T\) is the absolute temperature in Kelvin, and \(M\) is the molar mass of the gas.
From this formula, we can see the direct proportionality between \(v_{rms}\) and the square root of the temperature:
\[ v_{rms} \propto \sqrt{T} \] Step 3: Detailed Explanation:
Let the initial temperature be \(T_1\) and the initial rms velocity be \(v_1\).
Let the final temperature be \(T_2\) and the final rms velocity be \(v_2\).
We are given that the final temperature is 9 times the initial temperature:
\[ T_2 = 9T_1 \] Using the proportionality \(v_{rms} \propto \sqrt{T}\), we can set up a ratio:
\[ \frac{v_2}{v_1} = \frac{\sqrt{T_2}}{\sqrt{T_1}} \] Substitute the value of \(T_2\):
\[ \frac{v_2}{v_1} = \sqrt{\frac{9T_1}{T_1}} \] The \(T_1\) terms cancel out:
\[ \frac{v_2}{v_1} = \sqrt{9} = 3 \] This means the final velocity is 3 times the initial velocity:
\[ v_2 = 3v_1 \] So, the rms velocity of the gaseous molecule increases by 3 times.
Step 4: Final Answer:
When the absolute temperature is multiplied by 9, the rms velocity is multiplied by \(\sqrt{9} = 3\). The new velocity is 3 times the original. This corresponds to option (B).