Calculate the average kinetic energy of hydrogen molecule at `0^(@)C`. Given `k_(B)=1.38xx10^(-23)JK^(-1)`.

Explanation

To calculate the average kinetic energy of a hydrogen molecule at \(0^\circ C\), we can follow these steps:

### Step 1: Understand the formula for average kinetic energy
The average kinetic energy \(K\) of a molecule of a gas with \(F\) degrees of freedom is given by the formula:
\[ K = \frac{F}{2} k_B T \]
where:
- \(k_B\) is the Boltzmann constant,
- \(T\) is the absolute temperature in Kelvin.

### Step 2: Identify the degrees of freedom for hydrogen
Hydrogen (\(H_2\)) is a diatomic molecule. For diatomic gases, the degrees of freedom are:
- 3 translational degrees of freedom,
- 2 rotational degrees of freedom.

Thus, the total degrees of freedom \(F\) for hydrogen is:
\[ F = 3 + 2 = 5 \]

### Step 3: Convert the temperature to Kelvin
The temperature at \(0^\circ C\) is:
\[ T = 0 + 273 = 273 \text{ K} \]

### Step 4: Substitute the values into the formula
Now we can substitute \(F\), \(k_B\), and \(T\) into the kinetic energy formula:
\[ K = \frac{5}{2} \times (1.38 \times 10^{-23} \text{ J/K}) \times (273 \text{ K}) \]

### Step 5: Perform the calculation
Calculating the right-hand side:
\[ K = \frac{5}{2} \times 1.38 \times 10^{-23} \times 273 \]
Calculating \(1.38 \times 273\):
\[ 1.38 \times 273 \approx 376.14 \]
Now, substituting back:
\[ K = \frac{5}{2} \times 376.14 \times 10^{-23} \]
Calculating \(\frac{5}{2} \times 376.14\):
\[ K \approx 5 \times 188.07 \approx 940.35 \times 10^{-23} \]
Thus:
\[ K \approx 9.4035 \times 10^{-21} \text{ J} \]

### Final Answer
The average kinetic energy of a hydrogen molecule at \(0^\circ C\) is approximately:
\[ K \approx 9.41 \times 10^{-21} \text{ J} \]

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