States of Matter and Diffusion: Question 5

Worked solution

Step 1: Recall what diffusion depends on

Diffusion is the net movement of particles, by their own random motion, from a region of higher concentration to a region of lower concentration. At the same temperature, all gas particles have the same average kinetic energy, since kinetic energy depends on temperature.

Kinetic energy is related to mass and speed by: $E_k = \tfrac{1}{2}mv^2$

For two gases at the same temperature (the same $E_k$), a particle with a smaller mass, $m$, must have a greater speed, $v$, to have that same amount of kinetic energy. So, at a given temperature, particles with a lower relative molecular mass move faster on average, and gases with a lower $M_r$ diffuse more quickly.

Step 2: Compare compounds J and K

• Compound J: $M_r = 34$ (lower)
• Compound K: $M_r = 136$ (higher, exactly $4\times$ that of J)

Since compound J’s particles have the lower relative molecular mass, they travel faster, on average, than compound K’s particles at the same room temperature. Faster particles cover the same distance (from the refill to the sensor at the midpoint) in less time.

Compound J’s vapour therefore reaches the sensor first.

Why the other options are wrong

• B: this reverses the relationship. It is the lighter particles (lower $M_r$), not the heavier ones, that move faster and diffuse more quickly.
• C: it is true that both gases have the same average kinetic energy at the same temperature, but this does not mean they have the same speed. A lighter particle needs a higher speed to match a heavier particle’s kinetic energy, so their speeds (and diffusion rates) are different.
• D: diffusion is a physical process (particles simply spreading out through random motion); the two scent vapours mixing in the corridor does not amount to a chemical reaction that would stop either vapour from reaching the sensor.

Final answer

• Compound J (the lower $M_r$ vapour) reaches the sensor first, option A.