- #1
JesseC
- 251
- 2
Just learned about it recently, and I'm curious to understand more. The 'mixing' form of the paradox goes something like this:
Imagine a box partitioned in the middle. The two partitions are the same volume. In each there is a gas, on one side labelled A, and on the other B. The are held at exactly the same temperature and pressure.
Now suppose we get rid of the partition and allow the two to mix. Thus each gas can now occupy twice its original volume. Mathematically we can calculate the expected change of entropy to be
[tex] \Delta S = (n_A+n_B) ln(2) [/tex]
or something a bit like that anyway, the details of the mathematics are not particularly important I don't think.
But then suppose we had the same gas in each partition, A = B. So mathematically the entropy change would become
[tex] \Delta S = 2n_A ln(2) [/tex]
However when thought about carefully, it is impossible to achieve an entropy change in this situation. Say each side were to expand into the volume of the other, and we replaced the partition, we'd have exactly the picture as we started with thus no entropy change can have occurred.
What bothers me is, if the maths is predicting something incorrectly, it suggests there is something going wrong perhaps with the details, or the definitions, or the theory. Getting around the problem by saying the particles are indistinguishable is fine, but it doesn't make the mathematical answer correct! Has this been resolved?
Imagine a box partitioned in the middle. The two partitions are the same volume. In each there is a gas, on one side labelled A, and on the other B. The are held at exactly the same temperature and pressure.
Now suppose we get rid of the partition and allow the two to mix. Thus each gas can now occupy twice its original volume. Mathematically we can calculate the expected change of entropy to be
[tex] \Delta S = (n_A+n_B) ln(2) [/tex]
or something a bit like that anyway, the details of the mathematics are not particularly important I don't think.
But then suppose we had the same gas in each partition, A = B. So mathematically the entropy change would become
[tex] \Delta S = 2n_A ln(2) [/tex]
However when thought about carefully, it is impossible to achieve an entropy change in this situation. Say each side were to expand into the volume of the other, and we replaced the partition, we'd have exactly the picture as we started with thus no entropy change can have occurred.
What bothers me is, if the maths is predicting something incorrectly, it suggests there is something going wrong perhaps with the details, or the definitions, or the theory. Getting around the problem by saying the particles are indistinguishable is fine, but it doesn't make the mathematical answer correct! Has this been resolved?