Abstract

The paper presents the essential information for understanding the basic chemical thermodynamics for chemistry majors. The connection among the thermodynamic potentials, the thermodynamic potentials with their natural variables and the tools for navigating the Potential energy-Properties pairs are also presented. Several illustrations and examples to ease the understanding of this topic abound in this short paper.

Keywords:Laws of Thermodynamics, Thermodynamic potential Functions, System, Equilibrium State, Internal Energy, Enthalpy, Helmholtz free energy, Gibbs free energy, Equations of State, State functions, Chemical Potentials, Partial molar Energies, and Maxwell Relations.

Introduction

Thermodynamics (and particularly Chemical thermodynamics) is not a novel subject. However, innovation into the ways and means of presenting the subject to a first time physical chemistry student can be made interesting and captivating. The present day undergraduate chemistry students consider chemistry a very volatile subject. Regardless of how many times they read the topics it never registers into their faculty. By relating the various topics on this subject to objects that the students can visualize would help them to understand the subject a lot more. Moreover, simplifying the various technical features in the subject and supplementing with many examples could appeal to the students and present a good understanding and mastery of the subject.

After many years of teaching chemical thermodynamics, I have found an easier way to introduce the forest of equations in this subject. Every student in the physical chemistry class appears lost when we first introduce the equations of state, Maxwell relations, and the relationships among these thermodynamic equations. Students find it very difficult to connect the application of these equations to physical situations. For example, when we introduce the state functions, we emphasize that each function has a total differential and that a state function does not depend on the path taken between the initial and the final state. When we present an equation of state for the students and ask them to apply the definition of state function, there is a disconnection. For this reason, we should begin with the very basic equation of state. Equation of state has had enormous number of reviews and varying models [6-12] since 1873 when Johannes Diderik van der Waals (the 1910 Nobel Prize winner in Physics on the equation of states for gases and liquids) [2] defended his PhD thesis: On the Continuity of the Gaseous and Liquid States, Leiden University Netherlands [1-5].

State function

A state function f with two independent variables x and y can be represented as f(x, y) [13, 14]. We emphasize that the function must satisfy the following three equations (1a), (2a) and (3a):

Equation (1a) defines the state function, equation (2a) emphasizes that a state function f has a total (or exact) differential, and equation (3a) emphasizes that a state function does not depend on the path chosen between the initial state and the final state. Equation (3a) indicates that differentiating f with respect to x first and differentiating the result with respect to y will give the same result as differentiating f with respect to y first and differentiating the result with respect to x. The order of differentiation does not matter. For example, if we apply this definition to a fixed mole ideal gas equation of state, PV=RT, and make volume a function of temperature and pressure, then the equation becomes V = RT/P. This equation defines volume in terms of temperature and pressure; here R is the gas constant. The total or exact differential of V is carried out by first differentiating V with respect to T, holding P constant, and then adding the result to the differentiation of V with respect to P, holding T constant. Finally we carry out the path independence of a state function by differentiating the quantity (𝜕𝑉𝜕𝑇)𝑃 with respect to P to obtain 𝜕𝜕𝑃(𝜕𝑉𝜕𝑇)𝑃 and repeat the same by differentiating (𝜕𝑉𝜕𝑃)𝑇 with respect to T to obtain 𝜕𝜕𝑇(𝜕𝑉𝜕𝑃)𝑇. We present below the overall derivation for a fixed amount of gas that obeys the ideal gas equation.

𝑉 = (𝑇,) 𝑜𝑟 𝑉 = 𝑅𝑇𝑃 (1b)

d𝑉 = (∂𝑉∂𝑇)𝑃∂𝑇 + (∂𝑉∂𝑃)𝑇∂𝑃 = (𝑅𝑃) d𝑇 − (𝑅𝑇𝑃2 ) d𝑃 (2b)

∂𝜕𝑃 (𝜕𝑉𝜕𝑇)= − (𝑅𝑃2) =∂𝜕𝑇 (𝜕𝑉𝜕𝑃)𝑇= − (𝑅𝑃2) (3b)

We have demonstrated that ∂𝜕𝑃 (𝜕𝑉𝜕𝑇)=∂𝜕𝑇 (𝜕𝑉𝜕𝑃)𝑇= − (𝑅𝑃2), and for this reason, V is a state function,

particularly from equation (3b). Similarly, pressure P, is a state function by the analysis displayed in equations (1c), (2c) and (3c)

𝑃 = 𝑃(𝑇,𝑉) 𝑜𝑟 𝑃 =𝑅𝑇𝑉 (1c)

𝑑𝑃 = (𝑅𝑉)𝑑𝑇 − (𝑅𝑇𝑉2) 𝑑𝑉 (2c)

∂∂V (∂P∂T)V= ∂∂T (∂P∂V)T = − (RP2) (3c)

Temperature T, is also a state function by the same analysis and are displayed in equations (1d), (2d) and (3d).
T = T(V,P) or T =PVR (1d)
dT = (PR) dV + (VR) dP (2d)
∂∂P (∂T∂V)P= ∂∂V (∂T∂P)V =1R (3d)

This concept can be extended to other equations of state such as P(V−nb) = nRT, and the van der Waals equation of state.

First and Second Law of Thermodynamics

In the treatment that follows, we shall restrict our discussion to closed systems [13-14]. A closed system does not allow matter to enter or leave the system.

First Law of Thermodynamics

The first law of thermodynamics emphasizes the energy conversation law. The law relates the change of internal energy (ΔU) to the heat (q) added to the system and the work (w) done on the system. The law states that the change in internal energy of an isolated system is equal to the sum of heat added and the work done on the system [13].

ΔU = q + w (4)

The symbol w refers to all the work done on the system; it is the sum of gravitational, mechanical, surface, electrical, chemical work, etc., and so is the heat. For convenience we use just the mechanical pressure-volume work, and for this reason the first law equation is usually written as:

ΔU = q – PdV (5)

In this equation the pressure-volume work is the work done by the gas (in this case the system) on the surroundings. The negative sign in equation (5) indicates that the system does some work on the surroundings as a result of heat added to the system. In some cases we want to know the internal energy of a system when we hold the volume constant. For this we represent ΔU = qv, because when we keep the volume constant then there is no mechanical or pressure-volume work done by the system. qv represents the heat added to the system when volume is constant. The change in internal energy at constant volume raises the temperature of the system. In chemistry, we introduce enthalpy because we carry out several reactions at constant pressure. Enthalpy is the heat added to the system when we hold pressure constant. For this case we introduce enthalpy as yet another energy term, and we represent the enthalpy equivalent of the first law as:

ΔU = qp – PdV (6)

qp is the heat at constant pressure, and we give it the symbol ΔH. When we rearrange equation (6) we obtain

qp= ΔH = ΔU + PdV

We may represent equation (6) in differential form as:

dH = dU + PdV (7)

The internal energy U and the enthalpy H are state functions and they describe the equilibrium state of the system [13, 14]. These functions in turn describe the state variables like volume, pressure and temperature that depend on current equilibrium state of a system. The sum of w and q is a state function; however, neither q nor w is a state function. A state function depends only on the initial and the final state of the system. Work energy moves atoms or molecules in a uniform manner, but heat energy moves atoms or molecules in a disorganized or chaotic manner. Heat is a disordered form of energy, and when it is released it makes the universe more random (or disordered). While we can write a total differential for U we cannot write a total differential for either q or w because they depend on the path taken between the initial and the final state of the system. We shall illustrate this idea using the following examples.

Calculations to illustrate state functions and non-state functions

Calculate ΔU, ΔH, q, and w for each step and the entire cycle when one mole of neon gas initially at 1.0 bar pressure and a volume of 25.0 liters (call this point A) is slowly heated at constant volume until the pressure is tripled (call this point B). Next the gas is allowed to expand reversibly and at constant temperature until the pressure decreases to 0.5 bar (call this point C). The gas further undergoes reversible adiabatic compression to return to the initial state of 1.0 bar pressure and a volume of 25.0 liters. We observe that the gas undergoes some heating during the isochoric process from the initial state to the second step, and the temperature triples because under this condition pressure is directly proportional to the temperature. There is no pressure-volume work done at this step because there is no change in volume. For this reason, the internal energy is equal to the heat added to the gas. We call this path I. The isothermal expansion second step, the gas experiences some cooling and the temperature returns to the initial temperature of 300.7K. There is no change in the internal energy because there is no change in temperature. The work done by the gas is equal to the heat flow. We call this step path II. The last step returns the system to the initial state. This path III involves no heat flow whatsoever, and the work done on the gas is equal to the internal energy. We summarize the calculations in the three steps given below.

a) Path I: Constant Volume (or Isochoric ) step A to B
w = −p ΔV = 0 kJ and Δ𝑈=𝑞 Because ΔV = 0 and ΔU= q − p ΔV ΔU = nCvΔT= n(32)R ΔT,where 𝐶𝑣=32𝑅 ΔU= (1.0 mol)(32)(8.3145 JK−1mol−1)(902 − 301)K = 7.5 kJ q = ΔU = 7.5 kJ
ΔH = nCp ΔT =n(52)R ΔT, where Cp=52R
ΔH=(1.0 mol)(52) (8.3145 JK−1mol−1)(902 − 301)K = 12.5 kJ
b) Path II: Constant Temperature (or Isothermal) expansion step B to C
ΔU = q − p ΔV, and ΔU = nCv ΔT = 0 kJ Because ΔT = 0 .
w = −nRT ln(V3V2)= − (1.0 mol)(8.3145 JK−1mol−1)(301)K ln(15025)= − 4.5 kJ 𝑞=𝑝Δ𝑉 = − w = 4.5 kJ ΔH = nCpΔT = 0, because ΔT = 0.
c) Path III: Constant heat (or Adiabatic) compression step C to A
ΔU = q −p ΔV, ΔU = nCv ΔT Because q = 0 w =35(P1V1−P3V3) = − 7.5kJ ΔU=nCv ΔT=n(32)R ΔT= (1.0 mol)(32)(8.3145 JK−1mol−1)(301 − 902)K = −7.5 kJ
ΔH = nCp ΔT = n(52)R ΔT=(1.0 mol)(52)(8.3145 JK−1mol−1)(301−902)K=−12.5 kJ

Figure 1: Pressure-Volume plot of the three-step cycle (A→B→C→A)

Figure 2: Pressure-Volume plot of the four-step cycle (A→B→C→D→A) problem

Figure 3: TAVUS-HPG Memory diagram representation of the Thermodynamic Potentials (A, U, H, G) and thermodynamic Properties (T, V, S, P).

d) Paths (I + II + III) : Overall Cycle (ABCA)
ΔUcycle = ΔUI + ΔUII + ΔUIII = (7.5 + 0 −7.5) kJ =0 kJ ΔHcycle = ΔHI + ΔHII + ΔHIII = (12.5 + 0 −12.5)kJ =0 kJ qcycle = qI + qII + qIII = (7.5 + 4.5 + 0) kJ = 12 kJ wcycle = wI + wII + wIII = (0 − 4.5 − 7.5) kJ = −12 kJ
In this exercise, we notice that the total change in internal energy ΔU and the total change in enthalpy ΔH are each equal to zero. U and H are state functions and the value for the overall cycle is zero for each of them. The cycle in this case is that we start from a point and go round to return to the same point. On the other hand, the values for q and w are not equal to zero for the same cycle because they are not state functions. However, the sum of these values: q plus w is a state function, as we can see. In the above exercise and for neon (a monoatomic gas) we have used the values 1.5R and 2.5R for Cv and Cp respectively. Cv is the heat capacity at constant volume and Cp the heat capacity at constant pressure.
Let us repeat the above calculations by going through a different route to return to the initial state as follow: heat one mole neon gas initially at 1.0 bar and a volume of 25.0 liters (call this point A) at constant volume until the pressure is tripled (call this point B). Next we heat the gas at constant pressure until the volume is six times the initial volume (call this point C). For the next step we cool the gas at constant volume until the pressure is decreased to 1.0 bar (call this point D) and finally compress the gas to return to the initial state (point A).
There are four steps in this second exercise. The first step is an isochoric heating process; the second is isobaric expansion (another heating process); the third step is isochoric cooling; and the fourth step is isobaric compression.
a) Path I: Isochoric heating step A to B
w = −p ΔV = 0 kJ Because ΔV = 0
ΔU=nCvΔT= n(32)R ΔT= (1.0 mol)(32)(8.3145 JK−1mol−1)(902−301)K=7.5 kJ q = ΔU = 7.5 kJ ΔH = nCp ΔT =n(52)R ΔT=(1.0 mol)(52) (8.3145 JK−1mol−1)(902−301)K = 12.5 kJ
b) Path II: Constant Pressure (or Isobaric) expansion step B to C
w= −p ΔV = − (1.0 bar)×(150.0 – 25.0)L× (100) JL−1bar−1 = −12.5 kJ ΔU=nCVΔT= n(32)R ΔT=(1.0 mol)(32)(8.3145 JK−1mol−1)(5412−902)K = 56.25 kJ q=nCpΔT=n(52)R ΔT=(1.0 mol)(52)(8.3145 JK−1mol−1)(5412−902)K = 93.75 kJ ΔH=nCpΔT=n(52)R ΔT=(1.0 mol)(52)(8.3145 JK−1mol−1)(5412−902)K = 93.75 kJ
c) Path III: Isochoric cooling step C to D
w = −p ΔV = 0 kJ Because ΔV = 0 ΔU=nCvΔT= n(32)R ΔT=(1.0 mol)(32)(8.3145
JK−1mol−1)(1804−5412)K=−45.0kJ
q = ΔU = −45.0 kJ ΔH=nCp ΔT=n(52)R ΔT=(1.0 mol)(52)(8.3145 JK−1mol−1)(1804−5412)=−75.0 kJ
d) Path IV: Isobaric compression step D to A
w = −p ΔV= (1.0 bar)(25.0 – 150.0)L (100)JL−1bar−1 = −12.5 kJ
ΔU =nCVΔT = n(32)R ΔT=(1.0 mol)(1.5) (8.3145 JK−1mol−1)(301−1804)K = −18.75kJ q = nCp ΔT =n(52)R ΔT =(1.0 mol)(2.5) (8.3145 JK−1mol−1)(301−1804)K = − 31.25 kJ ΔH= nCp ΔT =n(52)R ΔT =(1.0 mol)(2.5) (8.3145 JK−1mol−1)(301−1804)K = − 31.25 kJ

e) Paths (I + II + III + IV): Overall Cycle (ABCDA) ΔUcycle = ΔUI+ΔUII + ΔUIII + ΔUIV = (7.5 + 56.25 – 45.0 – 18.75)kJ= 0 kJ ΔHcycle= ΔHI+ ΔHII+ΔHIII+ΔHIV= (12.5 + 93.75 – 75.0 −31.25) kJ = 0 kJ Δqcycle= ΔqI+ ΔqII+ΔqIII+ΔqIV= (7.5+93.75 −4 5.0 −31.25 ) kJ = 25.0 kJ Δwcycle= ΔwI+ ΔwII+ΔwIII+ΔwIV= (0 −12.5 + 0 −12.5) kJ = − 25.0 kJ

If we make a pressure-volume plot of the second exercise we would realize that work done is actually the area under plot. This is a rectangular area with the height equal to 2.0 bar and the length is 125.0 L. This gives an overall work for the cycle a 25.0 kJ value. The sign is negative because the gas does work on the surroundings. Also, if we compare the two exercises, we see that both ΔU and ΔH have overall values of 0 kJ for the cycle even though we took the gas through different paths. This is because in both cases we started from the same initial point, returned to the point, and U and H are state functions. On the other hand w and q are not state functions. They depend on the path taken for the cycle and since the paths are different w and q give different values for the cycles. However, w + q is a state function and that summed value is 0 kJ. Some thermodynamic texts attempt to write the total differential for the first law of thermodynamics to include heat and work. In order to do this they introduce “δ” instead of “d” to indicate that q and w are not state functions. They indeed depend on the path taken between the initial state and the final state as we confirmed this in the last calculations.

dU = δq + δw (4)

Second Law of Thermodynamics
The second law tells us that we cannot break even in the energy calculus. We cannot break even because absolute zero temperature is unattainable [13, 14]. For example, if we put a dollar’s worth of octane gas into an automobile, we will not obtain a dollar’s worth of work in return. One interpretation of the second law is that there is no machine that is one hundred percent efficient. When energy is in a form other than thermal energy, it can be transformed into another form of energy with good or excellent efficiency. The second law also implies that work can be converted totally to heat, but heat cannot be transformed totally to work. The second law of thermodynamics limits the efficiency of thermal energy. This law is usually interpreted in terms of entropy. The second law of thermodynamics states: “the entropy of the Universe always increases in a spontaneous process”.
The second law may be manipulated to the relate heat to the change in entropy.
q = TdS
The combined first and second laws become
dU = TdS − PdV (8)

The internal energy U(S,V), is a function of the entropy and volume, and it is a state function. We can apply the definition of state function to U by choosing the properties S and V. This equation defines U in terms of entropy and volume. The total or exact differential of U is carried out first by differentiating U with respect to S and holding V constant, and second by differentiating U with respect to V and holding S constant. Finally we differentiate the quantity (∂U/∂S)P with respect to V to get d/∂V (∂U/∂S)P and repeat the same process by differentiating (∂U/∂V)T with respect to S to obtain d/∂S (∂U/∂V)T. We present the entire process as before in the equations below.
U = U(S,V) (8a)
dU = (∂U∂S)V∂S + (∂U∂V)S∂V (8b)
∂∂V (∂U∂S)V = ∂∂S (∂U∂V)S (8c)
When we apply the definitions in equation (2a) we see that (∂U/∂S)V = T and (∂U/∂V)S = -V, and this is in agreement with the combined first and second laws (equation (8) above.

The equivalent equations for the enthalpy are:
dH = TdS + VdP (9)
H = H(S,P) (9a)
dH = (∂H∂S)P∂S + (∂H∂P)S∂P (9b)
∂∂P (∂H∂S)P = ∂∂S (∂H∂P)S (9c)
Two remaining thermodynamic potentials are the Helmholtz and Gibbs free energy functions. It is important to note that the equation of state U(S, V) for a closed system is not convenient to work with. This is because we do not usually carry out experiments at constant heat condition. From the second law of thermodynamics we see that heat is related to entropy; hence the internal energy function is represented as U(S,V). In many cases we carry out experiments at constant temperature by allowing heat exchange between the system and the thermostat at some temperature T. The Helmholtz function is an important consideration when keeping the temperature constant instead of the heat in an experiment. In order to allow a variation of temperature we switch to the Helmholtz and Gibbs free energy equations of state [9, 10]. They are represented respectively as A(T,V) and G(T,P) and defined as:
dA = − SdT − PdV (10)
dG = − SdT + VdP (11)

Physical chemistry students are confused after wading through these equations. We can make correlations among these equations and simplify the relationship by arranging the thermodynamic potentials, U, A, G and H, as well as the properties V, S, P and T, in a square diagram. Other texts used E (for the internal energy) and F (for the Helmholtz free energy) instead of U and A respectively. We place the thermodynamic potentials at the centers of each side of the square and place the properties at the corners of the square according to the TAVUS-HPG diagram presented below. TAVUS-HPG label starts with the property T at the top right corner of the square then the thermodynamic potential A at the top center and alternates between corners and centers as we go round the square in a counter-clockwise manner.

Next, we draw two diagonal arrows to connect the paired properties V, P and T, S. The tail of the first arrow starts from P and the head points to V. The second arrow connects S to T with the head of the arrow pointing to T. In this orientation we interpret the head of the arrow as the positive end and the tail the negative end. We can write several thermodynamic potential dependencies, thermodynamic partial differential equations, and the thermodynamic equations easily with this simple TAVUS-HPG diagram. In the vertical arrangement V and S sandwich the potential function U; we interpret this as U is a function of S and V, and we write U(S, V). Many texts interpret this as: the natural variables of U are S and V. Similarly, we write H(S, P), A(T, V) and G(T, P). We can derive several differential relations such as
(∂U∂S)V = T (8d)

We interpret this as the derivative of U with respect to S holding V constant is equal to “+T” because the arrow stretches from S and points to T. Similarly, when we differentiate U with respect to V and hold S constant, we obtain “–P” because this is the tail of the arrow.
(∂U∂V)S = −P (8e)

We can derive similar relations for the thermodynamic potentials: H, A, and G; the results are:
(∂H∂S)P = T (9d)
(∂H∂P)S= V (9e)
(∂A∂V)T = − P (10d)
(∂A∂T)V = −S (10e)
(∂G∂P)T = V (11d)
(∂G∂T)P = − S (11e)
Since U, H, A and G are state functions we can write the total or exact differentials for these functions. For the thermodynamic potential U the total differential is: 𝑑𝑈 = (𝜕𝑈𝜕𝑆)𝑉𝑑𝑆 + (𝜕𝑈𝜕𝑉)𝑆𝑑𝑉 (8b)
We recognize that the partial differentials (∂U/∂S)V and (∂U/∂V) are T and –P respectively from the TAVUS–HPG diagram. When we substitute these differentials into equation (8b), which is the combined first and second law, we obtain the expected dU equation (8).
𝑑𝑈 = 𝑇𝑑𝑆 – 𝑃𝑑𝑉 (8)
We can write similar equations for the thermodynamic potentials H, A, and G.

Using the TAVUS-HPG diagram we can write down all the thermodynamic potentials with their natural variables, write down all the derivatives of the thermodynamic potentials with respect to each natural variables, write down the total differential functions as well as the thermodynamic potential equations. We present these varieties below.

Thermodynamic potentials with their natural variables

𝑈(𝑆,𝑉); 𝐻(𝑆,𝑃); 𝐴(𝑇,𝑉) 𝑎𝑛𝑑 𝐺(𝑇,𝑃)

Derivatives of each thermodynamic potential: U, H, A, and G with respect to the natural variables

T, V, S, and P
(𝜕𝑈𝜕𝑆)𝑉 = 𝑇 (8d)
(𝜕𝑈𝜕𝑉)𝑆 = − 𝑃 (8e)
(𝜕𝐻𝜕𝑆)𝑃 = 𝑇 (9d)
(𝜕𝐻𝜕𝑃)𝑆= 𝑉 (9e)
(𝜕𝐴𝜕𝑉)𝑇 = − 𝑃 (10d)
(𝜕𝐴𝜕𝑇)𝑉= − 𝑆 (10e)
(𝜕𝐺𝜕𝑃)𝑇 = 𝑉 (11d)
(𝜕𝐺𝜕𝑇)𝑃 =−𝑆 (11e)

The total differentials of the thermodynamic potential equations

𝑑𝑈 = (𝜕𝑈𝜕𝑆) + (𝜕𝑈𝜕𝑉)𝑑𝑉 (8b)
𝑑𝐻 = (𝜕𝐻𝜕𝑆)𝑃𝑑𝑆 + (𝜕𝐻𝜕𝑃)𝑆𝑑𝑃 (9b)
𝑑𝐴 = (𝜕𝐴𝜕𝑇)𝑉𝑑𝑇 + (𝜕𝐴𝜕𝑉)𝑇𝑑𝑉 (10b)
𝑑𝐺 = (𝜕𝐺𝜕𝑇)𝑃𝑑𝑇 + (𝜕𝐺𝜕𝑃)𝑇𝑑𝑃 (11b)

The thermodynamic potential equations

𝑑𝑈 = 𝑇𝑑𝑆 – 𝑃𝑑𝑉 (8)
𝑑𝐻 = 𝑇𝑑𝑆 + 𝑉𝑑𝑃 (9)
𝑑𝐴 = − 𝑆𝑑𝑇 – 𝑃𝑑𝑉 (10)
𝑑𝐺 = − 𝑆𝑑𝑇 + 𝑉𝑑𝑃 (11)

Path independence relationships of the thermodynamic potential functions leading to Partial Differential Quotients called Maxwell Relations

𝜕𝜕𝑉 (𝜕𝑈𝜕𝑆)𝑉 = 𝜕𝜕𝑆 (𝜕𝑈𝜕𝑉)𝑆 𝑖𝑚𝑝𝑙𝑖𝑒𝑠 𝑡ℎ𝑎𝑡 (𝜕𝑇𝜕𝑉)𝑆 = − (𝜕𝑃𝜕𝑆)𝑉 𝜕𝜕𝑃(𝜕𝐻𝜕𝑆)𝑃 = 𝜕𝜕𝑆(𝜕𝐻𝜕𝑃)𝑆 𝑖𝑚𝑝𝑙𝑖𝑒𝑠 𝑡ℎ𝑎𝑡 (𝜕𝑇𝜕𝑃)𝑆 = (𝜕𝑉𝜕𝑆)𝑃 𝜕𝜕𝑉(𝜕𝐴𝜕𝑇)𝑉 = 𝜕𝜕𝑇(𝜕𝐴𝜕𝑉)𝑇 𝑖𝑚𝑝𝑙𝑖𝑒𝑠 𝑡ℎ𝑎𝑡 (𝜕𝑆𝜕𝑉)𝑇 = (𝜕𝑃𝜕𝑇)𝑉 𝜕𝜕𝑃(𝜕𝐺𝜕𝑇)𝑃 = 𝜕𝜕𝑇(𝜕𝐺𝜕𝑃)𝑆 𝑖𝑚𝑝𝑙𝑖𝑒𝑠 𝑡ℎ𝑎𝑡 −(𝜕𝑆𝜕𝑃)𝑇 = (𝜕𝑉𝜕𝑇)𝑃

The path independence relationship of a state function such as the thermodynamic potential function is very important in experimental applications [13, 14]. For example, if we desire to find the quantity (∂P/∂S)V for some equation of state experimentally, it is not obvious how we can go about measuring the change in pressure versus change in entropy at constant volume. However, the path independence relation indicates that (∂P/∂S)V is the same as the negative of the change in temperature versus change in volume at constant entropy, - (∂T/∂V)S. It is easier to design an experiment to measure the change in temperature versus change in volume. The (∂T/∂V)S approach provides an indirect method of measuring (∂P/∂S)V.

Multi-Component Systems, Partial Molar Quantities and Chemical Potentials

For two or more component systems for example, the Gibb’s free potential energy function is represented by G(T,P, n1, n2, n3,…) [13, 14]. In this representation, T and P retain their meanings and the ni’s are the moles of the component ith species in the mixture. We define chemical potential of component i as μi and this symbol is the change in G with respect to change in ni, by holding T, P and all other ni’s (ni ≠ nj) constant. 𝜇𝑖= (𝜕𝐺𝜕𝑛𝑖)(𝑇,𝑃,𝑛𝑖 ≠ 𝑛𝑗) 𝑤ℎ𝑒𝑟𝑒 𝜇1 = (𝜕𝐺𝜕𝑛1)(𝑇,𝑃,𝑛2,𝑛3,…); 𝜇2 = (𝜕𝐺𝜕𝑛2)(𝑇,𝑃,𝑛1,𝑛3,…) ; 𝜇3= (𝜕𝐺𝜕𝑛3)(𝑇,𝑃,𝑛1,𝑛2,𝑛4…) 𝑒𝑡𝑐.

Just like increasing temperature and pressure represent the capability of a system to give energy and take volume respectively, Chemical potential is the capability of a system to give particles. Temperature of a system spontaneously goes from high to low until thermal equilibrium is achieved. Pressure spontaneously flows from region of high to low until mechanical equilibrium is achieved. In the same manner particles move from the region of high to a region of low concentration until chemical equilibrium is achieved. It is important to note that when the chemical potential is defined for the Gibb’s free potential energy function we must hold the temperature and pressure of the system constant. The multi-component representations for the other three thermodynamic potential energy functions are: U(S,V, n1, n2, n3,…), H(S,P, n1, n2, n3,…) and A(T,V, n1, n2, n3,…), accordingly. Altogether, the following three equations are the remaining chemical potential representations:

𝜇𝑖= (𝜕𝑈𝜕𝑛𝑖)𝑆,𝑉,𝑛𝑖 ≠ 𝑛𝑗 𝜇𝑖 = (𝜕𝐻𝜕𝑛𝑖)𝑆,𝑃,𝑛𝑖 ≠ 𝑛𝑗 𝜇𝑖 = (𝜕𝐴𝜕𝑛𝑖)𝑇,𝑉,𝑛𝑖 ≠ 𝑛𝑗

In general, Chemical potentials are members of the partial molar quantities [13, 14]. All chemical potential functions are partial molar quantities but not all partial molar quantities are chemical potentials. For this reason the examples above: (∂G/∂ni)T,V, ni ≠ nj, (∂A/∂ni)T,P, ni ≠ nj, (∂U/∂ni)S,P, ni ≠ nj and (∂H/∂ni)S,V, ni ≠ nj are partial molar Gibb’s free energy, partial molar Helmholtz free energy, partial molar internal energy and partial molar enthalpy respectively and as well chemical potentials.