Physics 213
Homework 6
Atmospheric Pressure, Surfaces, Semiconductors, Phase Transition

 Home Homework 1 Homework 2 Homework 3 Homework 4 Homework 5 Homework 6 About Us problem 1 ATMOSPHERIC PRESSURE Given: Assume that the atmosphere has a temperature T that is independent of altitude. The pressure of the atmosphere is the sum of the partial pressures of the gases present (N2 , O2 , etc.). The partial pressure of each type of gas varies with altitude, h, according to the condition of constant chemical potential: µ(h)  =  kT ln(n(h)/nQ ) + mgh  =  constant where n(h) is the gas density (particles/volume) at altitude h, and nQ is the quantum density (which is independent of h). Find: At what altitude, h, does the partial pressure of nitrogen (molecular weight= 28) fall to 1/2 of its value at sea level? Give your answer in kilometers. Solution: µ(0)  =  kT ln(n(0)/nQ ) µ(h)  =  kT ln(n(h)/nQ ) + mgh and you know µ(h)  =  µ(h!= 0) in equilibrium, so set them equal and isolate mgh on one side kT ln(n(0)/nQ ) - kT ln(n(h)/nQ ) = mgh since ln(a) - ln(b) = ln(a/b), the nQs will cancel and that equation will become kT ln(n(0)/n(h)) = mgh. The question is asking for when the partial pressure falls to n(h) = .5n(0), so the equation can be further simplified to kT ln(.5 ) = mgh, where m is the grams/particle of nitrogen (aka molweight * 1/avagadro's#). Solve for h. A Given: Same Find: the same calculation for hydrogen (molecular weight = 2). Solution: Follow the same steps as in part A, don't forget to calculate the new m. B

 problem 2 ATOMS ON SURFACES Given: Consider a solid surface with two kinds of binding sites for a particular type of atom. There are M1 sites with binding energy delta1and M2 sites with binding energy delta2. A total of N atoms occupy these sites: N1 are on site 1 and N2 are on site 2. You will calculate the equilibrium ratio N1/N2 at temp T by minimizing the total free energy Find: The free energy if all of the atoms are on sites with delta1 (i.e., N 2 = 0). Solution: F = U - TS U = (N)(delta1) T is given S = k ln(omega) = k [ln(M^N) - ln(N!)] = k[(N)(lnM) - NlnN + N] using M1 Plug and chug. A Given: The same thing Find: The free energy if all 900 atoms were to occupy only the sites with delta2 (i.e., N 1 = 0). Solution: F = U - TS U = (N)(delta2) T is given S = k ln(omega) = k [ln(M^N) - ln(N!)] = k[(N)(lnM) - NlnN + N] using M2 Plug and chug. Again. B Given: Same Find: Calculate the relative number of atoms on sites 1 and 2 in equilibrium at Teq K Solution: µ1 = µ2. kT ln(N1/M1) - delta1 = kTeq ln(N2/M2 ) - delta2 kT ln[(N1M2)/(N2M1)] = delta1 - delta2 (N1/N2) = (M1/M2)e^[ (delta1-delta2)/kT] C Given: Same Find: The number (to the nearest integer) of atoms on sites 1 and 2. Solution: You know N1 + N2 = whatever N they gave you. From part C you have N1/N2 = some number. This is a system of equations. Solve for N1 and N2. D Given: Same Find: The free energy for the equilibrium state. Solution: Basically you have to find F1 and F2 like you did in parts A and B, then add them to get the total free energy F. F = U - Teq*k[(N)(lnM) - NlnN + N] For F1 make sure you use N1, M1, Teq and so on so forth for F2. E

 problem 3 ELECTRONS AND HOLES IN A SEMICONDUCTOR Given: nQ  = (2.6e+024)(T/300K)^(3/2) delta k = 8.617×10 -5  eV/K Find: The density of free electrons (number per cubic meter) in a pure crystal of GaAs at T1°C Solution: µe = - µh kT ln(ne /nQ ) + delta = -kT ln(nh /nQ ) where ne = nh (ne and nh are really n sub e and n sub h) ne/nQ = e^( -delta/ (2kT)) solve for the nQ for the T1 they give you sub it in and solve for ne A Given: Same Find: what factor is the density of free electrons increased when the temperature is raised to T2°C? Solution: nehot/nQ = e^( -delta/ (2kT)) solve for the nQ for the T2 they give you sub it in and solve for nehot nehot/ne_from_part_A = your answer B Given: Another ne. Find: the density of holes at T3°C Solution: It's time you experienced the joy of the Law of Mass Action, ne*nh = ni^ 2. You're given ne, and the temp they give you in this problem is probably the same you used in part B, so use nehot for ni. Solve for nh. C

 problem 4 WATER AND STEAM Given: At a temp T and pressure p, liquid water is in equilibrium with water vapor. The density of liquid water is 10^3 kg/m^3 . The molecular weight of water is 0.018 kg/mole. Consider a cylinder of water (liquid +vapor) held at temp T. There is m total mass of water (in all forms) in the cylinder. In this problem, T and p are held constant, but V may change. Assume that water vapor is an ideal gas, and use 2.256 × 10^6  J/kg for the latent heat of vaporization of water. Find: the volume when all of the water is liquid Solution: m/density A Given: same Find: As the volume is increased, at what volume does the last of the liquid water turn to vapor Solution: pV = nRT, make sure you're using the right units, mol not kg, etc B Given: Same Find: If heat is added (vaporizing the liquid) so that the water expands from the volume in part (a) to the volume in part (b), keeping T and p constant, slowly forcing a piston outwards, how much work is done on the piston? Solution: W = p(deltaV), where deltaV = partBanswer - partAanswer C Given: Same Find: How much heat flowed into the system in the process in part (c) Solution: Q = Lm where L is the latent heat of vap given to you and m is the total mass of water D Given: Same Find: How much the internal energy of the water changed in this process Solution: deltaU = W - Q = partDanswer - partCanswer E

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