## Isothermal Reversible and Irreversible (Free) Expansions

Volume: $\Delta S = nR\ln \frac{V_{2}}{V_{1}}$
Temperature: $\Delta S = nC\ln \frac{T_{2}}{T_{1}}$

Viraj B 3A
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Joined: Fri Sep 26, 2014 2:02 pm

### Isothermal Reversible and Irreversible (Free) Expansions

I wanted to know why $\Delta$S is the same for an isothermal reversible process and an isothermal irreversible (free expansion) process? What does it mean to find entropy for an isothermal irreversible system?

I also am struggling with the concept of free expansion. Free expansion (isothermal irreversible) occurs when the external pressure is 0 (in a vacuum). Hence w=0 because when there's an isothermal irreversible change w=-Pext$\Delta$V. Therefore system does no expansion work when it expands into vacuum because there's no opposing force for system to push against. But why would the gas want to expand in the first place? I guess the benefits of expansion would be increase in entropy but wouldn't this imply that the gas would expand indefinitely?

Thank you for any insight!

Niharika Reddy 1D
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Joined: Fri Sep 26, 2014 2:02 pm

### Re: Isothermal Reversible and Irreversible (Free) Expansions

ΔS is a state function, so the entropy of the system will be the same regardless of the path taken to get to the same final state from the same initial state. This is why ΔSsys is the same for both processes.

Justin Le 2I
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Joined: Fri Sep 26, 2014 2:02 pm

### Re: Isothermal Reversible and Irreversible (Free) Expansions

There is nothing stopping the gas from expanding so if it's thermal energy increases then the volume will increase.

Chem_Mod
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### Re: Isothermal Reversible and Irreversible (Free) Expansions

When you add more volume, the gas will expand as the entropy always increases. Thus, it is more favorable to fill up the larger volume when possible. It is like there are two people waiting for 1 treadmill, as soon as another treadmill is available, one of them will rush into that treadmill. So the gas will always fill up the volume you give it. Of course as you are increasing the volume of the system, the pressure will drop.

And as the discussion above mention, since entropy is the state function, you just need to calculate $\Delta S$ of the reversible process.