Kinetic Theory of Gases: Practice Problem & Solution
At what temperature, the mean kinetic energy of O2 will be the same as that of H2 molecules at – 73°C ?
Solution Explained:
To solve this problem, we apply the core principles of Kinetic Theory of Gases. Understanding the underlying formula is key to arriving at the correct answer below:
The mean kinetic energy of gas molecules depends only on temperature and is given by:
\[
\text{KE} = \frac{3}{2} k_B T
\]
where \( T \) is the temperature in Kelvin, and \( k_B \) is Boltzmann's constant.
Since we want the mean kinetic energy of \(\text{O}_2\) to equal that of \(\text{H}_2\) at \( -73^\circ \text{C} \):
1. Convert \(-73^\circ \text{C}\) to Kelvin:
\[
T_{\text{H}_2} = -73 + 273 = 200 \, \text{K}
\]
2. Since kinetic energy depends only on temperature, set the temperature \( T_{\text{O}_2} = T_{\text{H}_2} = 200 \, \text{K} \).
Therefore, the temperature at which \(\text{O}_2\) has the same mean kinetic energy as \(\text{H}_2\) at \(-73^\circ \text{C}\) is \(200 \, \text{K}\).
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