Two gases - argon (atomic radius $0.07\, nm$, atomic weight $40$) and xenon (atomic radius $0.1\, nm$, atomic weight $140$) have the same number density and are at the same temperature. The ratio of their respective mean free times is closest to :

Question

At which temperature will the r.m.s. velocity of a hydrogen molecule be equal to that of an oxygen molecule at $ 47^\circ \text{C} $?

Answer 1

To find the ratio of the mean free time of argon to xenon, we need to first understand the concept of mean free time ($\tau$), which is related to the mean free path ($\lambda$) and the average velocity ($v$) of the gas molecules. The formula for mean free time is given by:

$\tau = \frac{\lambda}{v}$

The mean free path is dependent on the number density ($n$) of the gas and the effective diameter ($d$) of the molecule:

$\lambda = \frac{1}{\sqrt{2} \pi d^2 n}$

Since the gases are given to have the same number density and temperature, we need to focus on the relationship between atomic radius and atomic weight, which affects the average velocity. The average velocity ($v$) of a gas is given by:

$v = \sqrt{\frac{8kT}{\pi m}}$

Here, $k$ is Boltzmann's constant, $T$ is temperature, and $m$ is the mass of the gas molecule. For atomic masses, the mass ($m$) can be expressed in kilograms. The mean free time for each gas can be expressed in terms of atomic weight and radius as:

For Argon ($A$): $\tau_A = \frac{1}{\sqrt{2} \pi (2 \times 0.07 \times 10^{-9})^2 n} \times \sqrt{\frac{\pi m_A}{8kT}}$
For Xenon ($X$): $\tau_X = \frac{1}{\sqrt{2} \pi (2 \times 0.1 \times 10^{-9})^2 n} \times \sqrt{\frac{\pi m_X}{8kT}}$

To find the ratio of mean free times, $\frac{\tau_A}{\tau_X}$:

\[ \frac{\tau_A}{\tau_X} = \frac{\sqrt{\frac{\pi m_A}{8kT}}}{\sqrt{\frac{\pi m_X}{8kT}}} \times \frac{(0.1)^2}{(0.07)^2} \]

Simplifying, we have:

\[ \frac{\tau_A}{\tau_X} = \sqrt{\frac{m_A}{m_X}} \times \left(\frac{0.1}{0.07}\right)^2 \]

Given $m_A = 40$, $m_X = 140$, calculate:

$\sqrt{\frac{40}{140}} = \sqrt{\frac{2}{7}}$

$\left(\frac{0.1}{0.07}\right)^2 = \left(\frac{10}{7}\right)^2 = \frac{100}{49}$

Thus,

\[ \frac{\tau_A}{\tau_X} = \sqrt{\frac{2}{7}} \times \frac{100}{49} \approx 1.09 \]

Thus, the correct answer is 1.09.

Answer 2