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Topic: dG of vaporisation at boiling point  (Read 20448 times)

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Offline mike

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dG of vaporisation at boiling point
« on: October 23, 2005, 10:32:21 PM »
Question: What is the standard molar Gibbs free energy of vaporisation at the normal boiling point of water?

Answer: Is it zero?
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Offline Yggdrasil

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Re:dG of vaporisation at boiling point
« Reply #1 on: October 24, 2005, 12:25:06 AM »
Yes.  The boiling point of a liquid is the temperature at which the gas phase and liquid phase will be at equilibrium.  At equilibrium, dG is zero.

GCT

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Re:dG of vaporisation at boiling point
« Reply #2 on: October 24, 2005, 12:52:29 PM »
Question: What is the standard molar Gibbs free energy of vaporisation at the normal boiling point of water?

Answer: Is it zero?


try finding out for yourself, at constant pressure and temperature

dG=dH-TdS, or assuming that neither the standard ethalpy of vaporization or standard entropy of vaporization changes significantly for a reasonable range of temperature

deltaGstd,m=delta u std = delta Hstdvap- T delta S stdvap, plug in the boiling point temperature

Offline mike

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Re:dG of vaporisation at boiling point
« Reply #3 on: October 24, 2005, 06:06:18 PM »
Quote
try finding out for yourself, at constant pressure and temperature

dG=dH-TdS, or assuming that neither the standard ethalpy of vaporization or standard entropy of vaporization changes significantly for a reasonable range of temperature

deltaGstd,m=delta u std = delta Hstdvap- T delta S stdvap, plug in the boiling point temperature

Well this is how I got my answer:

H2O   liquid <------> vapour

Keq = avapour/aliquid

and avapour = p/p0

and: aliquid = 1

therefore:  Keq = p/p0

for the normal boiling point of water p = p0 = 1

so Keq = 1

now:

dG = -RTlnKeq

and as Keq = 1 then lnKeq = 0

so therefore dG = 0

 :D
There is no science without fancy, and no art without facts.

GCT

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Re:dG of vaporisation at boiling point
« Reply #4 on: October 25, 2005, 07:39:23 PM »
dG also equals -TdS for an isothermal process.  Not quite sure what you're doin there.

Offline mike

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Re:dG of vaporisation at boiling point
« Reply #5 on: October 25, 2005, 07:45:10 PM »
Quote
Not quite sure what you're doin there.

this shows how the change in state has zero gibbs free energy change, ie using the Keq for the liquid to vapour change for water.

I am not sure how yours shows this?

 :) :)
There is no science without fancy, and no art without facts.

GCT

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Re:dG of vaporisation at boiling point
« Reply #6 on: October 26, 2005, 01:21:04 PM »
so are you assuming constant volume, wouldn't the hemholtz equation be more appropriate (assuming that you were to measure the vapor pressure of water in a constant volume container, water wouldn't boil; if you were to do it with a frictionless piston, the boiling point would be different).  If you were to say, have this occur in an open system, how would the vapor pressure of water apply?

The vaporization process still has changes of enthalpy and entropy associated with it, the chemical potential of liquid water and its vapor are not the same.

GCT

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Re:dG of vaporisation at boiling point
« Reply #7 on: October 26, 2005, 02:23:04 PM »
ok, I'm getting way ahead of myself here, so at the beginning I mentioned that

dG=dH-TdS, at constant temperature and pressure

since dS=dH/T at constant pressure

dG=dH-TdH/T=0

so yes, du seems to be zero, sorry for the confusion

Offline mike

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Re:dG of vaporisation at boiling point
« Reply #8 on: October 26, 2005, 05:49:30 PM »
Quote
since dS=dH/T at constant pressure

dG = dH - TdS

if dG = 0

dH - TdS = 0

therefore:

dH = TdS

dS = dH/T

 ??? ??? ???

aren't you just making the assumption that dG =0, this doesn't prove anything! What is wrong with my proof??
There is no science without fancy, and no art without facts.

GCT

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Re:dG of vaporisation at boiling point
« Reply #9 on: October 26, 2005, 08:43:33 PM »
alright, I did some internet searching and actually found discussions related to free energy of vaporization....with actual values.  Yes, I assumed that dS(system)=dH/T, assuming that at each stage of vaporization, the system is at equilbirum, making it a reversible process (which I thought was so).  Interesting.

You may actually want to plug in actual values, for instance, find the entropy and enthalpy of vaporization (CRC or NIST, scifinder) and with the respective temperatures, plug into the dG formula.

Another possibility is that at one atm, and at the corresponding boiling point, the free energy of vaporization is zero, but at temperatures otherwise, it may not be zero, since the chemical potential of the liquid and gases differ respectively.  

GCT

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Re:dG of vaporisation at boiling point
« Reply #10 on: November 04, 2005, 03:41:15 PM »
actually deltaGvap=u*(gas)-u*(liquid), where u* is the chemical potential of the pure gas/liquid, you can have a deltaGvap value but at the boiling point I think it's zero since the chemical potentials are equivalent for first order phase transitions.

Offline mike

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Re:dG of vaporisation at boiling point
« Reply #11 on: November 05, 2005, 03:08:07 AM »
Thanks GCT, well you really have been researching this haven't you ;)

Take a look at my working out further up this post and see what you think now. I think with my way you don't need to know any values and you only assume that that the vapour pressure equals atmospheric pressure at boiling (which is true). Let me know what you think.

mike
There is no science without fancy, and no art without facts.

GCT

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Re:dG of vaporisation at boiling point
« Reply #12 on: November 05, 2005, 12:47:24 PM »
I think that I can understand your though process, however, I usually find it more exact to work with the fundamental equations, because there are a lot of assumptions that are made from the derivations.  Another thing is that I'm not quite sure if you can apply free energy to an equilibrium process, that is whatever the proportion of vapor to liquid there may be, at the temperature, equivalent chemical potentials for gas and liquid for first order transitions, free energy doesn't really apply; so it seems from the fundamental equations.

For instance, say that we're talking about the situation besides the normal boiling point, another point of the equilibrium curve of the liquid-gas phase diagram, where the pressure is not 1 atm, but you're still at the boiling point for that respective pressure.  What will be the result of your method?

p.s. I'm not actually on the equilibrium chapter yet, just finished activity

Offline Donaldson Tan

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Re:dG of vaporisation at boiling point
« Reply #13 on: November 05, 2005, 08:54:55 PM »
consider boiling water in a close system

boiling occurs at constant temperature, so the change in molar gibb's free energy is given as: dg = dh - Tds

given that boiling occurs at saturated condition, then this process can be regarded as reversible. This implies that T.ds is therefore equivalent to the heat supplied (q = Tds)


boiling occurs at constant pressure:
du = q - p.dv
h = u + pv
dh = du + p.dv = (q - p.dv) + p.dv = q

given that q = T.ds = dh then it must be true that 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

GCT

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Re:dG of vaporisation at boiling point
« Reply #14 on: November 05, 2005, 09:35:09 PM »
consider boiling water in a close system

boiling occurs at constant temperature, so the change in molar gibb's free energy is given as: dg = dh - Tds

given that boiling occurs at saturated condition, then this process can be regarded as reversible. This implies that T.ds is therefore equivalent to the heat supplied (q = Tds)


boiling occurs at constant pressure:
du = q - p.dv
h = u + pv
dh = du + p.dv = (q - p.dv) + p.dv = q

given that q = T.ds = dh then it must be true that dg = dh - Tds = 0



first off, I don't think that boiling pertains to closed isolated systems, a supercritical state would be relevant where you have no distinction between the gas and liquid phase.

I've already mentioned this case in one of the above posts, but it's a bit insignificant ....you say that dG=dH-TdS, at constant temperature and pressure, the dS actually pertains to the entropy of the system, unfortunately dSsys is not always defined as dq/T, though it equals dq/T when you  assume an reversible process.

For instance, dS does not equal dq/T for an irreversible process.

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