Change in entropy in an isolated system

[Jenkz]

Homework Statement

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.

Homework Equations

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

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[itex]_{b}[/itex]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.

please help!

 

[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)

 

[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

 

[jambaugh]

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

 

[Jenkz]

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