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Homework 3
Homework 4
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Homework 6
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problem 1
VOLUME EXCHANGE 

Given:
A volume exchange problem with 6 cells, multiple occupancy, identical particles, and a movable partition.
Ml denotes the number of cells on the right, Mr denotes the number of cells on the right.
Find: The most probable value of Ml
Solution:
The easiest way is just to count the microstates, just as a general note, t he most probable value happens when the densities on each side of the partition are equal.

A

Given: The same thing.
Take a good look at the top right of your equation sheet at the chart with the formulas for bin statistics, the thing with the single/unlimited occupancy and the distinguishable/identical particles. You'll be using this chart a lot. M is the number of bins, N is the number of particles.
Find: The (dimensionless) entropy for the configuration with Ml = 2
Solution:
We were told that the particles are identical and the bins allow for multiple aka unlimited occupancy, so omega = (N+M1)!/ [(M1)!(N)!]. Calculate the omegas for the left and right side of the partition, and then plug and chug. sigma = dimensionless entropy.
sigma = ln(omega) = ln[(omegaLeft)*(omegaRight)]

B 
Given: Same
Find: The probability of finding the configuration with Ml= 2
Solution:
Depending on the number of cells you were given, there are (that number  1) places the partitions could be. So go down the line and calculate the number of microstates (also known as the variable omega) with the same formula as part B for each partition position. Then divide the number of microstates for the config Ml=2 by the total number of microstates.
For example, if you had 6 cells, there are 5 places the partition could be.
omega# = (microstates for the left side) * (microstates for the right side)
omega1 = 1 * [6!/(4!2!)] = 15
omega2 = [2!/(1!1!)] * [5!/(3!2!)] = 20 < this is when Ml=2
omega3 = [3!/(2!1!)] * [4!/(2!2!)] = 18
and so on
Then for the probability we do 20/(omega1 + omega2 + ... + omega6)

C 

problem 2
CARNOT CYCLE 

Given:
A Carnot engine and fridge.
The engine operates between a hot reservoir at Th and a cold reservoir at Tc for a time t. During this time, the engine absorbs Qh Joules of heat from the hot reservoir.
Find: The heat it expels to the cold reservoir, Qc
Solution:
epsilon = 1  (Qc/Qh), move some variables around a bit to get Qc = (1epsilon)*Qh. epsilon of a carnot engine is 1  Tc/Th.

A

Given: The same thing
Find: The Wby
Solution:
epsilon = Wby/Qh
Wby = epsilon(Qh)

B 
Given: Same
Find: How much heat flows from the nitrogen into the environment
Solution:
epsilon = 1  Tc/Th. Make sure the temps are in Kelvin. ALWAYS make sure.

C 
Given: A fridge, Wdone
Find: Qc
Solution:
Won = Wby
epsilon = 1  Tc/Th = Wby/Qh, solve for Qh. This is actually the answer to E.
Qc = Qh  Wby

D 
Given: Same
Find: Qh
Solution:
You solved for it already in D. 
E 
Given: Same
Find: K
Solution:
K = 1/epsilon  1 
F 

problem 3
PARTICLES IN A BOX 

Given:
x distinguishable particles in a box with two equal cells at equilibrium and that a microstate is specified by giving the position (left side or right side) of every particle.
Find: The probability of finding all the particles in the left side of the box
Solution:
1/M^N = 1/2^x

A

Given: Same
Find: entropy of the macrostate of part a
Solution: all the particles being on the left side of the box is actually one microstate. sigma = ln(1) = 0.

B 
Given: Same
Find: probability of finding y particles on the left side of the box
Solution: (x choose y)/2^x

C 

Given: Same
Find: what's the entropy of the configuration of part C
Solution: sigma = ln(x choose y)

D 

Given: Same
Find: probability of finding z particles on the left side of the box
Solution: (x choose z)/2^x

E 

Given: Same
Find: what's the entropy of the configuration of part E
Solution: sigma = ln(x choose z)

F 

Given: config A has a higher entropy than config B
Find: which config is more likely to occur
Solution: A

G 

problem 4
CARRIER DIFFUSION IN A SEMICONDUCTOR 

Given:
A semiconductor crystal at T1 K has one surface illuminated with light, so that free carriers (electrons and holes) are generated at the surface. The carriers diffuse away into the sample. You may assume for this problem that all the carriers have the same mass, that of a free electron, Me = 9.11e31 kg.
Find: avg thermal speed of the carrier
Solution:
(3/2)kT1 = .5mv^2, solve for v 
A

Given: same
Find: what is the mean free path, ?,
if the particle has a mean scattering time of tao seconds
Solution:
tao = l/v
l = (tao)v < this is your answer

B 
Given: Same
Find: diffusion constant
Solution:
D = vl/3
Thus, delta U = Q  Wby becomes deltaU = Q.
U = (alpha)nRT = (alpha)pV right? So deltaU = (alpha)(deltaP)(V). You're given a V, you know the alpha = 3/2 because it's a monatomic gas (particles move in the x, y, and z directions), and you have both p values. So solve for U already.

C 
Given: Same
Find: avg time for particle to diffuse x mm in the x direction
Solution:
x^2 = 2Dt^2
solve for t


