Question

The root mean square velocity of a gas molecule is $100\text{ ms}^{-1}$. The mass of the molecule is increased four times keeping the temperature constant. Then, the root mean square velocity is

• A. 25 ms$^{-1}$
• B. 50 ms$^{-1}$
• C. 75 ms$^{-1}$
• D. 2500 ms$^{-1}$
• E. 125 ms$^{-1}$

Hint:

Heavier gas molecules move slower than lighter ones at the same temperature. Increasing mass by 4 times reduces the rms velocity by a factor of $\sqrt{4} = 2$.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
The root mean square (rms) velocity of a gas describes the average speed of particles based on temperature and their molecular mass. It depends inversely on the square root of the mass.
Step 2: Key Formula or Approach:
The rms velocity formula is \(v_{\text{rms}} = \sqrt{\frac{3RT}{M}}\), where \(T\) is absolute temperature and \(M\) is molar mass.
If temperature \(T\) is constant, \(v_{\text{rms}} \propto \frac{1}{\sqrt{M}}\).
Step 3: Detailed Explanation:
Let the initial mass be \(M_1\) and initial velocity be \(v_1 = 100 \text{ ms}^{-1}\).
The new mass is \(M_2 = 4M_1\).
Using the proportionality relationship:
\[ \frac{v_2}{v_1} = \sqrt{\frac{M_1}{M_2}} \] Substitute the given relation:
\[ \frac{v_2}{100} = \sqrt{\frac{M_1}{4M_1}} \] \[ \frac{v_2}{100} = \sqrt{\frac{1}{4}} \] \[ \frac{v_2}{100} = \frac{1}{2} \] Solve for \(v_2\):
\[ v_2 = \frac{100}{2} = 50 \text{ ms}^{-1} \] Step 4: Final Answer:
The new root mean square velocity is 50 ms\(^{-1}\).