The average translational kinetic energy of a molecule in an ideal gas can be calculated using the formula:
\[ \text{KE} = \frac{3}{2} k_B T \]
where:
- \( \text{KE} \) is the average kinetic energy,
- \( k_B \) is the Boltzmann constant (\( k_B \approx 1.38 \times 10^{-23} , \text{J/K} \)),
- \( T \) is the absolute temperature in Kelvin.
First, we need to convert the temperature from Celsius to Kelvin. The conversion is done using the formula:
\[ T(K) = T(°C) + 273.15 \]
For 29°C:
\[ T = 29 + 273.15 = 302.15 , \text{K} \]
Now, substituting the values into the kinetic energy formula:
\[ \text{KE} = \frac{3}{2} (1.38 \times 10^{-23} , \text{J/K}) (302.15 , \text{K}) \]
Calculating it step by step:
- Calculate the product of \( k_B \) and \( T \):
\[ 1.38 \times 10^{-23} , \text{J/K} \times 302.15 , \text{K} \approx 4.16 \times 10^{-21} , \text{J} \]
- Now, calculate the average kinetic energy:
\[ \text{KE} = \frac{3}{2} (4.16 \times 10^{-21} , \text{J}) \approx 6.24 \times 10^{-21} , \text{J} \]
Therefore, the average translational kinetic energy of a molecule of an ideal gas at a temperature of 29°C is approximately:
\[ \text{KE} \approx 6.24 \times 10^{-21} , \text{J} \]