At what temperature will the average velocity of oxygen molecules be sufficient to escape from the earth. Given mass of oxygen molecule `= 5.34 xx 10^(-26) kg`. Boltzmann constant, `k = 1.38 xx 10^(-23) J "molecule"^(-1) K^(-1)`. Escape velocity of earth `= 11.0 km s^(-1)`.

Explanation

To find the temperature at which the average velocity of oxygen molecules is sufficient to escape from the Earth, we can use the relationship between kinetic energy and escape velocity.

### Step-by-Step Solution:

1. **Understand the relationship between kinetic energy and escape velocity**:
The kinetic energy of a molecule can be expressed as:
\[ KE = \frac{1}{2} m v^2 \]
where \( m \) is the mass of the molecule and \( v \) is its velocity.

The average kinetic energy of a gas molecule at temperature \( T \) is given by:
\[ KE = \frac{3}{2} k T \]
where \( k \) is the Boltzmann constant.

2. **Set the kinetic energy equal to the escape energy**:
For a molecule to escape the Earth's gravitational pull, its kinetic energy must be equal to or greater than the energy required to reach escape velocity \( v_E \):
\[ \frac{3}{2} k T = \frac{1}{2} m v_E^2 \]

3. **Rearrange the equation to solve for temperature \( T \)**:
\[ T = \frac{m v_E^2}{3 k} \]

4. **Substitute the known values**:
- Mass of oxygen molecule, \( m = 5.34 \times 10^{-26} \, \text{kg} \)
- Escape velocity, \( v_E = 11.0 \, \text{km/s} = 11.0 \times 10^3 \, \text{m/s} \)
- Boltzmann constant, \( k = 1.38 \times 10^{-23} \, \text{J molecule}^{-1} \, \text{K}^{-1} \)

Substitute these values into the equation:
\[ T = \frac{(5.34 \times 10^{-26}) (11.0 \times 10^3)^2}{3 (1.38 \times 10^{-23})} \]

5. **Calculate \( v_E^2 \)**:
\[ v_E^2 = (11.0 \times 10^3)^2 = 121 \times 10^6 = 1.21 \times 10^8 \, \text{m}^2/\text{s}^2 \]

6. **Substitute \( v_E^2 \) back into the equation**:
\[ T = \frac{(5.34 \times 10^{-26}) (1.21 \times 10^8)}{3 (1.38 \times 10^{-23})} \]

7. **Calculate the numerator**:
\[ 5.34 \times 10^{-26} \times 1.21 \times 10^8 = 6.46 \times 10^{-18} \]

8. **Calculate the denominator**:
\[ 3 \times 1.38 \times 10^{-23} = 4.14 \times 10^{-23} \]

9. **Now calculate \( T \)**:
\[ T = \frac{6.46 \times 10^{-18}}{4.14 \times 10^{-23}} \approx 1.56 \times 10^5 \, \text{K} \]

### Final Answer:
The temperature at which the average velocity of oxygen molecules will be sufficient to escape from the Earth is approximately:
\[ T \approx 1.56 \times 10^5 \, \text{K} \]