What is the temperature at which rms velocity of a gas is half its value at `0^(@)c`, if the pressure is kept constant ?

Explanation

To solve the problem of finding the temperature at which the root mean square (RMS) velocity of a gas is half its value at 0°C while keeping the pressure constant, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the RMS Velocity Formula**:
The RMS velocity (\(V_{rms}\)) of a gas is given by the formula:
\[ V_{rms} = \sqrt{\frac{3RT}{M}} \]
where \(R\) is the gas constant, \(T\) is the absolute temperature in Kelvin, and \(M\) is the molar mass of the gas.

2. **Set Up the Equation**:
According to the problem, the RMS velocity at temperature \(T\) is half of the RMS velocity at 0°C. Therefore, we can write:
\[ V_{rms}(T) = \frac{1}{2} V_{rms}(0°C) \]

3. **Calculate \(V_{rms}(0°C)\)**:
The temperature at 0°C in Kelvin is:
\[ T_{0°C} = 0 + 273 = 273 \text{ K} \]
Thus, the RMS velocity at 0°C can be expressed as:
\[ V_{rms}(0°C) = \sqrt{\frac{3R \cdot 273}{M}} \]

4. **Substitute into the Equation**:
Now substituting \(V_{rms}(0°C)\) into the equation for \(V_{rms}(T)\):
\[ V_{rms}(T) = \sqrt{\frac{3RT}{M}} = \frac{1}{2} \sqrt{\frac{3R \cdot 273}{M}} \]

5. **Square Both Sides**:
To eliminate the square root, we square both sides:
\[ \frac{3RT}{M} = \frac{1}{4} \cdot \frac{3R \cdot 273}{M} \]

6. **Cancel Out Common Terms**:
Since \(R\) and \(M\) are common on both sides, we can cancel them:
\[ 3T = \frac{1}{4} \cdot 3 \cdot 273 \]

7. **Simplify the Equation**:
This simplifies to:
\[ T = \frac{1}{4} \cdot 273 \]

8. **Calculate the Temperature**:
Now, calculating \(T\):
\[ T = \frac{273}{4} = 68.25 \text{ K} \]

9. **Convert to Celsius**:
To convert Kelvin to Celsius:
\[ T_{°C} = T - 273 = 68.25 - 273 = -204.75 °C \]

### Final Answer:
The temperature at which the RMS velocity of the gas is half its value at 0°C is:
\[ T = -204.75 °C \]