Kinetic Theory of Gases: Practice Problem & Solution
At room temperature, the r.m.s. speed of the molecules of certain diatomic gas is found to be 1930 m/s. The gas is :
Short Solution (Best for NEET):
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 root mean square (r.m.s.) speed \( v_{\text{rms}} \) of gas molecules is given by the formula:
\[
v_{\text{rms}} = \sqrt{\frac{3 k_B T}{m}}
\]
where:
- \( k_B \) is the Boltzmann constant,
- \( T \) is the temperature (assumed to be room temperature here),
- \( m \) is the mass of one molecule of the gas.
Rearranging to find \( m \):
\[
m = \frac{3 k_B T}{v_{\text{rms}}^2}
\]
For a diatomic gas like \( \text{H}_2 \), using the given \( v_{\text{rms}} = 1930 \, \text{m/s} \), we calculate that this speed corresponds to the molecular mass of hydrogen (\( \text{H}_2 \)). Therefore, the gas is identified as \( \text{H}_2 \).
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