1.9: First law of thermodynamics and thermodynamic potentials

Consider a system of \(n\) moles in a container of volume \(V\) with a fixed internal energy \(E\). The variables \(n\), \(V\), and \(E\) are all macroscopic thermodynamic quantities referred to as control variables. Control variables are simply quantities that characterize the ensemble and that determine other thermodynamic properties of the system. Different choices of these variables lead to different system properties. In order to describe the thermodynamics of an ensemble of systems with given values of \(n\), \(V\), and \(E\), we seek a unique state function of these variables. We will now show that such a state function can be obtained from the first law of thermodynamics, which relates the energy \(E\) of a system to a quantity \(Q\) of heat absorbed and an amount of work \(W\) done on the system:

\[E = Q + W \label{10.1}\]

The derivation of the desired state function begins by examining how the energy changes if a small amount of heat \(dQ\) is added to the system and a small amount of work \(dW\) is done on the system. Since \(E\) is a state function, this thermodynamic transformation may be carried out along any path, and it is particularly useful to consider a reversible path for which

\[dE = dQ_\text{rev} + dW_\text{rev} = dQ_\text{irrev} + dW_\text{irrev} \label{10.2}\]

Note that since \(Q\) and \(W\) are not state functions, it is necessary to choose whether to use a reversible or irreversible path. For simplicity, let us choose the reversible (’rev’) path. The amount of heat absorbed by the system can be related to the change in the entropy \(\Delta S\) of the system by

\[\Delta S = \int \frac{dQ_\text{rev}}{T}, \: \: \: \: \: \: \: dS = \frac{dQ_\text{rev}}{T} \label{10.3}\]

where \(T\) is the temperature of the system. Therefore, \(dQ_\text{rev} = T \: dS\). Work done on the system is measured in terms of the two control variables \(V\) and \(N\). Let \(P(V)\) be the pressure of the system at the volume \(V\). Mechanical work can be done on the system by compressing it from a volume \(V_1\) to a new volume \(V_2 < V_1\):

\[W_{12}^\text{(mech)} = -\int_{V_1}^{V_2} P(V) \: dV \label{10.4}\]

where the minus sign indicates that work is positive in a compression. A small volume change \(dV\) corresponds to an amount of work \(dW_\text{rev}^\text{(mech)} = -P(V) \: dV\). Although we will typically suppress the explicit volume dependence of \(P\) on \(V\) and write simply, \(dW_\text{rev}^\text{(mech)} = -P \: dV\), it must be remembered that \(P\) depends not only on \(V\) but also on \(n\) and \(E\). In addition to the mechanical work done by compressing a system, chemical work can also be done on the system by increasing the number of moles.

\[dW_\text{rev}^\text{(chem)} = \mu \: dn \label{10.5}\]

is done on the system. if there are \(M\) distinct chemical species, each with its own chemical potential \(\mu_k\), \(k = 1, \ldots, M\), then we would express the chemical work as

\[dW_\text{rev}^\text{(chem)} = \sum_{k=1}^M \mu_k \: dn_k \label{10.6}\]

\[dW_\text{rev} = dW_\text{rev}^\text{(mech)} + dW_\text{rev}^\text{(chem)} = -P \: dV + \sum_{k=1}^M \mu_k \: dn_k \label{10.7}\]

so that the total change in energy is

\[dE = T \: dS - P \: dV + \sum_{k=1}^M \mu_k \: dn_k \label{10.8}\]

By writing Equation 10.8 in the form

\[dS = \frac{1}{T} dE + \frac{P}{T} dV - \sum_{k=1}^M \frac{\mu_k}{T} dn_k \label{10.9}\]

it is clear that the entropy is both a state function and a natural function of \(V\), \(E\), and \(n_1, \ldots, n_M\). Since \(S\) is a function of \(n_1, \ldots, n_M\), \(V\), and \(E\), the change in \(S\) resulting from small changes in \(n_k\), \(V\), and \(E\) can also be written using the chain rule as

\[dS = \left( \frac{\partial S}{\partial E} \right)_{\left\{ n \right\}, V} dE + \left( \frac{\partial S}{\partial V} \right)_{\left\{ n \right\}, E} dV + \sum_{k=1}^M \left( \frac{\partial S}{\partial n_k} \right)_{V, E} dn_k \label{10.10}\]

Comparing Equation 10.10 with Equation 10.9 shows that the thermodynamic quantities \(T\), \(P\), and \(\mu_k\) can be obtained by taking partial derivatives of the entropy with respect to each of the three control variables:

\[\frac{1}{T} = \left( \frac{\partial S}{\partial E} \right)_{\left\{ n \right\}, V}, \: \: \: \: \: \: \: \frac{P}{T} = \left( \frac{\partial S}{\partial V} \right)_{\left\{ n \right\}, E}, \: \: \: \: \: \: \: \frac{\mu_k}{T} = -\left( \frac{\partial S}{\partial n_k} \right)_{V, E} \label{10.11}\]

We now recall that the entropy is a quantity that can be related to the number of microscopic states of the system. This relation was first proposed by Ludwig Boltzmann in 1877, although it was Max Planck who actually formalized the connection. Let \(\Omega\) be the number of microscopic states available to a system. The relation connecting \(S\) and \(\Omega\) states that

\[S(n_1, \ldots, n_M, V, E) = k_B \: \text{ln} \: \Omega(n_1, \ldots, n_M, V, E) \label{10.12}\]

Since \(S\) is a function of \(n_1, \ldots, n_M\), \(V\), and \(E\), \(\Omega\) must be as well. Since we can determine \(\Omega(n_1, \ldots, n_M, V, E)\) from a microscopic description of the system, Equation 10.12 then provides a connection between this microscopic description and macroscopic thermodynamic observables.

\[\frac{1}{k_B T} = \left( \frac{\partial \: \text{ln} \: \Omega}{\partial E} \right)_{\left\{ n \right\}, V}, \: \: \: \: \: \: \frac{P}{k_B T} = \left( \frac{\partial \: \text{ln} \: \Omega}{\partial V} \right)_{\left\{ n \right\}, E}, \: \: \: \: \: \: \: \frac{\mu_k}{k_B T} = -\left( \frac{\partial \: \text{ln} \: \Omega}{\partial n_k} \right)_{V, E} \label{10.13}\]

which shows that we can start with a microscopic description of the system and obtain the thermodynamic properties of the system.

On the other hand, we also see that \(E\) is a natural function of \(V\), \(S\), and \(n_1, \ldots, n_M\). Thus, from the chain rule, we also have

\[dE = \left( \frac{\partial E}{\partial V} \right)_{\left\{ n \right\}, S} dV + \left( \frac{\partial E}{\partial S} \right)_{\left\{ n \right\}, V} dS + \sum_{k=1}^M \left( \frac{\partial E}{\partial n_k} \right)_{S, V} dn_k \label{10.14}\]

and comparing to Equation 10.8, we have the additional thermodynamic relations

\[T = \left( \frac{\partial E}{\partial V} \right)_{\left\{ n \right\}, V}, \: \: \: \: \: \: \: P = - \left( \frac{\partial E}{\partial V} \right)_{\left\{ n \right\}, S}, \: \: \: \: \: \: \: \mu_k = \left( \frac{\partial E}{\partial n_k} \right)_{V, S} \label{10.15}\]

Changing Ensembles: Legendre Transforms

In thermodynamics, changing thermodynamic potentials between ensembles requires the use of a technique known as the Legendre transform. In order to see what a Legendre transform is, consider a simple function \(f(x)\) of a single variable \(x\). Suppose we wish to express \(f(x)\) in terms of a new variable \(s\), where \(s\) and \(x\) are related by

\[s = f'(x) \equiv g(x) \label{10.16}\]

with \(f'(x) = df/dx\). Can we determine \(f(x)\) at a point \(x_0\) given only \(s_0 = f'(x_0) = g(x_0)\)? The answer to this question, of course, is no. The reason, as Figure 10.1 makes clear, is that \(s_0\), being the slope of the line tangent to \(f(x)\) at \(x_0\), is also the slope of \(f(x) + c\) at \(x = x_0\) for any constant \(c\). Thus, \(f(x_0)\) cannot be uniquely determined from

Figure 10.1: Depiction of the Legendre transform.

\(s_0\). However, if we specify both the slope, \(s_0 = f'(x_0)\), and the \(y\)-intercept, \(b(x_0)\), of the line tangent to the function at \(x_0\), then \(f(x_0)\) can be uniquely determined. In fact, \(f(x_0)\) will be given by the equation of the line tangent to the function at \(x_0\):

\[f(x_0) = f'(x_0) \: x_0 + b(x_0) \label{10.17}\]

Equation 10.17 shows how we may transform from a description of \(f(x)\) in terms of \(x\) to a new description in terms of \(s\). First, since Equation 10.17 is valid for all \(s_0\), it can be written generally in terms of \(x\) as

\[f(x) = f'(x) \: x + b(x) \label{10.18}\]

Then, recognizing that \(f'(x) = g(x) = s\) and \(x = g^{-1} (s)\), and assuming that \(s = g(x)\) exists and is a one-to-one mapping, it is clear that the function \(b(g^{-1} (s))\), given by

\[b(g^{-1} (s)) = f(g^{-1} (s)) - sg^{-1} (s) \label{10.19}\]

contains the same information as the original \(f(x)\) but expressed as a function of \(s\) instead of \(x\). We call the function \(\overset{\sim}{f} (s) = b(g^{-1} (s))\) the Legendre transform of \(f(x)\). The function \(\overset{\sim}{f} (s)\) can be written compactly as

\[\overset{\sim}{f} (s) = f(x(s)) - sx(s) \label{10.20}\]

where \(x(s))\) serves to remind us that \(x\) is a function of \(s\) through the variable transformation \(x = g^{-1} (s)\).

The generalization of the Legendre transform to a function \(f\) of \(n\) variables \(n_1, \ldots, x_n\) is straightforward. In this case, there will be a variable transformation of the form

\[\begin{align} s_1 = \frac{\partial f}{\partial x_1} &= g_1 (x_1, \ldots, x_n) \\ &\vdots \\ s_n = \frac{\partial f}{\partial x_n} &= g_n (x_1, \ldots, x_n) \end{align} \label{10.21}\]

Again, it is assumed that this transformation is invertible so that it is possible to express each \(x_i\) as a function \(x_i (s_1, \ldots, s_n)\) of the new variables. The Legendre transform of \(f\) will then be

\[\overset{\sim}{f} (s_1, \ldots, s_n) = f(x_1(s_1, \ldots, s_n), \ldots, x_n (s_1, \ldots, s_n)) - \sum_{i=1}^n s_i x_i (s_1, \ldots, s_n) \label{10.22}\]

Note that it is also possible to perform the Legendre transform of a function with respect to any subset of the variables on which the function depends.

Transforming to the Canonical Ensemble

The Legendre transformation technique introduced in the previous section is the method by which thermodynamic potentials are transformed between ensembles. What we need to do is transform from a function of \(V\), \(S\), and \(\left\{ n \right\}\) to a function of \(V\), \(T\), and \(\left\{ n \right\}\). That is, we need to transform from \(E(n_1, \ldots, n_M, V, S)\) to a new energy that is a function of \(n_1, \ldots, n_M, V, T\). The Legendre transform is appropriate for this operation since

\[T = \left( \frac{\partial E}{\partial S} \right)_{\left\{ n \right\}, V} \label{10.23}\]

Applying the Legendre transform formula, we obtain a new energy

\[\begin{align} \overset{\sim}{E} (\left\{ n \right\}, V, T) &= E ( \left\{ n \right\}, V, S( \left\{ n \right\}, V, T)) - \frac{\partial E}{\partial S} S(\left\{ n \right\}, V, T) \\ &= E(\left\{ n \right\}, V, S( \left\{ n \right\}, V, T)) - TS(\left\{ n \right\}, V, T) \end{align} \label{10.24}\]

The function \(\overset{\sim}{E} (\left\{ n \right\}, V, T)\) is a new state function known as the Helmholtz free energy and is denoted \(A (\left\{ n \right\}, V, T)\) or \(F (\left\{ n \right\}, V, T)\). We will show shortly that when a thermodynamic transformation of a system from state \(1\) to state \(2\) is carried out on a system along a reversible path, then the work needed to effect this transformation is equal to the change in the Helmholtz free energy \(\Delta A\). From Equation 10.24, it is clear that \(A\) has both energetic and entropic contributions, and the delicate balance between these two contributions can sometimes have a sizeable effect on the free energy. Free energy is a particularly useful concept as it determines whether a process is thermodynamically favorable, indicated by a decrease in free energy, or unfavorable, indicated by an increase in free energy. It is important to note that although thermodynamics can determine if a process is favorable, it has nothing to say about the time scale on which the process occurs.

A process in which \(n_1, \ldots, n_M\), \(V\), and \(T\) change by small amounts \(dn_k\), \(dV\), and \(dT\) leads to a change \(dA\) in the Helmholtz free energy of

\[dA = \sum_{k=1}^M \left( \frac{\partial A}{\partial n_k} \right)_{V, T} dn_k + \left( \frac{\partial A}{\partial V} \right)_{\left\{ n \right\}, T} dV + \left( \frac{\partial A}{\partial T} \right)_{\left\{ n \right\}, V} dT \label{10.25}\]

via the chain rule. However, since \(A = E - TS\), the change in \(A\) can also be expressed as

\[\begin{align} dA &= dE - S \: dT - T \: dS \\ &= T \: dS - P \: dV + \sum_{k=1}^M \mu_k \: dn_k - S \: dT - T \: dS \\ &= -P \: dV + \sum_{k=1}^M \mu_k \: dn_k - S \: dT \end{align} \label{10.26}\]

where the second line follows from the first law of thermodynamics. By comparing the last line of Equation 10.26 with Equation 10.25, we see that the thermodynamic variables obtained from the partial derivatives of \(A\) are:

\[\mu_k = \left( \frac{\partial A}{\partial n_k} \right)_{V, T}, \: \: \: \: \: \: P = -\left( \frac{\partial A}{\partial V} \right)_{\left\{ n \right\}, T}, \: \: \: \: \: \: \: S = -\left( \frac{\partial A}{\partial T} \right)_{\left\{ n \right\}, V} \label{10.27}\]

Note that at constant temperature,

\[dA = -P \: dV + \sum_{k=1}^M \mu_k \: dn_k = dW_\text{rev} \label{10.28}\]

which shows that, physically, the change in free energy is simply the reversible work that we would need to perform on a system in order to change its thermodynamic state by changing the volume or the number of moles of the different species.

Note that we can also derive a relationship between \(A\) and the canonical partition function \(Q(\left\{ n \right\}, V, T)\) from the Legendre transform relation \(A = E - TS\). Since

\[E = -\frac{\partial \: \text{ln} \: Q}{\partial \beta} \label{10.29}\]

and

\[TS = -T \frac{\partial A}{\partial T} = \beta \frac{\partial A}{\partial \beta} \label{10.30}\]

we see that

\[A = -\frac{\partial \: \text{ln} \: Q}{\partial \beta} - \beta \frac{\partial A}{\partial \beta} \label{10.31}\]

This simple differential equation for \(A\) has the solution

\[A(\left\{ n \right\}, V, T) = -\frac{1}{\beta} \: \text{ln} \: Q(\left\{ n \right\}, V, T) \label{10.32}\]

which can be verified simply by substituting this solution back into the equation and showing that the equation is satisfied.

Transforming to the Isothermal-Isobaric Ensemble

The isothermal-isobaric ensemble uses \(\left\{ n \right\}\), \(P\), and \(T\) as the control variables. Thus, this ensemble results from performing a Legendre transform of the Helmholtz free energy from volume \(V\) to pressure \(P\). The volume in the Helmholtz free energy \(A(\left\{ n \right\}, V, T)\) is transformed into the external pressure \(P\) yielding a new free energy denoted \(\overset{\sim}{A} (\left\{ n \right\}, P, T)\):

\[\overset{\sim}{A} (\left\{ n \right\}, P, T) = A(\left\{ n \right\}, V(\left\{ n \right\}, P, T), T) - V(\left\{ n \right\}, P, T) \frac{\partial A}{\partial V} \label{10.33}\]

Using the fact that \(P = -\partial A/\partial V\), we obtain

\[\overset{\sim}{A} (\left\{ n \right\}, P, T) = A(\left\{ n \right\}, V(\left\{ n \right\}, P, T), T) + PV(\left\{ n \right\}, P, T) \label{10.34}\]

The function \(\overset{\sim}{A}(\left\{ n \right\}, P, T)\) is known as the Gibbs free energy and is denoted \(G(\left\{ n \right\}, P, T)\). Since \(G\) is a function of \(\left\{ n \right\}\), \(p\), and \(T\), a small change in each of these control variables yields a change in \(G\) given by

\[dG = \sum_{k=1}^M \left(\frac{\partial G}{\partial n_k} \right)_{P, T} dn_k + \left( \frac{\partial G}{\partial P} \right)_{\left\{ n \right\}, T} dP + \left( \frac{\partial G}{\partial T} \right)_{\left\{ n \right\}, P} dT \label{10.35}\]

However, since \(G = A + PV\), the differential change \(dG\) can also be expressed as

\[\begin{align} dG &= dA + P \: dV + V \: dP \\ &= -P \: dV + \sum_{k=1}^M \mu_k \: dn_k - S \: dT + P \: dV + V \: dP \\ &= \sum_{k=1}^M \mu_k \: dn_k + V \: dP - S \: dT \end{align} \label{10.36}\]

where the second line follows from Equation 10.26. Thus, equating Equation 10.36 with Equation 10.35, the thermodynamic relations of the isothermal-isobaric ensemble follow:

\[\mu_k = \left(\frac{\partial G}{\partial n_k} \right)_{P, T}, \: \: \: \: \: \: \: \langle V \rangle = \left( \frac{\partial G}{\partial P} \right)_{\left\{ n \right\}, T}, \: \: \: \: \: \: S = -\left( \frac{\partial G}{\partial T} \right)_{\left\{ n \right\}, P} \label{10.37}\]

As before, the volume in Equation 10.37 must be regarded as an average over instantaneous volume fluctuations.

Since \(G = A + PV = E - TS + PV = (E + PV) - TS\), \(G = H - TS\), where \(H\) is the enthalpy. We can also derive a relationship between \(G\) and the canonical partition function \(\Delta (\left\{ n \right\}, P, T)\) from the Legendre transform relation \(G = H - TS\). Since

\[H = -\frac{\partial \: \text{ln} \: \Delta}{\partial \beta} \label{10.38}\]

and

\[TS = -T \frac{\partial G}{\partial T} = \beta \frac{\partial G}{\partial \beta} \label{10.39}\]

we see that

\[G = -\frac{\partial \: \text{ln} \: \Delta}{\partial \beta} - \beta \frac{\partial G}{\partial \beta} \label{10.40}\]

This simple differential equation for \(A\) has the solution

\[G(\left\{ n \right\}, P, T) = -\frac{1}{\beta} \: \text{ln} \: \Delta (\left\{ n \right\}, P, T) \label{10.41}\]

which can be verified simply by substituting this solution back into the equation and showing that the equation is satisfied.