# Giuseppe Mensitieri » 8.State and equilibrium. Part 5

### Baric Derivatives

BARIC DERIVATIVES

Still referring to systems for which $\sigma \equiv\{V,T,x\}$, we introduce other two state functions which are also additive functions of mass, i.e.:

$\text{ENTHALPY: H=U+pV}$

$\text{GIBBS FREE ENERGY (or FREE ENTHALPY): G=H-TS=A+pV}$

Reminding that, for the state we are considering, $p=-\partial a/\partial V$ and , $S=-\partial a/\partial T$ we obtain the following relationships between the different mappings introduced so far:

$u(\cdot)=a(\cdot)-T\frac{\partial a}{\partial T};h(\cdot)=a(\cdot)-T\frac{\partial a}{\partial T}-V\frac{\partial a}{\partial V}; g(\cdot)=a(\cdot)-V\frac{\partial a}{\partial V}$

In many practical problems we know the value of pressure instead of that of volume. Hence, we would prefer to have p among the state variables in place of V. If we suppose that a function $f^{-1}(\cdot)$ exists such that:

$V=f^{-1}(p,T,x)$

Then, by substituting this function into the constitutive mappings, we can generate new mapping functions which map the triplet p, T and x into the value of the function of state.

### Baric Derivatives

We discuss here in some details the issue of invertibility. Consider, as an example, the p-V plot for the liquid-vapor equilibrium for a pure substance (see figure below). Here, the internal state variable, x, is the vapour phase fraction. With reference to the figure, we have that the equilibrium line where x = x*, is the line ‘0257′:

In the entire range of pressure p14- p36 the function is not invertible for the volume. In particular we have that:

• 0123‘ – corresponds to x = 0 (liquid phase) with the ‘23‘ line representing superheated liquid (non equilibrium condensed phase).
• 34‘ – is an hypothetical phase which cannot be realized; all the points on the curve correspond to absolutely unstable conditions.
• 4567‘ – corresponds to x = 1 (vapour phase) with the ‘45′ line and representing undercooled vapour (non equilibrium condition).

### Baric Derivatives

Even$f^*(\cdot)$ is not invertible from point 2 to point 5. However, the curves ‘0123′ (x = 0) and ‘4567′ (x = 1) have everywhere a finite derivative. We then conclude that, when T and x are fixed, $f(\cdot)$ is locally invertible quite everywhere. This implies that the inversion $f^{-1}(\cdot)$ can be done, but complying to the constraints imposed by local invertibility.
If we consider all the changes in the state of the system for which the system is located along that part of the p – V curve where there is invertibility at a certain pressure ( i.e. p = const.), we consequently have that for these changes df = 0, or, equivalently:

$\frac{\partial f}{\partial V}dV=-\left[\frac{\partial f}{\partial T}dT+\frac{\partial f}{\partial x}dx\right]\;\;\;\text{eq. (2)}$

From which we can get dV only if $\partial f/\partial V$is finite.
If we then consider a generic change in state $d\sigma\equiv\{dT, dx,dV\}$ at ρ = const. (dρ = 0), in the regions of local invertibility, we can evaluate dV from eq. (2). As a consequence, for this change in s we have:

$at\;\;p=const.\;\;\;dL=\left[\frac{\partial l}{\partial T}-\frac{(\partial l/\partial V){(\partial f/\partial T})}{\partial f/\partial V}\right]dT+\left[\frac{\partial l}{\partial x}-\frac{(\partial l/\partial V)(\partial f/\partial x)}{\partial f/\partial V}\right]dx$

### Baric Derivatives

From the previous expression we can define the following baric derivatives (i.e. derivatives at constant ρ):

$\frac{\partial L}{\partial T}\biggl |_{p,x}=\frac{\delta l}{\delta T}=\frac{\partial l}{\partial T}-\frac{(\partial l/\partial V)(\partial f/\partial T)}{\partila f/\partial V}$

$\text{eqs. (3a)}$

$\frac{\partial L}{\partial x}\Biggl |_{p,T}=\frac{\delta l}{\delta x}=\frac{\partial l}{\partial x}-\frac{(\partial l/\partial V)(\partial f/\partial x)}{\partial f/\partial V}$

Where

$\frac{\partial l}{\partial T}=\frac{\partial L}{\partial T}\Biggl |_{V,x}\;\;\;\;and\;\;\; \frac{\partial l}{\partial x}=\frac{\partial L}{\partial x}\Biggl |_{V,T}$

### Baric Derivatives

From eq. (3a) we obtain (see note 4 at the end of the chapter 2 of the lecture notes):

$\frac{\delta g}{\delta \bullet}=\frac{\partial a}{\partial \bullet}\text{or, equivalently}\;\; \frac{\partial G}{\partial \bullet}\Biggl |_{p. \bullet\bullet}=\frac{\partial A}{\partial \bullet}\Biggl |_{V, \bullet\bullet}$

We need finally to define only $\delta l/\delta p$. In fact, at constant T and x, we have:

$dp=\frac{\partial f}{\partial V}dV; dL=\frac{\partial l}{\partial V}dV\Rightarrow \frac{\delta l}{\delta p}=\frac{\partial l/\partial V}{\partial f/\partial V}$

### Maxwell Relations

We still consider the case of systems for which the state is:

$\sigma\equiv\{V,T,x\}$

The state functions (constitutive functions) we consider are: p, S, H, U, G and A. We have that for a differential change in state, :

$dA=\frac{\partial a}{\partial V}dV+\frac{\partial a}{\partial T}dT+\frac{\partial a}{\partial x}=-pdV-SdT+\theta dx\;\;\;\;\text{eq. (4)}$

dA is actually a differential since p, S and θ have finite values.
The invertibility assumption for the constitutive functions implies that, for the same differential change in state, the change of all the other functions of state are differential:

$dU=-pdV+TdS+\theta dx\;\;\;\text{eq.(5)}$

$dH=Vdp+TdS+\theta dx\;\;\;\text{eq. (6)}$

$dG=Vdp-SdT+\theta dx\;\;\;\text{eq. (7)}$

Equations (4) – (7) are called Maxwell relations. We underline here that we have used the following equalities:

$\theta=\frac{\partial A}{\partial x}\Biggl |_{V,T}=\frac{\partial U}{\partial x}\Biggl |_{V,S}=\frac{\partial H}{\partial x}\Biggl |_{p,S}=\frac{\partial G}{\partial x}\Biggl |_{p,T}$

### Maxwell Relations

In fact, as an example, from the definition of baric derivatives, we have:

$\frac{\delta g}{\delta x}=\frac{\partial G}{\partial x}\Biggl |_{p,T}=\frac{\partial A}{\partial x}\Biggl |_{V,T}$

While, from the definition of entropic derivatives, which will be illustrated later on, we have:

$\frac{\partial U}{\partial x}\Biggl |_{V,S}=\frac{\partial A}{\partial x}\Biggl |_{V,T}$

From the previous relationships we obtain several other relationships. For example:

a) in a direct way we obtain:

$-S=\frac{\partial G}{\partial T}\Biggl |_{p,x}=\frac{\partial A}{\partial T}\Biggl |_{V,x}$

b) putting equal to zero two differentials on the right hand side of eqs. (5) and (6):

$\frac{\partial H}{\partial T}\Biggl |_{p,x}=T\frac{\partial S}{\partial T}\Biggl |_{p,x}\;\;\;\text{and}\;\;\; \frac{\partial U}{\partial T}\Biggl |_{V,x}=T\frac{\partial S}{\partial T}\Biggl |_{V,x}$

### Maxwell Relations

c) by differentiation and substitution:

$\begin{array}{cc}\frac{\partial H}{\partial T}\Biggl |_{V,x}=T\frac{\partial S}{\partial T}\Biggl |_{V,x}+V\frac{\partial p}{\partial T}\Biggl |_{V,x}\\ \\ \frac {\partial H}{\partial T}\Biggl |_{V,x}=\frac{\partial U}{\partial T}\Biggl |_{V,x}
+V\frac{\partial p}{\partial T}\Biggl |_{V,x}\end{array}\right\}\Rightarrow T\frac{\partial S}{\partial T}\Biggl |_{V,x}=\frac{\partial U}{\partial T}\Biggl |_{V,x}$

d) by commutation of second derivatives:

$\frac{\partial p}{\partial T}\Biggl |_{V,x}=-\frac{\partial ^2 A}{\partial V\partial T}\Biggl |_{x}=\frac{\partial S}{\partial V}\Biggl |_{T,x}$

Other useful relationships are the following:

$C_V\equiv\frac{\partial U}{\partial T}\Biggl |_{V,x};C_p\equiv\frac{\partial H}{\partial T}\Biggl |_{p,x}\equiv\frac{\delta H}{\delta T}$

where CV and Cp are, respectively, the specific heat at constant volume and at constant pressure.

### Maxwell Relations

Consequently:

$T\frac{\partial S}{\partial T}\Biggl |_{V,x}=\frac{\partial U}{\partial T}\Biggl |_{V,x}\Rightarrow \frac{\partial S}{\partial T}\Biggl |_{V,x}=\frac{C_V}T$

$T\frac{\delta S}{\delta T}\equiv T \frac{\partial S}{\partial T}\Biggl |_{p,x}=\frac{\delta H}{\delta T}\equiv\frac{\partial H}{\partial T}\Biggl |_{p,x}\Rightarrow \frac{\delta S}{\delta T}\equiv\frac{\partial S}{\partial T}\Biggl |_{p,x}=\frac{C_p}T$

### Entropic Derivatives

We will show how, using procedures analogous to those adopted to obtain the expressions of baric derivatives, we can obtain expressions for entropic derivatives. In particular, we will obtain the expressions for:

$\frac{\partial U}{\partial V}\Biggl |_{x,S}~~;~~\frac{\partial U}{\partial x}\Biggl |_{v,S}~~;~~\frac{\partial U}{\partial S}\Biggl |_{V,x}$

We pass from a description of the state of the type $\sigma\equiv\{V,T,x\}$, to a description of the same state of the type $\sigma\equiv\{V,S,x\}$. If we consider a process at constant S, i.e. dS = 0, we have:

$\frac{\partial S}{\partial T}\Biggl |_{V,x}dT=-\left[\frac{\partial S}{\partial V}\Biggl |_{T,x}dV+\frac{\partial S}{\partial x}\Biggl |_{T,V}dx\right]$

### Entropic Derivatives

Since, for a generic differential change in state we have:

$dU=\frac{\partial U}{\partial V}\Biggl |_{T,x}dV+\frac{\partial U}{\partial T}\Biggl |_{V,x}dT+\frac{\partial U}{\partial x}\Biggl |_{T,V}dx$

For a differential change in state complying to the constraint that dS = 0, we obtain:

$dU_{at\;S=const}=\left[\frac{\partial U}{\partial V}\Biggl |_{T,x}-\frac{(\partial U/\partial T)_{V,x}(\partial s/\partial V)_{T,x}}{(\partial S/\partial T)_{V,x}}\right]dV+\left[\frac{\partial U}{\partial x}\Biggl|_{T,V}-\frac{(\partial U/\partial T)_{V,x}(\partial S/\partial x)_{T,V}}{(\partial S/\partial T)_{V,x}}\right] dx$

### Entropic Derivatives

From this expression we easily derive the expression for entropic derivatives we were seeking for:

$\left \begin{array}{ccc}\frac{\partial U}{\partial V}\Biggl |_{x,S}=\frac{\partial U}{\partial V}\Biggl |_{T,x}-\frac{(\partial U/\partial T)_{V,x}(\partial S/\partial V)_{T,x}}{(\partial S/\partial T)_{V,x}}\\ \\ \frac{\partial U}{\partial x}\Biggl |_{V,S}=\frac{\partial U}{\partial V}\Biggl |_{T,V}-\frac{(\partial U/\partial T)_{V,x}(\partial S/\partial x)_{T,V}}{(\partial S/\partial T)_{V,x}}\\ \\ \frac{\partial U}{\partial S}\Biggl |_{V,x}=\frac{(\partial U/\partial T)_{V,x}}{(\partial S/\partial T)_{V,x}} \end{array}\right \}$

### Entropic Derivatives

We can use these expressions, to demonstrate that (see note 5 at the end of chapter 2 of the notes):

$\theta=\frac{\partial A}{\partial x}\Biggl |_{V,T}=\frac{\partial U}{\partial x}\Biggl |_{V,S}$

and that, more in general:

$\frac{\partial U}{\partial \bullet}\Biggl |_{\bullet\bullet, S}=\frac{\partial A}{\partial \bullet}\Biggl |_{\bullet\bullet,T}$

### Equilibrium Conditions for ‘Internal’Systems

We have already seen in lecture 7 that, under proper smoothness hypothesis, if $\sigma\equiv\{V,T,x\}$, with x internal state variable, the equilibrium condition is:

$\frac{\partial A}{\partial x}\Biggl |_{V,T}=0$

We demonstrate now that, if we express the state as $\sigma\equiv\{V,S,x\}$, the equilibrium condition is:

$\frac{\partial U}{\partial x}\Biggl |_{V,S}=0$

In fact, from the 2nd law inequality:

$\left[\frac{\partial A}{\partial V}\Biggl |_{T,x}+p\right]\dot V+\left[\frac{\partial A}{\partial T}\Biggl |_{V,x}+S\right]\dot T+\frac{\partial A}{\partial x}\Biggl |_{V,T}\dot x\leq 0\;\;\;\text{(8)}$

### Equilibrium Conditions for ‘Internal’Systems

Moreover:

$\left \begin{array}{ccc}\frac{\partial A}{\partial V}\Biggl |_{T,x}=\frac{\partial U}{\partial V}\Biggl |_{S,x}\\ \\ \frac{\partial A}{\partial x}\Biggl |_{T,V}=\frac{\partial U}{\partial x} \Biggl |_{S,V}\\ \\ \frac{\partial A}{\partial T}\Biggl |_{V,x}=\frac{\partial U}{\partial T}\Biggl |_{V,x}-T\frac{\partial S}{\partial T}\Biggl |_{V,s}-S \end{array}\right \} \text{eqs . (9)}$

Now, since we have demonstrated that from the 2nd law inequality that:

$\frac{\partial A}{\partial T}\Biggl |_{V,x}=-S\;\;and\;\; \frac{\partial A}{\partial V}\Biggl |_{T,x}=-p$

we have:

$dU=-pdV+TdS+\theta dx$

### Equilibrium Conditions for ‘Internal’Systems

And we can write

$\left \begin{array}{ll}\frac{\partial U}{\partial V}\Biggl |_{S,x}=-p\\ \\ \frac{\partial U}{\partial T}\Biggl |_{V,x}=T\frac{\partial S}{\partial T}\Biggl |_{V,x}\end{array}\right \} \text{eqs . (10)}$

Substituting eqs. (9) into (8) we obtain:

$\left[\frac{\partial U}{\partial V}\Biggl |_{S,x}+p\right]\dot V+\left[\frac{\partial U}{\partial T}\Biggl |_{V,x}-T\frac{\partial S}{\partial T}\Biggl |_{V,x}-S+S\right]\dot T+\frac{\partial U}{\partial x}\Biggl |_{V,S}\dot x\leq 0$

That, using eqs. (10) becomes:

$[-p+p]\dot V+\left[T\frac{\partial S}{\partial T}\Biggl |_{V,x}-T\frac{\partial S}{\partial T}\Biggl |_{V,x}-S+S\right]\dot T+\frac{\partial U}{\partial x}\Biggl |_{V,S}\dot x \leq 0\Rightarrow \frac{\partial U}{\partial x}\Biggl |_{V,S}\dot x\leq 0\;\;\;\text{(11)}$

### Equilibrium Conditions for ‘Internal’Systems

Inequality (11) implies that U is minimum at x = x*, i.e. at equilibrium at constant Vand S. Using a similar procedure analogous equilibrium conditions can be found for G and H.
We can then summarize the equilibrium conditions for an ‘internal’system whose state is identified, equivalently (invertibility hypothesis holds true) by $\{V,T,x\};\{p,T,x\};\{V,S,x\}$, as $\{p,S,x\}$ follows:

S is at a maximum for a closed and isolated system
A is at a minimum at constant V and T
G is at a minimum at constant p and T
U is at a minimum at constant V and S
H is at a minimum at constant p and S

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Approfondimento

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