![]() ![]() ![]() But if we have a dilute gas, there is a lot of space, and very few of our boxes will have more than one atom in it. ![]() We should really divide by the number of permutations of the atoms in each box. Now this isn't quite right, since if we put more than one atom into a box it doesn't matter in which order we put it in. We can do this for each molecule, so the total number of ways we can put the molecules into the bins is $M \times M \times M. We can put one molecule into the left side of the box into the box in M ways. Let's first calculate the entropy of the initial state. (Small compared to the size of the box, but large compared to the size of an atom, so we don't have to worry about atoms "filling up one of the small volume".) One way to do this is to imagine breaking up the left side of the box into M small volumes. When the partition separating the two halves of the box is removed and the system reaches equilibrium again, how does the new entropy of the gas compare to the energy of the original system? An ideal gas occupies half of the container and the other half is empty. Presenting a sample problemĬonsider an example of an isolated box of volume $2V$ divided into two equal compartments. We choose a model that is simple enough that we can see all the details. (This example will be of particular interest when we start thinking about free energy.) Now, let's look at a toy model example that lets us see how the microstate counting form of entropy, $S = k_B \ln W$, can also give us insights into what happens spontaneously. In our analysis of the entropy change in heat flow, we analyzed a toy model to see how the macroscopic form of entropy, $S = Q/T$, gave us insight into what would happen spontaneously. ![]()
0 Comments
Leave a Reply. |
AuthorWrite something about yourself. No need to be fancy, just an overview. ArchivesCategories |