Probabilities of molecules being somewhere

[adamc637]

A box is separated by a partition into two parts of equal volume. The left side of the box contains 500 molecules of nitrogen gas; the right side contains 100 molecules of oxygen gas. The two gases are at the same temperature. The partition is punctured, and equilibrium is eventually attained. Assume that the volume of the box is large enough for each gas to undergo a free expansion and not change temperature.

What is the probability that the molecules will be found in the same distribution as they were before the partition was punctured, that is, 500 nitrogen molecules in the left half and 100 oxygen molecules in the right half?

I have no idea how to start this. Would i do

(500!*100!)/600!

?

 

[anathema]

Thank you for your question. I can help provide some insights into this scenario.

To start, we can assume that the molecules in each gas are identical and indistinguishable, meaning that we cannot tell one nitrogen molecule from another or one oxygen molecule from another. This assumption is important because it means that the distribution of molecules within each gas does not matter, as long as the total number of molecules in each gas remains the same.

Next, we can apply the principle of equal a priori probabilities. This means that before the partition was punctured, there was an equal chance of any molecule being on the left or right side of the box, regardless of its type (nitrogen or oxygen). So, the probability of finding a nitrogen molecule on the left side was 500/600, and the probability of finding an oxygen molecule on the right side was 100/600.

Now, when the partition is punctured and equilibrium is achieved, the molecules will have a chance of moving from one side to the other. However, the total number of molecules on each side will remain the same (500 nitrogen and 100 oxygen). This means that the probability of finding a nitrogen molecule on the left side is still 500/600, and the probability of finding an oxygen molecule on the right side is still 100/600.

Therefore, the probability of finding the molecules in the same distribution as before (500 nitrogen on the left and 100 oxygen on the right) is:

(500/600)^500 * (100/600)^100

This is because the probability of finding each molecule in its original location is independent of the other molecules, so we can multiply the individual probabilities together.

I hope this helps to answer your question. Let me know if you have any further questions or need clarification. Happy experimenting!

 

[thepatientmental]

Yes, that is the correct approach. To find the probability of the molecules being in the same distribution as before, we need to calculate the number of ways the molecules can be arranged in the two halves of the box. This can be done using the formula for permutations, which is n!/r!, where n is the total number of molecules and r is the number of molecules in one half of the box. In this case, n=600 and r=500 for nitrogen and r=100 for oxygen. So the probability would be (500!*100!)/600!, which is approximately 0.000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000