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### Topic: dG of vaporisation at boiling point  (Read 21800 times)

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#### Donaldson Tan

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##### Re:dG of vaporisation at boiling point
« Reply #15 on: November 06, 2005, 03:28:58 PM »
For instance, dS does not equal dq/T for an irreversible process.

Is assuming boiling a reversible process a bad assumption? I doubt so..

The conditions where boiling occur is such that both liquid and gas phases are at equilibrium with each other.

consider boiling in an open system:
dh = q + v.dp = q (dp = 0 at constant pressure)
q = T.ds (since boiling is a reversible process)
since q = T.ds = dh, then dg = dh - Tds = 0

"Say you're in a [chemical] plant and there's a snake on the floor. What are you going to do? Call a consultant? Get a meeting together to talk about which color is the snake? Employees should do one thing: walk over there and you step on the friggin� snake." - Jean-Pierre Garnier, CEO of Glaxosmithkline, June 2006

#### mike

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##### Re:dG of vaporisation at boiling point
« Reply #16 on: November 07, 2005, 10:10:48 PM »
dG = dGo + RTlnK

at equilibrium dG = 0

therefore:

0 = dGo + RTlnK

dGo = -RTlnK

K = equilibtium constant for this equilibrium:

H2O(l) <---->  H2O(g)

now:

K = avapour/aliquid right?

and aliquid = 1 ok?

therefore:

K = avapour = p/po

and:

po = p at the boiling point (vapour pressure equals atmospheric pressure)

so: K = 1

therefore:

dGo = -RTlnK
dGo = -RTln1
dGo = 0
There is no science without fancy, and no art without facts.

#### Donaldson Tan

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##### Re:dG of vaporisation at boiling point
« Reply #17 on: November 08, 2005, 11:24:12 AM »
according to mike, dG = 0

so dH = TdS must be valid too.

given at constant pressure, Q = dH, then Q = TdS must be valid.

This shows that boiling is a reversible process.

GCT: what's wrong with my method?
"Say you're in a [chemical] plant and there's a snake on the floor. What are you going to do? Call a consultant? Get a meeting together to talk about which color is the snake? Employees should do one thing: walk over there and you step on the friggin� snake." - Jean-Pierre Garnier, CEO of Glaxosmithkline, June 2006

#### GCT

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##### Re:dG of vaporisation at boiling point
« Reply #18 on: November 10, 2005, 12:42:29 PM »
according to mike, dG = 0

so dH = TdS must be valid too.

given at constant pressure, Q = dH, then Q = TdS must be valid.

This shows that boiling is a reversible process.

GCT: what's wrong with my method?

I don't really think there's anything explicitly wrong with your method, but the whole point of this thread was to find the basis of why and if dG (rather du, chemical potential), free energy of vaporization, is zero for a boiling point transition.  It seems to me that you have assumed that dG=0 for free energy of vaporization from the start.  All in all, you're right, the whole process of boiling is a reversible process,it's when the chemical potential of the gas and liquid are equivalent for a first order transition. I though that there was a more fundamental way to show this, for instance du(gas)-du(liquid)=dG vap (note that dG vaporization is not always zero at non-equivalent chemical potentials). It's always interesting though, when through a discussion similiar to this, one discovers exceptions, I was simply trying to find any exceptions and so far I've found none for the case at boiling point temperature free energy of vaporization for a first order transition.

mike, try working with a = gX , g is equivalent to the activity coefficient. g=1 for a pure substance.  a=X.  K=[gas vapor pressure]/1, the vapor pressure refers to constant volume, we can equate the gas vapor pressure for different respective atmospheric pressures to find our temperature of interest, boiling temperature.  Really doesn't prove anything though, if you do everything correctly though, you may find that dG=0, but this simply proves that the fundamental derivations of the equations were done correctly.  You're merely checking upon the validity of the derived equation.  And that's why we need to work with the fundamentals.

#### Donaldson Tan

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##### Re:dG of vaporisation at boiling point
« Reply #19 on: November 14, 2005, 08:24:18 AM »
I don't really think there's anything explicitly wrong with your method, but the whole point of this thread was to find the basis of why and if dG (rather du, chemical potential), free energy of vaporization, is zero for a boiling point transition.

The chemical potential of a pure substance is equivalent to the molar chemical potential.

Question: What is the standard molar Gibbs free energy of vaporisation at the normal boiling point of water?

We are indeed considering the case for pure substance - water.
« Last Edit: November 14, 2005, 08:25:38 AM by geodome »
"Say you're in a [chemical] plant and there's a snake on the floor. What are you going to do? Call a consultant? Get a meeting together to talk about which color is the snake? Employees should do one thing: walk over there and you step on the friggin� snake." - Jean-Pierre Garnier, CEO of Glaxosmithkline, June 2006