Question

A droplet grows by condensation from a radius of 2 um to 20 um in 10 min at a temperature of 0°C and a pressure of 70 kPa. Estimate the ambient supersaturation, neglecting the solution and curvature terms in the growth equation. 7. 2. S= 0. 45%

Answer 1

the estimated ambient supersaturation is approximately 18.7%.

Step-by-step explanation:

ln(R2/R1) = (4/3) * S * L * t

Where:

R1 is the initial radius of the droplet (2 um)

R2 is the final radius of the droplet (20 um)

S is the ambient supersaturation (what we need to find)

L is the Kelvin factor, which is dependent on temperature and pressure

t is the time duration (10 min)

Since we are neglecting the solution and curvature terms, we can assume the Kelvin factor (L) to be 1.

Now let's calculate the ambient supersaturation (S) using the given data:

ln(20/2) = (4/3) * S * 1 * 10

ln(10) = (4/3) * 10S

ln(10) = 40S/3

S = (3 * ln(10)) / 40

S ≈ 0.187

To convert the supersaturation to a percentage, we multiply it by 100:

S ≈ 0.187 * 100

S ≈ 18.7%