At constant pressure, calculate the root mean square velocity of a gas molecule at temperature `27^@ C` if its rms speed at `0^@ C`. Is 4 km/s.

Explanation

To solve the problem of calculating the root mean square (RMS) velocity of a gas molecule at 27°C given its RMS speed at 0°C is 4 km/s, we can follow these steps:

### Step-by-step Solution:

1. **Understand the Relationship**: The RMS speed of a gas is related to the temperature of the gas. At constant pressure, the RMS speed \( V_{rms} \) is proportional to the square root of the absolute temperature \( T \). This can be expressed as:
\[ V_{rms} \propto \sqrt{T} \]

2. **Set Up the Proportionality**: We can express the relationship between the RMS speeds at two different temperatures:
\[ \frac{V_{rms1}}{V_{rms2}} = \sqrt{\frac{T_1}{T_2}} \]
Where:
- \( V_{rms1} \) is the RMS speed at \( T_1 \) (0°C = 273 K)
- \( V_{rms2} \) is the RMS speed at \( T_2 \) (27°C = 300 K)

3. **Insert Known Values**: We know:
- \( V_{rms1} = 4 \, \text{km/s} \)
- \( T_1 = 273 \, \text{K} \)
- \( T_2 = 300 \, \text{K} \)
Thus, we can write:
\[ \frac{4 \, \text{km/s}}{V_{rms2}} = \sqrt{\frac{273}{300}} \]

4. **Calculate the Square Root**: Calculate the square root of the temperature ratio:
\[ \sqrt{\frac{273}{300}} = \sqrt{0.91} \approx 0.9539 \]

5. **Rearranging the Equation**: Rearranging the equation to solve for \( V_{rms2} \):
\[ V_{rms2} = 4 \, \text{km/s} \cdot \frac{1}{\sqrt{0.91}} \approx 4 \, \text{km/s} \cdot 1.0541 \]

6. **Final Calculation**: Now calculate \( V_{rms2} \):
\[ V_{rms2} \approx 4 \, \text{km/s} \cdot 1.0541 \approx 4.2164 \, \text{km/s} \]

7. **Rounding Off**: Rounding off the final answer, we get:
\[ V_{rms2} \approx 4.22 \, \text{km/s} \]

### Final Answer:
The root mean square velocity of the gas molecule at 27°C is approximately **4.22 km/s**.