Free Expansion is non-spontaneous?

In summary: 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.
  • #1
MacNCheese
7
0
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?
 
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  • #2
MacNCheese said:
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.)
 
  • #3
But isn't entropy change equal to heat supplied divided by the temperature?
 
  • #4
MacNCheese said:
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.
 
  • #5
MacNCheese said:
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!
 
  • #6
Just to add, the entropy change determines if a process is *reversible*. The (Gibbs) free energy change determines if a process is *spontaneous*.
 
  • #7
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
 
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What is free expansion?

Free expansion is a process in thermodynamics where a gas expands into a vacuum or a region with significantly lower pressure, without any external work being done on the gas.

Why is free expansion considered non-spontaneous?

Free expansion is considered non-spontaneous because it violates the second law of thermodynamics, which states that in a closed system, the total entropy (or disorder) cannot decrease. In free expansion, the gas molecules spread out and become more disordered, leading to a decrease in entropy, which goes against the law.

What factors affect the spontaneity of free expansion?

The spontaneity of free expansion is affected by the pressure, temperature, and volume of the gas, as well as the size of the vacuum or low-pressure region the gas is expanding into. A higher pressure or temperature of the gas will make free expansion more spontaneous, while a larger volume or smaller vacuum will make it less spontaneous.

Is free expansion ever observed in real-life situations?

Free expansion is rarely observed in real-life situations, as it is an idealized process that only occurs in perfectly isolated systems. In most real-life scenarios, there is some external work being done on the gas, making free expansion unlikely to happen.

What are the implications of free expansion being non-spontaneous?

The non-spontaneity of free expansion has implications for the efficiency of certain processes, such as heat engines, which utilize gas expansion to do work. It also affects our understanding of thermodynamics and the second law, as free expansion goes against the idea of increasing disorder in a closed system.

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