Thermodynamics at the Molecular Level

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Such relationships are particularly useful for systems where, for example, it might be impossible to keep the volume of the system constant. The constantvolume derivative can be expressed in terms of derivatives at constant temperature and constant pressure, two conditions that are easy to control in any laboratory setting.

1.9 Thermodynamics at the Molecular Level

A fundamental concept in chemistry is that matter is ultimately composed of atoms and molecules. As such, any model in chemistry must be consistent with that con

cept. Here, we will consider how thermodynamics is impacted by the atomic theory.

We begin by considering a completely unrelated concept: the pressure of the earth’s atmosphere versus altitude. The atmosphere, a gas, is subject to the force of gravity, as given by

F 5 mg

where m is the mass of an atmosphere particle and g is the acceleration due to grav

ity, which is about 9.81 m/s2 at the surface of the earth. Because g varies with dis

tance above the earth’s surface, so does the density, r, of the atmosphere. Consider a column of air that has area A, as illustrated in Figure 1.12. At a certain height h, we will mark an infinitesimal increase in height, dh. The infinitesimal volume of that infinitesimal increase in height is A # dh, and the mass of the infinitesimal volume is the density times the volume, or r# A # dh. We can now determine the infinitesimal force on this mass. However, in order for the gas to be staying at that altitude, it must be exerting a force against that of gravity, so we must include a minus sign with the expression. The force that the gas is exerting is thus

dF5 2dm # g5 2(r # A # dh) # g

Because pressure is force divided by area, the infinitesimal pressure of this sample of gas is

dp5 dF

A 5 2r # A # dh # g

A 5 2r # g # dh (1.29)

where in the last part we have rearranged the variables to put the infinitesimal last, as is customary.

To determine the total pressure due to height, we integrate both sides of this equation. The limits are 1 to p for the pressure variable and 0 to h for the height variable. Thus, we have

dp53

h 0

(2r # g # dh)

We can use the ideal gas law to determine an expression for the density of the gas.

Recall the definition of density:

r 5 m V 5 nM

where n is the number of moles of gas and M is its molar mass. Rearranging the ideal gas law:

pV 5nRT S n V5 p

RT Now we substitute for the n/V part of the density:

r 5 pM RT

A }dh

Figure 1.12 A model of a column of air, used to determine how the gas pres

sure varies with altitude.

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22 Chapter 1 | Gases and the Zeroth Law of Thermodynamics

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Substituting this expression into the integral on the right:

dp53

h 0

a2pM

RT #g#dhb

Bringing the pressure variable over to the other integral, we rearrange this to get 3

dp p 5 23

h 0

aMg RTb dh

Approximating that g and T are constant with height, each of these expressions can be integrated to yield

ln p

1 atm 5 2Mgh RT which can be rearranged to get

p5e2Mgh/RT atm (1.30)

Thus, atmospheric pressure follows, roughly, a negative exponential function.

Equation 1.30 is called the barometric formula.

Negative exponential functions show up many times in physical chemistry. The exponent itself is always a pure number; that is, it is unitless. Thus, the combined units of Mgh have the same combined units as RT. In this case, both units are units of energy (in standard units, joules). The RT term is important in thermodynamics, and it is often referred to as thermal energy: It represents an amount of energy due to temperature. The fact that Mgh also has units of energy (in fact, the product rep

resents the gravitational potential energy of the infinitesimal volume) suggests that we can replace the numerator in exponent with any kind of energy; hence, we see

property~e2E/RT

(where the “~” symbol means “is proportional to”) very commonly in science. This negative exponential expression is called the Boltzmann factor after Ludwig Boltz

mann, who first derived it.

Atoms and molecules contain energy. As such, some of their properties can be related to the Boltzmann factor. For example, gasphase atoms and molecules have translational energy, whose value is ẵmv2. Thus, properties of gases can, in part, be related as

property~e2ẵmvRT25e2mv2/2RT

One of the important properties that can be related to the Boltzmann factor is the probability of having a particular energy because of temperature. If a particular state has energy E, a fundamental concept of thermodynamics states that the probability of a particle being in that state is equal to

probability5e2DE/RT (1.31)

where DE represents the difference in energy from the energy minimum; if the energy minimum is zero, then DE equals the absolute energy, E. Using this idea, we can actually calculate the average energy of a sample of matter. It is not as simple, however, as adding up the energy values for each particle and then dividing by the number of particles—it is impossible to know the energy of each particle individually! However, there is another way to determine an average. If Pi represents the probability of having a certain value I of a property, then the average value 8I9 is given by

8I95 gipi#I

giPi (1.32)

This example illustrates how to use this formula.

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1.9 | Thermodynamics at the Molecular Level 23

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ExamplE 1.9

In a class of four students, scores on a quiz were 5, 5, 5, and 10 out of 10 points.

Calculate the average score using equation 1.32.

Solution

Before using equation 1.32, let us determine the average the normal way:

average551515110

4 5 25

4 56.25

We expect, therefore, that equation 1.32 should give the same result. First, we need the probabilities, Pi, for each value. Of the four values, there are three values of 5, so the probability of having a 5 is three out of four, or ắ. There is one score of 10 out of four, so the probability of having 10 as a score is ẳ. Using equation 1.32, there are only two terms in each sum. Thus, we have

average5 giPi #I giPi 5

4b#51 a1 4b#10 a3

4b 1 a1 4b

average5 a15

4b 1 a10 4b

1 5 a15

4b1 a10 4b5 25

4 56.25

This verifies that equation 1.32 is an appropriate way to calculate an average value.

This is the common way of calculating an average: Sum the values and divide by the number of values.

Each summation is only two terms in this example, because we only have two discrete scores, 5 and 10.

Let us calculate, then, the average translational energy of a gas due to tempera

ture, known as its thermal energy. Consider onedimensional translations, in which the gas particles are moving in a single dimension. We already know the value of the translational (that is, kinetic) energy, ẵmv2. Thus, using the Boltzmann factor as a probability, the average energy is given by

8E9 5 giPi # E

giPi 5 gie2mv2/2RT # ẵmv2 gie2mv2/2RT

where each summation is over each possible value of the velocity. While this expres

sion looks daunting, it can be simplified in two ways. First, because for all practical purposes the velocity can have any value, the discrete sums in the fractions can be substituted with integrals (which is, mathematically, what an integral is: a smooth, rather than discrete, sum). We have

8E95 3

0e2mv2/2RT# ẵmv2dv 3

e2mv2/2RT

Note how the sum of all probabilities in the denominator equals 1. This is com- monly, but not always, the case and sim- plifies the math.

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24 Chapter 1 | Gases and the Zeroth Law of Thermodynamics

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The second simplification is to recognize that both of these integrals have known solutions (see Appendix 1), so we can substitute these solutions into the fractions.

Substituting the solutions, we get

8E9 5 1 2m°1

4 ọ

p a m

2RTb3

1 2

ọ p a m

2RTb Although complicatedlooking, this fraction simplifies to

8E9 51

2RT for one dimension (1.33) Space has three dimensions, and each dimension contributes ẵRT to the translaư

tional energy of a gas. This concept is called the equipartition principle, the idea that each dimension (or “degree of freedom”) contributes equally to the overall energy of the particle. Thus, the total translational energy of a gas is

8E953/2RT (1.34)

For gases whose particles are just atoms (like helium, neon, and argon) that only have translational energies, this result is very useful and its implications can be veri

fied experimentally.

Molecular gases possess other forms of energy. For example, molecules can rotate in threedimensional space, rotating about three different spatial axes (only two different spatial axes for linear molecules, because a rotation about the molecular axis is not defined). For a rotating body, the energy of rotation is

E 5 ẵIv2

where I is a quantity called the moment of inertia (which depends on mass) and v is the rotational velocity. Note that the expression for the rotational energy is analo

gous to the expression for the translational energy: onehalf times a massrelated quantity times a squared velocity. Because of this, the analysis we performed above for translations (including the separation into one dimension at a time) is also anal

ogous, as is the result:

8E953/2RT for a nonlinear molecule

8E95RT for a linear molecule (1.35) This is another example of the equipartition principle, and its implications about the thermodynamic properties of molecules are also verifiable experimentally.

Energy can also be stored in motions that are described as vibrations: Atoms in a molecule “wiggle” back and forth with respect to each other. For a molecule with N atoms, there are 3N 2 6 different ways for the atoms to vibrate (3N 2 5 if the molecule is linear). The population of vibrational energies can be predicted by the Boltzmann factor, just like the other ways molecules have to store energy. However, there are two crucial differences. First, vibrational energies cannot be treated as con

tinuous. Rather, they must be treated as separate because allowable vibrational ener

gies can, in some instances, be much greater than RT. As a consequence, we cannot turn our summations for the average in equation 1.32 into integrals. Second, ideal vibrational energies are equally spaced. That is, if our first nonzero vibrational energy level is at E, our second nonzero vibrational energy level is at 2E, our third vibrational energy level is at 3E, and so forth. We will take advantage of this mathematically.

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1.9 | Thermodynamics at the Molecular Level 25

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To calculate the average thermal energy due to vibrations, we use the definition of an average:

8E95 giPi#E

giPi 5 0#e20/RT1E#e2RTE 12E#e22ERT13E#e2RT3E1 # # # e20/RT1e2RTE 1e22ERT1e23ERT1 # # #

Here, we are assuming that the lowest vibrational energy is zero (which, as we will find out later, is technically incorrect, but the error in our derivation will be un

noticeable). Note that the first term in the numerator will be exactly zero, while the first term in the denominator will be exactly one. If we substitute for x 5 e2E/RT, this average energy can be rewritten as

8E9 5 x#E#(112x 13x214x31###) 11x1x21x31###

In the limit of an infinite number of terms, each summation in the above fraction can be rewritten in a more simple form. The summation in the numerator is the expansion for (1 2 x)22, while the summation in the denominator is the expansion for (1 2 x)21. Substituting:

8E95 x#E#(12x)22 (12x)21 One of the (1 2 x) terms cancels, leaving

8E95x#E#(12x)21 8E95 x #E

(12x) Replacing for the value of x, this becomes

8E9 5 e2E/RT#E 112e2E/RT2 which simplifies to

8E95 E

eE/RT21 (1.36)

for each of the 3N 2 6 (or 3N 2 5) independent vibrations of an Natom molecule.

Thus, the average vibrational thermal energy of a molecule is a sum of 3N 2 6 (or 5) expressions of equation 1.36.

This expression can be considered at its two extremes. If the energy of the vibra

tion E is much larger than RT (which at 298 K equals 2480 J/mol), then e E/RT is very large, the denominator of the fraction in equation 1.36 is very large, and 8E9 < 0.

This implies that there will be no thermal energy stored by the vibrations of the molecule. However, if E is small compared to RT, then e E/RT is close to 1 and can be approximated as

eEyRT<11 E RT 1 1

2!a E

RTb21 1 3!a E

RTb31c

If we only keep the first two terms of this approximation, we can substitute into equation 1.36 to get

8E9< E 11 E

RT21

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26 Chapter 1 | Gases and the Zeroth Law of Thermodynamics which simplifies to

8E9 5RT for each of the lowenergy vibrations.

Many molecules have a mixture of lowenergy and highenergy vibrations, so experimental measurements of the thermal energies and other properties of vibrat

ing molecules were historically difficult to understand. Now that it is understood that vibrational energies cannot be modeled as continuous and have both small and large values, we know the reason why molecular gases have the thermodynamic properties that they do. We will look more into this topic in chapters on quantum mechanics and statistical thermodynamics, two additional important topics in physical chemistry.

K E Y E Q u a t i o n S

pV5nRT (ideal gas law)

Z5 pV

RT (definition of compressibility)

Z5 pV

RT511 B V1 C

V21c (virial equation in terms of volume) Z5 pV

RT511 Brp

RT 1Crp2

RT 1c (virial equation in terms of pressure) ap1 an2

V2b(V2nb)5nRT (van der Waals equation of state) TB5 a

bR (Boyle temperature estimation)

Gases are introduced first in a detailed study of thermodynamics because their behavior is simple. Boyle enunciated his gas law about the relationship between pressure and volume in 1662, making it one of the oldest of modern chemical principles. Although it is certain that not all of the “simple” ideas have been dis

covered, in the history of science, the more straightforward ideas were developed first. Because the behavior of gases was so easy to understand, even with more com

plicated equations of state, they became the systems of choice for studying other state variables. Also, the calculus tool of partial derivatives is easy to apply to the behavior of gases. As such, a discussion of the properties of gases is a fitting intro

ductory topic for the subject of thermodynamics. A desire to understand the state of a system of interest, which includes state variables not yet introduced and uses some of the tools of calculus, is at the heart of thermodynamics. We will proceed to develop such an understanding in the next seven chapters.

1 . 1 0 S u m m a r Y

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EXERCISES FOR CHAPTER 1 27

a'A 'Bb

5 a'A 'Db

Ea'D 'Bb

1 a'A 'Eb

Da'E 'Bb

C (chain rule for partial derivatives) a'A

'Cb

Ba'B 'Ab

Ca'C 'Bb

A5 21 (cyclic rule for partial derivatives) a 5 1

Va'V 'Tb

p (definition of expansion coefficient)

k 5 21 Va'V

'pb

T (definition of isothermal

compressibility)

P5e2Mgh/RT atm (barometric formula)

8I95 giPi#I

giPi (average value of I)

8E9 53/2RT (average translational energy of a

gas; average rotational energy for a nonlinear gas)

8E9 5RT (average rotational energy for a

linear gas) 8E9 5 E

eE/RT21 (average vibrational energy for each

vibration of a molecular gas)

1.2 System, Surroundings, and State

1.1. A bomb calorimeter is a sturdy metal vessel in which samples can be ignited and the amount of heat given off can be measured as the heat warms up surrounding water. Draw a rough sketch of such an experimental setup and label (a) the system and (b) the surroundings.

1.2. Differentiate between the system and the surroundings.

Give examples of both.

1.3. What does it mean to speak of a closed system? Give an example.

1.4. Use the equalities listed in Table 1.1 to convert the given values to the desired units. (a) 12.56 L to cm3 (b) 45°C to K (c) 1.055 atm to Pa (d) 1233 mmHg to bar (e) 125 mL to cubic centimeters (f) 4.2 K to °C (g) 25,750 Pa to bar

1.5. Which temperature is higher? (a) 0 K or 0°C (b) 300 K or 0°C (c) 250 K or 220°C

1.6. A standard drinking straw is 23 cm tall. What pressure difference is needed to raise a column of water that high? (When we drink, this pressure difference is supplied by one’s mouth.) Assume the density of water is 1.0 g/cm3. You will need to use F 5 mg, where g 5 9.80 m/s2.

E x E r C i S E S F o r C H a p t E r 1

1.7. Referring to the previous exercise: Without special pressurized equipment, the maximum pressure difference that can be generated is 1 atm, or 101,325 pascals. How high a column of water can this pressure difference support? What is this approximate height in feet?

1.3 & 1.4 Zeroth Law of Thermodynamics;

Equations of State

1.8. A pot of cold water is heated on a stove, and when the water boils, a fresh egg is placed in the water to cook.

Describe the events that are occurring in terms of the zeroth law of thermodynamics.

1.9. What difference is necessary for heat to flow between two systems? Can you think of an exception to your answer?

1.10. What is the value of F(T) for a sample of gas whose volume is 2.97 L and pressure is 0.0553 atm? What would the volume of the gas be if the pressure were increased to 1.00 atm?

1.11. What is the value of F(p) for a sample of gas whose temperature is 233.0°C and volume is 0.0250 L? What temperature is required to change the volume to 66.9 cm3?

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28 Chapter 1 | Gases and the Zeroth Law of Thermodynamics

1.12. Calculate the value of the constant in equation 1.9 for a 1.887-mol gas sample with a pressure of 2.66 bar, a volume of 27.5 L, and a temperature of 466.9 K. Compare your answer to the values in Table 1.2. Are you surprised with your answer?

1.13. Hydrogen gas is used in weather balloons because it is less expensive than helium. Assume that 5.57 g of H2 is used to fill a weather balloon to an initial volume of 67 L at 1.04 atm. If the balloon rises to an altitude where the pressure is 0.047 atm, what is its new volume? Assume that the temperature remains constant.

1.14. A weather balloon having an initial volume of 67 L and 1.04 atm is weighted down and dropped into the ocean.

Every 10.1 m, water pressure increases by 1 atm pressure.

What is the volume of the balloon after it sinks 64.0 m below the water’s surface? Assume that the temperature remains constant.

1.15. A 2.0-L soda bottle is pressurized with 4.5 atm of CO2 at 298 K. If the temperature is increased to 317 K, what is the pressure of the CO2?

1.16. The Mount Pinatubo volcano eruption in 1991 released an estimated 1.82 3 1013 g of SO2 into the atmosphere. If the gas had an average temperature of 217.0°C and filled the troposphere, whose approximate volume is 8 3 1021 L, what is the approximate partial pressure of SO2 caused by the eruption?

1.17. Show that one value of R, with its associated units, equals another value of R with its different associated units.

1.18. Scottish physicist W. J. M. Rankine proposed an absolute temperature scale based on the Fahrenheit degree, now called degrees Rankine (abbreviated R) and used by some engineering fields. If a degree Rankine is 5/9 of a degree Kelvin, what is the value of the ideal gas law constant in L # atm/mol # R?

1.19. Use the two appropriate values of R to determine a conversion between L # atm and J.

1.20. Calculations using STP and SATP use (the same? different?) value(s) of R. Choose one phrase to make the statement correct and defend your choice.

223.8 K With an initial temperature of 273.15 K, the final temperature is about 249 K

The Carnot Cycle and Efficiency