Kinetic Theory of Gases: Practice Problem & Solution
A mixture of 2 moles of helium gas (atomic mass = 4 amu) and 1 mole of argon gas (atomic mass = 40 amu) is kept at 300 K in a container. The ratio of the rms speeds \( \left( \frac{v_{rms}(helium)}{v_{rms}\left( argon \right)} \right) \) is:
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 root mean square (rms) speed \( v_{\text{rms}} \) of a gas is given by:
\[
v_{\text{rms}} = \sqrt{\frac{3RT}{M}}
\]
where \( M \) is the molar mass.
For helium (\( M = 4 \, \text{g/mol} \)) and argon (\( M = 40 \, \text{g/mol} \)) at the same temperature, the ratio of their rms speeds is:
\[
\frac{v_{\text{rms}(\text{He})}}{v_{\text{rms}(\text{Ar})}} = \sqrt{\frac{M_{\text{Ar}}}{M_{\text{He}}}} = \sqrt{\frac{40}{4}} = \sqrt{10} \approx 3.16
\]
Thus, the ratio \( \frac{v_{\text{rms}(\text{He})}}{v_{\text{rms}(\text{Ar})}} \) is approximately 3.16.
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