Question

The plot that depicts the behavior of the mean free time (time between two successive collisions) for the molecules of an ideal gas, as a function of temperature ($T$), qualitatively, is: (Graph are schematic and not drawn to scale)

• A.
• B.
• C.
• D.

Answer 1

The correct option is A

The explanation for Correct Option:

Step 1: Mean Free Path

We know that the mean free path is given as,

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

where $D$ is the particle diameter and $n$ is the particle number density.

Step 2: Average Speed

We know that average speed is given as,

$v=\sqrt{\frac{8RT}{{\mathrm{\pi M}}_{\mathrm{w}}}}$

where $R$ is Rydberg's constant, $T$ is the temperature and $M_w$ is the mass of each molecule of the gas.

Step 3: Mean Free Time

Therefore, the mean free time can be given as,

$$\begin{gathered} \tau =\frac{\lambda }{v} \\ =\frac{{\displaystyle \frac{1}{\sqrt{2}{\mathrm{\pi D}}^2\mathrm{n}}}}{\sqrt{{\displaystyle \frac{8RT}{{\mathrm{\pi M}}_{\mathrm{w}}}}}} \\ =\frac{\sqrt{{\mathrm{\pi M}}_{\mathrm{w}}}}{\sqrt{16R}{\mathrm{\pi D}}^2\mathrm{n}}\frac{1}{\sqrt{T}} \\ \Rightarrow \tau \propto \frac{1}{\sqrt{T}} \end{gathered}$$

Step 4: Explanation for Graph

1. From the above result, we can see that the mean free path increases linearly with $\frac{1}{\sqrt{T}}$.
2. Also, the temperature cannot be zero, otherwise, the term will become undefined.
3. Hence, the $\tau$ versus $\frac{1}{\sqrt{T}}$ graph must be linear and cannot pass through the origin.
   

Therefore, the correct answer is option (C).