The temperature of an ideal gas is increased from 120K to 480K. If at 120K the root-mean-squre velocity of the gas molecules is v, at 480K it becomes

Options

A. `4v`
B. `2v`
C. `v//2`
D. `v//4`

Explanation

To find the root-mean-square (RMS) velocity of the gas molecules at the new temperature of 480K, we can use the relationship between the RMS velocity and temperature for an ideal gas. The RMS velocity (v_rms) is given by the formula:

\[ v_{rms} = k \sqrt{T} \]

where \( k \) is a constant that depends on the gas and \( T \) is the absolute temperature in Kelvin.

### Step-by-Step Solution:

1. **Identify the initial and final temperatures:**
- Initial temperature \( T_1 = 120 \, K \)
- Final temperature \( T_2 = 480 \, K \)

2. **Write the expression for the RMS velocity at the initial temperature:**
- At \( T_1 = 120 \, K \), the RMS velocity is given as \( v_{rms1} = v \).

3. **Write the expression for the RMS velocity at the final temperature:**
- At \( T_2 = 480 \, K \), the RMS velocity will be:
\[ v_{rms2} = k \sqrt{T_2} \]

4. **Relate the two RMS velocities using the temperature:**
- Since \( v_{rms} \) is proportional to the square root of the temperature, we can express the relationship as:
\[ \frac{v_{rms2}}{v_{rms1}} = \sqrt{\frac{T_2}{T_1}} \]

5. **Substituting the known values:**
- Substitute \( v_{rms1} = v \), \( T_1 = 120 \, K \), and \( T_2 = 480 \, K \):
\[ \frac{v_{rms2}}{v} = \sqrt{\frac{480}{120}} \]

6. **Calculate the ratio:**
- Simplifying the fraction:
\[ \frac{480}{120} = 4 \]
- Therefore:
\[ \sqrt{4} = 2 \]

7. **Finding the final RMS velocity:**
- Now, substituting back into the equation:
\[ v_{rms2} = v \cdot 2 = 2v \]

### Final Answer:
The root-mean-square velocity of the gas molecules at 480K becomes \( 2v \).