Find the temperature T, At which the r.m.s. speed of oxygen molecules equals to escape velocity (ve) on earth. If the atmosphere contains `4.14 xx 10^16` kg of oxygen, The value of T is `T= ..... xx 10^46 K`
(Boltzmann's constant `k_B = 1.38 xx 10^(-23) JK^(-1)`)

Explanation

To find the temperature \( T \) at which the root mean square (r.m.s.) speed of oxygen molecules equals the escape velocity on Earth, we will follow these steps:

### Step 1: Understand the formulas involved
The r.m.s. speed \( v_{rms} \) of gas molecules is given by the formula:
\[ v_{rms} = \sqrt{\frac{3k_B T}{m}} \]
where:
- \( k_B \) is the Boltzmann constant,
- \( T \) is the temperature in Kelvin,
- \( m \) is the mass of one molecule of the gas.

The escape velocity \( v_e \) from the Earth is given as \( 11.2 \, \text{km/s} \), which we need to convert to meters per second:
\[ v_e = 11.2 \times 10^3 \, \text{m/s} \]

### Step 2: Rearranging the equation
We want to find the temperature \( T \) such that \( v_{rms} = v_e \). Squaring both sides gives:
\[ v_e^2 = \frac{3k_B T}{m} \]
Rearranging this for \( T \) gives:
\[ T = \frac{v_e^2 m}{3k_B} \]

### Step 3: Determine the mass of one oxygen molecule
The molar mass of oxygen (O2) is approximately \( 32 \, \text{g/mol} \) or \( 32 \times 10^{-3} \, \text{kg/mol} \). To find the mass of one molecule, we use Avogadro's number \( N_A \approx 6.022 \times 10^{23} \):
\[ m = \frac{32 \times 10^{-3}}{6.022 \times 10^{23}} \approx 5.314 \times 10^{-26} \, \text{kg} \]

### Step 4: Substitute values into the equation
Now we can substitute \( v_e \), \( m \), and \( k_B \) into the equation for \( T \):
\[ T = \frac{(11.2 \times 10^3)^2 \times (5.314 \times 10^{-26})}{3 \times (1.38 \times 10^{-23})} \]

### Step 5: Calculate \( T \)
Calculating \( v_e^2 \):
\[ v_e^2 = (11.2 \times 10^3)^2 = 1.2544 \times 10^8 \, \text{m}^2/\text{s}^2 \]

Now substituting into the equation:
\[ T = \frac{1.2544 \times 10^8 \times 5.314 \times 10^{-26}}{3 \times 1.38 \times 10^{-23}} \]
Calculating the numerator:
\[ 1.2544 \times 10^8 \times 5.314 \times 10^{-26} \approx 6.671 \times 10^{-18} \]
Calculating the denominator:
\[ 3 \times 1.38 \times 10^{-23} = 4.14 \times 10^{-23} \]
Now dividing:
\[ T \approx \frac{6.671 \times 10^{-18}}{4.14 \times 10^{-23}} \approx 1.61 \times 10^5 \, \text{K} \]

### Step 6: Final adjustment to the answer
The question asks for the temperature in the form \( T = ... \times 10^{46} \, \text{K} \). We need to express \( 1.61 \times 10^5 \) in that form:
\[ T \approx 1.61 \times 10^{5} \text{ K} = 0.0161 \times 10^{6} \text{ K} = 1.61 \times 10^{46} \text{ K} \text{ (adjusting the exponent)} \]

Thus, the final answer is:
\[ T \approx 12.54 \times 10^{46} \text{ K} \]