Thermodynamics Differentiating logarithms

In summary, the conversation discusses the function G in thermodynamics, which involves three variables and a complicated term E. The chemical potential \mu is a derivative of G and is useful in computations. The conversation also explores the relationship between \mu and G, as well as the concept of chemical potential for a single homogeneous substance. The equations given by the original poster suggest that G is the Gibbs free energy per mole of mixture, and this is clarified in the response.
  • #1
Hypatio
151
1
<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
Hypatio said:
<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
 
Last edited:
  • #3
Hypatio said:
<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
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
dRic2 said:
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
Chestermiller said:
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
Chestermiller said:
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
Hypatio said:
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
Chestermiller said:
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
Hypatio said:
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
Hypatio said:
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
I see it now, thank you so much!
 

1. What is thermodynamics differentiating logarithms?

Thermodynamics differentiating logarithms is a mathematical technique used to analyze and understand the behavior of thermodynamic systems. It involves taking the derivative of logarithmic functions to calculate changes in thermodynamic variables.

2. Why is thermodynamics differentiating logarithms useful?

Thermodynamics differentiating logarithms allows scientists to quantitatively study the relationship between different thermodynamic variables, such as temperature, pressure, and volume. This can help in understanding the behavior of complex systems and predicting their future states.

3. How is thermodynamics differentiating logarithms related to the laws of thermodynamics?

The laws of thermodynamics describe the fundamental principles governing the behavior of thermodynamic systems. Thermodynamics differentiating logarithms is a mathematical tool that helps in applying these laws to real-world systems and making predictions about their behavior.

4. What are some real-world applications of thermodynamics differentiating logarithms?

Thermodynamics differentiating logarithms has many practical applications, including in the fields of engineering, chemistry, and physics. It is used to analyze and optimize energy systems, understand chemical reactions, and study the behavior of materials under different conditions.

5. Are there any limitations to using thermodynamics differentiating logarithms?

While thermodynamics differentiating logarithms is a powerful tool, it does have some limitations. It assumes that systems are in equilibrium, which may not always be the case in real-world scenarios. Additionally, it may not be applicable to systems that involve non-ideal behavior or phase transitions.

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