Confused about thermodynamic quantities and their minimums.


by AntiElephant
Tags: confused, minimums, quantities, thermodynamic
AntiElephant
AntiElephant is offline
#1
Jan17-13, 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 constant-temperature, constant-pressure 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?
Phys.Org News Partner Physics news on Phys.org
The hemihelix: Scientists discover a new shape using rubber bands (w/ video)
Mapping the road to quantum gravity
Chameleon crystals could enable active camouflage (w/ video)
Studiot
Studiot is offline
#2
Jan17-13, 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).
AntiElephant
AntiElephant is offline
#3
Jan17-13, 04:35 PM
P: 15
Quote Quote by Studiot View Post
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).
Thanks for the reply. That actually made a lot of sense. So my problem was differenating between whether it was talking about the heat bath or the system. However, if the system were at constant temperature and pressure, Gibbs energy would never change (from that relation in my OP), right?

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?

Studiot
Studiot is offline
#4
Jan17-13, 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 + dwrev (Gibbs equation)

At constant temp TdS = d(TS) so we can rewrite the above as

d(U-TS) = dwrev (at constant T)

Let F = A-TS then dF = dwrev (at const T)

Now the expression wrev refers to all possible kinds of work.

If only pressure/volume work is envisaged

dwrev = 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 + dwrev + PdV + VdP

or at constant T and P

d(H-TS) = PdV + dwrev

This new function is called the Gibbs free energy and is G = H - TS = F+PV

So dG = PdV + dwrev

Again restricting this to PV work dwrev, dwrev = -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(H-TS) = 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?
facenian
facenian is offline
#5
Jan18-13, 04:02 AM
P: 205
Quote Quote by AntiElephant View Post
http://img842.imageshack.us/img842/7200/img0103iu.jpg


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?

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
DrDu
DrDu is offline
#6
Jan18-13, 06:02 AM
Sci Advisor
P: 3,376
Quote Quote by AntiElephant View Post


[itex] dG = -SdT + VdP [/itex]
holds only true if ##dU=TdS-pdV##. In that case, G cannot change at constant T and P, indeed.
However generally
##dU=TdS-pdV +\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.
Studiot
Studiot is offline
#7
Jan18-13, 06:03 AM
P: 5,462
facenian
In case the transformation is not reversIble the correct equation is dG≤0, 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
That's an interesting statement.

Under what irreversible process do you think ΔG = 0 ?

As a matter of interest your expression is an inequality not an equation.
facenian
facenian is offline
#8
Jan18-13, 04:52 PM
P: 205
Quote Quote by Studiot View Post
Under what irreversible process do you think ΔG = 0 ?
that's an interesting question and is related to the conceptual problem I'm having right now(see my post "Conceptual thermodynamics question" posted the same day as yours)
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
AntiElephant
AntiElephant is offline
#9
Jan18-13, 07:48 PM
P: 15
Quote Quote by facenian View Post
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.
Quote Quote by DrDu View Post
holds only true if ##dU=TdS-pdV##. In that case, G cannot change at constant T and P, indeed.
Quote Quote by studiot View Post
So for a change at constant temperature along a reversible path

dU = TdS + dwrev (Gibbs equation)
Thanks for the replies. Yes Studiot that did help, as long as I can understand one thing. My problem might be stemming from this;



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?
DrDu
DrDu is offline
#10
Jan19-13, 12:45 AM
Sci Advisor
P: 3,376
Quote Quote by AntiElephant View Post

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?
No, the last statement is in the text is surely correct and also your further thoughts about it.
AntiElephant
AntiElephant is offline
#11
Jan19-13, 09:21 AM
P: 15
Oh my mistake DrDu, I had misread your original post. But then;

Quote Quote by facenian View Post
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
How can

[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(E-TS+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?
Studiot
Studiot is offline
#12
Jan19-13, 03:54 PM
P: 5,462
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?
Yes this is true and I do not think anyone said anything to imply otherwise.
For a transition between equilibrium state 1 and equilibrium state 2.

ΔG is always G2-G1, 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=VdP-SdT 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.

G2-G1 = q - w +(P2V2-P1V1) -(T2S2-T1S1)

If T1 = T2 = T and P1 = P2 = P

Then by second law

q ≤ T(S2-S1)

Thus

w-P(V2-V1) ≤ (G2-G1)

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(V2-V1) 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(V2-V1)

This leads to the expression

w* ≤ - (G2-G1)
facenian
facenian is offline
#13
Jan20-13, 06:13 AM
P: 205
Quote Quote by AntiElephant View Post
How can

[itex] dE = TdS - PdV [/itex]

Hold for any process, but;

[itex] dG = -SdT + VdP [/itex]

Hold only for reversible ones?
The reason is this : [itex] dE = TdS - PdV [/itex] stated in this form is not valid for any procesess, what is valid for any process is [itex] dE = dQ - PdV [/itex] the problem stems from the inequation dQ≤ TdS in which the equality holds for reversible processes while for irreversible ones the strick inequality must be chosen
DrDu
DrDu is offline
#14
Jan21-13, 03:06 AM
Sci Advisor
P: 3,376
Quote Quote by AntiElephant View Post
Oh my mistake DrDu, I had misread your original post. But then;



How can

[itex] dE = TdS - PdV [/itex]

Hold for any process, but;

[itex] dG = -SdT + VdP [/itex]
Apparently you have not read my post #6, where I showed you that already the first equation does not hold for every system, especially for those systems where consideration of G becomes interesting, e.g. when there are other mechanical variables than V or when chemical reactions can occur.


Register to reply

Related Discussions
Thermodynamic processes and the relationships between the quantities they describe Classical Physics 3
Maximums and Minimums Calculus & Beyond Homework 6
Maximums and Minimums Calculus 4
maximums and minimums Calculus & Beyond Homework 5
maximums and minimums Calculus & Beyond Homework 7