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 follow these steps:

### Step 1: Understand the relationship between average kinetic energy and escape velocity
The average kinetic energy of a gas molecule is given by the formula:
\[ KE_{avg} = \frac{3}{2} k T \]
where \( k \) is the Boltzmann constant and \( T \) is the temperature in Kelvin.

The kinetic energy required to escape from the Earth is given by the formula for kinetic energy:
\[ KE_{escape} = \frac{1}{2} m v^2 \]
where \( m \) is the mass of the molecule and \( v \) is the escape velocity.

### Step 2: Set the average kinetic energy equal to the escape kinetic energy
To find the temperature at which the average kinetic energy equals the escape kinetic energy, we can set the two equations equal to each other:
\[ \frac{3}{2} k T = \frac{1}{2} m v^2 \]

### Step 3: Solve for temperature \( T \)
Rearranging the equation to solve for \( T \):
\[ T = \frac{m v^2}{3 k} \]

### Step 4: Plug in the values
Given:
- Mass of oxygen molecule, \( m = 5.34 \times 10^{-26} \) kg
- Escape velocity, \( v = 11.0 \) km/s = \( 11.0 \times 10^3 \) m/s
- Boltzmann constant, \( k = 1.38 \times 10^{-23} \) J/molecule·K

Substituting these values into the equation:
\[ T = \frac{(5.34 \times 10^{-26} \, \text{kg}) \times (11.0 \times 10^3 \, \text{m/s})^2}{3 \times (1.38 \times 10^{-23} \, \text{J/molecule·K})} \]

### Step 5: Calculate \( v^2 \)
Calculating \( v^2 \):
\[ v^2 = (11.0 \times 10^3)^2 = 121 \times 10^6 \, \text{m}^2/\text{s}^2 = 1.21 \times 10^8 \, \text{m}^2/\text{s}^2 \]

### Step 6: Substitute \( v^2 \) back into the equation for \( T \)
Now substituting \( v^2 \) back into the equation:
\[ T = \frac{(5.34 \times 10^{-26}) \times (1.21 \times 10^8)}{3 \times (1.38 \times 10^{-23})} \]

### Step 7: Calculate the numerator and denominator
Calculating the numerator:
\[ 5.34 \times 10^{-26} \times 1.21 \times 10^8 = 6.46 \times 10^{-18} \]

Calculating the denominator:
\[ 3 \times 1.38 \times 10^{-23} = 4.14 \times 10^{-23} \]

### Step 8: Final calculation for \( T \)
Now, divide the numerator by the denominator:
\[ T = \frac{6.46 \times 10^{-18}}{4.14 \times 10^{-23}} \approx 1.56 \times 10^5 \, \text{K} \]

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

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