Nucleation of Water Droplets Questions and Answers

Answer: C

The Kohler curve is simply the product of these two competing influences. Which one dominates depends on the radius, and the crossover is what produces the maximum on the curve. Solute dominates at small radius and curvature governs the approach to the peak.

Answer: D

A droplet driven past the peak grows without limit at the ambient supersaturation and is said to be activated. Below the peak it settles at a stable equilibrium size and remains a haze droplet. The peak is therefore the dividing line between haze and cloud.

Answer: A

Correct answer: Both A and R are true, and R is the correct explanation of A

Both are true and the reason gives the mechanism. The Kohler curve is the sum of the Kelvin term, which raises the required humidity, and the solute term, which lowers it; the peak is where the balance turns. Beyond it the required supersaturation falls as the droplet grows, so a droplet in air held at that supersaturation finds itself always above equilibrium and grows away without limit. That is what activation means.

Answer: B

Correct answer: 1 and 2 only

Statements 1 and 2 describe haze droplets correctly: below the Kohler peak the equilibrium is stable, so a droplet that grows finds the required humidity has risen and stops, and it tracks the ambient humidity up and down. Statement 3 contradicts that stability - unactivated droplets remain small, which is exactly why haze is a haze and not a cloud.

Answer: C

More solute lowers the Kohler curve and moves its peak to a larger radius and a smaller supersaturation. This is why the largest and most soluble particles in a rising parcel are the first to activate into cloud droplets.

Answer: D

Activation follows the critical supersaturation, which falls as size and solubility rise. The insoluble particle has no solute effect at all and requires the full curvature correction to be overcome. In practice such particles activate only in the strongest updraughts.

Answer: A

Correct answer: Both A and R are true, and R is the correct explanation of A

Both are true and the reason gives the ordering. A larger nucleus dissolves more solute into its droplet, which depresses the equilibrium vapour pressure further and lowers the peak of its Kohler curve. Its critical supersaturation is therefore smaller, and as a rising parcel’s supersaturation climbs it passes that threshold first. The same applies to more soluble nuclei of equal size.

Answer: B

Correct answer: True

This is the feedback that governs droplet number. As nuclei activate they grow rapidly and consume vapour, and once that consumption matches the supply from cooling the supersaturation stops rising and begins to fall. Nuclei whose critical supersaturation was never reached remain as haze. The cloud therefore selects its own droplet population, and adding more nuclei does not increase the number indefinitely.

Answer: C

An insoluble coating reduces the solute available and hinders wetting, raising rather than lowering the requirement. The other three each strengthen the solute effect that opposes curvature. Solubility and size are the two properties that matter most in practice.

Answer: D

Correct answer: a few tenths of one per cent

Peak supersaturation in a natural cloud is typically 0.1 to 1 per cent, and it is held there by the feedback described above. The figure matters because it is far below the several hundred per cent homogeneous nucleation would require - which is the whole reason condensation nuclei are indispensable - and because it decides which part of the aerosol population becomes cloud droplets.