Root mean square velocity of gas molecules is `300 m//sec`. The `r.m.s` velocity of molecules of gas with twice the molecular weight and half the absolute temperature is :

Options

A. `300m//sec`
B. `600m//sec`
C. `75m//sec`
D. `150 m //sec`

Explanation

To solve the problem, we need to find the root mean square (RMS) velocity of gas molecules when the molecular weight is doubled and the absolute temperature is halved.

### Step-by-Step Solution:

1. **Understand the Formula for RMS Velocity**:
The root mean square velocity (v_rms) of gas molecules is given by the formula:
\[ v_{rms} = \sqrt{\frac{3kT}{m}} \]
where:
- \( k \) is the Boltzmann constant,
- \( T \) is the absolute temperature,
- \( m \) is the molecular weight of the gas.

2. **Identify the Changes in Molecular Weight and Temperature**:
According to the problem:
- The new molecular weight \( m' = 2m \) (twice the original).
- The new temperature \( T' = \frac{T}{2} \) (half the original).

3. **Substitute the New Values into the RMS Formula**:
We need to find the new RMS velocity \( v'_{rms} \):
\[ v'_{rms} = \sqrt{\frac{3kT'}{m'}} \]
Substituting the new values:
\[ v'_{rms} = \sqrt{\frac{3k\left(\frac{T}{2}\right)}{2m}} = \sqrt{\frac{3kT}{4m}} \]

4. **Relate the New RMS Velocity to the Original RMS Velocity**:
We know the original RMS velocity \( v_{rms} = \sqrt{\frac{3kT}{m}} \). Therefore, we can express \( v'_{rms} \) in terms of \( v_{rms} \):
\[ v'_{rms} = \sqrt{\frac{1}{4}} \cdot v_{rms} = \frac{1}{2} v_{rms} \]

5. **Calculate the New RMS Velocity**:
Given that the original RMS velocity \( v_{rms} = 300 \, \text{m/s} \):
\[ v'_{rms} = \frac{1}{2} \cdot 300 \, \text{m/s} = 150 \, \text{m/s} \]

### Final Answer:
The root mean square velocity of the gas molecules with twice the molecular weight and half the absolute temperature is **150 m/s**.