# Stat Mech

1. Apr 20, 2009

### latentcorpse

What is the probability that at one instant, in one mole of an ideal gas, all $N=N_A$ molecules will be ofund in a volume $\frac{V}{2}$? Express your answer as a power of 10?

usually this type of question is just a simple combinatorics question but i don't know how to set it up. i.e. how do i make it "something choose something"

i thought there was some information missing to be honest but it's a past exam question so i'm assuming it's legit...

2. Apr 20, 2009

### dx

Divide the volume in your mind into half. How many ways can you distribute N particles into these two halfs? How many of these ways have all particles in one half? The ratio of these two numbers is the required probability. If you have one mole of gas, then the number of molecules N is the Avogadro number.

3. Apr 20, 2009

### latentcorpse

at a guess there's N! ways of arranging N distinguishable objects
only 1 of these will have all in one half so the answer is just N! yeah?

4. Apr 20, 2009

### dx

No it's not N! ways. N! is the number of ways of ordering N distinguishable objects in a line. We don't want to order the particles, but just to assign to each of them either "left side" or "right side". How many ways are there of doing that?

5. Apr 20, 2009

### latentcorpse

$\binom{N}{n_R} \binom{N-n_R}{n_L}$ but the second coeffeicient will just be one will it not as $n_L=N-n_R$

6. Apr 21, 2009

### latentcorpse

is this right? i just found the same question again and still haven't really grasped what's going on.

7. Apr 22, 2009

### dx

There are two ways to assign and L or an R to particle one, i.e. just L and R. Once you assign a side to particle one, there are two ways to assign a side to particle two, so for each of the two possible choices for particle one, there are two ways for particle two. This gives four ways for both: LL, LR, RL, RR. Can you see how to extend this reasoning to N particles?

8. Apr 22, 2009

### latentcorpse

2N possible combinations

i want 1 particular one (i.e. all on one side)

therefore $\binom{2N}{1}=2N$

9. Apr 22, 2009

### dx

Check your result for three particles:

LLR
LLL
LRR
LRL
RLR
RLL
RRR
RRL

There are 8 ways. But your formula 2N gives 2(3) = 6. Read post #7 again carefully and try to extend the logic to three particles, and you'll see the pattern.

10. Apr 22, 2009

### latentcorpse

ok 2^N then but i still only want to choose one of those combinations (i.e. all on one side) so i have 2^N choose 1=2^N ways of doing this.

11. Apr 22, 2009

### dx

There is only one way to assign "left" to all of them: (LLLL ... L). Be careful about what the "choose" operation means. N choose k is the number of ways of choosing a subset of k elements from N. Here, we don't want the number of ways to choose one element from 2^N, we just want the number of ways in which all of the N particles are on one side. They can all be either on the left side, or all on the right, so there are just two ways of doing it. The required probability is therefore 2/(2^N).

12. Apr 23, 2009

### latentcorpse

ok. i kind of follow that argument. but still a little hazy about why we can't use the choose argument - clearly it gives a high probability which is unphysical but surely there are 2^Npossible configurations and we want to choose 1 of them?

13. Apr 23, 2009

### dx

We want to count how many of them have all particles on one side. The technical meaning of "N choose k" is "the number of ways of selecting a subset of k elements from a set of N elements." Are we looking for the number of ways of choosing 1 element from a set of 2^N elements? No! We just want to count how many of them satisfy our condition "all on one side".