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What are conjugate variables

Perhaps I am too dumb, but this article doesn't really explain what conjugate variables really are... --Dan| 09:44, 15 March 2006 (UTC)

Also I read that 'Enthalpy' and 'inverse temperature' are conjugate thermodynamic variables. Can someone please explain or expand? --Dan| 09:51, 15 March 2006 (UTC)

What do you mean by really are? The article explains that conjugate variables form a pair - one is a generalized displacement, one is a generalized force, their product having dimensions of energy. Perhaps the article could go into Hamiltonian dynamics for a deeper explanation. Would that get closer to what they really are. As far as enthalpy and inverse temperature, they are not conjugate variables in this sense. Enthalpy has units of energy to begin with, so the product of the two will have dimensions of energy/temperature rather than energy. PAR 16:13, 15 March 2006 (UTC)

Bronsted’s work principle

I am still digging around, but I think Johannes Bronsted might have been the first to summarize his “work principle”, in his 1955 Principles and Problems in Energetics, that the over-all work ∆W performed by a system is the sum of contributions due to transport of extensive quantities ∆K_i across a difference of "conjugated potentials" P_i1 - P_i2 :

${\displaystyle \Delta W=\sum _{i=1}^{k}(P_{i1}-P_{i2})\Delta K_{i}}$

in which P_i1 - P_i2 may be T₁ - T₂ (thermal potential difference), μ₁ - μ₂ (chemical potential difference), or ψ₁ - ψ₂ (electric potential difference) and ∆K_i will be ∆S (quantity of entropy), ∆n (quantity of substance), or ∆e (quantity of electricity), respectively. These being relations loosely derived from Gibbs general internal energy expression.--Sadi Carnot 16:27, 23 July 2006 (UTC)

a serious misunderstanding

Quote:

${\displaystyle \mathrm {d} U\leq T\mathrm {d} S-P\mathrm {d} V+\mu _{1}\mathrm {d} N_{1}+\mu _{2}\mathrm {d} N_{2}\,.}$

More than that is true. Equality is always the case. So the correct formula is:

${\displaystyle \mathrm {d} U=T\mathrm {d} S-P\mathrm {d} V+\mu _{1}\mathrm {d} N_{1}+\mu _{2}\mathrm {d} N_{2}\,.}$

because the variables depend only on the state and not on the process.

Consider for simplicity the special case:

${\displaystyle \mathrm {d} N_{1}=\mathrm {d} N_{2}=0\,}$

If the process is reversible the term TdS is the heat absorbed and the term PdV is the work done.

TdS = heat_absorbed

PdV = work_done

dU = heat_absorbed − work_done = TdS − PdV

If the process is irreversible then

TdS > heat_absorbed

PdV > work_done

whether or not the BROWN is tested it will turn a blue-black colour in the sun. Consider an experiment where compressed air is expanded irreversibly and adiabatically by opening a valve. Then heat_absorbed=0 and work_done=0 and dU=0, but the entropy S has increased due to the irreversibility such that TdS>0, and the volume V has increased due to the expansion such that PdV>0, but still

dU = heat_absorbed − work_done = TdS − PdV

Please check the sources. Bo Jacoby (talk) 14:16, 5 February 2008 (UTC).

I agree! A very serious error indeed. Let's hope that wiki critics like Chris Hillman didn't see this. B.t.w. I violated wiki convention and corrected a typo: PdV is larger than work_done, not smaller. Count Iblis (talk) 14:21, 12 April 2008 (UTC)

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