The kinetic energy of translation of an oxygen molecule at a particular temperature of `6.27 xx 10^(-31) J`. Calculate the temperature . Boltzman's constant `= 1.38 xx 10^(-23) J//K`.

Explanation

To calculate the temperature of an oxygen molecule given its kinetic energy, we can use the formula for the kinetic energy of a molecule in terms of temperature:

\[ E = \frac{3}{2} k_B T \]

Where:
- \(E\) is the kinetic energy,
- \(k_B\) is Boltzmann's constant,
- \(T\) is the temperature in Kelvin.

### Step-by-Step Solution:

1. **Identify the given values**:
- Kinetic energy \(E = 6.27 \times 10^{-31} \, \text{J}\)
- Boltzmann's constant \(k_B = 1.38 \times 10^{-23} \, \text{J/K}\)

2. **Rearrange the formula to solve for temperature \(T\)**:
\[ T = \frac{2E}{3k_B} \]

3. **Substitute the values into the equation**:
\[ T = \frac{2 \times (6.27 \times 10^{-31} \, \text{J})}{3 \times (1.38 \times 10^{-23} \, \text{J/K})} \]

4. **Calculate the numerator**:
\[ 2 \times (6.27 \times 10^{-31}) = 1.254 \times 10^{-30} \, \text{J} \]

5. **Calculate the denominator**:
\[ 3 \times (1.38 \times 10^{-23}) = 4.14 \times 10^{-23} \, \text{J/K} \]

6. **Divide the numerator by the denominator to find \(T\)**:
\[ T = \frac{1.254 \times 10^{-30}}{4.14 \times 10^{-23}} \approx 303 \, \text{K} \]

### Final Answer:
The temperature \(T\) is approximately \(303 \, \text{K}\).

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