# Change in entropy in an isolated system

1. Aug 24, 2011

### Jenkz

1. The problem statement, all variables and given/known data

Derive an equation for the change in entropy that occurs in an isolated (micro-canonical) system containing N particles, if an adiabatic expansion from volume V1 to volume V1 takes place. Show that the number of microstates is given by V^N.

2. Relevant equations

Entropy S = K$_{b}$ ln $\Omega$
Where $\Omega$ is multiplicity, the number of microstates for distinguishable partciles= N!/$\Pi$$_{i}$n$_{j}$!

3. The attempt at a solution

Ok i'm not too sure where to start. I know that dQ = 0 as this is an adiabatic expansion.
Meaning dU = dW = - NK$_{b}$T ln (V2/V1), but i'm not sure if this helps anything.

I also know that a microcanonical system is thermally isolated and has a fixed N. So would thermally isolated mean dT = 0? in which case dU = 0 ... confused.

2. Aug 24, 2011

### lightgrav

thermally isolated, so Q=0, means all the Work goes into changing the Temperature.
How much Work is done during an adiabatic expansion? (as a function of Volumes)

3. Aug 25, 2011

### Jenkz

Is my answer in my first post incorrect?

"I know that dQ = 0 as this is an adiabatic expansion.
Meaning dU = dW = - NKbT ln (V2/V1)"

as dU= dW = (integrating from V1 to V2) - pdV
where P = NKbT / V

4. Aug 25, 2011

### jambaugh

Thermally isolated means no heat enters or leaves, thus the expansion will change the temperature.

5. Aug 25, 2011

### Jenkz

Ok thanks, but I still don't understand how what to do for this question. :S