Thermodynamics

We begin with the required thermodynamic formulae. We will assume that all processes are equilibrium processes which occur reversibly, and we consider here only thermal, mechanical, and chemical equilibrium. The fundamental relation of thermodynamics is:

$$E = E(S,V,{\vec N}),$$ (1.1)

where $E$ is the internal energy, $S$ the entropy, $V$ the volume, and ${\vec N}$ the set of all numbers of the distinguishable chemical components in the system. The same information content is contained in an equation expressing the entropy in terms of the internal energy, volume, and the numbers of particles of the chemical components. These expressions contain the maximal thermodynamic information about a system.

An infinitesimal change in the internal energy is given by:

$$dE = TdS - pdV + \sum\mu_idN_i.$$ (1.2)

Here $T$ is the temperature, $p$ the pressure, and $\mu_i$ the chemical potential of component $i$. The first term on the rhs of Eq. 1.2 corresponds to the heat change of the system, the second to mechanical work, and the third to changes in internal energy due to chemical transformations. Note that the heat transferred ($TdS$) at constant volume and composition is the same as the change in internal energy.

The internal energy is a first-order homogeneous function of its variables. This results from the paired combinations of intensive and extensive variables occurring in Eq. 1.2. Hence,

$$E = TS - pV + \sum\mu_iN_i.$$ (1.3)

We now take the derivative of both sides of Eq. 1.3, which leads to

$$SdT - Vdp + \sum N_i d \mu_i = 0.$$ (1.4)

This formula is known as the Gibbs-Duhem equation. It shows that at constant temperature and pressure, the chemical potential changes are not all independent of one another. This formula is useful in analysis of phase equilibria: the Clausius-Clapeyron equation can be derived from the Gibbs-Duhem relation for two phases at equilibrium.

The internal energy is a state function, meaning changes do not depend on the path taken. In contrast, the heat exchange and mechanical work do depend on the path. The path independence of state functions implies that the infinitesimal change in $E$ is an exact differential. This is equivalent to stating that the second derivatives of state functions are independent of the order of the differentiation. The Maxwell relations follow from this fact. An example is

$$\frac{\partial^2 E}{\partial S \partial V} = -\left( \frac{... ... S} = \left( \frac{\partial T}{\partial V}\right)_{S,{\vec N}}$$ (1.5)

An equation of state contains less than complete thermodynamic information. An example equation of state for the temperature is

$$T = T(S, V, {\vec N}) = \left(\frac{\partial E} {\partial S}\right)_{V, {\vec N}}$$ (1.6)

Since the temperature is intensive, it is homogeneous of order zero: scaling the variables $S$, $V$, or ${\vec N}$ by a constant does not change the value of the temperature. Therefore, the temperature of two connected subsystems at equilibrium must be the same as that for the separate subsystems; that is, the temperature is constant at equilibrium. The same holds, of course, for the pressure and all chemical potentials. The fact that the temperature is the derivative of the internal energy with respect to the entropy shows that Eq. 1.6 cannot contain the full thermodynamic information; the process of integrating it leaves an undetermined constant. (Knowledge of all equations of state does contain complete thermodynamic information.)

We now introduce the other state functions which are very useful in chemical thermodynamics. Each of these state functions may be obtained from the internal energy equation by a mathematical operation called a Legendre transformation. This transformation maintains the same information content, so each of the additional state functions (enthalpy, Helmholtz free energy, Gibbs free energy) contains the same complete thermodynamic information. The three functions listed above are

$$H = E + pV \hspace{0.5cm} (\mbox{Enthalpy})$$ (1.7)

$$A = E - TS \hspace{0.5cm} (\mbox{Helmholtz FE})$$ (1.8)

$$G = H - TS = \sum \mu_i N_i \hspace{0.5cm} (\mbox{Gibbs FE})$$ (1.9)

Taking the derivatives of each of the above, we find

$$dH = TdS + Vdp + \sum \mu_i dN_i$$ (1.10)

$$dA = -SdT - pdV + \sum \mu_i dN_i$$ (1.11)

$$dG = -SdT + Vdp + \sum \mu_i dN_i$$ (1.12)

The variables inside the differentials define the `natural' variables for each state function, that is $H = H(S,p,{\vec N})$, $A = A(T,V,{\vec N})$, $G = G(T,p,{\vec N})$. The heat exchange at constant pressure and composition is the enthalpy change, so the constant pressure heat capacity is

$$C_p = \left(\frac{\partial H}{\partial T}\right)_{p,{\vec N}... ...-T \left(\frac{\partial^2 G}{\partial T^2}\right)_{p,{\vec N}}$$ (1.13)

The heat capacity must be positive (LL, SPI, section 21, p. 63). The system Helmholtz and Gibbs free energies are minimized at equilibrium for constant ($T,V,{\vec N}$), ($T,p,{\vec N}$) respectively. The motivation for introducing the Helmholtz and Gibbs free energies is that they are natural functions of quantities which are easily handled experimentally. We can also solve Eq. 1.2 for $dS$ to obtain

$$dS = \frac{1}{T} dE + \frac{p}{T} dV - \sum \frac{\mu_i}{T} dN_i$$ (1.14)

From this relation it follows that the temperature can be defined as

$$\left( \frac{\partial S}{\partial E} \right)_{V,{\vec N}} = \frac{1}{T}$$ (1.15)

The chemical potential of component $i$, $\mu_i$, corresponds to the change in one of the state functions ($E, H, A, G$) with respect to the number of molecules of that component:

$$\mu_i = \left( \frac{\partial E}{\partial N_i} \right)_{S,V... ...)_{T,V} = \left( \frac{\partial G}{\partial N_i} \right)_{T,p}$$ (1.16)

Other thermodynamic quantities can be obtained by differentiating the free energies:

$$\left( \frac{\partial G}{\partial T} \right)_{p,{\vec N}} = -S$$ (1.17)

$$\left( \frac{\partial A}{\partial V} \right)_{T,{\vec N}} = -p$$ (1.18)

Consider a binary mixture. From the Gibbs-Duhem relation, we can obtain a relation between changes in chemical potentials at constant temperature and pressure:

$$d\mu_1 = -\frac{N_2}{N_1} d\mu_2$$ (1.19)

Taking the derivative with respect to $N_2$, we obtain

$$\left( \frac{\partial \mu_1}{\partial N_2} \right)_{T,p,N_1}... ... \left( \frac{\partial \mu_2}{\partial N_2} \right)_{T,p,N_1}$$ (1.20)

The derivative on the rhs is positive due to required thermodynamic inequalities for solutions (LL, SPI, section 96, p. 286). Thus the left hand side must be negative. That is, the chemical potential of component one must decrease if component two is added to the system.