By what factor does the average velocity of a gaseous molecule increase when the temperature (in Kelvin) is doubled ?

Options

A. `2.8`
B. `4.0`
C. `1.4`
D. `2.0`

Explanation

To determine by what factor the average velocity of a gaseous molecule increases when the temperature is doubled, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the relationship between average velocity and temperature**:
According to the kinetic molecular theory, the average velocity (V) of gas molecules is related to temperature (T) by the equation:
\[ V \propto \sqrt{T} \]
This means that the average velocity is directly proportional to the square root of the temperature.

2. **Define the initial and new temperatures**:
Let the initial temperature be \( T \). If the temperature is doubled, the new temperature \( T' \) will be:
\[ T' = 2T \]

3. **Express the new average velocity**:
Using the proportionality established earlier, the average velocity at the new temperature \( T' \) can be expressed as:
\[ V' \propto \sqrt{T'} \]
Substituting for \( T' \):
\[ V' \propto \sqrt{2T} \]

4. **Relate the new and old average velocities**:
Now, we can express the ratio of the new average velocity \( V' \) to the old average velocity \( V \):
\[ \frac{V'}{V} = \frac{\sqrt{2T}}{\sqrt{T}} = \sqrt{2} \]

5. **Calculate the factor of increase**:
Since \( \sqrt{2} \) is approximately equal to 1.414, we can conclude that:
\[ V' = V \cdot \sqrt{2} \]
Therefore, the average velocity increases by a factor of \( \sqrt{2} \) or approximately 1.414 when the temperature is doubled.

### Final Answer:
The average velocity of a gaseous molecule increases by a factor of approximately **1.414** (or \( \sqrt{2} \)) when the temperature is doubled.