Thermodynamic Potentials Applied for Equations of State (Post #3)

Equation of State of Ideal Gas

\[ A(T, V, \bar{n}) - A(T, V_R, \bar{n}) = \int_{V_R}^{V} \left( \frac{\partial A}{\partial V} \right)_{T, \bar{n}} dV = -\int_{V_R}^{V} P dV \tag{1} \]

In Equation 1, the term \( V_R \) acts as an arbitrary reference volume, which corresponds to a specific but unknown reference Helmholtz free energy, \( A_R \) (where \( A(T, V_R, \bar{n}) = A_R \)). To see how this plays out mathematically, let's look at the most basic example using the ideal-gas equation of state:

\[ P^{IG}(T, V, \bar{n}) = \frac{n R_g T}{V} \tag{2} \]

By substituting this into the fundamental volume derivative of the Helmholtz energy and performing the integration, we get:

\[ A^{IG}(T, V, \bar{n}) - A^{IG}(T, V_R, \bar{n}) = -\int_{V_R}^{V} P^{IG} dV \]

\[ = -\int_{V_R}^{V} \frac{n R T}{V} dV = -n R_g T \ln\left(\frac{V}{V_R}\right) \tag{3} \]

While Equation 3 allows us to calculate changes in Helmholtz free energy, it only provides the difference between states—we still do not know the absolute values of \( A \) or \( A_R \). If we differentiate Equation 3 with respect to temperature and apply the canonical entropy relationship, we arrive at a familiar expression for entropy:

\[ S^{IG}(T, V, \bar{n}) - S^{IG}(T, V_R, \bar{n}) = n R_g \ln\left(\frac{V}{V_R}\right) \tag{4} \]

Notice that in Equation 4, the temperature remains constant between the state of interest and the reference state.

If we were to differentiate Equation 3 with respect to volume instead, we would simply get the ideal gas law (\( P = n R_g T / V \)). However, differentiating with respect to the number of moles yields the change in chemical potential:

\[ \mu_i^{IG}(T, V, \bar{n}) - \mu_i^{IG}(T, V_R, \bar{n}) = -R_g T \ln\left(\frac{V}{V_R}\right) \tag{5} \]

These relationships highlight a major limitation: an equation of state (EOS) is not very helpful for calculating absolute thermodynamic properties. For instance, Equation 4 doesn't give us the actual overarching formula for entropy; it only shows how entropy differences relate to changes in volume and composition. The same limitation applies to Equation 5 regarding chemical potential.

For now, we can set this limitation aside. Moving forward, we will apply this exact same mathematical framework to much more powerful equations of state. To make those models work, we will introduce the concept of a residual property—which is simply the mathematical difference between a property calculated from our advanced real-gas model and that same property calculated for an ideal gas.

Soave–Redlich–Kwong Equation of State

Now, let's apply the integration specified in Equation 3 to a more advanced model: the Soave-Redlich-Kwong equation of state (SRK EOS).

\[ P = \frac{n R_g T}{V - nb} - \frac{n^2 a}{V(V + nb)} \tag{6} \]

In this expression, the parameters \( a \) and \( b \) are typically dependent on both temperature and composition. Notice that we have structured the SRK EOS here as a function of \( T \), \( V \), and the total number of moles (\( \bar{n} \)), rather than using mole fractions. We have also kept the unit choices generalized, consistent with our earlier discussions.

Finally, because it will be necessary for future calculations, we can determine the absolute minimum possible volume for the SRK EOS simply by inspecting the denominator of the first term:

\[ V_{min}^{SRK} = nb \tag{7} \]

Before we dive into the integration, it helps to algebraically rewrite the SRK EOS into a form that is easier to integrate:

\[ P = \frac{n R_g T}{(V - nb)} - \frac{n^2 a}{nb} \left( \frac{1}{V} - \frac{1}{V + nb} \right) \tag{8} \]

Applying the integration from Equation 3 to this new form yields:

\[ A(T, V, \bar{n}) - A(T, V_R, \bar{n}) = \left[ -n R_g T \ln(V - nb) + \frac{n^2 a}{nb} \ln \frac{V + nb}{V} \right]_{V_R}^{V} \]

\[ = -n R_g T \ln \frac{(V - nb)}{(V_R - nb)} + \frac{n^2 a}{nb} \ln \frac{(V + nb)V_R}{V(V_R + nb)} \tag{9} \]

At this point, we could try to push forward, but we run into the exact same roadblock we saw with Equation 3: we still don't know the absolute value of \( A(T, V_R, \bar{n}) \) for an SRK fluid at an arbitrary reference volume.

However, we can bypass this issue by comparing our SRK fluid calculations directly to those of an ideal gas. If we subtract the ideal gas result (Equation 3) from our SRK result (Equation 9), we get:

\[ A(T, V, \bar{n}) - A(T, V_R, \bar{n}) - A^{IG}(T, V, \bar{n}) + A^{IG}(T, V_R, \bar{n}) \]

\[ = -n R_g T \ln \frac{(V - nb)V_R}{(V_R - nb)V} + \frac{n^2 a}{nb} \ln \frac{(V + nb)V_R}{V(V_R + nb)} \tag{10} \]

At first glance, it might look like we just mashed two underdefined equations together. But remember two crucial facts: first, the conditions \( (T, V, V_R, \bar{n}) \) apply equally to both the SRK fluid and the Ideal Gas. Second, we have complete freedom to choose our reference volume, \( V_R \), however is most convenient.

If we define \( V_R \) to approach infinity (\( V_R \to \infty \)), we are simulating an infinitely expanded gas. Because any valid, useful equation of state must trend toward ideal gas behavior at infinite volume, we can safely conclude that as \( V_R \to \infty \), the real fluid properties converge with the ideal gas properties (\( A(T, V_R, \bar{n}) \to A^{IG}(T, V_R, \bar{n}) \)). Additionally, under this mathematical limit, terms like \( \frac{V_R}{V_R \pm nb} \) simply approach 1. 

Note: As \( V_R \to \infty \), the \( nb \) term becomes negligible compared to the infinitely expanding volume, causing the fraction to approach unity:

\[ \lim_{V_R \to \infty} \frac{V_R}{V_R \pm nb} = 1 \tag{11} \]

Alternatively, expressed as a simple directional trend:

\[ \frac{V_R}{V_R \pm nb} \to 1 \tag{12} \]

By applying this infinite-volume reference state, Equation 10 beautifully simplifies to:

\[ A(T, V, \bar{n}) - A^{IG}(T, V, \bar{n}) = -n R_g T \ln \left[ \frac{V - nb}{V} \right] + \frac{n^2 a}{nb} \ln \left[ \frac{V + nb}{V} \right] \tag{13} \]

The left-hand side of Equation 13 provides us with the formal definition of the Residual Helmholtz Free Energy:

\[ A^{Res} \equiv A(T, V, \bar{n}) - A^{IG}(T, V, \bar{n}) \tag{14} \]

It is crucial to understand the conceptual leap that led us to Equation 13, particularly how specific terms algebraically cancel out as our reference volume approaches infinity (\( V_R \to \infty \)). Furthermore, the core definition of any Residual Property strictly requires that both the real fluid and the Ideal Gas (IG) reference state share the exact same temperature (\( T \)), volume (\( V \)), and composition (\( \bar{n} \)). By substituting residual properties into our earlier definitions, we obtain the following relationships:

\[ \left( \frac{\partial A^{Res}}{\partial T} \right)_{V, \bar{n}} = -S^{Res} = -S(T, V, \bar{n}) + S^{IG}(T, V, \bar{n}) \tag{15} \]

\[ \left( \frac{\partial A^{Res}}{\partial V} \right)_{T, \bar{n}} = -P^{Res} = -P(T, V, \bar{n}) + \frac{n R_g T}{V} \tag{16} \]

\[ \left( \frac{\partial A^{Res}}{\partial n_i} \right)_{T, V, n_{k \neq i}} = \mu_i^{Res} = \mu_i(T, V, \bar{n}) - \mu_i^{IG}(T, V, \bar{n}) \tag{17} \]

However, Equations 15 through 17 do not magically solve all our problems. While we can differentiate Equation 13 with respect to \( T \), \( V \), or \( \bar{n} \) to generate residual properties, we still cannot determine the absolute property of the SRK fluid without knowing the corresponding property of the ideal gas.

Fortunately, we do know the pressure for both the ideal gas and the SRK fluid, which allows us to completely solve for the volume derivative, \( \left( \frac{\partial A^{Res}}{\partial V} \right)_{T, \bar{n}} \).

Regarding composition (\( \bar{n} \)), we will see in the next section that if we can successfully formulate the ideal gas chemical potential (\( \mu_i^{IG} \)), we can then derive a highly useful expression for the chemical potential (\( \mu_i \)) of the SRK fluid (and other equations of state).

Finally, when it comes to temperature (\( T \)), we will discover that for a massive category of thermodynamic calculations, we actually do not need to know the absolute ideal gas entropy. In practice, those terms frequently end up canceling out algebraically when we assemble the final equations.

The Chemical Potential of an Ideal Gas

Let's start from the canonical derivative relation for the ideal gas Helmholtz free energy (Equation 18):

\[ \left( \frac{\partial A^{IG}}{\partial V} \right)_{T, \bar{n}} = -P^{IG} \tag{18} \]

Substituting the ideal gas law, \( P^{IG} = \frac{n R_g T}{V} \) (Equation 2), yields:

\[ \left( \frac{\partial A^{IG}}{\partial V} \right)_{T, \bar{n}} = -\frac{n R_g T}{V} \tag{19} \]

Integrating this expression with respect to volume (\( V \)), while keeping temperature (\( T \)) and the composition vector (\( \bar{n} \)) constant, gives us:

\[ A^{IG} = \int \left( -\frac{n R_g T}{V} \right) dV = -n R_g T \ln(V) + C \tag{20} \]

In multivariable calculus, the "constant of integration" (\( C \)) is not simply a static number; it can depend on any variables that were deliberately held constant during the partial integration. Because \( T \) and \( \bar{n} \) were treated as constants, \( C \) is actually an arbitrary function, \( F(T, \bar{n}) \). This leaves us with:

\[ A^{IG} = -n R_g T \ln(V) + F(T, \bar{n}) \tag{21} \]

Derivation of Equation 30

Equation 30 resolves the structure of this unknown integration function \( F(T, \bar{n}) \) by leveraging Euler's theorem for homogeneous functions alongside the Legendre transform.

1. First Line of Equation 30: Euler Expansion of Helmholtz Energy

From classical thermodynamics, we know the fundamental potentials are related by:

\[ A \equiv U - TS = (H - PV) - TS = (H - TS) - PV = G - PV \tag{22} \]

Applying Euler's theorem for first-order extensive thermodynamic potentials tells us that the total Gibbs free energy is simply the sum of molar chemical potentials weighted by their respective mole numbers:

\[ G = \sum_r \mu_r n_r \tag{23} \]

By substituting this definition of \( G \) back into our relation \( A = G - PV \) specifically for an ideal gas mixture, we arrive at the first line of Equation 30:

\[ A^{IG} = -(PV)^{IG} + \sum_r \mu_r^{IG} n_r \tag{24} \]

2. Second Line of Equation 30: Matching the Volume and Composition Dependencies

Applying the ideal gas equation of state, we know that \( (PV)^{IG} = n R_g T \). Substituting this into our previous expression yields:

\[ A^{IG} = -n R_g T + \sum_r \mu_r^{IG} n_r \tag{25} \]

Recalling Equation 17, the chemical potential is defined as the partial derivative of the Helmholtz free energy with respect to the mole number:

\[ \mu_r^{IG} = \left( \frac{\partial A^{IG}}{\partial n_r} \right)_{T, V, n_{k \neq r}} \tag{26} \]

Now, if we differentiate Equation 21 with respect to \( n_r \)—keeping in mind that the total moles \( n = \sum_j n_j \), meaning \( \frac{\partial n}{\partial n_r} = 1 \)—we get:

\[ \mu_r^{IG} = -R_g T \ln(V) + \left( \frac{\partial F(T, \bar{n})}{\partial n_r} \right)_{T, n_{k \neq r}} \tag{27} \]

To simplify this, we can define a dimensionless component function, \( f_r(T, \bar{n}) \):

\[ f_r(T, \bar{n}) \equiv \frac{1}{R_g T} \left( \frac{\partial F}{\partial n_r} \right)_{T, n_{k \neq r}} \tag{28} \]

This allows us to rewrite the chemical potential cleanly as:

\[ \mu_r^{IG} = R_g T \left[ -\ln(V) + f_r(T, \bar{n}) \right] \tag{29} \]

Finally, substituting this chemical potential back into our earlier expression (\( A^{IG} = -(PV)^{IG} + \sum_r \mu_r^{IG} n_r \)) brings us to the complete Equation 30:

\[ A^{IG} = -n R_g T + R_g T \sum_r \left[ -\ln(V) + f_r(T, \bar{n}) \right] n_r \tag{30} \]

1. The Extensivity Requirement (1st Order Homogeneity)

Helmholtz free energy (\( A \)) is fundamentally an extensive property. This means that if we scale the entire size of our system by some factor \( \lambda > 0 \) while holding the temperature (\( T \)) constant, both the total volume (\( V \)) and the individual mole numbers (\( n_r \)) will scale by that exact same factor \( \lambda \):

\[ V \to \lambda V, \quad n_r \to \lambda n_r \]

For the ideal gas Helmholtz energy (\( A^{IG} \)) to scale proportionally alongside the system (\( A^{IG} \to \lambda A^{IG} \)), it must mathematically behave as a function that is homogeneous of degree 1 with respect to volume and composition \( (V, \bar{n}) \). 

Let's look back at our newly derived Equation 30 to see how this applies:

\[ A^{IG} = -n R_g T + R_g T \sum_r \left[ -\ln(V) + f_r(T, \bar{n}) \right] n_r \]

The leading term, \( -n R_g T \), is naturally first-order homogeneous because the total moles scale directly with \( \lambda \) (where \( n = \sum n_r \), meaning \( n \to \lambda n \)).

Inside the summation, the trailing factor \( n_r \) also scales perfectly as \( \lambda^1 \).

Because the \( n_r \) outside the brackets already satisfies the scaling requirement, the entire bracketed quantity, \( \left[ -\ln(V) + f_r(T, \bar{n}) \right] \), must be homogeneous of degree 0. In other words, it must act as an intensive quantity that remains completely unchanged when the system is scaled by \( \lambda \):

\[ -\ln(\lambda V) + f_r(T, \lambda \bar{n}) = -\ln(V) + f_r(T, \bar{n}) \tag{31} \]

By expanding the logarithm on the left side of that relationship, we can isolate the function \( f_r \) to understand exactly how it must behave:

\[ -\ln(\lambda) - \ln(V) + f_r(T, \lambda \bar{n}) = -\ln(V) + f_r(T, \bar{n}) \tag{32} \]

\[ f_r(T, \lambda \bar{n}) - f_r(T, \bar{n}) = \ln(\lambda) \tag{33} \]

2. Derivation of Equation 37 via Euler's Theorem

Euler's homogeneous function theorem states that for any function \( g(x_1, x_2, \dots, x_m) \) that is homogeneous of degree \( k \), the following relationship holds true:

\[ \sum_j x_j \frac{\partial g}{\partial x_j} = k \cdot g \tag{34} \]

We can apply this theorem directly to our bracketed function, \( g(V, \bar{n}) = -\ln(V) + f_i(T, \bar{n}) \), which we just established is homogeneous of degree zero (\( k = 0 \)):

\[ V \frac{\partial \left[ -\ln(V) + f_i \right]}{\partial V} + \sum_r n_r \frac{\partial \left[ -\ln(V) + f_i \right]}{\partial n_r} = 0 \cdot \left[ -\ln(V) + f_i \right] = 0 \tag{35} \]

Next, let's quickly evaluate those two partial derivatives individually:

1. With respect to volume: \( \frac{\partial \left[ -\ln(V) + f_i \right]}{\partial V} = -\frac{1}{V} \)

2. With respect to mole numbers: \( \frac{\partial \left[ -\ln(V) + f_i \right]}{\partial n_r} = \frac{\partial f_i}{\partial n_r} \)

Substituting these evaluated derivatives back into our Euler relation gives us the first line of Equation 37:

\[ -\frac{1}{V} V + \sum_r \frac{\partial f_i}{\partial n_r} n_r = 0 \tag{36} \]

Finally, by simplifying the first term (\( -\frac{1}{V} V = -1 \)) and moving that negative one to the right-hand side of the equation, we arrive perfectly at Equation 37:

\[ \sum_r \frac{\partial f_i}{\partial n_r} n_r = 1 \tag{37} \]

1. Understanding the Behavior of \( f_i \)

Equation 37 reveals a couple of fascinating mathematical quirks about our component function, \( f_i \):

\( f_i \) is not homogeneous in \( \bar{n} \): Because the right-hand side of Equation 37 equals 1 rather than \( k \cdot f_i \), the function itself does not follow standard homogeneity rules.

Universal sum to 1: For that sum to reliably equal 1 regardless of the mixture's actual composition, the partial derivative must perfectly isolate the specific component's mole number. This means:

\[ \frac{\partial f_i}{\partial n_i} = \frac{1}{n_i} \quad \text{and} \quad \frac{\partial f_i}{\partial n_r} = 0 \quad (r \neq i) \tag{38} \]

If we take that relationship and directly integrate \( \frac{1}{n_i} dn_i \), we arrive at Equation 39:

\[ f_i = \ln(n_i) + k_i(T) \tag{39} \]

In this expression, \( k_i(T) \) serves as an integration constant that is exclusively dependent on the temperature, \( T \).

2. The Resulting Chemical Potential (Equation 40)

We are finally ready to put the pieces together. By substituting our newly derived \( f_i = \ln(n_i) + k_i \) back into the ideal gas chemical potential expression we built earlier (\( \mu_i^{IG} = R_g T \left[ -\ln(V) + f_i \right] \)), we get Equation 40:

\[ \mu_i^{IG} = -R_g T \ln(V) + R_g T \ln(n_i) + R_g T k_i \tag{40} \]

By algebraically grouping the volume and mole terms together, the underlying physics finally become visible. This arrangement reveals a direct, tangible connection to the molar concentration (\( c_i = n_i / V \)) or the partial pressure (\( P_i = y_i P \)):

\[ \mu_i^{IG} = R_g T \ln\left(\frac{n_i}{V}\right) + R_g T k_i(T) \tag{41} \]

Determination of \( k_i \):

1. Matching \( \mu_i^{IG} \) to the Pure Component Limit

In Equation 40, the ideal gas chemical potential is given by:

\[ \mu_i^{IG} = -R_g T \ln(V) + R_g T \ln(n_i) + R_g T k_i = R_g T \ln\left(\frac{n_i}{V}\right) + R_g T k_i \tag{42} \]

For a pure substance (\( n_i = n \)), the chemical potential is identical to the molar Gibbs free energy:

\[ \mu_i^{IG} = g_i^{IG} \tag{43} \]

Using the ideal gas relation (\( V = \frac{n R_g T}{P} \)), the molar concentration becomes \( \frac{n_i}{V} = \frac{P}{R_g T} \). Substituting this into Equation 42 gives us an expression for the pure ideal gas:

\[ g_i^{IG}(T, P) = R_g T \ln\left(\frac{P}{R_g T}\right) + R_g T k_i = R_g T \ln(P) - R_g T \ln(R_g T) + R_g T k_i \tag{44} \]

2. Physical Origin of \( g_i^\circ(T) \) and Equation 48

To assign an actual numerical value to \( g_i^{IG}(T, P) \), thermodynamics anchors the property to standard formation data:

Reference State: Data for standard Gibbs energy of formation (\( \Delta g_{f, i}^\circ \)) or standard enthalpy of formation (\( \Delta h_{f, i}^\circ \)) and absolute entropy (\( s_i^\circ \)) are tabulated at \( T_0 = 298.15\text{ K} \) and \( P_0 = 1\text{ bar} \).

Temperature Integration: Using pure-component ideal-gas heat capacity polynomials (\( C_{p, i}^0(T) \)), the enthalpy and entropy are integrated from \( T_0 \) to the system temperature \( T \):

\[ h_i^\circ(T) = \Delta h_{f, i}^\circ(T_0) + \int_{T_0}^T C_{p, i}^0(T') \, dT' \tag{45} \]

\[ s_i^\circ(T) = s_i^\circ(T_0, P_0) + \int_{T_0}^T \frac{C_{p, i}^0(T')}{T'} \, dT' \tag{46} \]

\[ g_i^\circ(T) \equiv h_i^\circ(T) - T s_i^\circ(T) = g_i(T, P_0) \tag{47} \]

Pressure Isotherm (Equation 48): At a constant temperature, the change in Gibbs free energy with respect to pressure is given by \( dg_i = v \, dP = \frac{R_g T}{P} \, dP \). Integrating this from the standard pressure \( P_0 \) to the system pressure \( P \) gives:

\[ g_i^{IG}(T, P) = g_i(T, P_0) + R_g T \ln\left(\frac{P}{P_0}\right) = g_i^\circ(T) + R_g T \ln(P) - R_g T \ln(P_0) \tag{48} \]

3. Solving Explicitly for \( k_i(T) \)

We now have two different expressions for \( g_i^{IG}(T, P) \). Equating them directly isolates our unknown integration constant, \( k_i(T) \):

\[ R_g T \ln(P) - R_g T \ln(R_g T) + R_g T k_i = g_i^\circ(T) + R_g T \ln(P) - R_g T \ln(P_0) \tag{49} \]

Notice that the \( R_g T \ln(P) \) term beautifully cancels out on both sides:

\[ -R_g T \ln(R_g T) + R_g T k_i = g_i^\circ(T) - R_g T \ln(P_0) \tag{50} \]

Finally, by dividing everything through by \( R_g T \), we arrive at our explicit solution for the temperature-dependent integration constant:

\[ k_i(T) = \frac{g_i^\circ(T)}{R_g T} + \ln\left(\frac{R_g T}{P_0}\right) \tag{51} \]

Let's start by revisiting Equation 40 and factoring out the \( R_g T \) term:

\[ \mu_i^{IG} = -R_g T \ln(V) + R_g T \ln(n_i) + R_g T k_i \tag{52} \]

\[ \mu_i^{IG} = R_g T \left[ \ln\left(\frac{n_i}{V}\right) + k_i(T) \right] \tag{53} \]

From our previous derivation (Equation 51), we solved for the explicit expression of \( k_i(T) \):

\[ k_i(T) = \frac{g_i^\circ(T)}{R_g T} + \ln(R_g T) - \ln(P_0) \tag{54} \]

Step 2: Substitution of \( k_i(T) \)

Now, we substitute that expression for \( k_i(T) \) directly into our bracketed term:

\[ \mu_i^{IG} = R_g T \left[ \ln\left(\frac{n_i}{V}\right) + \frac{g_i^\circ(T)}{R_g T} + \ln(R_g T) - \ln(P_0) \right] \tag{55} \]

By distributing \( R_g T \) back through the equation, we can pull the standard Gibbs energy out front:

\[ \mu_i^{IG} = g_i^\circ(T) + R_g T \left[ \ln\left(\frac{n_i}{V}\right) + \ln(R_g T) - \ln(P_0) \right] \tag{56} \]

Step 3: Combining the Logarithmic Terms

Using standard logarithmic identities (\( \ln(x) + \ln(y) = \ln(xy) \) and \( \ln(a) - \ln(b) = \ln(a/b) \)), we can compress the terms inside the bracket:

\[ \ln\left(\frac{n_i}{V}\right) + \ln(R_g T) - \ln(P_0) = \ln\left( \frac{n_i R_g T}{V \cdot P_0} \right) \tag{57} \]

Next, we express the total mixture pressure using the ideal gas law (\( P = \frac{n R_g T}{V} \)), which rearranges to \( \frac{R_g T}{V} = \frac{P}{n} \). Substituting this into our logarithmic argument simplifies it beautifully:

\[ \frac{n_i R_g T}{V \cdot P_0} = \frac{n_i}{n} \cdot \frac{P}{P_0} \tag{59} \]

Step 4: The Final Form of Equation 61

Let \( c_i \equiv y_i = \frac{n_i}{n} \) represent the mole fraction of species \( i \) in our ideal gas mixture. We can split the product inside the logarithm one last time:

\[ \ln\left( \frac{P}{P_0} \cdot c_i \right) = \ln\left(\frac{P}{P_0}\right) + \ln(c_i) \tag{60} \]

Multiplying this by \( R_g T \) and adding it back to \( g_i^\circ(T) \) yields the final, highly useful Equation 61:

\[ \mu_i^{IG}(T, P, \bar{n}) = g_i^\circ(T) + R_g T \ln\left(\frac{P}{P_0}\right) + R_g T \ln(c_i) \tag{61} \]

Physical Interpretation & Real Fluid Extension

Equation 61 is incredibly elegant because every single term tells a specific physical story about the gas:

Reference State (\( g_i^\circ \)): This captures the pure substance's caloric properties at temperature \( T \) and standard pressure \( P_0 \) (typically 1 bar).

Isothermal Pressure Effect (\( R_g T \ln(P/P_0) \)): This mathematically compresses or expands the pure ideal gas isothermally from standard pressure to the actual system pressure \( P \).

Ideal Mixing Effect (\( R_g T \ln c_i \)): This accounts for the entropy of mixing at constant temperature and pressure, representing Dalton's partial pressure contribution.

For a real fluid, we simply add the residual chemical potential (\( \mu_i^{Res} = \left(\frac{\partial A^{Res}}{\partial n_i}\right)_{T, V, n_{k \neq i}} \)) to define the total chemical potential:

\[ \mu_i = \mu_i^{IG} + \mu_i^{Res} = g_i^\circ(T) + R_g T \ln\left(\frac{P}{P_0}\right) + R_g T \ln(c_i) + \mu_i^{Res} \tag{62} \]

Because the chemical potential of a real component is formally defined via its fugacity (\( \hat{f}_i \)) as:

\[ \mu_i \equiv g_i^\circ(T) + R_g T \ln\left(\frac{\hat{f}_i}{P_0}\right) \tag{63} \]

We can equate the two forms. By defining the fugacity coefficient as \( \phi_i \equiv \frac{\hat{f}_i}{c_i P} \), we directly recover the fundamental link between residual properties and phase equilibrium:

\[ \ln \phi_i = \frac{\mu_i^{Res}}{R_g T} - \ln Z \tag{64} \]

(Note: The \( \ln Z \) term, where compressibility factor \( Z = \frac{PV}{n R_g T} \), acts as the required mathematical adjustment for the coordinate transformation from the \( (T, V) \) Helmholtz domain over to the \( (T, P) \) Gibbs domain).