The most probable velocity of the molecules of a gas is 1 km/sec. The R.M.S velocity of the molecules is

Options

A. 1.128 km/sec
B. 1.224 km/sec
C. 1.5 km/sec
D. 1.086 km/sec

Answer

To find the R.M.S (Root Mean Square) velocity of the molecules of a gas when the most probable velocity is given, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the relationship between most probable velocity and R.M.S velocity**:
- The most probable velocity (\(v_{mp}\)) is given by the formula:
\[ v_{mp} = \sqrt{\frac{2RT}{M}} \]
- The R.M.S velocity (\(v_{rms}\)) is given by the formula:
\[ v_{rms} = \sqrt{\frac{3RT}{M}} \]

2. **Given data**:
- The most probable velocity (\(v_{mp}\)) is given as 1 km/sec.

3. **Relate \(v_{mp}\) to \(v_{rms}\)**:
- From the equations, we can express \(v_{rms}\) in terms of \(v_{mp}\):
\[ v_{rms} = \sqrt{\frac{3}{2}} \cdot v_{mp} \]

4. **Substitute the value of \(v_{mp}\)**:
- Substitute \(v_{mp} = 1 \text{ km/sec}\):
\[ v_{rms} = \sqrt{\frac{3}{2}} \cdot 1 \text{ km/sec} \]

5. **Calculate \(v_{rms}\)**:
- Calculate \(\sqrt{\frac{3}{2}}\):
\[ \sqrt{\frac{3}{2}} = \frac{\sqrt{3}}{\sqrt{2}} \approx 1.224 \]
- Therefore:
\[ v_{rms} \approx 1.224 \text{ km/sec} \]

6. **Final Answer**:
- The R.M.S velocity of the molecules is approximately:
\[ v_{rms} \approx 1.224 \text{ km/sec} \]