Undergrad Thermodynamics Differentiating logarithms

[Hypatio]

<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 \mu 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 \mu_x 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

 

[fresh_42]

<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 \mu 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 \mu_x 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

 

[Chestermiller]

<mentor: change title>
$\mu_x = \frac{d}{dn_x}G = ?$

I think $\mu_x$ should be $$\mu_x=\frac{\partial [(n_x+n_y+n_z)G]}{\partial n_x}$$

 

[dRic2]

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 $G(T, P, N) = \mu(T, P) N$.

 

[Chestermiller]

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.

 

[dRic2]

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

 

[Hypatio]

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

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

 

[Chestermiller]

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?

 

[Hypatio]

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 3(2n_x+1)n_y

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 3n_xn_y, with respect to n_x, is 3(2n_x+1)n_y??

 

[Hypatio]

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 3(2n_x+1)n_y

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 3n_xn_y, with respect to n_x, is 3(2n_x+1)n_y??

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 3(1-n_x)n_y. I do not know where the (1-n_x) term comes from.

Perhaps a better example is to find the derivative of

$G = Bn_xn_y(n_x-n_y)$

with respect to n_x, because I can't even figure it out graphically.

 

[Chestermiller]

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 $$x\mu_x+y\mu_y=3y^2x+3x^2y=3(x+y)(xy)=3xy=G$$

 

[Hypatio]

I see it now, thank you so much!