The root mean square velocity of the gas molecule is 300 m/s. What will be the root mean square speed of he molecule if the atomic weight is doubled and absolute temperature is halved ?

Options

A. 300 m/s
B. 150 m/s
C. 600 m/s
D. 75 m/s

Explanation

To solve the problem, we need to find the new root mean square (RMS) speed of gas molecules when the atomic weight is doubled and the absolute temperature is halved. We will use the formula for RMS speed:

\[ V_{rms} = \sqrt{\frac{3RT}{M}} \]

Where:
- \( V_{rms} \) is the root mean square speed,
- \( R \) is the universal gas constant,
- \( T \) is the absolute temperature,
- \( M \) is the molecular weight of the gas.

### Step 1: Identify the initial conditions
Given:
- Initial RMS speed, \( V_{rms1} = 300 \, \text{m/s} \)
- Let the initial temperature be \( T \) and the initial molecular weight be \( M \).

### Step 2: Write the expression for the initial RMS speed
Using the formula for RMS speed, we have:

\[ V_{rms1} = \sqrt{\frac{3RT}{M}} \]

### Step 3: Define the new conditions
According to the problem:
- The atomic weight is doubled: \( M_2 = 2M \)
- The absolute temperature is halved: \( T_2 = \frac{T}{2} \)

### Step 4: Write the expression for the new RMS speed
Now, we can write the expression for the new RMS speed:

\[ V_{rms2} = \sqrt{\frac{3RT_2}{M_2}} = \sqrt{\frac{3R\left(\frac{T}{2}\right)}{2M}} \]

### Step 5: Simplify the expression for \( V_{rms2} \)
Substituting \( T_2 \) and \( M_2 \) into the equation:

\[ V_{rms2} = \sqrt{\frac{3R\left(\frac{T}{2}\right)}{2M}} = \sqrt{\frac{3RT}{4M}} = \frac{1}{2} \sqrt{\frac{3RT}{M}} = \frac{1}{2} V_{rms1} \]

### Step 6: Calculate the new RMS speed
Since \( V_{rms1} = 300 \, \text{m/s} \):

\[ V_{rms2} = \frac{1}{2} \times 300 \, \text{m/s} = 150 \, \text{m/s} \]

### Conclusion
The new root mean square speed of the gas molecule when the atomic weight is doubled and the absolute temperature is halved is:

\[ \boxed{150 \, \text{m/s}} \]