Meteo 437  Atmospheric Physics II  (Cloud physics and chemistry)

Problem set 2

 

 

Assigned:        26 January 2004 

Due:                2 February 2004

 

 

 

Use the constants in the Appendix in Bohren and Albrecht or in Rogers and Yau.  You can work with others, but be sure that you understand how to do the problems.

 

  1. A reasonable value for the surface area density of a tropospheric aerosol is 10-7 cm2 cm-3, where the units are cm2 of the total surface area of aerosol to cm3 volume of air ( a surface area density).  Assume that the temperature is 20o C and the relative humidity is 50%.  Find the total number of water vapor molecules that impact the surface of the aerosols per unit volume of air in units of molecules cm-3 s-1.

The flux of molecules is given by F = ¼ n∙v, where n is the number density of molecules and v is the mean velocity of those molecules.  Thus, the total number of molecules hitting the aerosols is F∙A, where A is the total aerosol surface area.

 

  1. With what energy per second per m2 do “air” molecules strike you skin when T = 295 K and p = 960 hPa?

The mass of air, Mair, is about 0.029 kg mole-1.  The energy of each molecule is thus about ½ mair∙v2.  Thus, we need only multiply the molecular flux times the energy per each molecule to get the energy per unit surface area.

To do this problem absolutely correctly, we should look at the mean velocity and mass of each component of air and add them together; we should also use the root mean squared velocity when thinking about the energy and not the mean velocity.  Ideally, we would even do the integral over the Boltzmann distribution.  However, the differences caused  by using the mass of air and the mean velocity of air are small.

 

 

 

This equals about 1.4x107 J.  A kilowatt-hr is equal to 3.6x106 J, so that molecules strike you with the energy of about 4 kW-hr all the time.  Why doesn’t this hurt?

 

 

  1. The H-O bond strength in water vapor is 5.12 eV = 460 kJ mole-1.  What wavelength must a photon have if it is to break the O-H bond?

 

Use the equation:

E = hc/l = (1.98x10-16/ (nm)) J photon-1,

where l is in units of nm.

 

Note that 1 Joule = 6.242x1018 eV, 1 cal = 4.184 J; 1 eV = 23.06 kcal mol-1 = 96.48 kJ mol-1 = 1.60x10-19 J molecule-1. 

To break a single bond requires the number of Joules per molecule, which is equal to:

 

Eb = (96.5x103 J/eV 5.12 eV)/mole x (1 mole/6.02x1023 molecules) = 8.2x10-19 J/molecule (one photon per molecule)

 

l = 1.98x10-16/8.21e-19 = 242 nm

 

In reality, because of other quantum mechanical considerations, the photodissociation of water vapor occurs with a reasonable cross section only for wavelengths below 200 nm.