The average translational kinetic energy (KE) of a molecule of an ideal gas can be calculated using the formula:
\[ \text{KE} = \frac{3}{2} k T \]
where:
- \( k \) is the Boltzmann constant, approximately \( 1.38 \times 10^{-23} , \text{J/K} \)
- \( T \) is the absolute temperature in Kelvin.
To convert the temperature from degrees Celsius to Kelvin, you can use the following conversion:
\[ T(K) = T(°C) + 273.15 \]
For a temperature of 29 degrees Celsius:
\[ T = 29 + 273.15 = 302.15 , \text{K} \]
Now, we can plug this value into the formula for average translational kinetic energy:
\[ \text{KE} = \frac{3}{2} \times (1.38 \times 10^{-23} , \text{J/K}) \times (302.15 , \text{K}) \]
Calculating this:
\[ \text{KE} = \frac{3}{2} \times 1.38 \times 10^{-23} \times 302.15 \]
\[ \text{KE} \approx \frac{3}{2} \times 4.169 \times 10^{-21} , \text{J} \]
\[ \text{KE} \approx 6.253 \times 10^{-21} , \text{J} \]
Thus, the average translational kinetic energy of a molecule of an ideal gas at a temperature of 29 degrees Celsius is approximately \( 6.25 \times 10^{-21} , \text{J} \).