How Can I Find the Entropy of the Universe Using Sterling's Approximation?

[t_n_p]

Homework Statement

A model universe comprises 100 atoms in system 1 and 1500 atoms in system 2. Compute the entropy for the universe when there are 3 atoms on in system 2 and 97 atoms on in system 1 (using sterlings approximation).

The Attempt at a Solution

I am able to find the entropy of systems 1 and 2 by initially finding the number of microstates and then the equation:

entropy = boltzmanns * ln(microstates)

Just wondering how I would get the entropy of the universe though.

 

[t!m]

Well, if you know how many microstates there are for systems 1 and 2, then how many microstates are possible for the combined system (1+2)?

 

[t_n_p]

I was unsure of the terminology. So taking universe as systems 1 and 2:

I can find number of microstates using Ω = C(N,n)
(i.e. the number of ways of moving n atoms from N sites)

Thus for system 1:
Ω = C(1500,3)?
This is a massive number!

Am I on the right track?

 

[Ygggdrasil]

Yup that's the correct way of doing the calculation. The calculation may be made easier by finding a formula for C(N,n) in terms of factorials, then applying Stirling's approximation.

 

[t_n_p]

sterlings approximation only helps once I get a value of Ω though correct?
I cannot even compute C(1500,3)=1500!/3!1497! due to overflow!

 

[Ygggdrasil]

Try to calculate ln(Ω) instead.

 

[t_n_p]

Ok, so I managed to compute C(1500,3) and C (100,97).
I got 561375500 and 646800 respectively.

(I wish to keep in terms of boltzmanns)
Therefore for system 1:
entropy = K*ln(561375500) = 20.15K

and system 2:
entropy = K*ln(646800) = 13.38K

total = 33.53K?

Where does Sterling's Approximation come into this?

 

[Ygggdrasil]

ln(Ω) = ln(1500!/3!1497!) = ln(1500!) - ln(3!) - ln(1497!) = 1500ln(1500) - 1500 - ln(6) - 1497ln(1497) + 1497

Here's where sterling's approximation saves you from evaluating horrible factorials.

 

[t_n_p]

yeah i completed it, thanks for help