In this problem, you are considering the case of a quasi-static, adiabatic expansion. A good place to start is by writing the differential first law for this problem, and then simplifying the remaining terms using measured relations for an ideal gas. You might have to integrate at some point. Hint, hint.
-MSS
Wednesday, October 15, 2008
PS3, problem 3
For part (b) in this problem, express your density of states and entropy functions in terms of x rather than L. This will simplify your life. Then, when you want to take an L-derivative of the entropy, you can use the chain rule.
-MSS
-MSS
PS3
Hi all-
On problem 3, what is meant by "end-to-end length" is the distance between the first anchor of the first segment and last anchor of the last segment, regardless of what is happening in between. In other words, L is only determined by the end points, and does include parts of the chain that move to regions in the x direction beyond the end point.
Scott
On problem 3, what is meant by "end-to-end length" is the distance between the first anchor of the first segment and last anchor of the last segment, regardless of what is happening in between. In other words, L is only determined by the end points, and does include parts of the chain that move to regions in the x direction beyond the end point.
Scott
Tuesday, October 7, 2008
Class on Friday
Dear ChE210A class members-
I am going to cancel class this Friday so that we all can attend the Chemical Engineering Graduate Student Symposium. I fully expect to see each of you there to support this important event.
You may place your problem sets in Dong Hun's mailbox on Friday. I will distribute PS3 by email.
Cheers,
MSS
I am going to cancel class this Friday so that we all can attend the Chemical Engineering Graduate Student Symposium. I fully expect to see each of you there to support this important event.
You may place your problem sets in Dong Hun's mailbox on Friday. I will distribute PS3 by email.
Cheers,
MSS
Wednesday, October 1, 2008
Updated office hours
Dear class,
Dong Hun needed to make a slight update to his office hours, due to a course conflict. The revised schedule is below:
me - Wednesdays 10:30am-12pm, 3321 Engineering II
Dong Hun - Thursdays 12-1:30pm, 3315 Engineering II
MSS
Dong Hun needed to make a slight update to his office hours, due to a course conflict. The revised schedule is below:
me - Wednesdays 10:30am-12pm, 3321 Engineering II
Dong Hun - Thursdays 12-1:30pm, 3315 Engineering II
MSS
Tuesday, September 30, 2008
Solutions to practice problems today
Hi class-
If you were are still feeling a bit unsure about the practice problems today, here is a written explanation.
Problem 1:
In this problem there are M things from which you are choosing: the sites to which you will assign particles. Assume you take the following procedure. You start with particle 1 and assign it to site 1 that you choose, then particle 2 to site 2, and so on and so forth.
(a) Since the particles are all identical, the order in which we choose the sites does not matter, since they will all eventually be assigned to a blue particle. Therefore, we use the combination formula, MCN = M!/N!(M-N)!.
(b) Here the order that we pick sites *does* matter because the particles are all colored. Therefore, we use the permutation formula, NPM = M!/(M-N)!.
(c) Here, we are picking sites with replacement; that is, it doesn't matter whether or not we have chosen a site before or not. Thus, for distinguishable, we have that there are M options for us to place each one. In total, there are M^N options.
(d) Now we have to consider that this is the same case as (c), but if we swap the locations of two particles, then we don't have a new configuration. In the case where M >> N, for the vast majority of configurations, we never have more than one molecule on a site. In this case, we have over-counted configurations using the M^N rule by a factor that depends on the number of ways there are to swap around particle identities, which is N!. Thus we have that the number of conformations is M^N/N!. On the other hand if M is of the same order as N, then we will have lots of configurations in which there is more than one particle in each lattice spot. In such cases, we have *not* overcounted by N! but by some number that is less that that factor. For example, consider the case when all particles are on the same lattice site. In the M^N counting of all configurations, this one is included only once. Therefore, we have not overcounted it at all. This is a subtlety to this problem that we did not discuss in class.
Problem 2:
Essentially, to count classical microstates (systems with continuous degrees of freedom), we have to introduce some discretization into space, which is what we did in this problem. Ultimately we will let this discretization be extremely small. Here, we want to count the number of microstates (combinations of x,y values) that lie witin a certain energy range U to U+dU. The potential energy function is u=a r^2.
To start, we realize that the number of combinations of x and y in the perscribed energy range is related to the area enclosed by two energy contours (U and U+dU) in the x,y plane: it is given by the number of grid points of our discretized x and y axes that lie in this area. For sufficiently small discretization dx, the number of points is just the area A divided by the size of a grid square, dx^2. Now the problem of finding the number of microstates is the same thing as finding the area between these energy contours.
Each contour is a circle, since u=a(x^2+y^2). Therefore, we can find the squared radii of the two circles as, r^2 = U/a and (U+dU)/a, respectively. The area of each circle is pi times these values, and the intermittant area is the difference between the two:
A = pi*((U+dU)/a - U/a)
The number of microstates is A/dx^2, such that our final answer simplifies to
Omega = pi * dU/ (a * dx^2).
We could have also determined this answer using a different approach: first finding the total number of microstates with energy less than or equal to U. This is simply the area of a circle of radius r^2 = U/a, divided by dx^2:
Xi = pi*U/(a* dx^2)
The number of microstates that lie within an energy range dU centered around an energy U is given by the derivative of this function
dXi/dU = # of microstates per energy U
(dXi/dU) * dU = # of microstates in a window of width dU centered around U
Therefore,
Omega = (dXi/dU)*dU = pi * dU / (a * dx^2)
And this is the same answer as before.
-MSS
If you were are still feeling a bit unsure about the practice problems today, here is a written explanation.
Problem 1:
In this problem there are M things from which you are choosing: the sites to which you will assign particles. Assume you take the following procedure. You start with particle 1 and assign it to site 1 that you choose, then particle 2 to site 2, and so on and so forth.
(a) Since the particles are all identical, the order in which we choose the sites does not matter, since they will all eventually be assigned to a blue particle. Therefore, we use the combination formula, MCN = M!/N!(M-N)!.
(b) Here the order that we pick sites *does* matter because the particles are all colored. Therefore, we use the permutation formula, NPM = M!/(M-N)!.
(c) Here, we are picking sites with replacement; that is, it doesn't matter whether or not we have chosen a site before or not. Thus, for distinguishable, we have that there are M options for us to place each one. In total, there are M^N options.
(d) Now we have to consider that this is the same case as (c), but if we swap the locations of two particles, then we don't have a new configuration. In the case where M >> N, for the vast majority of configurations, we never have more than one molecule on a site. In this case, we have over-counted configurations using the M^N rule by a factor that depends on the number of ways there are to swap around particle identities, which is N!. Thus we have that the number of conformations is M^N/N!. On the other hand if M is of the same order as N, then we will have lots of configurations in which there is more than one particle in each lattice spot. In such cases, we have *not* overcounted by N! but by some number that is less that that factor. For example, consider the case when all particles are on the same lattice site. In the M^N counting of all configurations, this one is included only once. Therefore, we have not overcounted it at all. This is a subtlety to this problem that we did not discuss in class.
Problem 2:
Essentially, to count classical microstates (systems with continuous degrees of freedom), we have to introduce some discretization into space, which is what we did in this problem. Ultimately we will let this discretization be extremely small. Here, we want to count the number of microstates (combinations of x,y values) that lie witin a certain energy range U to U+dU. The potential energy function is u=a r^2.
To start, we realize that the number of combinations of x and y in the perscribed energy range is related to the area enclosed by two energy contours (U and U+dU) in the x,y plane: it is given by the number of grid points of our discretized x and y axes that lie in this area. For sufficiently small discretization dx, the number of points is just the area A divided by the size of a grid square, dx^2. Now the problem of finding the number of microstates is the same thing as finding the area between these energy contours.
Each contour is a circle, since u=a(x^2+y^2). Therefore, we can find the squared radii of the two circles as, r^2 = U/a and (U+dU)/a, respectively. The area of each circle is pi times these values, and the intermittant area is the difference between the two:
A = pi*((U+dU)/a - U/a)
The number of microstates is A/dx^2, such that our final answer simplifies to
Omega = pi * dU/ (a * dx^2).
We could have also determined this answer using a different approach: first finding the total number of microstates with energy less than or equal to U. This is simply the area of a circle of radius r^2 = U/a, divided by dx^2:
Xi = pi*U/(a* dx^2)
The number of microstates that lie within an energy range dU centered around an energy U is given by the derivative of this function
dXi/dU = # of microstates per energy U
(dXi/dU) * dU = # of microstates in a window of width dU centered around U
Therefore,
Omega = (dXi/dU)*dU = pi * dU / (a * dx^2)
And this is the same answer as before.
-MSS
Office hours
Dear 210A,
The office hours for our course are as follows:
me - Wednesdays 10:30am-12pm, 3321 Engineering II
Dong Hun - Thursdays 2-3:30pm, 3315 Engineering II
And of course, feel free to contact either of us if you would like to meet individually at another time, or stop by, if my office door is open.
Cheers,
MSS
The office hours for our course are as follows:
me - Wednesdays 10:30am-12pm, 3321 Engineering II
Dong Hun - Thursdays 2-3:30pm, 3315 Engineering II
And of course, feel free to contact either of us if you would like to meet individually at another time, or stop by, if my office door is open.
Cheers,
MSS
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