# Giuseppe Mensitieri » 7.State and equilibrium. Part 4

### 2nd law and ‘Internal’ Systems

2nd LAW AND INTERNAL SYSTEMS
SYSTEM WITH INTERNAL STATE VARIABLES
(e.g. CHEMICAL REACTION, RELAXING SYSTEMS, PHASE TRANSITION)

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

The irreversibility we have encountered in the viscous systems is in some way ’special’: in fact, it can be made as little as one desires if the transformation is slow enough. If we think, instead, of the case of a reacting mixture, the composition x of the system cannot be assigned at will and independently of all the other state variables. There are some important questions:

a) what is the irreversibility related to the time evolution of x?
b) what are the differences with the irreversibility we have seen for the case of viscous systems?
c) what is the concept of equilibrium for the case of ‘internal’systems?

On the basis of the assumed state, we have:

$A=a(V,T,x); U=u(V,T,x); p=f(V,T,x); S=s(V,T;x)$

We need also an expression for:

$\dot x=r(V,T,x)$

### 2nd law and ‘Internal’ Systems

By adopting procedures similar to those already used for the other analysed systems, we obtain starting from the 2nd law for the case of uniform T:

$\left(\frac{\partial a}{\partial V}+p\right)\dot V+\left(\frac{\partial a}{\partial T}+S\right)\dot T+\frac{\partial a}{\partial x}\dot x\leq 0$

From which we conclude that for this system:

$p=-\frac{\partial a}{\partial V}\;\;\;\;S=-\frac{\partial a}{\partial T}\;\;\;\text{and}\;\;\;\frac{\partial a}{\partial x}\dot x\leq0\Rightarrow Z=-\frac{\partial a}{\partial x}\dot x \geq 0\;\;\;\;(1)$

and

The previous expression indicates that A never increases for such system. In fact if we imagine a generic behaviour of A as a function of x:

• at x = xa and $\partial a/\partial x\leq 0$, hence from (1) $\dot x>0$ and  xa evolves toward x*;
• at x = xb and $\partial a/\partial x\geq 0$, hence from (1) $\dot x<0$ and xb evolves toward x*.

### 2nd law and ‘Internal’ Systems

Sooner or later, x will become equal to x* and it will happen when $\dot x=0$. We make here a smoothness hypothesis, i.e. that $\dot x$ is invertible for x at x* (see figure on the right). Consequently, x = x* when $\dot x=0$, i.e. when the system is at equilibrium. Hence, for such case, the system is at equilibrium when A is at minimum at constant V and T (under smoothness hypothesis: see note in the following slide).

### 2nd law and ‘Internal’ Systems

Note on the smoothness hypothesis
From the figure on the right in the previous slide, reporting a possible dependence of $\dot x$ on x, we see that the assumption of smooth behaviour consists in the invertibility of $\dot x$ as a function of x.

$\frac{\partial\dot x}{\partial x}\right\biggl |_{x^*}\neq 0$

$\frac{\partial \dot x}{\partial x}\biggl |_{x^*}\neq 0$

Now, from (1):

$\frac {\partial a}{\partial x}\dot x \leq 0$

We have $(\partial a/\partial x)\dot x$ that is maximum at equilibrium. Hence, at equilibrium we have:

$\left(\frac{\partial a}{\partial x}\dot x\right)^*=0\;\;and\;\; d\left[\left(\frac{\partial a}{\partial x}\dot x\right)^*\right]=0$

Consequently:

$\dot x d\left(\frac{\partial a}{\partial x}\right)+\frac{\partial a}{\partial x}d(\dot x)=0$

### 2nd law and ‘Internal’ Systems

But, at equilibrium, since $\dot x=0$, we have:

$\dot x d\left(\frac{\partial a}{\partial x}\right)=0$

We then conclude that:

$\left[\frac{\partial a}{\partial x}d(\dot x)\right]^*=0\;\;\;\text{but}\;\;\; d(\dot x)=\frac{\partial \dot x}{\partial x}dx$

but

Hence, in the smoothness hypothesis (i.e. if  $\frac{\partial \dot x}{\partial x}\biggl|_{x^*}\neq 0$) we have:

$\left(\frac{\partial a}{\partial x}\right)^*=0$

### 2nd law and ‘Internal’ Systems

Now we can compare the viscous dissipation with the dissipation related to internal state variables. In fact, in the case of uniform temperature, we have seen that the viscous local velocity of energy dissipation per unit mass is:

$Z=\left(-\frac{\partial a}{\partial V}-p\right)\dot V\geq 0$

It can be made small at will by decreasing $\dot V$.

On the other hand, the local velocity of energy dissipation per unit mass due to evolution of internal state variables is:

$Z=-\frac{\partial a}{\partial x}\dot x\geq 0$

It cannot be made small at will.
For example, if we consider the isothermal transformation of a gas from volume V1 to volume V2, the total dissipation during the process is, assuming an order 1 approximation for p dependence on $\dot V$:

$\int Zdt=B\int\dot V^2dt$

If we slow down the process, the integration interval increases linearly with time while the integrand decreases to the second power. Hence, the dissipation can be made small at will. On the contrary, if an ‘internal’system evolves from x1 to x2, the total dissipation will display a finite value which cannot be reduced.

### 2nd law and ‘Internal’ Systems

We now introduce the definition of affinity of the system, θ:

$\theta=\frac{\partial a}{\partial x}$

Since, by the 2nd law we have that $-\theta\dot x\geq 0$, we have also seen that, under the hypothesis of smooth behaviour, we have θ = 0 when $\dot x=0$, i.e. when $\dot x=r(V,T,x)=0$.

$\dot x=0$ is an equilibrium condition and the value of x at equilibrium, i.e. x*, is the solution of the equation $r(V,T, x^*)=0$. Consequently:

$x^*=K(V,T)\rightarrow\theta (V,T,K)V,T))=\theta^*(V,T)=0$

It is worth noting that for a viscous system the equilibrium condition is instantaneously attained as soon as one starts to keep the site constant. For systems with internal state variables, instead, the rate of evolution toward the equilibrium can be so slow that one does not detect any change of the system with time, although the system is not at equilibrium.

### I materiali di supporto della lezione

Approfondimento

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