Answers · Diffusion

Diffusion — final check key

Check meaning, not exact wording. These are teaching points, not official Cambridge marks.

1. SK1 · Final check

Model answer. (a) Soon after, the gas is concentrated near the left end (35 at A) and has hardly reached C and D. Later, the concentration at A has fallen to 16, while at B, C and D it has risen (to 12, 6 and 4). The gas has spread from the left end along the tube, and the readings are evening out. (b) The scented gas particles move at random, colliding with air particles and changing direction each time. There are many more of them near the left end, so at every point more particles cross from the left than from the right. This net movement from higher to lower concentration lowers the concentration at A and raises it at D. No particle aims for D. (c) In the end all four sensors read the same, about 10: the same amount of gas, spread evenly along the sealed tube, so the reading is the spread-out average. It lies between the later readings at D (4) and A (16), and it is not zero. The particles are still moving, but as many cross each way, so the readings stop changing.

A full answer includes

(a) I said that soon after, the concentration is high near the left end and almost zero further along; later it is lower at A (35 down to 16) and higher at B, C and D, so the gas has spread along the tube and the readings are evening out.
(b) I said the scented gas particles move at random and collide with air particles; there are more of them near the left end, so more cross from that side than back, giving a net movement from higher to lower concentration.
(b) I did not give the particles an aim, such as “they move to where there is more space”.
(c) I predicted readings that are equal at every sensor, between the later D and A readings and not zero (about 10), and said the particles are still moving: as many cross each way, so the readings stop changing.

Also accept

(a) A falls; B, C and D rise; the gas spreads from the left end and evens out. (b) Random motion and collisions; more particles cross from the high-concentration end than back: a net movement from high to low concentration. (c) Equal at every sensor, about 10; the particles are still moving, but there is no net movement.
(c) “The same at all four, somewhere between 4 and 16, not zero” meets the point without the number 10.

Watch for

“The gas particles move towards D because there is more space there.” — No particle aims anywhere. They move at random; more cross towards D because there are more of them on the A side.
“In the end all the readings are 0, because the gas has spread out.” — The tube is sealed, so all the gas is still there, spread evenly. Every sensor reads the same value, not zero.

2. SK2 · Final check

Model answer. (a) Hydrogen chloride. In the same time it spread 30 cm along the tube, while Z spread only 20 cm. (b) Z diffuses more slowly, so its M_r is higher than that of hydrogen chloride: more than 36.5. (c) At the same temperature, the particles of every gas have the same average kinetic energy, so lighter particles (lower M_r) move faster on average. A gas with a lower M_r therefore spreads further in the same time: it has a higher rate of diffusion. (d) If one plug went in earlier, that gas would have longer to spread, and the ring would form further from its end whatever its M_r. Putting both in at the same moment gives both gases the same time, so the distances compare their rates fairly. Using solutions of the same concentration helps too, so that neither gas starts with more particles.

A full answer includes

(a) Hydrogen chloride: it spread 30 cm in the time Z spread only 20 cm.
(b) Z has a higher M_r than hydrogen chloride: more than 36.5.
(c) At the same temperature, particles with a lower M_r (lighter particles) move faster on average, so the gas spreads further in the same time: it has a higher rate of diffusion. (Saying that both gases have the same average kinetic energy is good, but not needed.)
(d) So that both gases spread for the same time; then the distances compare their rates of diffusion fairly. (Adding that both solutions should have the same concentration is also a good fair-test point.)

Also accept

(a) Hydrogen chloride: 30 cm against 20 cm in the same time. (b) The M_r of Z is greater than 36.5. (c) Lower M_r, faster particles on average at the same temperature, so faster diffusion. (d) The same time for both gases, so the distances compare their rates fairly.
(d) “Otherwise one gas gets a head start” is fine, and so is adding “the same concentration of each solution”.

Watch for

“Z diffused faster, because the ring is nearer the Z end.” — The ring forms where the gases meet. Nearer the Z end means Z covered less of the tube, so Z was slower.
“Z is lighter, because it travelled a shorter distance.” — A shorter distance in the same time means slower diffusion, and that means heavier particles: a higher M_r.