The absolute temperature of the gas is increased 3 times. What will be the increases in root mean square velocity of the gas molecules?

Options

A. `upsilonsqrt(2)`
B. `upsilon//sqrt(3)`
C. `sqrt(3) upsilon`
D. same

Explanation

To solve the problem, we need to understand the relationship between the root mean square (RMS) velocity of gas molecules and the absolute temperature of the gas.

### Step-by-step Solution:

1. **Understand the Relationship**:
The root mean square velocity (V_rms) of gas molecules is directly proportional to the square root of the absolute temperature (T). This relationship can be expressed as:
\[ V_{rms} \propto \sqrt{T} \]

2. **Initial Temperature**:
Let the initial absolute temperature of the gas be \( T \). Therefore, the initial RMS velocity can be expressed as:
\[ V_{rms, initial} = k \sqrt{T} \]
where \( k \) is a proportionality constant.

3. **Increased Temperature**:
According to the problem, the absolute temperature is increased three times. Thus, the new temperature \( T' \) is:
\[ T' = 3T \]

4. **Calculate New RMS Velocity**:
Now, we can calculate the new RMS velocity at the increased temperature:
\[ V_{rms, new} = k \sqrt{T'} = k \sqrt{3T} = k \sqrt{3} \sqrt{T} \]
This can be rewritten as:
\[ V_{rms, new} = \sqrt{3} V_{rms, initial} \]

5. **Determine Increase in RMS Velocity**:
The increase in RMS velocity can be calculated as:
\[ \Delta V_{rms} = V_{rms, new} - V_{rms, initial} = \sqrt{3} V_{rms, initial} - V_{rms, initial} \]
Simplifying this gives:
\[ \Delta V_{rms} = (\sqrt{3} - 1) V_{rms, initial} \]

6. **Conclusion**:
The increase in the root mean square velocity of the gas molecules when the absolute temperature is increased three times is given by:
\[ \Delta V_{rms} = (\sqrt{3} - 1) V_{rms, initial} \]

### Final Answer:
The increase in root mean square velocity of the gas molecules is \( (\sqrt{3} - 1) V_{rms, initial} \).

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