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Author: Subject: Thermodynamics - Heat of Vaporization change with pressure
FluoroPunch
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Thermodynamics - Heat of Vaporization change with pressure

I'm a little unsure about how it's physically possible for the heat of vaporization of a substance to change as the pressure is reduced.

As I understand, this is the case so I'm not doubting that, I just don't understand how this is possible. From what I know, enthalpy is a state function, so the path taken should not matter, only the end state. This doesn't seem to agree with the heat of vaporization changing as the pressure is reduced. Shouldn't the energy to turn a liquid into a gas be constant since it's a path function? Perhaps its simply because the intermolecular forces do not prevent vaporization as much when the pressure is reduced. It just doesn't make sense thermodynamically...

Could anyone clarify? Maybe I'm just tired or going insane...or missing something simple?
DraconicAcid
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The enthalpy of vaporization will change slightly with pressure. Enthalpy is a state function, but a low-pressure end state isn't the same as a high-pressure end state. I don't know how to calculate the deltaH for gas going from one pressure to another, but I wouldn't be surprised if it's non-zero.

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FluoroPunch
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 Quote: Originally posted by DraconicAcid The enthalpy of vaporization will change slightly with pressure. Enthalpy is a state function, but a low-pressure end state isn't the same as a high-pressure end state. I don't know how to calculate the deltaH for gas going from one pressure to another, but I wouldn't be surprised if it's non-zero.

So if I'm understanding correctly, what you're implying is that there's some enthalpy change due to the pressure change? That makes sense... So its kind of like how after a reaction you have to bring the reactants to the same temperature at the start of reaction to find out the total heat of reaction, only in this case you would have to bring the pressure back to what it was initially after vaporization/condensation?
DraconicAcid
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Quote: Originally posted by FluoroPunch
 Quote: Originally posted by DraconicAcid The enthalpy of vaporization will change slightly with pressure. Enthalpy is a state function, but a low-pressure end state isn't the same as a high-pressure end state. I don't know how to calculate the deltaH for gas going from one pressure to another, but I wouldn't be surprised if it's non-zero.

So if I'm understanding correctly, what you're implying is that there's some enthalpy change due to the pressure change? That makes sense... So its kind of like how after a reaction you have to bring the reactants to the same temperature at the start of reaction to find out the total heat of reaction, only in this case you would have to bring the pressure back to what it was initially after vaporization/condensation?

Yes. If you want standard delta(H), all the gases have to be at 1 atm before and after.

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FluoroPunch
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Alright that was the missing piece of the puzzle. Thanks a bunch
Magpie
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The definition of enthalpy is "internal energy" + PV. So why wouldn't the enthalpy change with pressure.

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FluoroPunch
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 Quote: Originally posted by Magpie The definition of enthalpy is "internal energy" + PV. So why wouldn't the enthalpy change with pressure.

Oof...you're right. Guess I'm getting rusty on the basics I think that's explains it best, it is a function of pressure.
CharlieA
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From a theoretical view (definitely not my forte!), I would think that a substance would vaporize more easily (require less energy) at a lower pressure because the liquid molecules would have to "fight against" fewer vapor molecules to enter the gas phase (vaporize).
(Is that pretty deep for me, or what?) Regards, Charlie A
FluoroPunch
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