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:

(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:
(0)  =  kT ln(n(0)/nQ )
(h)  =  kT ln(n(h)/nQ ) + mgh
and you know
(h)  =  (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:
1 = 2.
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:
e = - h
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

 


Got comments? Corrections? Or even better, donations? You can try us at LexiconIndustries@gmail.com,
but just to warn you, we might not be home.