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### Course: Physical Chemistry (Essentials) - Class 11>Unit 6

Lesson 12: Spontaneity

# 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.

## Want to join the conversation?

• aren't there an infinite number of states in a volume since there are an infinite places a particle can be in the volume?
• In classical mechanics, the position and momentum variables of a particle CAN vary continuously, so the number of possible microstates is actually infinite. Integrals are therefore used. For more, see the following wikipedia article on the partition function: http://en.wikipedia.org/wiki/Partition_function_%28statistical_mechanics%29
• 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.
• 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?
• Yes. He acquired the two because his volume was doubled. You see something similar in the integral of 1/V from V1 to 2V1. If his original assumption had been 3 times the volume we would've integrated to 3V1 instead. But ultimately the 2 is just the ratio of volumes.
• When Sal blew away the partition wall at , he went on to say that the temperature of the system stayed the same. This makes sense seeing as the particles didn't have to knock against the container to expand it, that is to say they didn't use any of their kinetic energy to increase the volume as in previous examples. I'm happy with that, but when we look at the equation he finishes with: S=Q/T , if T has remained the same but S has changed (there are more possible states so S increases, that's easy) , Q must change... But this isn't intuitive, why should the heat increase just because there's more room for the particles to bounce around in!?
• In the example at , there is no work or heat, so he solves for S by molecular positioning (no mention of Q).

In the "nearly identical but more realistic" example at , the system does work (and receives heat) so he solves for S by Q/T.

He then shows these are the same, and S is a state variable regardless of the path taken. As we saw in the previous video, solving for S can be done using the most convenient manner possible.
• @,
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?
• But the states are changing as the molecules have velocity. If they wouldn't have had a velocity then your proposition would have been correct
• 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
• How can we assume that those both constants are the same(i.e Boltzmann constant)?? Is there some proof or is it just a pure assumption?
• I have the same question; he stated the first equation probably has a constant, but he never gets into a mathematical derivation of that that constant should be - so saying it must be Boltzmann's constant is a bit of a stretch.
(1 vote)
• Shortly after it's said that "we get to define" our formula for S and so choose to state it as a log because the number of states will be "really large". However, if the log were not included then the statistical (Boltzmann) formula for entropy would not be equivalent to the classical formula ∆S = Q/T. So including the log is required. It's not actually choice that we make. Can you offer another explanation for why the log is included in the statistical (Boltzmann) formula? Does it have something to do with the fact that the increase in entropy slows as temperature increases? What are your thoughts?