Question

Ratio between rms speed of Ar to the most probable speed of \(O^2\) at \(27\degree C\) is

• A. \(\frac{\sqrt8}{\Pi}\)
• B. \(\sqrt{\frac{8}{3}}\)
• C. \(\frac{\sqrt4}{\Pi}\)
• D. \(\frac{\sqrt4}{3}\)

Answer 1

The Correct Option is B

To solve this problem, we need to understand how to calculate the root mean square (rms) speed and the most probable speed of gases and then compare these for Argon (Ar) and Oxygen (\(O_2\)).

First, let's recall the formulae for these speeds:

• The root mean square speed (\(v_{rms}\)) of a gas is given by: 

\[v_{rms} = \sqrt{\frac{3kT}{m}}\]

• where \(k\) is the Boltzmann constant, \(T\) is the temperature in Kelvin, and \(m\) is the molecular mass of the gas.
• The most probable speed (\(v_{mp}\)) is given by: 

\[v_{mp} = \sqrt{\frac{2kT}{m}}\]

We need to find the ratio of the rms speed of Ar to the most probable speed of \(O_2\).

The molecular mass of Ar is approximately 40 u, and for \(O_2\), it is approximately 32 u.

Since both gases are at the same temperature (27°C = 300 K), we use these speeds:
The formula for the ratio becomes:

\[\text{Ratio} = \frac{v_{rms}(\text{Ar})}{v_{mp}(O_2)} = \frac{\sqrt{\frac{3kT}{M_{\text{Ar}}}}}{\sqrt{\frac{2kT}{M_{\text{O}_2}}}}\]

Plugging in the values and simplifying, we get:

\[\text{Ratio} = \frac{\sqrt{\frac{3}{40}}}{\sqrt{\frac{2}{32}}} = \sqrt{\frac{3 \cdot 32}{2 \cdot 40}} = \sqrt{\frac{96}{80}} = \sqrt{\frac{8}{3}}\]

Thus, the correct answer is \(\sqrt{\frac{8}{3}}\).

This proves that the option \(\sqrt{\frac{8}{3}}\) is indeed correct, based on the calculation of speeds using standard formulae of kinetic theory of gases.