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Old Jul 16th 2016, 09:11 AM   #1
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chemical potential

Hello,
I am having serious problems with the concept/definition of chemical potential:
By definition mu=partial derivative U(S,V,N) with respect to N.

Now
1) If S does not change,and the process of removing a particle is reversible,
dQ=0.
2)If V does not change, no work is applied.
3) Form the first law => dU=0

By consequence the chemical potential is always zero for any system.

This is simple proof is obviously nonsense, but what am I missing?

Thanks,

Wolfgang
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Old Jul 16th 2016, 02:11 PM   #2
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Originally Posted by wolfgang6444 View Post
3) Form the first law => dU=0
My Thermo is a bit spotty these days but this one leaps out a me. The internal energy of a system is conserved only if we have a closed system. If the number of particles in the system can change then we have particles entering or leaving the system. Thus if the system is closed to conserve internal energy then the chemical potential must be zero. If it is not closed then we could have a chemical potential.

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Old Jul 16th 2016, 03:37 PM   #3
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Thre is more to it than that since n or N can change in a closed system undergoing a phase change.
This subject sounds like the introduction to the phase rule.
The point is that you don't need chemical potential for single component systems and Topsquark's comments applies to these
(he is a physicist and physicists usually deal with single component systems and use N (number of molecules). Chemists more usually deal with mixtures and work in moles or n. (chemistry is after all largely about mixtures and reactions)
For a multiphase system and/or multicomponent system you equation is incomplete since it should include the summation of the chemical potentials of each component or phase.
I do not have time tonight to write more but I will see what I can do tomorrow (Sunday) if you are still interested.

Does this help?

Last edited by studiot; Jul 16th 2016 at 03:54 PM.
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Old Jul 16th 2016, 11:45 PM   #4
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Thanks to both of you,

yes indeed meanwhile I realized: the first law in the form dU=dQ+dA only applies to closed systems. This tends to be left out in most basic text books.
Meanwhile I have tried to calculate the contribution for a homogenous system:
U(S,V,N)=N*Up(S/N,V/N) =>dU/dN=1/N*U-S/N*dUp/dSp - V/*dUp/dVp
As dUp(Sp,Vp) = dU(S,V,N)/dS for a homogenous system
U(S,V,N) = U/N-S/N*dU/dS - V/N dU/dV =U/N -1/N S*T +1/N p*V
This is the well known form of the chemical potential for a homogenous system as Gibbs-Potential per particle.
So the chemical potential is NOT - as I thought - just the average internal Energie per particle but really the average Gibbs-potential per particle. Not very intuitive though.
I am aware that the main use for chemical potential are mixtures of phases and equlibrium of chemical reactions. Right now I was trying to use it to calculate p(h) and T(h) for the adiabatic atmosphere made up of a single ideal gas by considering particle exchange between to volumes at different height (yes, my background is physics, not chemistry ;-)

Thanks,

Wolfgang
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Old Jul 17th 2016, 03:06 AM   #5
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yes indeed meanwhile I realized: the first law in the form dU=dQ+dA only applies to closed systems.
What about (steady) flow systems?

Physicists don't usually consider these, but engineers do.
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Old Jul 17th 2016, 04:22 AM   #6
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Note also that the analysis you are using refers to ideal gases.

Ideal gases have a simplified thermodynamics in that the internal energy depends only on the absolute temperature of the gas. This is in fact one way to define an ideal gas.

For most of physics, chemistry and engineering this is adequate, but chemical engineers in particular need to deal with lots of real world gases and substitute a quantity called 'fugacity' for pressure.
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Old Jul 17th 2016, 04:29 AM   #7
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Why?

My hope is that my analysis applies to ANY homogenous system.

Wolfgang
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