Helium atom is two times heavier than a hydrogen molecule. At 289 K, the average kinetic energy of a helium atom is

Options

A. two times that of a hydrogen molecule
B. same as that of a hydrogen molecule
C. four times that of a hydrogen molecule
D. half that of a hydrogen molecule.

Explanation

To find the average kinetic energy of a helium atom at 289 K, we can use the formula for the average kinetic energy of a gas molecule:

1. **Understanding the Formula**: The average kinetic energy (KE) of a gas molecule is given by the formula:
\[ KE = \frac{3}{2} RT \]
where \( R \) is the universal gas constant and \( T \) is the temperature in Kelvin.

2. **Identifying Constants**: The value of the universal gas constant \( R \) is approximately \( 8.314 \, \text{J/(mol·K)} \).

3. **Substituting the Temperature**: We are given the temperature \( T = 289 \, \text{K} \).

4. **Calculating the Kinetic Energy**:
\[ KE = \frac{3}{2} \times 8.314 \, \text{J/(mol·K)} \times 289 \, \text{K} \]
\[ KE = \frac{3}{2} \times 8.314 \times 289 \]
\[ KE = \frac{3}{2} \times 2400.426 \]
\[ KE = 3600.639 \, \text{J/mol} \]

5. **Average Kinetic Energy per Atom**: Since we want the average kinetic energy per atom, we need to convert this energy from per mole to per atom. We can do this by dividing by Avogadro's number (\( N_A \approx 6.022 \times 10^{23} \)):
\[ KE_{\text{per atom}} = \frac{3600.639 \, \text{J/mol}}{6.022 \times 10^{23} \, \text{atoms/mol}} \]
\[ KE_{\text{per atom}} \approx 5.98 \times 10^{-21} \, \text{J} \]

6. **Conclusion**: The average kinetic energy of a helium atom at 289 K is approximately \( 5.98 \times 10^{-21} \, \text{J} \).