Chapter 4
Heat and Thermodynamics

All things are an exchange for fire.
Heraclitus (500 BCE)

Off[General::spell]

4.0 Introduction

Chemistry deals with two questions. Where is it going, and how fast is it getting there? Thermodynamics addresses the first question and chemical kinetics addresses the second.

Thermodynamics is the science that grew out of the industrial revolution as engineers and scientists tried to understand the principles of the driving force of the revolution, the steam (or heat) engine. We will take a historical approach and begin with a discussion of macroscopic or phenomenological thermodynamics and then turn to a microscopic or molecular treatment.

Strictly speaking the subject that address where a process goes is interested only in the inital state and the eventual final state. States are static and its study should be called thermostatics. However, systems change with time as they move between inital and final states. Hence the term thermodynamics.

Because thermodynamics is considered such a fundamental subject, a great deal of effort has been invested in understanding it. Consequently, there are numerous approaches and analyses of thermodynamics. They may divided into two areas, phenomenological macroscopic thermodynamics and theoretical microscopic thermodyanmics. We will present a roughly histrorical presentation proceding from macroscopic energy conservation through applications to heat engines and calorimeters to theoretical microscopic (molecular) statistical mechanics.

4.1 Conservation Laws

4.1.1 Symmetry

Quantities or properties of systems that could change but remain constant are said to be conserved (in time). Conservation is one manifestation of a broader aspect of systems called symmetry, referring to the property of redundancy. Geometrical symmetry occurs when objects have similar parts that could be superposed by some rearrangement. Humans have an external bilateral symmetry, snowflakes have six-fold symmetry, and spheres have continuous (infinite) rotatinal symmetry. Symmetry is considered to be one of the most fundamental properties of objects in the Universe and symmetry laws among the most basic properties of structure and behavior. For example, Newtonian mechanics is symmetric in time. If time is reversed in a dynamical system, the system returns to its former state.

4.1.2 Viva la Vis

Historically, Newton resolved a great debate that lasted half a century. At issue was the elucidation of something called vis viva, or the living force, that was supposed to distinguish organic matter from inanimate matter, motivated by the vis mortua, or inamite (dead) force. In 1637 Descartes showed that momentum is conserved for simple mechanical systems and identified it with the vis viva. When two otherwise isolated objects collide there is an exchange of momentum. Consider a moving billiard ball colliding with a stationary ball. If they have the same mass, the first ball will stop upon collision and the second will move away with the velocity of the first. If they have unequal masses, the relationship between the parameters (masses) and variables (velocites) is observed to be (using subscripts to distinguish the two balls):

(0 - ) = - 0)

or

=

where Δv ≡ -

In other words, the change momentum = mΔv between the subsystems is constant, and since the total momentum of the combined isolated system is zero (momentum is conserved in time.

On the other hand, from his newly discovered mechanics of falling objects, Galileo found another quantity that was conserved. Since

v =

f = mg =

and the same force operating over a given distance produces the constant quantity

f Δh =

Galileo identified this quantity with the vis viva, now called the kinetic energy. Today we use the abbreviation KE or the symbol T to represent kinetic energy.

In 1686 Leibnitz sided with Galileo, but in 1687 Newton showed that both Descartes and Galileo were right! For a system with no external forces, Newton's Second Law says

f = = 0

and by integrating over time, the momentum p is a dt = Δp = 0 ⇒ p = constant.

Integrate[D[p[t],t],t]

On the other hand, integrating force over distance yields

f = ma = m = m = mv

and ∫ f(r) dr = m∫ v dr = m∫ v dv = - = ΔT = 0

and the kinetic energy remains constant.

Integrate[m*v, v]

The integral of force exerted on a system over distance is now called the work done on the system.

w ≡ ∫ f(r) dr

4.1.3 Conservation of Mechanical Energy

After the invention of steam engines, the engineer L.M.N. Carnot in 1803 defined a quantity representing the stored work as potential energy, or the energy with the potential of doing work, which he called the "latent (hidden) vis viva,". We give it the abbreviation PE or the symbol V, defined in terms of force such that

f(r) ≡ -

When the potential energy is released to a system, the system acquires moving kinetic energy. The conversion between the two forms of energy suggests that the loss of potential energy causes a gain in kinetic energy, or

-ΔV = +ΔT

consistent with the definition of change as the final value minus the initial value:

- ) = - )

or

+ = +

This was the first statement of the conservation of total energy, where the total energy E is the sum of the kinetic and potential eneries of a given system:

E ≡ T + V = constant, or ΔE = 0

A familiar example illustrates the interchange between potential and kinetic energy and the conservation of their sum. Consider a child on a swing. The child on the swing is lifted up, causing an increase in potential enegy in the presence of the gravitational attraction to the Earth. On release, the child falls and the potential energy decreases, while the kinetic energy increases, as noticed by the increase in speed. Eventually the child reaches the minimum height allowed by the swing and no more potential energy is converted into kinetic energy. At this point the child is moving at maximum speed, whereupon the child begins to ascend causing kinetic energy to be converted back into potential energy. At the top of the swing the child is motionless for an instant before descending again and the maximum potential energy is realized when the kinetic energy is zero. In the absence of interactions with the surroundings, the child swings back and forth continuously while potential and kinetic energy exchange. The sum of the potential and kinetic enrgies at any point, however, remains constant, equal to the total energy of the system.

4.2 Energy and the First Law

4.2.1 Work

Intuitively, we think of raising an object against the gravity as doing work (on the object). This increases the potential energy of the object. Generalizing this concept leads to the definition of work as an increase in potential energy due to a force causing a displacement. Since f(r) ≡ - , we define work as

≡ ∫ (r) dr

There are many ways to perform work, depending on the force involved and the path of the displacement. For example, using the definition of pressure as the ratio of force to area, we can describe the force of a piston f moving an incremental distance dr inside a cylinder of area A as producing an amount of mechanical expansion work resulting from the expansion of gas in the cylinder by amount ΔV:

≡ ∫ (r) dr = -∫ A dr = -∫ dV

where is the pressure exerted on the piston by the gas, and the integral is evaluated from an initial volume to some final volume. Expansion work is not limited to gases and, in general, P depends on volume V, so we write the general formula

  = - dV

How pressure depends on volume determines the path taken in the (P,V,T) state space of the gas, and thus the value of the integral. Note that pressure is an intensive state variable (independent of amount) whereas volume is an extensive state variable (proportional to amount). Other forms of work involve the product of intensive and extensive state variables analogous to pressure and volume. The analogues for two-dimensional surfaces are surface tension and area, and for one-dimensional wires are tension and length. For electochemical cells, the analogues are voltage EMF and charge Q, so

  = - dQ

To illustrate how work depends on path, let's calculate the formulas for expansion work for various processes (paths). For our first case, consider the work of expansion against a constant pressure (an isobaric process). This general result applies to any phase, solid, liquid or gas.

Isobaric Expansion Work

ClearAll["`*"]
Print["Ideal gas isobaric pressure = ", P]
Print["Ideal gas isobaric expansion work = ", Simplify[Integrate[P, {V,Vi,Vf}]]]

For the next case, we consider a constant temperature path (isothermal process). Here we need an equation of state to relate pressure to volume. We consider the ideal gas equation of state. We add some replacement rules for logarithms to coax Mathematica into displaying the result in a familiar form.

Isothermal Ideal Gas Expansion Work

ClearAll["`*"]
Print["logrule: ", logrule = {Log[a_]-Log[b_] -> Log[a/b]}]

Print["Ideal gas isothermal pressure = ", Pig = P/.Solve[P*V == n*R*T, P][[1]]]

Print["Ideal gas isothermal expansion work = ", Simplify[Integrate[Pig, {V,Vi,Vf}]]/.logrule]

Literally hundreds of empirical equations of state have been proposed for substances. Some of them have limited ranges and many are advertised to have a given form for wide varieties of substances. One of the most famous is that due to van der Waals in 1873. It is empirical but attempts to account for real gas behavior by adapting idea gas behavior to include attractive and repulsive terms.

(P+a/)(V-nb) = nRT

The parameters a and b reflect the intermolecular attractive and repulsive forces, respectively. We can compare the state variables for a van der Waals gas with those for an ideal gas for, say, dioxygen graphically. Since the van der Waals gas has the same temperature dependence as an ideal gas, both follow Charles' Law, but the van der Waals equation of state is a cubic in volume (why?), and therefore has different isotherms from the hyperbolas of the ideal gas. At high temperature and low pressure the van der Waals gas approaches the ideal gas, but at low temperature and high pressure, the van der Waals gas exhibits condensation. This is manifested in the Boyle Law plots where the volume decreases to the point that the isotherm goes into a supercooled region that pases through a superheated condensed phased into a condensed phase iostherm. Portions of the isotherm having negative slope have no physical reality, however, and are artifacts of a continuous empirical equation of state.

van der Waals Gas vs. Ideal Gas

Clear[P,V,n,R,T,a,b]
P[V_,T_] := n*R*T/(V - n*b) - n^2*a/V^2
n = 1;
R = .082059;
a[O2] = 5.464; b[O2] = 0.03049;

Phase Diagram

Plot3D[P[V,T]/.{a->0,b->0}, {V,0.01,3}, {T,0,500},
       PlotPoints->{30,30}, ViewPoint->{2,-10,2},
       AxesLabel->{"V","T","P"}, PlotRange->{0,110},
       PlotLabel->"ideal gas"];
Plot3D[P[V,T]/.{a->a[O2],b->b[O2]}, {V,0.01,3}, {T,0,500},
       PlotPoints->{30,30}, ViewPoint->{2,-10,2},
       AxesLabel->{"V","T","P"}, PlotRange->{0,110},
       PlotLabel->"van der Waals gas"];

Charles' Law Isochores

Show[Table[Plot[P[V,T]/.{a->0,b->0}, {T,0,650},
     DisplayFunction->Identity], {V,0.01,100,10}],
     DisplayFunction->$DisplayFunction, AxesLabel->{"T","P"},
     PlotLabel->"ideal gas"];
Show[Table[Plot[P[V,T]/.{a->a[O2],b->b[O2]}, {T,0,650},
     DisplayFunction->Identity], {V,0.01,100,10}],
     DisplayFunction->$DisplayFunction, AxesLabel->{"T","P"},
     PlotLabel->"van der Waals gas"];

Boyle's Law Isotherms

Show[Table[Plot[P[V,T]/.{a->0,b->0}, {V,1,.04},
     DisplayFunction->Identity], {T,0,1000,100}],
     DisplayFunction->$DisplayFunction, AxesLabel->{"V","P"},
     PlotLabel->"ideal gas"];
 Show[Table[Plot[P[V,T]/.{a->a[O2],b->b[O2]}, {V,1,.04},
     DisplayFunction->Identity], {T,0,1000,100}],
     DisplayFunction->$DisplayFunction, AxesLabel->{"V","P"},
     PlotLabel->"van der Waals gas"];

To compute the isothermal work for a van der Waals gas, we use a more general logarithm replacement rule to simplify the result.

Isothermal van der Waals Gas Expansion Work

ClearAll["`*"]
Print["logrule: ", logrule = {m_*Log[a_]-m_*Log[b_] -> m*Log[a/b]}]

Print["van der Waals gas isothermal pressure = ", Pvdw = P/.Solve[(P+a*n^2/V^2)*(V-n*b) == n*R*T, P][[1]]]

Print["van der Waals isothermal gas expansion work = ", Simplify[Integrate[Pvdw, {V,Vi,Vf}]]/.logrule]

4.2.2 The Mechanical Equivalent of Heat

Newtonian mechanics, based on the concept of force, is the foundation of modern science. When energy became identified an alternative to force as the motivating quantity of change (e.g. in the form of Hamilton's  equations), people looked for energy analogues to the fundamental properties of force, such as conservation (symmetry in time). Conservation of force is expressed in Newton's  Third Law, every action force has an equal and opposite force. Conservation of energy is more subtle; energy consists of components the sum of which remains constant. The First Law of Thermodyanmics, which became recognized as a general principle in the middle of the 19th Century, may be stated as energy can change form but the sum of all forms remains constant for any system that is energetically isolated such that it does not exchange energy with any other system.

Empirically, Benjamin Thompson (Count Rumford) in 1798 and (Sir) Humpfrey Davy in 1803 observed that rubbing produces heat. In 1834 James Watt observed  that heat expands a gas, and in 1843 James Prescott Joule reported that stirring produces heat. Rumford was in charge of boring cannons and noticed that a dull boring tool produced heat instead of work (cutting). Davy was President of the Royal Society and reported that rubbing two pieces of ice together melted the ice. Quantitative measurements could be made in the form of proportionalities or constant ratios (water to cool the cannon proportional to the boring time, mass of melted ice proportional to rubbing time, etc.)

Watt was a steam engineer designer who explored improving the efficiency of  expanding steam to drive a piston in a chamber. He invented a governor to moderate the motion of the steam engine to improve it's performenance. Joule was the son of a Manchester brewer put in charge of tending the vats where he observed that stirring generated heat. He conducted a variety of very careful experiments and is credited with being the first to recognize the implications of the observations as conservation. His famous "paddle wheel" experiment was designed to change the state of water, as measured by temperature, two different ways. The first was simply to heat the water with, say, a flame. The second way was to stirr the water using a falling weight connected to a stirring paddle through a cord over a pulley.

For both processes quantitative amounts of energy could be calculated from existing definitions. By comparing the amounts of energy generated to heat a given amount of water through a given temperature change, Joule deduced what he called the "mechanical equivalent of heat." A temperature rise of 1 Fahrenheit degree for 1 pound of water produces an amount of heat defined as 1 BTU (British Thermal Unit). The heat produced is the product of the mass of the heated object with the temperature change and a characteristic parameter called the specific heat, SH, which has the value unity for water. Given the definition of a calorie (cal) in terms of a BTU we have

q = m SH Δt = 1 BTU () = 252 cal

On the other hand, Joule measured the work to heat the same amount of water the same amount. From the formula for the amount of work w done by an object of mass m falling a distance Δh, and the conversion between ft-lb (foot pounds) and Joules J, we have

w = mgΔh = 772 ft-lb () = 1054 J

The mechanical equivalent of heat is a conversion factor which converts between units of heat and work. Simple conversion factors are ratios derived from equalities involving different units. Thus, since 1 ft = 12 in,  = = 1 (why?), and we can multiply a quantity with one unit of length by unity to "convert" it to the same quantity but with different units, as in 3 ft * = 36 in. Here we have w = q, or ΔE(q = 0) = ΔE(w = 0) for the two ways of heating water, so by equating these two quantities we have

= = 4.18

This conversion factor is the mechanical equivalent of heat. The modern value focuses on the basic SI unit of the Joule as a Newton meter and defines the heat quantities such as the calorie in terms of the Joule as in

1 cal  ≡ 4.1845 J

Rubbing, expansion and stirring are forms of work (force moving through a distance). Since the change in the state of the water (as measured by a temperature change) is the same for both the heating and the mechanical work processes, the foundation was laid for the general recognition of the interconversion between work and heat. Furthermore, the same temperature change could be produced by either heat or work or a combination of heat and work, the sum of the heat and work applied to the system is a measure of the change in energy of the water (why the sum and not some other function?).

ΔE = q + w

Joule was only nineteen year old when he submited his first report to the Royal Society in 1838. He was ignored because of his youth. In 1844 he submitted a second report that predicted on the basis of his observations that falling water should heat up due to the conversion of potential energy into kinetic energy. This too was rejected, but the senior scientist William Thompson, who became Lord Kelvin, recognized the young genius. Two weeks later on a vacation to France he recorded that he met young Joule, who had just married, on his honeymoon, with his bride waiting in a carriage while Joule ascended the Alpine Mount Blanc (bordering Switzerland and Italy, the highest peak in Europe at 4708 m) with a long thermometer on his way to test the elevation of water fall temperatures!

Joule's reasoning is as follows. For an isolated system

ΔE = q + w = m SHΔt + mgΔh = 0

Δt = , independent of amount (m)

Because SH and g are positive but Δh = - < 0 for a falling object, we see the that the temperature increases (Δt > 0).

Kelvin was instrumental in getting published in 1849, eleven years after he first submitted his first paper. His suggestion that friction results in the conversion of work into heat was deleted by the reviewers.

4.2.3 Heat as Motion

Robert Boyle (1627-1691) was a friend of Newton who helped define the transition from alchemy to chemistry. He was a founding member of the Royal Society and, like Newton, an atomist. In 1661 he wrote The Sceptical Chymist, in which he presented the modern working definition of an element as the irreducible component of compound substances. A typical renaissance man his interests ranged widely and he is remembered for discovering the first law of gas behavior, which bears his name.

Antoine Laurent Lavoisier (1743-1794) wrote the first book on chemistry that is intelligible to modern readers, Traité élémentaire de Chimie in 1789, summarizing research by himself and others based on quantitative analysis of compounds and reactions (stoichiomery - measure of mass). He was aware of the Laws of Conservation of Mass and Constant Combining Proportions. Considered to be the father of modern chemistry, he probably would have been led to the atomic theory explanation of chemical reactions had his work not been cut short by the French Revolution invention of Dr. J. J. Guillotine.

John Dalton (1766-1844) was a Manchester schoolteacher who picked up where Lavoisier left off in the discovery of the explanation of chemical reaction stoichiometry in terms of atomic masses, called the Atomic Theory. The chapters of his first volume of A New System of Chemical Philosophy (1080, 1811) are titled On Heat of Caloric (111 pages), On the Constitution of Bodies (49 pages) and On Chemical Synthesis (6 pages). He became the leading chemical authority in part because he not only explained the Law of Constant Combining Proportions but because he also predicted the Law of Multiple Combining Proportions for the formation of compounds with multiple chemical formulas, and found evidence for it in the literature going back a half century.

Since the use of fire for cooking and manufacturing in prehistoric times, man has taken an interest in the heat attending chemical reactions. Thermochemistry is the application of thermodynamics to chemical reactions, particularly the heat energy of reactions. Since the amount of heat attending a chemical reaction is proportional to the amount of mass involved, quantitative measurements and calculations benefit from considering to be an additional substance involved in the reaction. This was called caloric and is listed as the second element (after light) in Laviosier's Table of Simple Substances. Dalton did not consider heat to be a material element, but rather a bonding agent not unlike an electron cloud. A half century later Maxwell's Kinetic Molecular Theory explained heat as molecular motion (to be discussed below). This provides a connection between macroscopic heat microscopic energy. As we shall see this suggest an interpretation of work as useful, ordered energy and heat as useless, disordered energy.

ΔE = w + q = ordered energy + random energy

4.2.4 Generalization of the First Law

Although many scientists participated in the development of thermodynamics, in England, Kelvin is given the credit for the first clear statement of the thermodynamic conservation of energy in 1851:

= w + q = 0, or E = constant for an isolated system

An alternative statement recognizes that a system can change its state in a variety of ways, but the change in energy is independent of the way the system changes, called the path from the initial to the final state. (This would be called the trajectory in dynamical system terminology.) Thus, Joule observed that one of the state variables of water, namely the temperature, could be modified by a given amount either by direct heating or doing mechanical work on the water.

= 0

To appreciate these forms of the First Law, consider climbing a mountain. As far as the change in potential energy needed to ascend the mountain, it doesn't  matter which path one takes, or even which method of transportation is used, Galileo's  formula assures us the gravitational energy acquired is

= mgΔh

Furthermore, all this energy will be released on descending to the initial altitude, so as far as thermodynamics is concerned, there has been no net expenditure (or gain) in energy in the round trip: = 0. A rather curious result when one considers that a mountain hike involves the expenditure of calories (or does it; explain)?

There are many ways to change the energy of a system. For example, work done compressing a gas at constant pressure increases its energy by PΔV (since ΔV < 0) and equals the work done by a gas in an equivalent expansion, which decreases its energy by -PΔV (ΔV > 0). In general,

ΔE = +

The remarkable thing about the First Law is that, while work and heat depend on the way they are transferred (the "path"), their sum does not.

4.3 Applications of the First Law

4.3.1 Calorimetry

There are two common devices for measuring heats of reactions, called calorimeters ("heat measure"). Both are based on the First Law of Thermodynamics since they consist of a chemical reaction system thermally connected to a surrounding medium (usually water) which measures the energy released (or absorbed) by the reaction. The total apparatus is thermally isolated (as much as possible) so that by conservation of energy, the energy lost (or gained) by the reaction is gained (or lost) by the surrounding mediuim. Practical considerations lead to calibrating the calorimeter using standard chemical reactions with known heats of reaction to determine the total heat capacity of the apparatus in terms of an equivalent amount of water of known heat capacity, the so-called calorimetry constant.

The first type of calorimeter consists of a closed container of constant volume ("bomb"), where the reaction takes place, immersed in a water bath to measure heat released or absorbed by the reaction. The second type is open to the atmosphere and most closely represents the typical chemical reaction taking place in an aqueous solution in a beaker ("beaker chemistry").

There are two type of heat capacity, corresponding to their historical measurement using calorimeters. The first is called the heat capacity at constant volume, and provides a direct measure of the energy of a reaction:

≡

from which we have by integration,   

ΔE = ∫ dT = ∫ dT = ∫ m dT

meaning and (calorimeter). n is the number of mols and is the molar heat capacity, or heat capacity per mol; m is the mass and the specific heat, or heat capacity per gram. (The various ways to express the energy change emphasize the importance of the quantity.) Temperature dependence has been explicitly shown to remind us that heat capacity varies with temperature in the general case. Constant volume processes are called isochoric processes.
As usual, Δ means change from inital to final state, so

= -

At constant volume, the work w = ∫ PdV = 0 and ΔE = (why?). If the heat capacity is constant, which will be approximately true for small temperature changes,

ΔE = ΔT = n ΔT = m ΔT

and

ΔT = (after reaction) - (before reaction)

A quantity related to energy is called enthalpy (or sometimes heat), symbolized by H, defined by

H ≡ E + PV

Enthalpy is associated naturally with heating and cooling at constant pressure:

P = constant  =>  w = ∫ PdV = PΔV  =>  ΔH = ΔE + Δ(PV) = ΔE + PΔV = w + q - w =

Constant pressure processes are called isobaric processes. Analogous to the heat capacity at constant volume used in a bomb calorimeter, there is heat capacity at constant pressure, which provides a direct measure of the enthalpy of a reaction in a beaker calorimeter:

≡

from which we have by integration,

ΔH = ∫ dT = ∫ n dT = ∫ m dT

meaning and (calorimeter). If the heat capacity is constant, which will be approximately true for small temperature changes,

ΔH = ΔT = n ΔT = m ΔT

Isobaric and isochoric heat capacities for dense substances, like solids and liquids are about the same.

≃ for condensed phases

On the other hand, isobaric and isochoric heat capacities differ for compressible gases. For example, an ideal gas has PV = nRT, so

H = E + PV  =>  = + => = + = + nR,  or = + R

and the isobaric and isochoric heat capacities of gases differ by about 8.3145 J/Mol-K.

For chemical reactions, the energy or enthalpy of reaction may be determined by measuring temperature changes:

Δ = - = = ∫ dT

Δ = - = = ∫ dT

C is the heat capacity of the total system, reaction + any solvent + calorimeter + calorimeter water + thermometer + etc. It is usually obtained from a standard reaction with a known reaction energy or enthalpy. Once C is known, temperature measurements of arbitrary reactions yield energies or enthalpies of reaction. Thus, calorimetry provides experimental procedures for the direct measurement of energies and enthalpies of reactions, interpreted as heats of reaction at constant volume and constant pressure, respectively.

The heat capacity of water at one time was operationally defined to be 1 calorie/g -degree °C at  15 °C and 1 atm pressure. Today the calorie is defined to be 4.184 J so the heat capacity of water at constant pressure is 4.184 J/g-degree °C at 15 °C, or = 75.376 J/mol-K at 25 °C. In gereral, heat capacity is a function of the state of the system, temperature, pressure and phase.

   •  Heat capacities are tabulated for many substances. Let's look up the heat capacity of water as a function of temperature and evaluate values at various temperatures and find the heat required to raise water from its freezing temperature to its boiling temperature.

Heat Capacity of Water

NIST Shomate Equation (http://webbook.nist.gov)

ClearAll["`*"]
Print["Heat required to raise water from T1 to T2 = ", Integrate[n*Cp[T],{T,T1,T2}]]

Cp[T_] := A+B*(T/10^3)+C*(T/10^3)^2+D*(T/10^3)^3+E/(T/10^3)^2

ΔH = Integrate[n*Cp[T],{T,T1,T2}]//Simplify

NIST Data

Unprotect[C,D,E];
A = -203.6060;
B = 1523.290;
C = -3196.413;
D = 2474.455;
E = 3.855326;

Variables

n = 1;
T1 = 273.15;
T2 = 373.15;

Heat capacity of water at various temperatures

Print["Molar heat capacity of water at 298.15 K = ", Cp[298.15]]

Print["Molar heat capacity of water at ", T1, " K = ", Cp[T1]]

Print["Molar heat capacity of water at ", T2, " K = ", Cp[T2]]

Plot[Cp[T],{T,T1,T2}, AxesLabel->{"T/Kelvin","Cp[H2O]/Joule/Mol-Kelvin"}];

Heat required to raise temperature

Print["Joules required to raise ",n," mol of water from ",T1," to ",T2, " = ", ΔH = Integrate[n*Cp[T],{T,T1,T2}]]

   •  This notebook shows the calculations involved in calibrating a bomb calorimeter using the known value of the heat (enthalpy) of combustion of the standard benzoic acid (-3226.7 KJ/mol). When 0.9802 g of benzoic acid is combusted in the calorimeter, the temperature increases by 3.83 C.

Benzoic acid combustion: + →  7 + 3 O(l)

Calorimetry Calibration

ClearAll["`*"]
<<Miscellaneous`Units`

Equations

eq1 = ΔH == ΔE+R*T*Δn
eq2 = n == (m/M)
eq3 = n*ΔE+Cp*ΔT == 0

Calibration Equation

Solve[{eq1,eq2,eq3},{n,ΔE,Cp}]//Simplify

Heat of Combustion Equation

Solve[{eq1,eq2,eq3},{n,ΔE,ΔH}]//Simplify

Constants

R = 8.3145 Convert[Joule, Kilo Joule]/(Mol*Kelvin)
T = 298.15 Kelvin

Calibration Calculation

Clear[Cp]
Δn = -.5
m = .9802 Gram
M = 122.118 Gram/Mol
ΔT = 3.83 Kelvin
ΔH = -3226.7 Kilo Joule/Mol

soln = Solve[{eq1,eq2,eq3},{n,ΔE,Cp}]//Simplify
Print["Calorimeter heat capacity = ", Cp = Cp/.soln[[1]]]

   •  We now use the calibrated calorimeter to determine the heat (enthalpy) of combustion of glucose. The combustion of 0.4654 g of glucose raises the temperature of the calorimeter by 1.07 °C.

Glucose combustion: + →  6 + 6 O(l)

Heat of Combustion Calculation

Clear[ΔH,Δn,m,M,ΔT,n]

Δn = 0
m = .4654 Gram
M = 180.156 Gram/Mol
ΔT = 1.07 Kelvin

soln2 = Solve[{eq1,eq2,eq3},{n,ΔE,ΔH}]//Simplify
Print["Heat of combustion = ", ΔHc[glucose] = ΔH/.soln2[[1]]]

4.3.2 Thermochemistry

The First Law of Thermodynamics provides a mechanism for summarizing vast amounts of information in relatively small tables of values. Since E and H are state functions, ΔE and ΔH are independent of path, or, equivalently, = = 0. This means a total ΔE or ΔH value is the sum of the partial ΔE and ΔH values that comprise the total. If X stands for a state function such as E or H, we have

  = - =

or

-   = 0

In terms of paths, a total change for a direct path is equivalent to the sum of changes for an indirect path

  =

In the case of gravitational potential energy, the total potential energy is the sum of any contributing parts. Thus in hiking over mountains, the total increase in potential energy is the sum of the increases of climbs and decreases of descents and is equal to the final potential energy at the end of the hike minus the initial potential energy at the begining. Near the surface of the earth this would be

= - =

In the case of a chemical reaction, the principle applies and the total energy (or enthalpy) of a reaction that is comprised of component parts is the sum of the energies (or enthalpies) of the parts:

  =

  =

This was observed experimentally for heats of reaction for beaker reactions at constant pressure) by Hess in 18?? for chemical reactions that added together to form an observed totral reaction. The energy (enthalpy) change for any chemical reaction can be calculated as the arithmetic sum of the energy (enthalpy) changes for any alternative set of reactions that sum to the given reaction, whether or not the subreactions can be observed experimentally!

   •  From the molar heat of combustion of glucose determined calorimetrically in the last section, we can find the molar heat(enthalpy) of formation of glucose.

Glucose formation:   →  

Glucose combustion: + →  6 + 6 O(l)

Heat of Formation from Heat of Combustion

ΔHf[H2O] = -285.830 Kilo Joule/Mol;
ΔHf[CO2] = -393.509 Kilo Joule/Mol;
ΔHf[O2] = 0 Kilo Joule/Mol;

Stoichiometric Coefficients

Glucose combustion: C6H12O6 + 6O2 = 6CO2 + 6H2O

n[CO2] = 6 Mol;
n[H2O] = 6 Mol;

n[glucose] = 1 Mol;
n[O2] = 6 Mol;

Heat of Formation

Solve[n[glucose]*ΔHc[glucose] == n[CO2]*ΔHf[CO2] + n[H2O]*ΔHf[H2O] - n[O2]*ΔHf[O2] - n[glucose]*ΔHf[glucose], ΔHf[glucose]]

This process may be generalized to calculate th eenergy and enthalpy of any reaction given energies and heats (enthalpies) of formation and mols of reactants and products (stoichiometric coefficients in the balanced chemical reaction.) If the reaction Reactants = Products is rearranged into the algebraically equivalent form Products - Reactants = 0,  product coefficients  have positive values and reactants have negative values, and the general formula is:

  = n(i)

where X stands for energy or enthalpy,

  = n(i)

  = n(i)

This formula could be implemented as a mathematica package that stores heats of formation of a (large) number of compounds and accepts stoichiometric coefficients as input and produces heat of reaction as output.

4.3.3 Bond Energies

We may envision a chemical reaction as taking place by breaking the bonds of all the reactant molecules, thereby generating all the atoms of the reactants, and then combining all the atoms to form new bonds of the product molecules. Note that the reaction thus conserves atoms in accordance with Dalton's model of chemical reactions. The same reasoning used to calculate the energy or enthalpy of reaction from formation energies or enthalpies yields similar formulas, with a few significant differences. Formation refers to formation of compound (final state) from elements (initial state), whereas bond enegy refers to breaking molecules (initial state) into atoms (final state).  This means opposite signs are involved in the sumations. Further, all the bonds of each molecule must be broken, adding a multiplicative factor m(i) to the bond energies for equivalent bonds.

  ≃ n(i)m(i)

≃ n(i)m(i)

where for the ith type of bond in a molecule (i) is the molar bond energy  and is the molar bond enthalpy. m(i) is the number of bonds of type i per molecule. In practice, bond energies are easier to obtain than bond enthalpies and are more commonly tabulated. Note that these estimates assume gaseous species. If other phases are involved, they must be corrected by the appropriate energies or enthalpies of phase changes to gase phases.

These formulas could be implemented as a mathematica package that stores bond energies and enthapies of a (large) number of compounds and accepts stoichiometric coefficients as input and produces energies and enthalpies of reaction as output. One hundred entries could be used to compute reaction energies for one hundred million reactions involving two reactants and two products! There is a tradeoff, however. Nonequivalent but similar bonds do not have identical bond energies. For example, NMR spectra confirm that the terminal hyrogen atoms of CCChave equivalent environments and bonding which differs from the central hydrogen atoms. But even in the case of equivalent bonds, as they are removed one at a time, different amounts of energy are required. Thus the energies to remove th efour hydrogen atoms sequentially in Care 438, 465, 422 and 338 kJ/mol. However, bond energies are approximately the same for a pair of atoms, independent of the environment. This is especially true of large molecules where one end is unaware of the other. This remarkable fact is what allows us to write Lewis chemical formulas representing atoms with letters and bonds with dots and lines. In fact, using a single type of line (or lines for multiple bonds) is a gross oversimplification, suggesting all bond of all types are equivalent. The truth is, all bonds in molecules are sensitive to their chemical envornment (other bonds), as described by wave mechanics.

In practice, bond energies are tabulated as averages over many examples of a given bond in different molecules. For example, the average value of the four bond energies in methane is 416 kJ/mol. Either using this value or combining it with C-H bond energies in other molecules yields an average bond energy that can be entered into a general table. The consequences are that not only is there no single C-H bond energy, but differnt tables may have different values for their average C-H bond energy. Nevertheless, approximate values of energies of reaction for a great variety of reactions can be obtained from simple average bond energy tables.

Typical Bond Energies (kJ/mol): Single Bonds

ClearAll["`*"]
DisplayForm[FrameBox[GridBox[Table[{
   {"","Li","Na","Mg","C","N","P","O","S","H","F","Cl","Br","I"},
   {"Li","106","","","","","","","","","","","",""},
   {"Na"," 90"," 77","","","","","","","","","","",""},
   {"Mg","-","-","9","","","","","","","","","",""},
   {"C","-","-","-","347","","","","","","","","",""},
   {"N","-","-","-","305","160","","","","","","","",""},
   {"P","-","-","-","264","209","213","","","","","","",""},
   {"O","341","256","362","358","201","351","146","","","","","",""},
   {"S","-","-","234","259","464","444","-","266","","","","",""},
   {"H","238","184","126","414","391","321","467","344","436","","","",""},
   {"F","577","519","462","485","270","439","197","327","570","155","","",""},
   {"Cl","469","410","328","397","200","289","203","253","429","256","240","",""},
   {"Br","423","370","327","280","243","-","201","218","362","237","219","190",""},
   {"I","352","301","285","209","159","-","201","-","295","<272","211","179","149"}
    }] ,RowLines->{True,False},ColumnLines->{True,False}]]]

Typical Bond Energies (kJ/mol): Double Bonds

DisplayForm[FrameBox[GridBox[Table[{
   {"","C","N","P","O","S"},
   {"C","613","","","",""},
   {"N","615","418","","",""},
   {"P","513","617","490","",""},
   {"O","725*","627","597","498",""},
   {"S","699","464","444","522","425"}
    }] ,RowLines->{True,False},ColumnLines->{True,False}]]]

*799 for C

Typical Bond Energies (kJ/mol): Triple Bonds

DisplayForm[FrameBox[GridBox[Table[{
   {"","C","N","O"},
   {"C","839","",""},
   {"N","891","946",""},
   {"O","1070","1063",""}
    }] ,RowLines->{True,False},ColumnLines->{True,False}]]]

Let's use bond energies to estimate the enthalpy for the reaction + = O  (the enthalpy of formation of water, or equivalently the enthalpy of combustion of dihydrogen). Assuming all species are gases, we have

  ≃ n(i)m(i)   =    + -2

BE = {436 Kilo Joule/Mol(* H-H *),
      498 Kilo Joule/Mol (* O=O *), 
      467 Kilo Joule/Mol(* H-O *)}

nspecies = 3;
n = {1,1/2,-1};
m = {1,1,2};

ΔEf[H2O] = Sum[n[[i]]*m[[i]]*BE[[i]], {i,1,nspecies}]
R = 8.3145*10^-3 Kilo Joule/(Mol*Kelvin);
T = (25 + 273.15) Kelvin;
Print["Estimated enthalpy of formation of gaseous water is ", ΔHf[H2Ogas] = ΔEf[H2O] + R*T*Sum[n[[i]],{i,1,3}]]

This may be compared with the experimental value of -241.818 kJ/mol, a 2% difference.  Typically,

When species are not in the gas phase, known enthalpies of condensation may be used to convert bond energy calculations to the appropriate phases. The molar vaproization of water can be obtained from the heats (entalpies) of formation of liquid and gas water:

ΔHf[H2O-g] = -241.818 Kilo Joule/Mol;
ΔHf[H2O-l] = -285.830 Kilo Joule/Mol;
Print["Molar enthalpy of vaporization of water = ", ΔHv[H2O] = ΔHf[H2O-g] - ΔHf[H2O-l]]

Print["Estimated enthalpy of formation of gaseous water is ", ΔHf[H2Ogas] - ΔHv[H2O]]

4.3.4 Kirchoff's Law

Thermodynamic tables often contain thermodynamic values for a single temperature (standard temperature is 25 °C). Kirchoff's Law provides a formula for computing the enthalpy of reaction at other temperatures. It is a simple extension of the definition of heat capacity.

Δ ≡

from which we have by integration,

ΔH( - ΔH(  = ∫ Δ dT ΔT

This formula applies to molar amounts as well:

Δ( - Δ(  = ∫ Δ(n dT ΔT

Heat of vaporization of water at 500 Kelvin assuming heat capacities are constant

<<Miscellaneous`Units`
Cp[H2O-g] = 33.6 Joule/(Mol*Kelvin);
Cp[H2O-l] = 76.0951 Joule/(Mol*Kelvin);
ΔCp = Cp[H2O-g] - Cp[H2O-l]
T1 = 298.15 Kelvin;
T2 = 500 Kelvin;
ΔT = T2 - T1
Print["Molar enthalpy of vaporization of water at ", T2, " = ", ΔHv[H2O] + Convert[ΔCp*ΔT, Kilo Joule/Mol]]

Heat of vaporization of water at 500 Kelvin given heat capacities as functions of temperature

NIST Shomate Equation (http://webbook.nist.gov)

Clear[T]
Cp[i_,T_] := A[[i]]+B[[i]]*(T/10^3)+C[[i]]*(T/10^3)^2+D[[i]]*(T/10^3)^3+E[[i]]/(T/10^3)^2

NIST Data

Unprotect[C,D,E];
A = {-203.6060,30.0920};
B = {1523.290,6.832514};
C = {-3196.413,6.793435};
D = {2474.455,-2.534480};
E = {3.855326,0.082139};

T1 = 298.15;
T2 = 500;
ΔCp = (Cp[2,T]-Cp[1,T]) Joule/(Mol)
Print["Molar enthalpy of vaporization of water at ", T2, " = ", ΔHv[H2O] + Convert[Integrate[ΔCp,{T,T1,T2}], Kilo Joule/Mol]]

CpLiquid[T_] := A[l]+B[l]*(T/10^3)+C[l]*(T/10^3)^2+D[l]*(T/10^3)^3+E[l]/(T/10^3)^2
CpGas[T_] := A[g]+B[g]*(T/10^3)+C[g]*(T/10^3)^2+D[g]*(T/10^3)^3+E[g]/(T/10^3)^2

4.4 Larger Examples

4.4.1 Adiabatic Expansion Work

Later on we will be interested in the work attending processes without any transfer of heat to or from the system. Such processes are called adiabatic processes. Starting from the differential form of the First Law, we can derive a law analogous to Boyle's Law for ideal gas expansion and compression processes that applies to adiabatic conditions.

dE = q + w = ndT - PdV = ndT - dV = ndT - = 0

This differential equation can be solved by collecting ("separating") the variables to opposite sides of the equation:

  =   = (1 - γ) ,  where γ

Assuming the ratio of heat capacities γ is a constant and integrating both sides of the equation,

    = (1 - γ) =

Exponentiating both sides of the equation and applying the Ideal Gas Law, we have

     = =  

Collecting like terms gives

=   = constant at constant temperature.

Air, for example, has a γ value equal to about 1.4.

We can compare the isobaric, isothermal and adiabatic work of expansion of an ideal gas to demonstrate how work depends on path.

Adiabatic Ideal Gas Expansion Work

Reduction rules

Ideal gas adiabatic expansion work

ClearAll["`*"]
Print["adiabatic rule: ", adiabaticRule = {Pin*Vi^γ->Pf*Vf^γ}]
Print["Ideal gas adiabatic pressure = ", Pad = P/.Solve[P*V^γ == Pin*Vi^γ, P][[1]]]
Print["Ideal gas adiabatic expansion work = ", Simplify[Integrate[Pad, {V,Vi,Vf}]/.adiabaticRule]]

Let's show numerically and graphically how the work differs for the case of expansion of one mole of ideal gas (air) isobarically, isothermally and adiabatically from an initial volume of 1 Liter at 298.15 Kelvin to a final volume of 5 Liters.

Ideal Gas Expansion Work

ClearAll["`*"]
<<Miscellaneous`Units`
n = 1 Mol;
R = .082059 Liter Atmosphere/(Mol Kelvin);
Ti = 298.15 Kelvin;
Print["Initial pressure = ", Pinitial = 5 Atmosphere]
Print["Final pressure = ", Pfinal = 1 Atmosphere]
Print["γ = ", γ = 1.4]
Print["Initial volume = ", Vi = V/.Solve[Pinitial*V == n*R*Ti, V][[1]]]

Clear[P,V]
Print["Final temperature = ", Tf = Ti]
Print["Final volume = ", Vf = V/.Solve[Pfinal*V == n*R*Tf, V][[1]]]
Print["Ideal gas isobaric pressure = ", Pib = Pfinal]

Print["Isobaric expansion work = ", Convert[Integrate[Pib, {V,Vi,Vf}],Joule]]
isobaricPlot = Plot[Pib/Atmosphere, {V,Vi/Liter,Vf/Liter}, DisplayFunction->Identity];

Clear[P]
Print["Final temperature = ", Tf = Ti]
Print["Final volume = ", Vf = V/.Solve[Pfinal*V == n*R*Tf, V][[1]]]
Print["Ideal gas isothermal pressure = ", Pig = P/.Solve[P*V == n*R*Ti, P][[1]]]

Print["logrule: ", logrule = {m_*Log[a_]-m_*Log[b_] -> m*Log[a/b]}]
Print["Ideal gas isothermal expansion work = ", Convert[Integrate[Pig, {V,Vi,Vf}]//Simplify/.logrule//Chop,Joule]]
isothermalPlot = Plot[Pig/Atmosphere/Liter, {V,Vi/Liter,Vf/Liter}, DisplayFunction->Identity];

Clear[P,T,V]
Print["Final temperature = ", Tf = T/.Solve[T/Ti == (Pfinal/Pinitial)^((γ-1)/γ), T][[1]]]
Print["Final volume = ", Vf = V/.Solve[Pfinal*V == n*R*Tf, V][[1]]]

Print["Ideal gas adiabatic pressure = ", Pad = P/.Solve[P*V^γ == n*R*Ti*Vi^(γ-1), P][[1]]]

Print["Ideal gas adiabatic expansion work = ", Convert[Integrate[Pad, {V,Vi,Vf}],Joule]]
adiabaticPlot = Plot[Pad/Atmosphere/Liter^γ, {V,Vi/Liter,Vf/Liter}, DisplayFunction->Identity];

Show[isobaricPlot, isothermalPlot, adiabaticPlot, PlotRange->{0,5},DisplayFunction->$DisplayFunction];

Note that the maximum work is obtained isothermally, about double the minimum work done adiabatically in this case.

Exercises

  1.  How high can a 70 Kg hiker climb using the energy from a 100 g candy bar?

  2.  How much water must a human perspire in one day to maintain normal body temperature (98.6 F), given that their body generates 6 kJ/kg-hr?

  3.  Confirm Joule's prediction that the water at the base of Niagra Falls in America should be about 1/5 degree Fahrenheit warmer than that at the top 50 meters higher.

  4.  What final temperature results from adding 20 g of ice at 0 Celsius to 200 g or water at 20 celsius?

  5.  What is the percentage difference in the work done by one mole of dinitrogen at 3000 C expanding from 1 Liter to 5 Liters as an ideal gas the work done as a van der Waals gas?  (a[N2] = 1.390 atm-L^2/mol^2, b[N2] = 0.03913 L/mol)

  6.  Obtain a formula for the work of an ideal gas undergoing an adiabatic expansion, for which P= constant.

  7. Explain how the expenditure of calories during a mountain hike is consistent with the First Law of Thermodymaics.

  8. Show that for an ideal gas adiabatic process P = constant .