Calculate the K.E. per mole of oxygen at `27^(@)C`. Given `N = 6.02 xx 10^(23), k = 1.38 xx 10^(-23) JK^(-1)`.

Explanation

To calculate the kinetic energy per mole of oxygen at \(27^\circ C\), we will follow these steps:

### Step 1: Convert the temperature from Celsius to Kelvin
The temperature in Kelvin (T) can be calculated using the formula:
\[ T(K) = T(°C) + 273 \]
Given \(T = 27^\circ C\):
\[ T = 27 + 273 = 300 \, K \]

### Step 2: Identify the degrees of freedom for oxygen
For diatomic gases like oxygen, the degrees of freedom (f) are:
- 3 translational degrees of freedom
- 2 rotational degrees of freedom
Thus, the total degrees of freedom for oxygen is:
\[ f = 5 \]

### Step 3: Use the formula for kinetic energy
The average kinetic energy per molecule for a gas can be expressed as:
\[ KE = \frac{f}{2} k T \]
Where:
- \(k\) is the Boltzmann constant (\(1.38 \times 10^{-23} \, J/K\))
- \(T\) is the temperature in Kelvin

Substituting the values:
\[ KE = \frac{5}{2} \times (1.38 \times 10^{-23}) \times 300 \]

### Step 4: Calculate the kinetic energy per molecule
Calculating the above expression:
\[ KE = \frac{5}{2} \times 1.38 \times 10^{-23} \times 300 \]
\[ KE = \frac{5 \times 1.38 \times 300}{2} \times 10^{-23} \]
\[ KE = \frac{2070}{2} \times 10^{-23} = 1035 \times 10^{-23} \, J \]

### Step 5: Calculate the kinetic energy per mole
To find the kinetic energy per mole, we multiply the kinetic energy per molecule by Avogadro's number (\(N = 6.02 \times 10^{23}\)):
\[ KE_{per \, mole} = KE \times N \]
\[ KE_{per \, mole} = (1035 \times 10^{-23}) \times (6.02 \times 10^{23}) \]
\[ KE_{per \, mole} = 1035 \times 6.02 \, J \]
\[ KE_{per \, mole} = 6220.7 \, J \]

### Final Answer
The kinetic energy per mole of oxygen at \(27^\circ C\) is approximately:
\[ KE_{per \, mole} \approx 6220.7 \, J \]

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