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 4km/s

Explanation

To calculate the root mean square (RMS) velocity of a gas molecule at a temperature of \(27^\circ C\) given that its RMS speed at \(0^\circ C\) is \(4 \, \text{km/s}\), we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Relationship**: The RMS speed of a gas molecule is related to temperature by the formula:
\[ V_{\text{rms}} \propto \sqrt{T} \]
where \(V_{\text{rms}}\) is the root mean square velocity and \(T\) is the absolute temperature in Kelvin.

2. **Convert Temperatures to Kelvin**:
- The temperature at \(0^\circ C\) is:
\[ T_1 = 0 + 273 = 273 \, \text{K} \]
- The temperature at \(27^\circ C\) is:
\[ T_2 = 27 + 273 = 300 \, \text{K} \]

3. **Set Up the Ratio**: Since the RMS speed is proportional to the square root of the temperature, we can write:
\[ \frac{V_{\text{rms},1}}{V_{\text{rms},2}} = \sqrt{\frac{T_1}{T_2}} \]
where \(V_{\text{rms},1} = 4 \, \text{km/s}\) (at \(0^\circ C\)) and \(V_{\text{rms},2}\) is what we want to find (at \(27^\circ C\)).

4. **Substitute Known Values**:
\[ \frac{4 \, \text{km/s}}{V_{\text{rms},2}} = \sqrt{\frac{273}{300}} \]

5. **Calculate the Square Root**:
\[ \sqrt{\frac{273}{300}} = \sqrt{\frac{91}{100}} = \frac{\sqrt{91}}{10} \]

6. **Rearranging the Equation**:
\[ V_{\text{rms},2} = 4 \, \text{km/s} \cdot \frac{10}{\sqrt{91}} \]

7. **Calculate \(V_{\text{rms},2}\)**:
- First, calculate \(\sqrt{91} \approx 9.54\).
- Then:
\[ V_{\text{rms},2} \approx 4 \cdot \frac{10}{9.54} \approx 4.19 \, \text{km/s} \]

### Final Answer:
The root mean square velocity of the gas molecule at \(27^\circ C\) is approximately \(4.19 \, \text{km/s}\).