Confused about thermodynamic quantities and their minimums.by AntiElephant Tags: confused, minimums, quantities, thermodynamic 

#1
Jan1713, 01:56 PM

P: 15

http://img842.imageshack.us/img842/7200/img0103iu.jpg
Mainly concerned with the first half of the page. You'll see that it derives for a system in contact with a constanttemperature, constantpressure heat bath that the change in Gibbs entropy is always less than or equal to zero. Equality being for reversible transformations. A is the availability of the system AND its environment. However, I've honestly never understood something about this at all, and I've just tried to ignore it but it's come cropping up again for a different module for mine. Maybe it's something really simple. The relation [itex] dG = SdT + VdP [/itex] Is derived by considering the defition of Gibbs free energy. However, this formula is based on state variables, and hence works for all transformations. Right? It even confirms that for me in a different textbook of mine. So, in a constant temperature, constant pressure heat bath [itex] dT = 0, dP = 0 [/itex] always applies surely? There is no change in pressure of the system, there is no change in temperature. But then surely [itex] dG = 0 [/itex] for no matter what process? What is wrong in my reasoning? 



#2
Jan1713, 03:04 PM

P: 5,462

A heat bath is another name for a heat reservoir. It is not part of the system undergoing the process.
http://en.wikipedia.org/wiki/Thermal_reservoir It has the property of remaining at constant temperature, however much heat it accepts or provides. However the system may or may not remain at constant temperature. Heat is transferred from/to the heat bath by the system process. If pressure remains constant no work is done so all the heat transferred increases/decreases the internal energy of the system. As an example, you melt some ice at constant pressure by adding latent heat from the surroundings (heat bath). 



#3
Jan1713, 04:35 PM

P: 15

If you wouldn't mind, would you clarify something for me again? There's another source for this topic which basically emulates a different proof; Note: I believe he was meant to write [itex] ΔW < ΔF [/itex] and [itex] ΔW = PΔV[/itex]. In this case work done is the work done on to the system. The theorem states that for a system kept at constant temperature and constant pressure, Gibbs never increases. True. But, by that thermodynamic relation in my OP, surely it will never decrease either? Its change is always zero. In this case there seems to be no reference to the system being connected to a constant pressure/temperature heat bath. Maybe that is automatically implied? Also I've always learnt that [itex] ΔW = PΔV [/itex] is only valid for reversible processes. P being the pressure of the system. He seems to make the equivalence without reference as to whether or not the process is reversible. The only way I can make sense of this is if by; [itex] W = PΔV [/itex], P is actually referring to the applied pressure, rather than the pressure of the system. The applied pressure P is always greater than system pressure due to frictional effects, and in the case of a reversible process [itex] P_{applied} = P_{system} [/itex]? Edit: The fact that it says that relation is only infinitesimal reversible processes also makes no sense to me. Infinitesimal? Okay. But the process depends on state variables and thus it works for all processes, not just reversible ones? 



#4
Jan1713, 06:47 PM

P: 5,462

Confused about thermodynamic quantities and their minimums.
Not quite sure of your symbols here. E is not normally used for enthalpy. Assuming you are using F for what is normally called the Helmoltz free energy (or work function) and H for enthalpy, equation 38 should read.
G = H  TS = F + PV. since there is some doubt as to the meaning of E I shall avoid it and use U for internal energy. So for a change at constant temperature along a reversible path dU = TdS + dw_{rev} (Gibbs equation) At constant temp TdS = d(TS) so we can rewrite the above as d(UTS) = dw_{rev} (at constant T) Let F = ATS then dF = dw_{rev} (at const T) Now the expression w_{rev} refers to all possible kinds of work. If only pressure/volume work is envisaged dw_{rev} = PdV At const volume dV = 0 so dF = 0 so F is a minimum, at equilibrium. So the condition for equilibrium at constant temp and volume is that the Helmholtz free energy is a minimum and dF = 0. Now to develop a similar discussion for constant pressure Note that H = U + PV dH = dU + PdV + VdP Substitute from the above dH = TdS + dw_{rev} + PdV + VdP or at constant T and P d(HTS) = PdV + dw_{rev} This new function is called the Gibbs free energy and is G = H  TS = F+PV So dG = PdV + dw_{rev} Again restricting this to PV work dw_{rev}, dw_{rev} = PdV So at const T and P dG = 0, Thus G is a minimum, at equilibrium. Now applying all that to my melting ice example. dG= 0 = d(HTS) = dH  d(TS) = dH  TdS dS = dH / T The entropy of melting (fusion) can be calculated by dividing the enthalpy (=latent heat) by the melting point. You might like to look at this thread and in particular post#11 (Which no longer contains a typo) http://www.physicsforums.com/showthr...lmholtz&page=2 does this help? 



#5
Jan1813, 04:02 AM

P: 205

Don't know if you already solved your problem but it seems simple to me. There is an error in your statement "and hence works for all transformations", it does not work for all transformations it only works for reversible transformations. In case the transformation is not reversIble the correct equation is [itex] dG ≤ 0 [/itex], where P is the external preasure and T is the source temperature that are supposed to be constant, the preasure and temperature of the system not even defined por a non equiblibrium state 



#6
Jan1813, 06:02 AM

Sci Advisor
P: 3,375

However generally ##dU=TdSpdV +\sum_i Y_i dX_i +\sum_j \mu_j dN_j ##, where ##X_i## are other extensive variables of the system and ##Y_i## the conjugate intensive variables, e.g. charge and voltage of a battery, ##\mu_j## is the chemical potential of component j and ##N_j## the ammount of component j. Then ##dG=\sum_i Y_i dX_i +\sum_j \mu_j dN_j SdT +VdP##, so even at constant T and P, G can change. 



#7
Jan1813, 06:03 AM

P: 5,462

Under what irreversible process do you think ΔG = 0 ? As a matter of interest your expression is an inequality not an equation. 



#8
Jan1813, 04:52 PM

P: 205

It seems that for irreversible processes the inequality must hold strickly but I yust can't find a correct demostration in the context of macroscopic thermodynamics Obs: in my previous post I said "which are suposed to be constant" and this is not correct because in such case dT=dP=0, the only restriction is that the process be reversible in which case as you already correctly noted dG=0 also, what you missed is that when the transformation is not reversible you can not use your equation as you stated it 



#9
Jan1813, 07:48 PM

P: 15

Is the statement at the bottom wrong then? It says since it only involves state variables, that equation must be for all changes. That has actually made sense to me. If a system is taken through an infinitesimal irreversible change from an equilibrium state A of energy, entropy, volume [itex] (E_1 ,S_1, V_1) [/itex] to an equilirium state B of [itex] (E_2, S_2, V_2) [/itex] then the energy change; [itex] dE = dQ^{irrev} + dW^{irrev} [/itex] However, energy is a state variable and we can construct a reversible process from the same initial state to the same final state and the energy, entropy and volume change will be the same; [itex] dE = dQ^{rev} + dW^{rev} [/itex] [itex] dE = TdS PdV [/itex] Now although this was for a reverisble process, all the quantities are state variables and so this formula should surely hold for any process? I know you have all said that's not true, but why mathematically is it wrong? 



#10
Jan1913, 12:45 AM

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P: 3,375





#11
Jan1913, 09:21 AM

P: 15

Oh my mistake DrDu, I had misread your original post. But then;
[itex] dE = TdS  PdV [/itex] Hold for any process, but; [itex] dG = SdT + VdP [/itex] Hold only for reversible ones? As far as I can tell, the infinitesimal change equation for Gibbs free energy is derived directly from that first equation, which holds for any process. Here is how I'd do it; Gibbs free energy is defined [itex] G = E  TS + PV [/itex] So then; [itex] dG = d(ETS+PV) = dE TdS SdT + PdV + VdP [/itex] Now this must be for any process, surely. As we have the equation for [itex] dE [/itex] for any process, we can sub that in; [itex] dG = TdS  PdV  TdS  SdT + PdV + VdP [/itex] [itex] dG = SdT + VdP [/itex] Which must be true for any process? 



#12
Jan1913, 03:54 PM

P: 5,462

For a transition between equilibrium state 1 and equilibrium state 2. ΔG is always G_{2}G_{1}, reversible or irreversible. However if you calculate in terms of other variables you have to be able to put values to these in states 1 and 2, and between. dG=VdPSdT can also be written [tex]dG = {\left( {\frac{{\partial G}}{{\partial P}}} \right)_T}dP + {\left( {\frac{{\partial G}}{{\partial T}}} \right)_P}dT[/tex] It is an exact differential and the second formulation shows more clearly what variables have to be held constant in the partials for it to hold correctly. For a transition between state 1 and state 2. G_{2}G_{1} = q  w +(P_{2}V_{2}P_{1}V_{1}) (T_{2}S_{2}T_{1}S_{1}) If T_{1} = T_{2} = T and P_{1} = P_{2} = P Then by second law q ≤ T(S_{2}S_{1}) Thus wP(V_{2}V_{1}) ≤ (G_{2}G_{1}) Where the expression inherits the same property that equality refers to the reversible and inequality refers to the irreversible. This is where the difference arises. Now Dr Du was referring to the following. The term P(V_{2}V_{1}) refers to the work done in moving the boundary at a steady pressure P. This is not necesarily all the work performed. For instance galvanic work may be performed. The w = w^{*} + P(V_{2}V_{1}) This leads to the expression w^{*} ≤  (G_{2}G_{1}) 



#13
Jan2013, 06:13 AM

P: 205





#14
Jan2113, 03:06 AM

Sci Advisor
P: 3,375




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