States of Matter and Diffusion: Question 10

Worked solution

Setting up the comparison

Two coloured gases are released into identical gas jars at the same moment, at the same temperature: bromine ($\text{Br}_2$) into jar R, and nitrogen dioxide ($\text{NO}_2$) into jar S. Both gases spread upward through the still air by their own random particle motion, moving from the region of high concentration where they were released towards the region of lower concentration further up each jar. Timing how long each colour takes to reach the same fixed mark gives a direct comparison of how quickly the two gases diffuse.

Part (a): Defining diffusion

Diffusion is the net movement of particles from a region of higher concentration to a region of lower concentration, as a result of the random motion of the particles.

Part (b): Comparing the average particle speeds

Bromine has $M_r = 160$ and nitrogen dioxide has $M_r = 46$, so nitrogen dioxide’s particles have a smaller mass than bromine’s particles.

At the same temperature, particles of every gas have the same average kinetic energy. Kinetic energy is related to mass and speed by: $E_k = \tfrac{1}{2}mv^2$

For two gases with the same $E_k$ (same temperature), a particle with a smaller mass, $m$, must have a greater speed, $v$, to have that same amount of kinetic energy.

Since nitrogen dioxide’s particles have the lower $M_r$, they must therefore travel at the faster average speed, compared with the heavier bromine particles.

Part (c): Checking the data against the prediction

A gas with a faster average particle speed diffuses more quickly, so it should cover the $20\,\text{cm}$ distance to the marked line in less time.

This is exactly what the table shows: nitrogen dioxide (the lower-$M_r$, predicted-faster gas) reaches the mark in only $23\,\text{s}$, while bromine (the higher-$M_r$, predicted-slower gas) takes much longer, $48\,\text{s}$, to travel the same $20\,\text{cm}$. The shorter recorded time for nitrogen dioxide is therefore consistent with it having the faster average particle speed, as found in part (b).

Part (d): Predicting the result for chlorine

Chlorine, $\text{Cl}_2$, has $M_r = 71$. This is:

• greater than nitrogen dioxide’s $M_r$ of 46, so chlorine’s particles move more slowly, on average, than nitrogen dioxide’s. Chlorine should take longer than $23\,\text{s}$ to reach the mark;
• but less than bromine’s $M_r$ of 160, so chlorine’s particles still move faster, on average, than bromine’s. Chlorine should still take less than $48\,\text{s}$ to reach the mark.

Predicted time for chlorine: somewhere between $23\,\text{s}$ and $48\,\text{s}$, since its $M_r$ (71) lies between the $M_r$ values of nitrogen dioxide (46) and bromine (160).

Final answers

• (a) Diffusion is the net movement of particles from an area of higher concentration to an area of lower concentration, due to the particles’ random motion.
• (b) Nitrogen dioxide has the faster average particle speed: its lower $M_r$ means, at the same temperature (same average kinetic energy), its particles must move faster than the heavier bromine particles.
• (c) Nitrogen dioxide’s shorter recorded time ($23\,\text{s}$ vs $48\,\text{s}$ for bromine) matches the prediction that the lower-$M_r$ gas diffuses faster.
• (d) Chlorine’s time would be between $23\,\text{s}$ and $48\,\text{s}$, since its $M_r$ (71) is between those of nitrogen dioxide (46) and bromine (160).