Energy dissipation in Thermodynamics

Often, laws are reformulated to change variables in order to simplify the analysis of the system. For example, Newton's laws of motion can be rewritten in completely equivalent Lagrange equations for appropriate coordinate systems in classical mechanics. The Lagrange equations, in turn, can undergo a change of variable to produce the related Hamilton equations. Change of variables, is known as Legendre differential transformations. If the state of the system is described by a function of two variables f(x,y), which satisfies the equation

$$\mbox{d}f=u\mbox{d}x+v\mbox{d}y$$ (28)

and we wish to change the description to one involving a new function g(u,y), satisfying a similar equation in terms of du and dy and leaving the information content intact (it is a contact or a linear bijective transform), then it is necessary to define the Legendre transform g(u,y) as

$$g(\partial_xf~,~y)= f-ux~=~f~-~x\partial_xf~~.$$ (29)

It is easily verified that g satisfies the equation [1]

$$\mbox{d}g=-x\mbox{d}u + v\mbox{d}y~~.$$ (30)

So if we take the first law of thermodynamics, namely,

$$\mbox{d}E = -P\mbox{d}V + T\mbox{d}S$$ (31)

where E, the internal energy is a function of volume V and entropy S. Therefore, E is convenient for situations involving changes in volumes and entropy. One could see the similarity between $\mbox{d}U$ and $\mbox{d}g$ in equation (35) above, so we could generate a new function $\mbox{d}f$ from $\mbox{d}E$ and call it $\mbox{d}H$, the enthalpy,

$$\mbox{d}H = V\mbox{d}P + T\mbox{d}S$$ (32)

and the enthalpy which is a function of P and S is

H = E + PV  . (33)

From (31) we get a new function called Helmholtz function

$$\mbox{d}F = -S\mbox{d}T - P\mbox{d}V$$ (34)

and

F = E - TS (35)

The last chararacteristic function one could generate from the enthalpy differential form namely

$$\mbox{d}H = T\mbox{d}S + V\mbox{d}P$$

which is also a state function and is called the Gibbs function [1],

$$\mbox{d}G = -S\mbox{d}T + V\mbox{d}P$$ (36)

and

G = H -TS  . (37)

In terms of the state functions (potential functions) so far defined, we have written four differential equations that are formulations of the first law, namely,

$$\begin{array}{c@{\:=\:}c@{\:{}\:}c} \mbox{d}E & -P\mbox{d}V +... ...\mbox{d}T\\ \mbox{d}G & V\mbox{d}P - S\mbox{d}T~~. \end{array}$$

The above equations are for a closed system, that is, the number of moles n of the system is constant.

If we permit matter to pass through a permeable membrane, that is, the number of moles can change, and thus becoming a thermodynamic variable, and the system will be called an open system. Now, for an open system E will depend not only on V and S, but also on the number of moles n₁ , n₂ ,..., n_N of N different substances. The expansion of U in this case will look like:

$$\mbox{d}E(V,S,n) = ~\partial_V^{S,n}E~~\mbox{d}V~+~\partial_S... ...~\mbox{d}n_1~+...+\partial_{n_N}^{V,S,n\prime}E~~\mbox{d}n_N~.$$ (38)

where the subscript $n\prime$ on all but the first two partial derivitives means that all n, except the one in the partial derivitive, are being held constant during the differentiation. From equations (31) and (38) we see that

$$\partial_V^{S,n}E~=~-P \quad\mbox{~and~}\quad \partial_S^{V,n}E~=~T$$

and we could write the remaining derivitives as

$$\partial_{n_N}^{V,S,n\prime}E~=~\mu_j$$

where $\mu_j$ is defined as the chemical potential of the jth substance.

Hence, the above set of equations could be written as in the following [1]

$$\mbox{d}E(V,S,n)~ = -P\mbox{d}V + T\mbox{d}S+\sum_{j=1}^N \mu_j\mbox{d}n_j$$ (39)

$$\mbox{d}H(P,S,n) =~ V\mbox{d}P + T\mbox{d}S+\sum_{j=1}^N \mu_j\mbox{d}n_j$$ (40)

$$\mbox{d}F(V,T,n) ~= -P\mbox{d}V - S\mbox{d}T+\sum_{j=1}^N \mu_j\mbox{d}n_j$$ (41)

$$\mbox{d}G(P,T,n) =~ V\mbox{d}P - S\mbox{d}T+\sum_{j=1}^N \mu_j\mbox{d}n_j~~.$$ (42)

If we take a single substance then the last term in the above equations will become $\mu \mbox{d}n$. We need only equation (39) to see the form of the energy dissipation in thermodynamics, but it would be useful here to illustrate how one could generate many different potential functions .

Theoretically, there must always exist a way from any set of three variables to achieve any other variables in a three step-type way . The three steps are writing the Legendre transformation, rearranging the terms and defining new monotonic variables till one reaches the desired potentials [9].

For example we would like to start from the energy function E(S,V,N), and reach other potential functions like A:=E/S^2/3, $B:=\mu/PT$, and C:=N/E, where these three potential functions could be measured experimentally (in this particular example).

We start by defining s:=S^2/3 $\Longrightarrow$s^3/2=S $\Longrightarrow$ S^-1/3=s^-1/2.

Differentiation, $\mbox{d}_S s=\frac{2}{3}S^{-1/3}=\frac{2}{3}s^{-1/2}$ $\Longrightarrow$ $\mbox{d}=\frac{3}{2}s^{1/2}\mbox{d}s$.

Upon substitution in (39) we get $\mbox{d}E=T.\mbox{d}s.\frac{3}{2}.s^{1/2}-P\mbox{d}V+\mu \mbox{d}N$ . Defining the term as $\tau:=\frac{3}{2}Ts^{1/2}$, we get the new relation

$$\mbox{d}E(s,V,N)=\tau \mbox{d}s-P\mbox{d}V+\mu \mbox{d}N$$

Rearranging the terms and defining new variables, we get

$$\begin{array}{c@{\:=\:}c@{\:-\:}c@{\:+\:}c} P\mbox{d}V & \ta... ...box{d}V & a\mbox{d}s & b\mbox{d}E & c\mbox{d}N \\ \end{array}$$

with $a:=\tau/P$, b:=1/P, and $c:=\mu/P$ . To interchange a with $\mbox{d}s$ and b with $\mbox{d}E$ we need two new Legendre transformations, as follows

$$Z:=V-sa \qquad\mbox{,}\qquad \mbox{d}Z=-s\mbox{d}a-b\mbox{d}E+c\mbox{d}N$$

and

$$Y:=Z+bE \qquad\mbox{,}\qquad \mbox{d}Y=-s\mbox{d}a+E\mbox{d}b+c\mbox{d}N~~.$$

Rearranging the terms and defining new variables

$$\begin{array}{c@{\:=\:}c@{\: \:}c@{\: \:}c} \mbox{d}b & -\fr... ...\mbox{d}b & -f\mbox{d}Y + X\mbox{d}a + e\mbox{d}N \end{array}$$

Defining W:=b-aX and $\mbox{d}W=f\mbox{d}Y-a\mbox{d}X+e\mbox{d}N$. We have chosen the last definition of the Legendre transform to target the definition X:=A^-1 and exchange $\mbox{d}X$ with the wanted differential $\mbox{d}A$ as follows

$$\mbox{d}_AX=-\frac{1}{A^2}=-X^2\Longrightarrow \mbox{d}X=-X^2\mbox{d}A$$

Substituting, we get

$$\mbox{d}W = f\mbox{d}Y+aX^2 \mbox{d}A+e\mbox{d}N$$

The potential function (W) now is a variable of Y, A, and N. It is clear that the first step is accomplished, which is achieving a potential function in terms of the three variables A, B, and C. Similarly, we could go one with this process till we exchange $\mbox{d}Y$ with $\mbox{d}B$ and $\mbox{d}N$ with $\mbox{d}C$ [9].