Free Expansion is non-spontaneous?

[MacNCheese]

I searched in the fora, and I did find https://www.physicsforums.com/showthread.php?t=292278" asking pretty much the same thing.

Internal energy change and work done are zero, so no heat is absorbed. Thus entropy is zero, hence it isn't spontaneous. What am I doing wrong?

Could you also explain why internal energy changes/work are zero?

 

[Doc Al]

Internal energy change and work done are zero, so no heat is absorbed. Thus entropy is zero, hence it isn't spontaneous. What am I doing wrong?

The entropy change is not zero--it increases as the gas expands.

Could you also explain why internal energy changes/work are zero?

No heat flows and no work is done. (Nothing pushes on it.)

 

[MacNCheese]

But isn't entropy change equal to heat supplied divided by the temperature?

 

[D H]

But isn't entropy change equal to heat supplied divided by the temperature?

You are talking about dS = δQ/T. This expression is true only for a reversible process. Free expansion is not a reversible process.

That said, you can use this expression, dS = δQ/T, to compute the change in entropy due to free expansion. Entropy is a "state variable": The entropy of a system is a function of the system's state only. Compare that to work and heat transfer. The amount of work done by a system in getting from some initial state to a final state, along with the heat transferred to a system, depend on the path taken as the system changes from that initial state to the final state.

Because entropy is path-independent, if you can find a path along which you can compute the change in entropy then the computed entropy will be the entropy of the system no matter what path is taken.

The final temperature of the gas in free expansion is the same as the initial temperature: It is an isothermal process. So, find a reversible isothermal process that has the same initial and final states as does the free expansion, compute the change in entropy for this process, and you will have the change in entropy for free expansion.

 

[Doc Al]

But isn't entropy change equal to heat supplied divided by the temperature?

For a reversible process, yes. But free expansion is not reversible.

If you can imagine a reversible process taking the gas from one volume to another, you can use it to calculate the change in entropy. (Entropy, being a state variable, doesn't depend on how you go from initial to final state.)

Edit: D H beat me to it!

 

[Andy Resnick]

Just to add, the entropy change determines if a process is *reversible*. The (Gibbs) free energy change determines if a process is *spontaneous*.

 

[MacNCheese]

Wow, thanks a bunch. That helped clear my concept of entropy a lot.

Does this mean that if I'm given the initial and final volumes and the moles of the gas, the change in entropy will be nR * ln(V_2/V_1)?

Also, is the change in entropy of the surroundings zero?

EDIT: Why can't I use an irreversible process? As long it gets to and from the same (corresponding) states, it should work fine, right? And why is dS = dQ/T defined for only reversible processes?

Sorry, I just cannot wrap my mind around most of thermodynamics. Wish I'd listened in class :P