Problem 4 Consider a classical gas with non-interacting molecules in the atmosph

Question

Problem 4 Consider a classical gas with non-interacting molecules in the atmosphere each with mass m in thermal equilibrium at absolute temperature T. The only external influence on the system is the Earth's uniform gravitational field. It is well known from elementary physics that the gravitational potential energy, Uh, of a particle of mass m at altitude h (near the Earth's surface) relative to some arbitrary reference level (level of zero gravitational energy) is Uh = mgh. Assume that h=0 is the reference level and that particle distribution in the atmosphere obey the Boltzmann distribution. (a) Determine the partition function of the gas. (8 points) (b) Show that the number of particles, n(0), at the zero level is related to the number of particles, n(h), at altitude h by n(h) = n(0)e^-mgh/(kT) where k is the Boltzmann constant. (8 points) (c) At 300 K, the gas composition at sea level in the Earth's atmosphere is primarily composed of 78% nitrogen and 22% oxygen, with concentrations of other trace gases neglected. What will be the nitrogen and oxygen atmospheric gas composition at a height 9 km above sea level at 300 K? Assume that the mass of molecular oxygen Is 32 amu and the mass of molecular nitrogen is 28 amu, where amu is the atomic mass unit given by 1 amu = 1.66053892 = 10^-27 kg. (12 points)

Explanation / Answer

non interacting gas which obeys Grand cannonical ensembles, by transfering energy and no of particles with respect to temperature T.

                          Pn ( En , nr)

a) probability of the finding particles

   Pn = C e-BEn

                 summation Pn = 1

                     C x summation e-BEn =1

                                  C = summation 1/ e-BEn

                                Pn =       e-BEn / sum e-BEn

                          Pn = e-BEn/ Z

                 partition function, Z = sum e-BEn

                                                               En = mgh

                                           Z = sum e-mgh/kT

b) average value of number of particles

                       n(h) = C e-mgh/kT

                at height h = 0

                        n(0) = C e0

                           C = n(0)

                    n(h) = n(0) e-mgh/kT

c) at earth's atmosphere, N2 =78% and O2 = 22%

                           n(0)N = 0.78                   n(0)O = 0.22

                 mN = 28 amu = 4.65x10-26 kg                mO = 32 amu = 5.31x10-26 kg

             n(h) = n(0) e-mgh/kT

                 T = 300 K,    h = 9 km = 9000 m

                 n(9 km)N = 0.78 x e- 4.65x10-26 x 9.8 x 9000 / 1.38x10-23 x 300

                                          = 0.78 x 0.371

                         n(9 km)N = 0.289

                                  n( 9 km)N     = 28.9 %

                    n(9 km)O = 0.22 x e -5.31x10-26 x 9.8x9000/ 1.38x10-23 x300

                                              = 0.22 x 0.323

                                    = 0.071

                             n( 9 km)O = 7.1 %

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