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Thermodynamic Properties :: Fixing the State

by Saladsamurai
Tags: fixing, properties, state, thermodynamic
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Saladsamurai
#1
Jul14-10, 07:03 PM
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I am reading through a thesis paper and it says that for a combustion process in which we have separated the chamber into 2 zones: burned and unburned ideal gas mixtures, if we know the pressure and equivalence ratio (ER) for the unburned ideal gas mixture, we can find all of its thermodynamic properties.

Doesn't the ER just allow us to find the number of moles of each species? And thus knowing pressure and ER (moles) we are still left with two unknowns (Tunburned and Vunburned) in the ideal gas relation:
pV = nRT.

Am I missing the obvious here? Any thoughts are appreciated.
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stewartcs
#2
Jul15-10, 12:50 PM
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Quote Quote by Saladsamurai View Post
I am reading through a thesis paper and it says that for a combustion process in which we have separated the chamber into 2 zones: burned and unburned ideal gas mixtures, if we know the pressure and equivalence ratio (ER) for the unburned ideal gas mixture, we can find all of its thermodynamic properties.

Doesn't the ER just allow us to find the number of moles of each species? And thus knowing pressure and ER (moles) we are still left with two unknowns (Tunburned and Vunburned) in the ideal gas relation:
pV = nRT.

Am I missing the obvious here? Any thoughts are appreciated.
Well, I'm not inclined to give an opinion without reading the paper. However, I will say that the thermodynamic state of a simple compressible system is completely specified (i.e. fixed) by two independent, intensive properties.

CS
Saladsamurai
#3
Jul15-10, 04:24 PM
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Thanks stewartcs. I believe that I have sorted it out. It was poorly worded and I took the statement as a "standalone" statement when I should not have.


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