1. Use the Kohler curve handout to estimate:

a) The radius of the droplet that will form on a sodium chloride particle of mass 10^-16 g when the air has a supersaturation of 0.1%.

radius of 0.11 mm.

b) the relative humidity of air adjacent to a droplet of radius 0.1 mm if there is 10^-15 g of salt dissolve in it.

about 87%

c) The supersaturation which has to be exceeded for cloud droplets to grow on cloud condensation nuclei consisting of 10^-15 g of salt.

about 0.13%

d) The radius at which a droplet becomes activated if it has 10^-15 g of salt in it.

0.6 mm

2 a) Check your answer to 1b with a calculation of relative humidity using the Kohler curve formula. Use A=3.25 x 10^-7 mK, B = (4.3 x 10^-6 m³ mol^-1) n_ion/M_s, where for NaCl n_ion = 2 and M_s = 58.4 g mol^-1, and temperature of T = 273 K.

NOTE - I derived the following equation in class:

e'\e_s = 1 + A/r - B/r³

where
A = 2s/nkT

B = imM_w/M_s(4/3pr³r)

The values for A and B above are for the equation as expanded below:

e_r\e_s = 1 + A/rT - B M_s/r³

Note that these values for A and B are based on terms that don't change with the nature of the solute or with the temperature, whereas mine were not really 'constants'.

Sorry for the confusion - the idea is that

B = 1.47 x 10^-7 m³ g^-1

m_s = 1x 10^-15 g

RH = 1 + 0.012 - 0.147 = 86.4%
 
 

b) What is the relative humidity above the droplet if the same mass CCN was instead made of ammonium sulfate (NH₄)₂SO₄ (M_s = 132 g mol^-1 and n_ion=3)?

B = 0.977 x 10^-7 m³ g^-1 m_s=1x 10^-15 g

RH = 1 + 0.012 - 0.098 = 91.4%