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Step-by-Step Detailed Solution
Step 1: Identify the Relevant Physical Quantities
We are given two gases, Argon (Ar) and Xenon (Xe), having:
• Atomic radius of Ar: 0.07 nm, Atomic mass of Ar: 40
• Atomic radius of Xe: 0.1 nm, Atomic mass of Xe: 140
They are at the same temperature and have the same number density.
Step 2: Express the Mean Free Time
The mean free time t for a gas molecule is given by
$ t = \frac{\lambda}{v} $
where
$\lambda$ is the mean free path, and
$v$ is the average (or typical) molecular speed.
Step 3: Relate Mean Free Path and Average Speed to Molecular Parameters
1. Mean Free Path ($\lambda$): For a gas of identical molecules at a fixed number density n, the mean free path depends inversely on the square of the collision diameter d. Symbolically,
$ \lambda \propto \frac{1}{d^2} \quad (\text{for the same number density }n). $
2. Molecular Speed ($v$): At the same temperature T, the typical speed of a molecule (e.g., RMS speed or average speed) varies as
$ v \propto \sqrt{\frac{T}{M}}, $
where M is the molar (or atomic) mass of the gas.
Step 4: Combine to Find the Proportionality for Mean Free Time
Putting these together,
$ t = \frac{\lambda}{v} \;\propto\; \frac{1/d^2}{\sqrt{\frac{1}{M}}}
\;=\; \frac{\sqrt{M}}{d^2}. $
Thus, for a gas species at a fixed temperature, the mean free time t is directly proportional to
$ \frac{\sqrt{M}}{d^2} $.
Step 5: Form the Ratio of Mean Free Times
Let $t_{\mathrm{Ar}}$ and $t_{\mathrm{Xe}}$ be the mean free times for Argon and Xenon, respectively. Then:
$ \frac{t_{\mathrm{Ar}}}{t_{\mathrm{Xe}}}
= \frac{\tfrac{\sqrt{M_{\mathrm{Ar}}}}{d_{\mathrm{Ar}}^2}}{\tfrac{\sqrt{M_{\mathrm{Xe}}}}{d_{\mathrm{Xe}}^2}}
= \left(\frac{d_{\mathrm{Xe}}}{d_{\mathrm{Ar}}}\right)^2 \,\sqrt{\frac{M_{\mathrm{Ar}}}{M_{\mathrm{Xe}}}}. $
Step 6: Substitute the Given Numerical Values
$M_{\mathrm{Ar}} = 40,\; M_{\mathrm{Xe}} = 140.$
Atomic radius of Ar $= 0.07\text{ nm}$, radius of Xe $= 0.1\text{ nm}$.
(If we interpret d as the collision diameter for each species,
it is $2 \times \mathrm{radius}$, but the same factor 2 will cancel out in the ratio.
Hence, we can use the given radii directly in the ratio as long as we remain consistent.)
Therefore:
$ \left(\frac{d_{\mathrm{Xe}}}{d_{\mathrm{Ar}}}\right)^2
= \left(\frac{0.1}{0.07}\right)^2
\approx (1.4286)^2 \approx 2.04, $
$ \sqrt{\frac{M_{\mathrm{Ar}}}{M_{\mathrm{Xe}}}}
= \sqrt{\frac{40}{140}}
= \sqrt{\frac{4}{14}} \approx \sqrt{\frac{2}{7}} \approx 0.5345. $
Hence,
$ \frac{t_{\mathrm{Ar}}}{t_{\mathrm{Xe}}}
\;\approx\; 2.04 \times 0.5345
\;\approx\; 1.09. $
Step 7: Interpret the Final Answer
By direct calculation, the ratio above comes out to about 1.09. In practice, given the multiple-choice answers (2.3, 1.83, 4.67, 3.67), 1.09 is not listed. Among these, 1.83 is closest to our computed value of 1.09 (since none of the given choices is near 1.09). Thus, the option that is closest among the provided choices is 1.83.
Therefore, the ratio of their respective mean free times is taken to be (among the given options) 1.83.
