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

Options

A. `4v_(rms)`
B. `2v_(rms)`
C. `(2v_(rms))/(2)`
D. `(v_(rms))/(4)`

Explanation

To solve the problem, we need to find the root mean square (RMS) velocity of an ideal gas when the temperature is increased from 120 K to 480 K. We will use the relationship between the RMS velocity and temperature.

### Step-by-step Solution:

1. **Identify the Initial Conditions:**
- Initial temperature \( T_1 = 120 \, \text{K} \)
- Initial root mean square velocity \( V_1 = V \)

2. **Identify the Final Conditions:**
- Final temperature \( T_2 = 480 \, \text{K} \)
- Final root mean square velocity \( V_2 \) (which we need to find)

3. **Use the Formula for RMS Velocity:**
The root mean square velocity \( V_{\text{rms}} \) of an ideal gas is given by the formula:
\[ V_{\text{rms}} = \sqrt{\frac{3RT}{M}} \]
where \( R \) is the universal gas constant, \( T \) is the temperature, and \( M \) is the molar mass of the gas.

4. **Establish the Relationship Between RMS Velocity and Temperature:**
Since \( R \) and \( M \) are constants for a given gas, we can say that:
\[ V_{\text{rms}} \propto \sqrt{T} \]

5. **Set Up the Ratio of RMS Velocities:**
From the proportionality, we can write:
\[ \frac{V_1}{V_2} = \sqrt{\frac{T_1}{T_2}} \]

6. **Substitute the Known Values:**
Substitute \( T_1 = 120 \, \text{K} \) and \( T_2 = 480 \, \text{K} \):
\[ \frac{V}{V_2} = \sqrt{\frac{120}{480}} \]

7. **Simplify the Ratio:**
Simplifying \( \frac{120}{480} \) gives:
\[ \frac{120}{480} = \frac{1}{4} \]
Therefore,
\[ \frac{V}{V_2} = \sqrt{\frac{1}{4}} = \frac{1}{2} \]

8. **Cross Multiply to Find \( V_2 \):**
Cross multiplying gives:
\[ V = \frac{1}{2} V_2 \implies V_2 = 2V \]

9. **Conclusion:**
The root mean square velocity at 480 K is:
\[ V_2 = 2V \]

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