A gas molecule at the surface of earth happens to have the rms speed for the gas at `0^(@)C` . Suppose it went straight up without colliding with other molecules, how high it would rise? Mass of the molecule is `4.65 xx 10^(-26) kg`. Boltzmann's constant `= 1.38 xx 10^(-23) JK^(-1)`.

Explanation

To solve the problem of how high a gas molecule would rise if it were to ascend straight up without colliding with other molecules, we can follow these steps:

### Step-by-Step Solution:

1. **Identify Given Values**:
- Mass of the molecule, \( m = 4.65 \times 10^{-26} \, \text{kg} \)
- Boltzmann's constant, \( k = 1.38 \times 10^{-23} \, \text{J/K} \)
- Temperature, \( T = 0^\circ C = 273 \, \text{K} \)

2. **Calculate the RMS Speed**:
The root mean square (RMS) speed of the gas molecules is given by the formula:
\[ v_{\text{rms}} = \sqrt{\frac{3kT}{m}} \]
Substituting the known values:
\[ v_{\text{rms}} = \sqrt{\frac{3 \times (1.38 \times 10^{-23} \, \text{J/K}) \times (273 \, \text{K})}{4.65 \times 10^{-26} \, \text{kg}}} \]

3. **Perform the Calculation**:
First, calculate the numerator:
\[ 3 \times (1.38 \times 10^{-23}) \times (273) = 1.131 \times 10^{-20} \, \text{J kg} \]
Now, divide by the mass:
\[ \frac{1.131 \times 10^{-20}}{4.65 \times 10^{-26}} \approx 2.43 \times 10^{5} \, \text{m}^2/\text{s}^2 \]
Taking the square root gives:
\[ v_{\text{rms}} \approx \sqrt{2.43 \times 10^{5}} \approx 493.1 \, \text{m/s} \]

4. **Calculate the Maximum Height**:
When the molecule rises to a height \( h \), its kinetic energy is converted into potential energy. The relationship is given by:
\[ mgh = \frac{1}{2} mv_{\text{rms}}^2 \]
Canceling \( m \) from both sides:
\[ gh = \frac{1}{2} v_{\text{rms}}^2 \]
Rearranging for \( h \):
\[ h = \frac{v_{\text{rms}}^2}{2g} \]
Substituting \( g = 9.8 \, \text{m/s}^2 \):
\[ h = \frac{(493.1)^2}{2 \times 9.8} \]

5. **Perform the Calculation for Height**:
First, calculate \( (493.1)^2 \):
\[ (493.1)^2 \approx 243,100.61 \, \text{m}^2/\text{s}^2 \]
Now, calculate \( 2g \):
\[ 2g = 2 \times 9.8 = 19.6 \, \text{m/s}^2 \]
Now, calculate \( h \):
\[ h = \frac{243,100.61}{19.6} \approx 12,400.03 \, \text{m} \]

6. **Convert Height to Kilometers**:
\[ h \approx 12.4 \, \text{km} \]

### Final Answer:
The height to which the gas molecule would rise is approximately **12.4 kilometers**.