Thermodynamics Differentiating logarithms

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<mentor: change title>

In thermodynamics, there is a function which, for the three variables x, y, and z, can be given as

##G = xG_x+yG_y+zG_z + N[x\ln(x) + y\ln(y)+ z\ln(z)]+E(x,y,z)##

where G_x, G_y, G_z, and N are some constants and E is some arbitrarily complicated term.

There is a derivative of G called the chemical potential [itex]\mu[/itex] which is useful because

##G=x\mu_x+y\mu_y+z\mu_z## [Equation A]

I am trying to understand how this is satisfied, but I am having trouble even when E=0. To my mind, it should be the case that

##\mu_x = \frac{d}{dx}G = G_x+N(\ln(x)+1)+\frac{d}{dx}E## [Equation B]

but if this is true then Equation A doesn't make sense unless the value "1" in Equation B is removed. There must be something basic that I am not getting because I am also failing to satisfy Equation A for any function for E.. The problem is probably related to the fact that [itex]\mu_x[/itex] is actually

##\mu_x = \frac{d}{dn_x}G = ?????##

so that

##x = n_x/(n_x+n_y+n_z)##

but I don't get how to evaluate the derivatives even with this "change".

Thanks
 
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  • #2
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<mentor: change title>

In thermodynamics, there is a function which, for the three variables x, y, and z, can be given as

##G = xG_x+yG_y+zG_z + N[x\ln(x) + y\ln(y)+ z\ln(z)]+E(x,y,z)##

where G_x, G_y, G_z, and N are some constants and E is some arbitrarily complicated term.

There is a derivative of G called the chemical potential [itex]\mu[/itex] which is useful because

##G=x\mu_x+y\mu_y+z\mu_z## [Equation A]
I assume this is one version of ##G## which is easy enough to make computations. So in this case it would be an Ansatz for ##G##. However, "there is" demands an explanation. Where did it come from and what should be achieved by it. Is it given, an assumption, a derivation of something? What?
I am trying to understand how this is satisfied, ...
I guess it is an additional condition. A certain version of possible ##G##. If this is the case, then you cannot "understand" it. If it can be understood, then tell us where did it come from.
... but I am having trouble even when E=0. To my mind, it should be the case that

##\mu_x = \frac{d}{dx}G = G_x+N(\ln(x)+1)+\frac{d}{dx}E## [Equation B]

but if this is true then Equation A doesn't make sense unless the value "1" in Equation B is removed. There must be something basic that I am not getting because I am also failing to satisfy Equation A for any function for E.. The problem is probably related to the fact that [itex]\mu_x[/itex] is actually

##\mu_x = \frac{d}{dn_x}G = ?????##

so that

##x = n_x/(n_x+n_y+n_z)##

but I don't get how to evaluate the derivatives even with this "change".

Thanks
 
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  • #3
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<mentor: change title>
##\mu_x = \frac{d}{dn_x}G = ?????##
I think ##\mu_x## should be $$\mu_x=\frac{\partial [(n_x+n_y+n_z)G]}{\partial n_x}$$
 
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  • #4
dRic2
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The chemical potential for a single homogeneous substance is defined as ##\mu = \left( \frac {\partial G} {\partial N} \right)_{T, P}##. ##G = G(T, P, N)## is a function of temperature, pressure and the number of particle (or moles, as you prefer). It is assumed that ##G## is an addictive quantity w.r.t. the number of particles ##N##, that is ##G(T, P, \alpha N) = \alpha G(T, P, N)##.
Now let's compute the chemical potential both for ##G(T, P, \alpha N)## and ##G(T, P, N)##. So, for ##G(T, P, N)## I simply have:
$$\mu = \left( \frac {\partial G(T, P, N)} {\partial N} \right)_{T, P}$$
For ##G(T, P, \alpha N)## I have:
$$ \mu' = \left( \frac {\partial G(T, P, \alpha N)} {\partial (\alpha N)} \right)_{T, P}$$
Exploiting the property ##G(T, P, \alpha N) = \alpha G(T, P, N)## it is easy to see that ##\mu' = \mu## for every choice of ##\alpha##, this implies that the chemical potential can not be a function of the number of particles for a pure 1-component system. Hence ##\mu = \mu(T, P) = \left( \frac {\partial G} {\partial N} \right)_{T, P}##. Integrating the last equation yields ##G(T, P, N) = \mu(T, P) N##.
 
  • #5
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The chemical potential for a single homogeneous substance is defined as ##\mu = \left( \frac {\partial G} {\partial N} \right)_{T, P}##. ##G = G(T, P, N)## is a function of temperature, pressure and the number of particle (or moles, as you prefer). It is assumed that ##G## is an addictive quantity w.r.t. the number of particles ##N##, that is ##G(T, P, \alpha N) = \alpha G(T, P, N)##.
Now let's compute the chemical potential both for ##G(T, P, \alpha N)## and ##G(T, P, N)##. So, for ##G(T, P, N)## I simply have:
$$\mu = \left( \frac {\partial G(T, P, N)} {\partial N} \right)_{T, P}$$
For ##G(T, P, \alpha N)## I have:
$$ \mu' = \left( \frac {\partial G(T, P, \alpha N)} {\partial (\alpha N)} \right)_{T, P}$$
Exploiting the property ##G(T, P, \alpha N) = \alpha G(T, P, N)## it is easy to see that ##\mu' = \mu## for every choice of ##\alpha##, this implies that the chemical potential can not be a function of the number of particles for a pure 1-component system. Hence ##\mu = \mu(T, P) = \left( \frac {\partial G} {\partial N} \right)_{T, P}##. Integrating the last equation yields ##G(T, P, N) = \mu(T, P) N##.
In the equations given by the OP, it appears clear that his G is the Gibbs free energy per mole of mixture. So your G is equal to his G times the total number of moles. That's what I tried to point out in my response. In Introduction to Chemical Engineering Thermodynamics by Smith and van Ness (I think Chapter 11), they offer a different derivation of the desired relationship.
 
  • #6
dRic2
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So your G is equal to his G times the total number of moles
My derivation is for a 1-component system. The generalization to a mixture is left as an exercise to the reader :oldbiggrin:
 
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  • #7
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I think ##\mu_x## should be $$\mu_x=\frac{\partial [(n_x+n_y+n_z)G]}{\partial n_x}$$
This helped me correct my computation of the derivatives with finite differences, but I'm still having trouble with the analysis. For example, I found that this special derivative (whatever it is called) of

##G = 3n_xn_y##
predicts
##\frac{\partial (n_x+n_y+n_z)G}{\partial n_x}= 3(1-n_x)n_y##

but I'm not seeing how this is proved analytically. How does this happen?

Algebraically the finite difference is

##\frac{1}{dn_x}[3(n_x+dn_x)n_y(1+dn_x)-3n_xn_y]##
 
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  • #8
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This helped me correct my computation of the derivatives with finite differences, but I'm still having trouble with the analysis. For example, I found that this special derivative (whatever it is called) of

##G = 3n_xn_y##
predicts
##\frac{\partial (n_x+n_y+n_z)G}{\partial n_x}= 3(1-n_x)n_y##

but I'm not seeing how this is proved analytically. How does this happen?

Algebraically the finite difference is

##\frac{1}{dn_x}[3(n_x+dn_x)n_y(1+dn_x)-3n_xn_y]##
I have no idea what you did here. How does this relate to the original problem?
 
  • #9
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I have no idea what you did here. How does this relate to the original problem?
Sorry the actual finite-difference is

##\left[3\frac{n_y}{1+dn_x}\left(\frac{n_x+dn_x}{1+dn_x}\right)(1+dn_x)-3n_xn_y\right]\frac{1}{dn_x}##

which is [itex]3(2n_x+1)n_y[/itex]

I'm trying to show that I don't know what I'm doing by finding this derivative for an even simpler equation. I can do it with finite differences, but I don't know how to do it plainly through analysis. I don't understand what the correct operations are.

How is it true that the derivative of [itex]3n_xn_y[/itex], with respect to [itex]n_x[/itex], is [itex]3(2n_x+1)n_y[/itex]??
 
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  • #10
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Sorry the actual finite-difference is

##\left[3\frac{n_y}{1+dn_x}\left(\frac{n_x+dn_x}{1+dn_x}\right)(1+dn_x)-3n_xn_y\right]\frac{1}{dn_x}##

which is [itex]3(2n_x+1)n_y[/itex]

I'm trying to show that I don't know what I'm doing by finding this derivative for an even simpler equation. I can do it with finite differences, but I don't know how to do it plainly through analysis. I don't understand what the correct operations are.

How is it true that the derivative of [itex]3n_xn_y[/itex], with respect to [itex]n_x[/itex], is [itex]3(2n_x+1)n_y[/itex]??
Sorry, my inability to do these derivatives is resulting in frustrating mistakes and I can't delete this post: the actual equation consistent with this finite-difference is [itex]3(1-n_x)n_y[/itex]. I do not know where the [itex](1-n_x)[/itex] term comes from.

Perhaps a better example is to find the derivative of

##G = Bn_xn_y(n_x-n_y)##

with respect to [itex]n_x[/itex], because I can't even figure it out graphically.
 
  • #11
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This helped me correct my computation of the derivatives with finite differences, but I'm still having trouble with the analysis. For example, I found that this special derivative (whatever it is called) of

##G = 3n_xn_y##
predicts
##\frac{\partial (n_x+n_y+n_z)G}{\partial n_x}= 3(1-n_x)n_y##

but I'm not seeing how this is proved analytically. How does this happen?

Algebraically the finite difference is

##\frac{1}{dn_x}[3(n_x+dn_x)n_y(1+dn_x)-3n_xn_y]##
Let's take a simpler example, where there are only two chemical components, and with G=3xy. So $$(n_x+n_y)G=\frac{3n_xn_y}{(n_x+n_y)}$$So, $$\mu_x=\frac{\partial}{\partial n_x}\left(\frac{3n_xn_y}{(n_x+n_y)}\right)=3\frac{n_y^2}{(n_x+n_y)^2}=3y^2$$and, then, $$\mu_y=3x^2$$

So $$x\mu_x+y\mu_y=3y^2x+3x^2y=3(x+y)(xy)=3xy=G$$
 
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  • #12
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I see it now, thank you so much!
 

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