# Homework Help: Statistical Thermodynamics/I need a physics genius

1. May 22, 2013

### Remy34

Here are to challenge questions given to the class to answer to be turned for a free slice of pizza. We have to days, and he said we can get help, and I'm taking him at his word. Here they are.

1). Suppose a room is divided into the front fifth, the middle fifths, and the back two fifths. What is the ratio of the probabilities of finding a 20%:40%:40% split in the particles (the most likely distribution), as opposed to a 19.9%:40.2%:39.9% split.( Assume all the particles are identical. . (Leave answer in terms of N, the total number of particles. But recognize that N is a
large number)

2). Consider the case of a gas in the atmosphere. Assume that the temperature is constant. Basd on the maxwell Boltzman distribution, at sea level the atmosphere contains 78.1% N2m 21% O2, 0.9% argon and 0.036 CO2. What are the ratios at the top of Everest?

relevant equations
Boltzmann-Maxwell.

3. The attempt at a solution

For The first one. Probability of finding a 20%:40%:40% split: 1
Probability of finding a 19.9%:40.2%:39.9% split= 19.9/20*39.9/40= 0.9925125

For the second, I don't even know where to begin. I know the probability is supposed to be lower (right?), but that is about it.

I know we're still going to get the pizza, but this bothering me to no end. I have to know it, and this why I joined, and will hopefully contribute in turn.

2. May 23, 2013

### clamtrox

How did you come up with these numbers?

Consider a simpler case, where the room is split into two equal parts. Then the probability of a random particle finding itself on either side is 1/2. Use binomial distribution to write the probability of having N1 particles on one side and N-N1 on the other, then somehow evaluate this for a large N (use the Stirling approximation, or just raw power and wolfram alpha)

What do you mean probability is supposed to be lower? Of course air pressure is lower on top of Everest, but you're asked for relative fractions. So they still sum up to 1. Would you expect the ratios of different gases to be different? What physical properties might affect the ratio?

3. May 23, 2013

### Remy34

Ok I see what you did. I gather from what you said is that I should do the following:

N!/(N1!(N-N1)!)=20%

(N-N1)!/(N1!(N-N1-N2)!)=40%

Right?

As for the second one, I have no idea how to approach, but you are right. The probabilities would still have to add to one. I was only thinking about the amount, which will change, right?

And thanks for bothering to respond.

4. May 23, 2013

### clamtrox

No no, you want to calculate those expressions, for when N1 = 0.2N or 0.199N, and N2 = 0.4N or 0.399N. And you also need to remember to include the intrinsic probabilities for each slot;
So
$$P(N, N_1, N_2) = \left(\frac{1}{5}\right)^{N_1} \left(\frac{4}{5}\right)^{N-N_1} \frac{N!}{N_1!(N-N_1)!} \left(\frac{1}{2}\right)^{N-N_1} \frac{(N-N_1)!}{N_2!(N-N_1-N_2)!}$$

right? (check I didn't make any stupid mistakes there) Then you "just" need to calculate $$\frac{P(N, 0.199N,0.399N)}{P(N,0.2N,0.4N)}$$

5. May 23, 2013

### Remy34

Thanks for that. But I suppose you can't calculate, seeing as you don't know what N is. I know that because N is big, we can use Sterlings Aprox. but why bother going the extra step?

As for the second question, how do you even begin?

Last edited: May 23, 2013