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# Reconciling thermodynamic and state definitions of entropy

Long video explaining why entropy is a measure of the number of states a system can take on (mathy, but mind-blowing). Created by Sal Khan.

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• aren't there an infinite number of states in a volume since there are an infinite places a particle can be in the volume? • if entropy is a state variable then what is the entropy of the gas at a point where pressure is P volume V and temperature T and internal energy U ?? if we can't say from these how does it becomes a state variable as per def S=Q/T where Q is the heat required to reach there which can be done in many ways so from these it seems that S depends on the process it is done ??? • Good question. We only see delta S = Q / T, i.e., the INCREMENTAL entropy from adding heat at a certain temperature. At absolute zero, S is also 0 and the stuff is a crystal. When you now start heating, the crystal will start vibrating, then it will crack into pieces, individual molecules will come off, etc. The heat capacity will change over these different phases. When you stop heating, the stuff will go into an equilibrium state at a certain temperature. And it won't matter how it got there, i.e., if first molecule 1 broke off and then m 2. I would believe that S not only depends on P, V, T, and U but also on the type of gas we are talking about here, i.e., what type of crystals it forms at very low temparature. So to answer your question: the reason we cannot compute the absolute value of S from macrostate variables is not that it would depend on how we reached that state. It is that the calculation would be extremely complex and would require information (e.g., about crystal properties) that we might not have. Hopefully somebody else can comment on this as well.
• Is change in entropy in adiabatic process zero?? as Q is zero?? • At an equation is started that is going to list all possible combination's for the location and momentum of a system with N particles. Sal says that X is equal to all possible combination's of those two variables for each particle, and to figure out how many possible combination's for all particles you take X^N. Does that then imply that two particles can share the same location? Just curious not in need of explanation for a test or anything.
Thank you Sal!
-Tyler • YES: the equation under discussion tacitly assumes that particles CAN inhabit the same point in space. This seems like a bizarre assumption, but it's actually not that bad!

In these videos, Sal makes repeated use of the ideal gas law, PV=nRT or PV=NkT. Now an ideal gas is the simplest kind of gas so it makes sense to start here; but an important approximation for the ideal gas is that the particles of the gas do not interact with one another, in which case there is no reason for two particles to avoid each other and they can sit in the very same spot.

A different way of looking at it is to say that the particles of an ideal gas occupy zero volume.
As I said, this is okay-ish because many gases we encounter are low density. If you added up the volumes of all the gazillions of air molecules in the room, the result would be tiny compared to the volume of the room itself. So each molecule can be said to take up approximately zero space.

If you want to go deeper, I suggest you take a look at the "van der Waals gas", which tries to account for finite-volume effects (as well as long-range interactions between molecules).
• Did not the particles in the first box expend heat expanding into the doubled volume? It appears similar to an abstract piston. I may be wrong in assuming that the expansion happens because particles push each other around. • The expansion is in a vacuum. The gas pushes against nothing, so no energy is spent by the gas during the expansion. The expansion happens not because the particles push against each other but because there is no wall that prevents a particle from moving into the available empty volume.
• At about Sal introduced the natural logarithm to describe the entropy. I didn't really get why a logarithm has to used. Can someone give me an explanation why the natural logarithm has to be used in this exact example? • In the beginning of the video, he introduces us to the idea of the state of the system, S. As you saw, the number of possible states for the total number of particles can be an exceedingly large number. In such cases, we often resort to the use of a logarithm to reduce the scale of the problem. Specifically, while the natural logarithm of 5 is about 1.6, the natural logarithm of 50 million is only about 18. In this way, large numbers (such as the number of molecules in a volume of gas) can more easily be handled.
The other motivation that we see in in employing the laws of logarithm. These include clearing variable exponents, etc.
Hope that helps.
• Sal's used "ln 2" throughout, but can that be generalized to "ln (v2/v1)", as in natural log of the ratio of the ending volume to the beginning volume? • @,
shouldn't it be x(x-1)(x-2)(x-3).....
because a already ocupied 1 state so b has only x-1 states to go to? • Watching this video, I realized I do not know how a vacuum works! I get the "big idea" in that the natural tendency of disparate pressures when exposed to each other is to equalize, but how does this work on the atomic level?

Sal implies that once the wall in this video gets eliminated, molecules that were to bounce off of it continue on filling the void, but other molecules not currently headed towards the wall don't "know" about the new volume. It doesn't seem right that it just happens that molecules bounce around enough to eventually equal out. If you poke a hole only large enough for a few molecules to pass through, the chance that they would happen to bounce through is slim, but intuitively there would be a constant stream. Are the molecules somehow dragging along others through? • No, they do not drag each other. What happens is that at the start, all molecules are in one side of the hole. If P is the probability of a molecule to go through the hole, the number of molecules that pass to the other side is n (number of molecules) * P. When some of them did already pass to the other side, n decreases, so there will be less molecules going through. Furthermore, there is also a chance of the molecules that already passed to come back. This eventually balances out when the pressure is the same on both sides.

This looks kinda weird at first, because intuitively (at least for me) it looks like that vacuum is sucking everything. It's NOT! Don't forget that the Earth is in the space and it's not exploding :D 