Animated Solution for Chemistry - States of Matter: Comprehension Passage

A Race in a Tube

Imagine a classic chemistry experiment: a long glass tube, $24\text{ cm}$ in length, acting as a racetrack for two volatile liquids. On the left side, we have a cotton plug soaked in liquid X, a relatively light molecule with a molar mass of $10\text{ g/mol}$. On the right side, we have liquid Y, a heavier contender weighing in at $40\text{ g/mol}$.

As these liquids evaporate, their vapors begin a race towards the center of the tube. The moment they meet, they react to form a visible product. The question is: exactly where will this product first appear?

Graham's Law

To predict the meeting point, we first turn to Graham's Law of Diffusion. This fundamental law states that under identical conditions of temperature and pressure, the rate of diffusion of a gas is inversely proportional to the square root of its molar mass. Mathematically, this is expressed as:

Let's plug in our given values. The molar mass of Y ($M_Y$) is $40$, and the molar mass of X ($M_X$) is $10$.

This elegant result tells us that gas X, being lighter, travels exactly twice as fast as gas Y.

Because both gases start their journey at the exact same time and meet at the exact same instant, the distance they cover is directly proportional to their speeds. If gas X travels a distance $d$, then gas Y must travel the remaining distance, which is $24-d$. Setting up the ratio:

Solving this simple linear equation:

So, theoretically, the product should form $16\text{ cm}$ away from the left plug. This perfectly answers the first part of our comprehension.

Why Did X Slow Down?

But physics and chemistry are rarely that simple in the real world. The problem throws a curveball: experimentally, the product forms at a distance less than $16\text{ cm}$. This means gas X didn't travel as far as Graham's Law predicted. It was somehow delayed.

To understand why, we must look at the environment inside the tube. It isn't a vacuum; it is filled with an inert gas at $1\text{ atm}$ pressure. The diffusing molecules aren't flying in a straight line; they are constantly bumping into the inert gas molecules, undergoing a random walk.

Could the mean free path ($\lambda$) be the culprit? The mean free path is the average distance a molecule travels between collisions:

The problem explicitly states that X and Y have equal molecular diameters ($\sigma$). Since they are diffusing through the same inert gas, the number density ($N^{\ast }$) of the background gas is also identical for both. Therefore, ${\lambda }_X={\lambda }_Y$. The mean free path cannot explain the discrepancy.

Collision Frequency

If the distance between collisions is the same, what about the number of collisions per second? This is the collision frequency ($Z_{12}$), which depends on the relative velocity ($U_{rel}$) between the diffusing gas and the inert gas:

The relative velocity is governed by the temperature and the reduced mass ($\mu$) of the colliding pair:

Where the reduced mass is calculated as:

Let's compare the reduced masses for our two systems. Because gas X ($10\text{ g/mol}$) is much lighter than gas Y ($40\text{ g/mol}$), the reduced mass of the X-inert gas system is significantly smaller than that of the Y-inert gas system (${\mu }_X<{\mu }_Y$).

A smaller reduced mass leads to a higher relative velocity. Consequently, gas X experiences a much higher collision frequency with the inert gas molecules compared to gas Y ($Z_{X,inert}>Z_{Y,inert}$).

Imagine two runners navigating a crowded street. The faster runner (gas X) will bump into pedestrians (inert gas) much more frequently than the slower runner (gas Y). These relentless collisions act as a microscopic drag force, impeding the progress of the lighter gas more severely than the heavier one. As a result, gas X covers less ground than Graham's ideal law predicts, shifting the meeting point closer to its starting line.