Question

A haze droplet is activated to form a cloud droplet when the ambient saturation ratio exceeds the critical value S* and the droplet radius exceeds r*. At this point the original condensation nucelous may be assumed to be completely dissolved in the droplet. Prove that, for a given hygroscopic material and a fixed temperature, the concentration of the solution when r=r* is stronger for small condensation nuclei than for large ones. Evaluate the concentration, in units of mass of salt per unit mass of water, for sodium chloride nuclei ranging from 10^-19 to 10^-14g.

Answer 1

Prove show that given hygroscopic material and a fixed temperature ,the concentration of the solution when r=r*is stronger for small condensation nuclei than for large ones.

The activation of cloud condensation nuclei to form cloud drops can be a kinetically limited process under some circumsrancces .Here the sensitivity of particles activation to parameters that are related to kinetic processes ,in particular condensational growth and ambient supersaturation,is examined .Specifically the parameters studied are the thermal and mass accommodation coefficients the rate of solute dissolution for a particle and the updraft velocity .Result are based on the equatuion for condensation growth and simulations using adiabetic parcel models ,We conclude that activation is not very sensitive to the thermal accommmodation coefficient but is sensitive to the mass accommodation coeffiecient below some critical value of 0.1 to 0.001 .

Dissolution rates (defined for 100 nm dry particles ) sodium choloride concentration .

In the range of 10 min to 1 day can significantly affect activation .Sensitivity to updraft velocity uncertainities can be significant and are likely to be more important for polluted (larger one )rather than for small condensation nuclei clean condition(r=r*)

Calculate what mass of sodium would be needed to get to that concentration

10^-15 g = 5.10^-3 g =5mg

So ,you need the solution to contain 5 mg of sodium cations .Use sodium and sodium chloride molar mass to help you find sodium chlorides percent composition

22.989770 gmol/.58.44277 gmol *100

=39.34% Na

This means that very 100 g of sodium chloride will contain 39.34 g of sodium .In your case ,you will get 5 mg of sodium from

5 mg Na . 100 mg Nacl/39.34 mg Na

=12.7097 Nacl

The anser will be

mNacl = 12.7097 mg

So,to get a solution that is 50 ppm in sodium cations ,you need to dissolve 12.7097 mg of sodium chloride in 100 mL of water .