Questions/comments about thermodynamics/statistical mechanics unit

Tue, Nov 14, 4:01 a.m. - For last year test for number 5, When I tried to solve it, I got 1/2mv^2+1/2mv^2+1/2ksp^2+1/2ksp^2 and then got 2KBT. But in the solutions the answers was just KBT. Why is spring constant not included.

First, to whoever is posting this question at 4:01 a.m. on the night before the exam -- you really need to be getting some sleep. In almost every case, any studying the night before that results in you not getting sleep reduces your exam score because you will be exhausted and won't be able to think straight. If you are looking at this on Tuesday morning, stop studying at least one-half hour before the exam and rest (but don't sleep through the exam).

As for your question, the reason why we don't include the potential energy terms is that the problem is asking only for the average kinetic energy. If the problem asked for the average thermal energy per molecule, then you would be including the potential energy terms.

So, for instance, if a problem said, ``A 2 kg ball is moving with a speed of 5 m/s at a height 10 meters above the ground. Determine the kinetic energy of the ball.'' You would write $K = (1/2)mv^2 = (1/2)(2 kg)(5 m/s)^2 = 25 J$. You would not write ``$K = (1/2)mv^2 + mgh$ = ...''

Mon, Nov 13, 9:11 p.m. - Hello, I want to ask about example 7.2 in Chapter 7. What is the speed of sound if we had a monatomic gas and a diatomic gas in the air?

The calculation in Example 7.2 is for a diatomic gas, so you already have the answer there. If the gas is monatomic, do the same calculation but instead of $\gamma = 7/5$, use $\gamma = 5/3$. If your question is what happens if you have a gas that is a mixture of the two, the answer is beyond the scope of this class.

Sun, Nov 12, 11:01 p.m. - When do you use E(therm)=nC(delta)T vs E(therm)=(f/2)nRT? Would you only use the latter equation when dealing with gases?

$E_{therm} = nC\Delta T$ is always valid (as long as there is no phase change -- remember, if there is a phase change, the $\Delta E_{therm} = nL$ comes into play). The issue here is ``what is C?'' If the problem specifies a particular substance, e.g., lead or water, then you look up the value of $C$ in the table (and remember that specific heats are different for solid vs liquid states).

If the problem says an ideal gas or an ideal solid, then you use the equation for $E_{therm}$ that you can derive from the Equipartition Theorem. (And remember: you are responsible for being able to do that derivation -- we have done this several times in lecture and in the homework.) That's where the $(f/2)nRT$ stuff comes from. Remember: $f$ is simply the number of terms that you get when you write out $E_{molecule}$ and since Equipartition says that when you take an average, you replace each of those terms with $(1/2)k_BT$, when you multiply by the number of molecules, you get $E_{therm} = Nf(1/2)k_BT = (f/2)nRT$.

Sun, Nov 12, 9:48 p.m. - I just asked the question about the use of the equation E=nRT instead of E=nCT, and I understood why we used it. You don't need to answer the question, thank you very much.

I actually decided to answer it anyway, because it is a good question and someone else might wonder about the same thing.

Sun, Nov 12, 9:37 p.m. - Isn't the equation E=nCT, C=3R? Can you please explain why you used E=nRT instead?

C isn't 3R in this case because it is a 1-D ideal solid. The on-line example shows how this work using the Equipartition Theorem.

Sun, Nov 12, 3:01 p.m. - How are entropy and other calculations relevant to heat engines different for heat flow between two reservoirs, heat flow between a reservoir a non-reservoir, and heat flow between two non-reservoirs? Change in entropy for a reservoir is Q/T because T is constant. For non-reservoirs, temperature changes...

... which means that you need to use the version of the $\Delta S$ equation for situations with changing T, i.e., "For changing T (but no phase transition), $\Delta S = ncln(T_f/T_i)$.

So, the procedure is the same, regardless of whether the heat source and dump are reservoirs or not. $\Delta S_{total} = \Delta S_{hot} + \Delta S_{cold}$ (and maybe $+ \Delta S_{cookie oven}$ or whatever. Then substitute in for each of the $\Delta S$ terms with whatever formula is appropriate for that piece of the system.

Sun, Nov 12, 12:35 p.m. - Can a heat engine operate between a reservoir and a non-reservoir?

Sure. Several of the homework problems that you did involve non-reservoirs, e.g., any problem involving a "brick" as a heat source. The Sterling engines that I demonstrated in lecture (those engines that sat on top of a mug with hot water) use the hot water as a source of heat, but that is clearly not a reservoir.

Sun, Nov 12, 12:21 p.m. - "Notice that the bound on how much heat we have to dump becomes smaller for very smaller Th or small Tc. Evidently, the more extreme the difference in temperature between the reservoir, the more work we will be able to extract." (pg. 211). This does not intuitively make sense to me. Do you mind conceptually explaining this? Thank you!

Well, I can explain the converse. Clearly, you can get less and less work the smaller the temperature difference because in the limit where Th = Tc, you can't get any work at all. (That should be intuitively reasonable that you can't get any work out of an engine where the reservoirs have the same temperature.) So, if the work gets less when Th - Tc is small, then it makes sense that it gets larger when Th - Tc is large.

Sun, Nov 12, 10:06 a.m. - Do you mind explaining the paragraph "you may be wondering why doing work..." on pg. 205? It doesn't make sense why doing work doesn't affect the entropy. Squeezing an object increases its thermal energy. We established in lab that energy units can be compared to thermal energy. If we increase thermal energy, we should increase energy units, which increases the entropy. But, according to this paragraph, the energy units don't increase if thermal energy increases. I am thoroughly confused.

This is a subtle point, and I would encourage you not to get too bogged down on this. (And if you really want a full discussion, take some of our more advanced physics courses.)

But the short version is that you actually can increase the thermal energy without increasing the number of energy units if you increase the spacing between the energy levels. I.e., if the unsqueezed system has energy levels of, say,n= 1, 2 and 3 corresponding to 2, 4, and 6 eV and we have, say, 5 units of energy (where each "unit" is 2 eV), we would have 10 eV total. But if the energy levels are 4, 8 and 12 eV, then 5 units of energy would correspond to 20 eV total. More energy overall, but still the same number of ways of distributing them.

Again, this is a subtle point and I would encourage you not to get too bogged down on this point.

Sat, Nov 11, 7:12 p.m. - How does the dunking bird work (step by step)? I understand that putting your hand on the bottom increases the temperature, which vaporizes the volatile liquid. This higher pressure of the gas (vaporized liquid), then, pushes the liquid up. I do not understand what happens next. Also, I do not understand the vaporized cooling scenario. I've talked to a few other people who are also thoroughly confused.

And another dunking bird question! This must be "dunking bird day" on the PHYS 211 questions board.

The step-by-step explanation is in a post below (from 11:24 am on Saturday). The "what happens next" part of your question starts with the bullet point "Eventually, ..."

Sat, Nov 11, 1:16 p.m. - How does the "drinking bird" work? Why/ what makes it drop then rise at a different pace?

I just responded to a question about the dunking bird. See comments below.

As for the pace, do you mean, "why do different birds" dunk at a different rate? If so, the answer is all about the balance of the bird; i.e., how high does the fluid have to go to cause them to tip over. Also, the felt on the head might be slightly different, which might cause evaporation to happen at a different rate, affecting the cooling rate. And there might be some slight contamination of the air in one or another.

As for the explanation of how it works, see the post below.

Sat, Nov 11, 11:59 a.m. - For the third test, I was wondering whether the whole table of thermodynamic properties of solids would be included ( including the ideal solid data) or whether we needed that for our notecard. Similarly, constants like avagadro's # and Kb will they be given to us too.

We won't put the entire table on the test, but we will provide you with anything from those tables that you need (along with several that you don't need).

The ideal solid stuff isn't really ``data'' -- it is a formula (i.e., specific heat = 3R). I'd put that on your 3x5 card.

Constants like Avogadro's number and $k_B$ will be provided.

Sat, Nov 11, 11:24 a.m. - What would be a concise explanation as to how the dunking bird works?

Concise ... concise ... Let's see how this goes.

Sun, Nov 5, 11:41 a.m. - For SUPP CH 9 #16b, I thought that delta-S total should be conserved. Why, then, is it 0.6 J/K if no energy is being added to the system as a whole?

Ugh. Here is again the fallacy that ``conservation'' has anything to do with entropy. It doesn't. Something is conserved if its value doesn't change. That's not the case with entropy. Entropy can and often does increase during a process.

In 9-16, you are taking a high temperature object and putting it in contact with a low temperature object, in which case heat will flow from the hot to the cold object, resulting in a decrease in the entropy of the hot object, and even larger increase in the entropy of the cold object, and an overall increase in entropy of the entire system. An increase in entropy means that the system is going to a more likely state. THAT'S THE WHOLE POINT!!! Heat flows from hot to cold because it is overwhelmingly, mind-bogglingly more likely that it will do so, rather than having no energy flow or (even less likely) a flow from cold to hot.

Sun, Nov 5, 11:34 a.m. - For SUPP CH 9 #14, the relationship is given to us in 9.10. Where might we want to look to begin deriving that relationship? Does it have to do something with how thermal energy is conserved, perhaps?

Please don't answer 9-14 by saying, ``$S_{AB} = S_A + S_B$ because equation 9.10 says so.'' You can prove this by (a) recognizing that the total multiplicity $\Omega_{AB} = \Omega_A \Omega_B$, from Boltzmann's relation for entropy $S = k_Bln\Omega$, and from the mathematics of natural logs. Specifically, remember that $ln(ab) = ln(a)+ln(b)$.

Sun, Nov 5, 11:27 a.m. - For SUPP CH 9 #7c, should the explanation be something along the lines of the fact that 7b has more energy units than 7a and an incremental unit of energy added to 7b yields a lower increase in entropy than an incremental unit of energy added to 7a? Do we have to explain anything on a molecular level? Even if we don't, though, how could one explain this phenomenon on a molecular and probabilistic level (rather than just a computational one)?

Yes to your first question. As for your second question, any explanation should involve probabilities/multiplicities and that is what you are doing in parts a and b. So, no -- once you have completely explained something, you do not have to come up with another explanation.

As for your third question: you are explaining this on a probabilistic level. That's what the computation in (a) and (b) are doing. Remember: entropy is a measure of probability (actually, likelihood, and the natural log of that).

Sun, Nov 5, 11:13 a.m. - For SUPP CH 9 #5, would the approach to answering this question be that the temperature of both can't go up (one has to go up and one has to go down per dS/dE = 1/T)? How do we prove entropy is conserved, though?

Answer to your first question: no. Answer to your second question: why would you want to prove the entropy is conserved? There is no ``conservation of entropy'' principle and if you ever try to prove that there is one, you will be wasting your time.

In class last week, we said that for dice, there is only one microstate that gives a total of 2 for a roll of two dice, but there are 6 microstates for a total of 7. So, rolling a 7 is more likely than rolling a 2, which means that ``7'' has a higher entropy. Does that mean that you will never roll at 2? Why is it that you can end up with smaller entropy states when rolling 2 dice, but you never end up with lower entropy states in thermodynamic systems?

Hint number 1: if you were working with an Einstein solid that had only two particles in it, you would often see the system entropy decreasing. Why don't you ever see an entropy decrease for a solid with, say, a mole of atoms?

Hint number 2: you might want to use the words, ``really'', ``unbelievably'' and ``mind-bogglingly'' in your answer when discussing probabilities/likelihoods when discussing systems with large numbers of molecules in the Einstein solid.

Sun, Nov 5, 10:56 a.m. - For SUPP CH 9 #3c, why can we not simply use the multiplicity formula (subbing q=6 and N=3)? What may be a better approach to solving this problem instead (to get 10/216 instead of the answer that the multiplicity formula gives with the approach I outlined above: 28).

The multiplicity formula that we use $\Omega = \frac{(q+N-1)!}{q!(N-1)!}$ is only valid for the Einstein solid. Please make sure that you have that on your 3x5 card and that you only use this formula if the problem specifically states that you are dealing with an Einstein solid.

For this problem, by far the easiest way to do it is to simply write down any combination of the three dice that will give you a sum of 6. I'll give you the first one: 1 on the first die, 1 on the second and 4 on the third. Keep going.

Sun, Nov 5, 10:55 a.m. - For SUPP CH 7 #15e, why do you have to use force = (NkT)/L instead of density * volume * average force per molecule (derived in part 15d)?

You aren't using NkT/L -- I'm not sure where that came from. I think that you are overthinking part (e). Once you have figured out in part (d) how much force a single molecule is exerting on the wall, simply multiply by the number of molecules in the box that are hitting the wall. Note the second assumption at the top of the problem that says pretend that a third of the particles are moving in each direction.

And the good news is that you will know if you have done this correctly because when you take the force determined in part (e) and divide by the area of the wall (in part f), you should get a pressure of 1 atm (i.e., 101 kPa).

Sun, Nov 5, 10:53 a.m. - For SUPP CH 7 #12, is the answer that you are adding thermal energy by dragging the molecule? What I thought is that if you add enough thermal energy to a few atoms/molecules, the thermal energy is dissipated across all neighboring molecules.

This is something that you should be noticing when you actually do the simulation. (By the way, if your response to this question doesn't explicitly discuss what you did and what you saw in the simulation, then don't expect to get full credit -- you actually have to do the simulation.) Yes, the energy that you put into dragging a single molecule gets transferred to other molecules (not just the neighboring ones). Can you explain why? And why does that result in melting of the solid? For that, refer back to 7-1(c).

Sun, Nov 5, 10:52 a.m. - For SUPP CH 7 #1c, is the answer that if the thermal energy is high enough that the molecules are too far from their minimum of their potential well, the pairs of molecules are stretched out far enough (at least d/10 per the Lindemann criteria) that causes the interaction of molecules to be weaker (the model of the spring goes away)?

It isn't that the interaction between the molecules goes away -- that's what happens when the stuff turns into a gas. It's only in the gas state that the molecules are far enough so that the interaction is negligible. See Fig. 6.1 for more about that.

A solid doesn't melt when the interactions become negligible. Rather, a solid melts when the atoms in the solid can move around (due to thermal jiggling) so that they can start exchanging places, rather than basically jiggling in place. That's the key difference between a solid and a liquid -- in both of them, the atoms are still bound together, but in a liquid, the atoms are still able to move around.

And that is the key to understanding why comparing $x_{therm}$ to the equilibrium spacing $d$ for a solid is the key. I'll let you figure out the rest of the argument.

Sat, Nov 4, 3:59 p.m. - For Supp CH 7 #12, where can we find the molecular dynamics applet?

It can be found on the calendar page for Lecture 16 or 17. You can also get to it by clicking here:

Sun, Oct 29, 2:36 p.m. - I noticed that there are no answers to CH8 on the back of the supplementary reading. Would you please post it on the website?

Thanks for catching this. We have posted the answers for Ch 8 on the calendar page for the October 31 lecture.

Sun, Oct 29, 11:29 a.m. - I have a question from chapter 6 in the supplementary book. First, it says that the gas constant R=8.31 J/mol.k. and then it says that the molar specific heat C for an ideal solid is 3*R=24.9 J/mol.k. But then in Table 6.1 on the book, there is a different C for every material. How can C be different if it is always 3*R and R is a constant?

C = 3R for the ideal solid, which is a model that is an approximation for real solids. It works quite well, as you can see in the table -- the real, experimentally-measured values of C end up quite close to 24.9 J/mol*K, but they aren't exactly the same.

So, C is not always 3*R. The 3R result is an idealized approximation which works quite well, but isn't the exact value for any of the solids.