Thermodynamics Molecular Model of a Gas Molar Heat...

ThermodynamicsMolecular Model of a Gas

Molar Heat Capacities

Lana Sheridan

De Anza College

May 3, 2016

Last time

• heat transfer mechanisms

• modeling an ideal gas at the microscopic level

• pressure, temperature, and internal energy from microscopicmodel

Overview

• modeling an ideal gas at the microscopic level

• equipartition of energy

• rms speed of molecules

• heat capacities for ideal gases

• adiabatic processes

• the Boltzmann (Gibbs) distribution

Kinetic Theory of Gases

Previously, we studied what happens in thermodynamic systems tobulk properties in various transformations.

Now we want to understand how these macroscopic quantitiesarise from the microscopic behavior of particles, on average.

We cannot model every the motion of every single particle in asubstance, but we can say a lot about the ensemble of particlesstatistically.

Molecular Model of an Ideal Gas

21.1 Molecular Model of an Ideal Gas 627

21.1 Molecular Model of an Ideal GasIn this chapter, we will investigate a structural model for an ideal gas. A structural model is a theoretical construct designed to represent a system that cannot be observed directly because it is too large or too small. For example, we can only observe the solar system from the inside; we cannot travel outside the solar system and look back to see how it works. This restricted vantage point has led to different historical structural models of the solar system: the geocentric model, with the Earth at the center, and the heliocentric model, with the Sun at the center. Of course, the latter has been shown to be correct. An example of a system too small to observe directly is the hydrogen atom. Various structural models of this system have been devel-oped, including the Bohr model (Section 42.3) and the quantum model (Section 42.4). Once a structural model is developed, various predictions are made for experimen-tal observations. For example, the geocentric model of the solar system makes pre-dictions of how the movement of Mars should appear from the Earth. It turns out that those predictions do not match the actual observations. When that occurs with a structural model, the model must be modified or replaced with another model. The structural model that we will develop for an ideal gas is called kinetic the-ory. This model treats an ideal gas as a collection of molecules with the following properties:

1. Physical components: The gas consists of a number of identical molecules within a cubic con-

tainer of side length d. The number of molecules in the gas is large, and the average separation between them is large compared with their dimensions. Therefore, the molecules occupy a negligible volume in the container. This assumption is consistent with the ideal gas model, in which we imagine the molecules to be point-like.

2. Behavior of the components: (a) The molecules obey Newton’s laws of motion, but as a whole their motion

is isotropic: any molecule can move in any direction with any speed. (b) The molecules interact only by short-range forces during elastic colli-

sions. This assumption is consistent with the ideal gas model, in which the molecules exert no long-range forces on one another.

(c) The molecules make elastic collisions with the walls.

Although we often picture an ideal gas as consisting of single atoms, the behavior of molecular gases approximates that of ideal gases rather well at low pressures. Usu-ally, molecular rotations or vibrations have no effect on the motions considered here. For our first application of kinetic theory, let us relate the macroscope variable of pressure P to microscopic quantities. Consider a collection of N molecules of an ideal gas in a container of volume V. As indicated above, the container is a cube with edges of length d (Fig. 21.1). We shall first focus our attention on one of these molecules of mass m0 and assume it is moving so that its component of velocity in the x direction is vxi as in Figure 21.2. (The subscript i here refers to the ith mol-ecule in the collection, not to an initial value. We will combine the effects of all the molecules shortly.) As the molecule collides elastically with any wall (property 2(c) above), its velocity component perpendicular to the wall is reversed because the mass of the wall is far greater than the mass of the molecule. The molecule is mod-eled as a nonisolated system for which the impulse from the wall causes a change in the molecule’s momentum. Because the momentum component pxi of the molecule is m0vxi before the collision and 2m0vxi after the collision, the change in the x com-ponent of the momentum of the molecule is

Dpxi 5 2m0vxi 2 (m0vxi) 5 22m0vxi (21.1)

d

d dz x

y

m 0

vxi

viS

One molecule of the gas moves with velocity v on its way toward a collision with the wall.

S

Figure 21.1 A cubical box with sides of length d containing an ideal gas.

Figure 21.2 A molecule makes an elastic collision with the wall of the container. In this construc-tion, we assume the molecule moves in the xy plane.

vyi

vxi

vyi

–vxi

viS

viS

The molecule’s x component of momentum is reversed, whereas its y component remains unchanged.

We modeled the particles of gas as small, identical, and obeyingNewton’s laws, with no long range interactions.

We assumed all collisions are elastic.

1Figure from Serway & Jewett.

Pressure from the Molecular Model

Pressure, P, can be related to the average translational kineticenergy, K̄trans, of the molecules of a gas.

P =2

3

N

VK̄trans

More K.E., or more molecules in less volume ⇒ higher pressure.

Relation to Macroscopic view of an Ideal Gas

Ideal gas equation:PV = nRT

or equivalently:PV = NkBT

If we put our new expression for pressure into this equation:

2

3NK̄trans = NkBT

We can cancel N from both sides and re-arrange:

K̄trans =1

2m0v2 =

3

2kBT

Temperature from the Molecular Model

We can also relate temperature to molecular motion!

T =2

3kbK̄trans

Temperature is directly proportional to the translational kineticenergy of the particles.

RMS Speed and Temperature

K̄ =1

2m0v2 =

3

2kBT

It would also be useful to express the average molecular speed interms of the temperature.

Since the motion of the gas molecules are isotropic, the averagevelocity is zero.

However, we can instead consider the root-mean-square (rms)speed.

That is convenient here because we have the average of thesquares of the speed, not the average speed itself.

RMS Speed and Temperature

root-mean-square (rms) speed:

vrms =√

v2 =

√3kBT

m0

Alternatively, it can be expressed

vrms =

√3RT

M

where M is the molar mass.

rms speed is higher for less massive molecules for a giventemperature.

RMS Speed Question

An ideal gas is maintained at constant pressure. If the temperatureof the gas is increased from 200 K to 600 K, what happens to therms speed of the molecules?

(A) It increases by a factor of 3.

(B) It remains the same.

(C) It is one-third the original speed.

(D) It is√

3 times the original speed.

1Serway & Jewett, page 644, question 2.

Equipartition of EnergyIn physics, the number of degrees of freedom a system has is thenumber of real number variables we need to specify to describe asystem.

For our purposes here, we can say that each degree of freedomcounts another way that a particle can possess energy.

It can move in the x-direction, having kinetic energy, but also inthe y and z directions. That’s 3 ways. 3 degrees of freedom.

Degrees of freedom count rotational and vibrational motion as wellas translational K.E.

Equipartition of energy theorem

Each degree of freedom for a molecule contributes an andadditional 1

2kBT of energy to the system.

Equipartition of Energy


Each degree of freedom for a molecule contributes an additional12kBT of energy to the system.

Strictly, this only is proven to hold for systems in thermalequilibrium that are ergodic, meaning all microstates (states of allthe particles) are equally probable over long periods of time.

It can be proven starting from the Boltzmann distribution ofenergies (to come).

This assumes a continuum of possible energies, so we expectproblems when we are in settings where the thermal energy kBT ismuch less than the energy spacing between energy levels predictedby quantum mechanics.

Kinetic Energy and Internal Energy

The total kinetic energy of an ideal monatomic gas of N particlesis the total translational K.E.

Ktot,trans =3

2NkBT =

3

2nRT

In a monatomic gas these are the three translational motions arethe only degrees of freedom. We can choose

Eint = Ktot,trans

(This is the thermal energy, so the bond energy is zero – if weliquify the gas the bond energy becomes negative.)

Question

Quick Quiz 21.11 Two containers hold an ideal gas at the sametemperature and pressure. Both containers hold the same type ofgas, but container B has twice the volume of container A.

(i) What is the average translational kinetic energy per molecule incontainer B?

(A) twice that of container A

(B) the same as that of container A

(C) half that of container A

(D) impossible to determine

1Serway & Jewett, page 631.

Question

Quick Quiz 21.11 Two containers hold an ideal gas at the sametemperature and pressure. Both containers hold the same type ofgas, but container B has twice the volume of container A.

(ii) From the same choices, describe the internal energy of the gasin container B.

(A) twice that of container A

(B) the same as that of container A

(C) half that of container A

(D) impossible to determine


Another Look at Heat CapacityWe already studied specific heat, c = Q

m∆T , particularly for solidsand liquids.

Now we must revisit this concept, because for gases there aremany ways to change the temperature of a gas by ∆T , withdifferent ways requiring different amounts of heat, Q.

520 CHAPTE R 19 TH E KI N ETIC TH EORY OF GAS E S

HALLIDAY REVISED

Molar Specific Heats at Constant Volume

CV

Molecule Example (J/mol ! K)

MonatomicIdeal R " 12.5

RealHe 12.5Ar 12.6

DiatomicIdeal R " 20.8

Real N2 20.7O2 20.8

PolyatomicIdeal 3R " 24.9

Real NH4 29.0CO2 29.7

52

32

Table 19-2

A change in the internal energy Eint of a confined ideal gas depends on only thechange in the temperature, not on what type of process produces the change.

Fig. 19-10 Three paths representingthree different processes that take an idealgas from an initial state i at temperature Tto some final state f at temperature T #$T. The change $Eint in the internal energyof the gas is the same for these threeprocesses and for any others that result inthe same change of temperature.

Pres

sure

Volume

i

f

T + T ∆

T

f

f

1

2

3

The paths are different,but the change in theinternal energy is thesame.

As examples, consider the three paths between the two isotherms in the p-V dia-gram of Fig. 19-10. Path 1 represents a constant-volume process. Path 2 represents aconstant-pressure process (that we are about to examine). Path 3 represents a processin which no heat is exchanged with the system’s environment (we discuss this in Section19-11). Although the values of heat Q and work W associated with these three pathsdiffer, as do pf and Vf , the values of $Eint associated with the three paths are identicaland are all given by Eq. 19-45, because they all involve the same temperature change$T.Therefore,no matter what path is actually taken between T and T # $T,we can al-ways use path 1 and Eq.19-45 to compute $Eint easily.

energy to the gas as heat Q by slowly turning up the temperature of the thermalreservoir. The gas temperature rises a small amount to T # $T, and its pressurerises to p # $p, bringing the gas to final state f. In such experiments, we wouldfind that the heat Q is related to the temperature change $T by

Q " nCV $T (constant volume), (19-39)

where CV is a constant called the molar specific heat at constant volume. Substi-tuting this expression for Q into the first law of thermodynamics as given by Eq. 18-26 ($Eint " Q % W ) yields

$Eint " nCV $T % W. (19-40)

With the volume held constant, the gas cannot expand and thus cannot do anywork.Therefore, W " 0, and Eq. 19-40 gives us

(19-41)

From Eq. 19-38, the change in internal energy must be

(19-42)

Substituting this result into Eq. 19-41 yields

(monatomic gas). (19-43)

As Table 19-2 shows, this prediction of the kinetic theory (for ideal gases) agreesvery well with experiment for real monatomic gases, the case that we haveassumed. The (predicted and) experimental values of CV for diatomic gases(which have molecules with two atoms) and polyatomic gases (which have mole-cules with more than two atoms) are greater than those for monatomic gases forreasons that will be suggested in Section 19-9.

We can now generalize Eq. 19-38 for the internal energy of any ideal gas bysubstituting CV for R; we get

Eint " nCVT (any ideal gas). (19-44)

This equation applies not only to an ideal monatomic gas but also to diatomicand polyatomic ideal gases, provided the appropriate value of CV is used. Just aswith Eq. 19-38, we see that the internal energy of a gas depends on the temper-ature of the gas but not on its pressure or density.

When a confined ideal gas undergoes temperature change $T, then from ei-ther Eq. 19-41 or Eq. 19-44 the resulting change in its internal energy is

$Eint " nCV $T (ideal gas, any process). (19-45)

This equation tells us:

32

CV " 32R " 12.5 J/mol!K

$Eint " 32nR $T.

CV "$Eint

n $T.

halliday_c19_507-535v2.qxd 28-10-2009 15:56 Page 520

1Diagram from Halliday, Resnick, Walker, 9th ed, page 520.

Another Look at Heat Capacity


HALLIDAY REVISED


CV



RealHe 12.5Ar 12.6


Real N2 20.7O2 20.8


Real NH4 29.0CO2 29.7

52

32

Table 19-2



Pres

sure

Volume

i

f

T + T ∆

T

f

f

1

2

3






$Eint " nCV $T % W. (19-40)


(19-41)


(19-42)










32

CV " 32R " 12.5 J/mol!K

$Eint " 32nR $T.

CV "$Eint

n $T.


For processes 1, 2, and 3:

Q1 = c1m∆T

Q2 = c2m∆T

Q3 = c3m∆T

Each process has a different value of c!

Molar Specific Heat of an Ideal Gas

For solids and liquids, heat capacity, C , and specific heat capacity,c , are defined to be for constant pressure processes.

This is purely because it is very difficult to stop a solid or liquidfrom expanding and maintain it at a constant volume!(∆V = βVi∆T )

However, it is possible to measure the heat capacities of solids andliquids at constant volume instead of constant pressure.

When that is done, small differences in the values of the heatcapacity are obtained.


In gases, the variation of the heat capacity obtained for differentpaths is quite big.

Define:

Molar heat capacity at constant volume, CV

Along an isovolumetric process (constant volume):

Q = nCV ∆T

Molar heat capacity at constant pressure, CP

Along an isobaric process (constant pressure):

Q = nCP ∆T

(These are both intensive quantities, like specific heat.)


We have defined molar heat capacities (cap. per mole) here andnot specific heat capacities (cap. per mass).

Why?

Using “per mole” as the reference for heat capacity allows us totalk about many different gases with the same relationships, sincewe will always be talking about the same number of molecules.

It is just more convenient.

Question

21.2 Molar Specific Heat of an Ideal Gas 633

P

VT

i

f

f !Isotherms

T " #T

For the constant-volume path, all the energy input goes into increasing the internal energy of the gas because no work is done.

Along the constant-pressure path, part of the energy transferred in by heat is transferred out by work.

Figure 21.4 Energy is trans-ferred by heat to an ideal gas in two ways.

Let’s now apply the results of this discussion to a monatomic gas. Substituting the internal energy from Equation 21.25 into Equation 21.28 gives

CV 5 32R 5 12.5 J/mol # K (21.29)

This expression predicts a value of CV 5 32R for all monatomic gases. This predic-

tion is in excellent agreement with measured values of molar specific heats for such gases as helium, neon, argon, and xenon over a wide range of temperatures (Table 21.2). Small variations in Table 21.2 from the predicted values are because real gases are not ideal gases. In real gases, weak intermolecular interactions occur, which are not addressed in our ideal gas model. Now suppose the gas is taken along the constant-pressure path i S f 9 shown in Figure 21.4. Along this path, the temperature again increases by DT. The energy that must be transferred by heat to the gas in this process is Q 5 nCP DT. Because the volume changes in this process, the work done on the gas is W 5 2P DV, where P is the constant pressure at which the process occurs. Applying the first law of thermodynamics to this process, we have

DE int 5 Q 1 W 5 nCP DT 1 (2P DV) (21.30)

In this case, the energy added to the gas by heat is channeled as follows. Part of it leaves the system by work (that is, the gas moves a piston through a displacement), and the remainder appears as an increase in the internal energy of the gas. The change in internal energy for the process i S f 9, however, is equal to that for the pro-cess i S f because E int depends only on temperature for an ideal gas and DT is the same for both processes. In addition, because PV 5 nRT, note that for a constant- pressure process, P DV 5 nR DT. Substituting this value for P DV into Equation 21.30 with DE int 5 nCV DT (Eq. 21.27) gives

nCV DT 5 nCP DT 2 nR DT

CP 2 CV 5 R (21.31)

This expression applies to any ideal gas. It predicts that the molar specific heat of an ideal gas at constant pressure is greater than the molar specific heat at constant vol-ume by an amount R, the universal gas constant (which has the value 8.31 J/mol ? K). This expression is applicable to real gases as the data in Table 21.2 show.

Table 21.2 Molar Specific Heats of Various GasesMolar Specific Heat ( J/mol ? K)a

Gas CP CV CP 2 CV g 5 CP/CV

Monatomic gasesHe 20.8 12.5 8.33 1.67Ar 20.8 12.5 8.33 1.67Ne 20.8 12.7 8.12 1.64Kr 20.8 12.3 8.49 1.69

Diatomic gasesH2 28.8 20.4 8.33 1.41N2 29.1 20.8 8.33 1.40O2 29.4 21.1 8.33 1.40CO 29.3 21.0 8.33 1.40Cl2 34.7 25.7 8.96 1.35

Polyatomic gasesCO2 37.0 28.5 8.50 1.30SO2 40.4 31.4 9.00 1.29H2O 35.4 27.0 8.37 1.30CH4 35.5 27.1 8.41 1.31

a All values except that for water were obtained at 300 K.

Quick Quiz 21.22 (i) How does the internal energy of an ideal gaschange as it follows path i → f ?

(A) Eint increases.

(B) Eint decreases.

(C) Eint stays the same.

(D) There is not enough information to determine how Eint

changes.2Serway & Jewett, page 631.

Question

21.2 Molar Specific Heat of an Ideal Gas 633

P

VT

i

f

f !Isotherms

T " #T

For the constant-volume path, all the energy input goes into increasing the internal energy of the gas because no work is done.

Along the constant-pressure path, part of the energy transferred in by heat is transferred out by work.

Figure 21.4 Energy is trans-ferred by heat to an ideal gas in two ways.

Let’s now apply the results of this discussion to a monatomic gas. Substituting the internal energy from Equation 21.25 into Equation 21.28 gives

CV 5 32R 5 12.5 J/mol # K (21.29)

This expression predicts a value of CV 5 32R for all monatomic gases. This predic-

tion is in excellent agreement with measured values of molar specific heats for such gases as helium, neon, argon, and xenon over a wide range of temperatures (Table 21.2). Small variations in Table 21.2 from the predicted values are because real gases are not ideal gases. In real gases, weak intermolecular interactions occur, which are not addressed in our ideal gas model. Now suppose the gas is taken along the constant-pressure path i S f 9 shown in Figure 21.4. Along this path, the temperature again increases by DT. The energy that must be transferred by heat to the gas in this process is Q 5 nCP DT. Because the volume changes in this process, the work done on the gas is W 5 2P DV, where P is the constant pressure at which the process occurs. Applying the first law of thermodynamics to this process, we have

DE int 5 Q 1 W 5 nCP DT 1 (2P DV) (21.30)

In this case, the energy added to the gas by heat is channeled as follows. Part of it leaves the system by work (that is, the gas moves a piston through a displacement), and the remainder appears as an increase in the internal energy of the gas. The change in internal energy for the process i S f 9, however, is equal to that for the pro-cess i S f because E int depends only on temperature for an ideal gas and DT is the same for both processes. In addition, because PV 5 nRT, note that for a constant- pressure process, P DV 5 nR DT. Substituting this value for P DV into Equation 21.30 with DE int 5 nCV DT (Eq. 21.27) gives

nCV DT 5 nCP DT 2 nR DT

CP 2 CV 5 R (21.31)

This expression applies to any ideal gas. It predicts that the molar specific heat of an ideal gas at constant pressure is greater than the molar specific heat at constant vol-ume by an amount R, the universal gas constant (which has the value 8.31 J/mol ? K). This expression is applicable to real gases as the data in Table 21.2 show.

Table 21.2 Molar Specific Heats of Various GasesMolar Specific Heat ( J/mol ? K)a

Gas CP CV CP 2 CV g 5 CP/CV

Monatomic gasesHe 20.8 12.5 8.33 1.67Ar 20.8 12.5 8.33 1.67Ne 20.8 12.7 8.12 1.64Kr 20.8 12.3 8.49 1.69

Diatomic gasesH2 28.8 20.4 8.33 1.41N2 29.1 20.8 8.33 1.40O2 29.4 21.1 8.33 1.40CO 29.3 21.0 8.33 1.40Cl2 34.7 25.7 8.96 1.35

Polyatomic gasesCO2 37.0 28.5 8.50 1.30SO2 40.4 31.4 9.00 1.29H2O 35.4 27.0 8.37 1.30CH4 35.5 27.1 8.41 1.31

a All values except that for water were obtained at 300 K.

Quick Quiz 21.22 (ii) From the same choices, how does theinternal energy of an ideal gas change as it follows path f → f ′

along the isotherm labeled T + ∆T?

(A) Eint increases.

(B) Eint decreases.

(C) Eint stays the same.

(D) There is not enough information to determine how Eint

changes.2Serway & Jewett, page 631.


Paths with the same ∆T ,∆Eint:


HALLIDAY REVISED


CV



RealHe 12.5Ar 12.6


Real N2 20.7O2 20.8


Real NH4 29.0CO2 29.7

52

32

Table 19-2



Pres

sure

Volume

i

f

T + T ∆

T

f

f

1

2

3






$Eint " nCV $T % W. (19-40)


(19-41)


(19-42)










32

CV " 32R " 12.5 J/mol!K

$Eint " 32nR $T.

CV "$Eint

n $T.


We have related internal energyto temperature through KE:

Eint = Ktot,trans =3

2nRT

and from the first law ofthermodynamics:

∆Eint = Q +W

Since W is different for thedifferent processes shown, so isQ.

The ratio Q∆T will be different

also.

Heat Capacity for Constant Volume ProcessesIn a constant volume process, no work is done: ∆Eint = Q

51919-8 TH E MOLAR S PECI FIC H EATS OF AN I DEAL GASPART 2

HALLIDAY REVISED

19-8 The Molar Specific Heats of an Ideal GasIn this section, we want to derive from molecular considerations an expression forthe internal energy Eint of an ideal gas. In other words, we want an expression for theenergy associated with the random motions of the atoms or molecules in the gas.Weshall then use that expression to derive the molar specific heats of an ideal gas.

Internal Energy EintLet us first assume that our ideal gas is a monatomic gas (which has individualatoms rather than molecules), such as helium, neon, or argon. Let us also assumethat the internal energy Eint of our ideal gas is simply the sum of the translationalkinetic energies of its atoms. (As explained by quantum theory, individual atomsdo not have rotational kinetic energy.)

The average translational kinetic energy of a single atom depends only onthe gas temperature and is given by Eq. 19-24 as . A sample of nmoles of such a gas contains nNA atoms.The internal energy Eint of the sample is then

(19-37)

Using Eq. 19-7 (k ! R/NA), we can rewrite this as

(monatomic ideal gas). (19-38)Eint ! 32nRT

Eint ! (nNA)Kavg ! (nNA)(32kT ).

Kavg ! 32 kT

KEY I DEA

Additional examples, video, and practice available at WileyPLUS

(c) What is the most probable speed vP at 300 K?

Speed vP corresponds to the maximum of the distributionfunction P(v), which we obtain by setting the derivativedP/dv ! 0 and solving the result for v.

Calculation: We end up with Eq. 19-35, which gives us(Answer)

This result is also plotted in Fig. 19-8a.

! 395 m/s.

! A 2(8.31 J/mol "K)(300 K)0.0320 kg/mol

vP ! A 2RTM

The internal energy Eint of an ideal gas is a function of the gas temperature only; itdoes not depend on any other variable.

Fig. 19-9 (a) The temperature of anideal gas is raised from T to T # $T in aconstant-volume process. Heat is added,but no work is done. (b) The process on a p-V diagram.

T Q

Pin

(a)

(b)

Pres

sure

Volume

V

i

p

f p + p ∆

T + T ∆

T

Pin

Thermal reservoir

The temperature increase is done without changingthe volume.

With Eq.19-38 in hand,we are now able to derive an expression for the molar spe-cific heat of an ideal gas.Actually,we shall derive two expressions.One is for the case inwhich the volume of the gas remains constant as energy is transferred to or from it asheat.The other is for the case in which the pressure of the gas remains constant as en-ergy is transferred to or from it as heat.The symbols for these two molar specific heatsare CV and Cp, respectively. (By convention, the capital letter C is used in both cases,even though CV and Cp represent types of specific heat and not heat capacities.)

Molar Specific Heat at Constant VolumeFigure 19-9a shows n moles of an ideal gas at pressure p and temperature T,confined to a cylinder of fixed volume V.This initial state i of the gas is marked onthe p-V diagram of Fig. 19-9b. Suppose now that you add a small amount of


Therefore,

∆Eint = nCV ∆T

and

CV =1

n

(∂Eint

∂T

)V

(the subscipt V means const volume)

Heat Capacity for Constant Volume Processes

CV =1

n

(∂Eint

∂T

)V

(the subscipt V means const volume)

Putting in our value for internal energy:

CV =1

n

(∂

∂T

(3

2nRT

))V

CV =3

2R

This is the value for CV for all monatomic gases.

Heat Capacity for Constant Pressure ProcessesIn a constant pressure process, the work done on the gas is:W = −P ∆V

52119-8 TH E MOLAR S PECI FIC H EATS OF AN I DEAL GASPART 2

HALLIDAY REVISED

Fig. 19-11 (a) The temperature of anideal gas is raised from T to T ! "T in aconstant-pressure process. Heat is addedand work is done in lifting the loaded pis-ton. (b) The process on a p-V diagram.Thework p "V is given by the shaded area.

T Q

(a)

(b)

Pres

sure

Volume

V

if

V + V∆

T + T∆

T

W

p

p V∆

Thermal reservoir

The temperature increase is done without changingthe pressure.

Monatomic Diatomic

nR ∆T7__2

nR ∆T5__2

nR ∆T Q @ con V

Q @ con p

W

W ∆Eint trans

3__2

∆Eint trans

Q @ con V

Q @ con p

W∆Eint trans

rotation

transrotation

∆Eint

WFig. 19-12 The rela-tive values of Q for amonatomic gas (left side)and a diatomic gas under-going a constant-volumeprocess (labeled “con V”)and a constant-pressureprocess (labeled “con p”).The transfer of the energyinto work W and internalenergy ("Eint) is noted.

pressure process . Note that for the latter, the value of Q is higherby the amount W, the work done by the gas in the expansion. Note also that forthe constant-volume process, the energy added as Q goes entirely into the changein internal energy "Eint and for the constant-pressure process, the energy addedas Q goes into both "Eint and the work W.

(Q # 52nR "T )

Molar Specific Heat at Constant PressureWe now assume that the temperature of our ideal gas is increased by the samesmall amount "T as previously but now the necessary energy (heat Q) is addedwith the gas under constant pressure. An experiment for doing this is shown inFig. 19-11a; the p-V diagram for the process is plotted in Fig. 19-11b. From suchexperiments we find that the heat Q is related to the temperature change "T by

Q # nCp "T (constant pressure), (19-46)

where Cp is a constant called the molar specific heat at constant pressure. ThisCp is greater than the molar specific heat at constant volume CV, because energymust now be supplied not only to raise the temperature of the gas but also forthe gas to do work—that is, to lift the weighted piston of Fig. 19-11a.

To relate molar specific heats Cp and CV, we start with the first law of ther-modynamics (Eq. 18-26):

"Eint # Q $ W. (19-47)

We next replace each term in Eq. 19-47. For "Eint, we substitute from Eq.19-45. For Q, we substitute from Eq. 19-46. To replace W, we first note that sincethe pressure remains constant, Eq. 19-16 tells us that W # p "V. Then we notethat, using the ideal gas equation (pV # nRT), we can write

W # p "V # nR "T. (19-48)

Making these substitutions in Eq. 19-47 and then dividing through by n "T, we find

CV # Cp $ Rand then

Cp # CV ! R. (19-49)

This prediction of kinetic theory agrees well with experiment, not only formonatomic gases but also for gases in general, as long as their density is lowenough so that we may treat them as ideal.

The left side of Fig. 19-12 shows the relative values of Q for a monatomic gasundergoing either a constant-volume process or a constant-(Q # 3

2nR "T )


From the ideal gas equation:

P ∆V = nR ∆T

So,W = −nR ∆T

Heat Capacity for Constant Pressure ProcessesFirst law:

∆Eint = Q +W

rearranging:

∆Eint − Q = W

nCV ∆T − nCP ∆T = −nR ∆T

because Q = nCP ∆T . Dividing by −n∆T :

CP − CV = R

For a monatomic gas:

CP =5

2R

Equipartition Consequences in Diatomic Gases

Reminder:


Each degree of freedom for each molecule contributes an andadditional 1

2kBT of energy to the system.

A monatomic gas has 3 degrees of freedom: it can havetranslational KE due to motion in 3 independent directions.

A diatomic gas has more ways to move and store energy.

It can:

• translate

• rotate

• vibrate

Equipartition Consequences in Diatomic Gases

Contribution to internal energy:

3

2kBT −→

(3 directions of motion)

2

2kBT −→

(rotations about x and z axes)

2

2kBT −→

(KE and PE of harmonic oscillator)

21.3 The Equipartition of Energy 635

21.3 The Equipartition of EnergyPredictions based on our model for molar specific heat agree quite well with the behavior of monatomic gases, but not with the behavior of complex gases (see Table 21.2). The value predicted by the model for the quantity CP 2 CV 5 R, however, is the same for all gases. This similarity is not surprising because this difference is the result of the work done on the gas, which is independent of its molecular structure. To clarify the variations in CV and CP in gases more complex than monatomic gases, let’s explore further the origin of molar specific heat. So far, we have assumed the sole contribution to the internal energy of a gas is the translational kinetic energy of the molecules. The internal energy of a gas, however, includes contributions from the translational, vibrational, and rotational motion of the mol-ecules. The rotational and vibrational motions of molecules can be activated by collisions and therefore are “coupled” to the translational motion of the molecules. The branch of physics known as statistical mechanics has shown that, for a large num-ber of particles obeying the laws of Newtonian mechanics, the available energy is, on average, shared equally by each independent degree of freedom. Recall from Section 21.1 that the equipartition theorem states that, at equilibrium, each degree of freedom contributes 1

2 kBT of energy per molecule. Let’s consider a diatomic gas whose molecules have the shape of a dumbbell (Fig. 21.5). In this model, the center of mass of the molecule can translate in the x, y, and z directions (Fig. 21.5a). In addition, the molecule can rotate about three mutually perpendicular axes (Fig. 21.5b). The rotation about the y axis can be neglected because the molecule’s moment of inertia Iy and its rotational energy 1

2 Iyv2 about

this axis are negligible compared with those associated with the x and z axes. (If the two atoms are modeled as particles, then Iy is identically zero.) Therefore, there are five degrees of freedom for translation and rotation: three associated with the translational motion and two associated with the rotational motion. Because each degree of freedom contributes, on average, 1

2kBT of energy per molecule, the inter-nal energy for a system of N molecules, ignoring vibration for now, is

E int 5 3N 112kBT 2 1 2N 11

2kBT 2 5 52 NkBT 5 5

2nRT

We can use this result and Equation 21.28 to find the molar specific heat at con-stant volume:

CV 51n

dE int

dT5

1n

ddT

152nRT 2 5 5

2R 5 20.8 J/mol ? K (21.33)

From Equations 21.31 and 21.32, we find that

CP 5 CV 1 R 5 72 R 5 29.1 J/mol ? K

g 5CP

CV5

72R52R

575

5 1.40

These results agree quite well with most of the data for diatomic molecules given in Table 21.2. That is rather surprising because we have not yet accounted for the possible vibrations of the molecule. In the model for vibration, the two atoms are joined by an imaginary spring (see Fig. 21.5c). The vibrational motion adds two more degrees of freedom, which cor-respond to the kinetic energy and the potential energy associated with vibrations along the length of the molecule. Hence, a model that includes all three types of motion predicts a total internal energy of

E int 5 3N 112kBT 2 1 2N 11

2kBT 2 1 2N 112kBT 2 5 7

2Nk BT 5 72nRT

and a molar specific heat at constant volume of

CV 51n

dE int

dT5

1n

ddT

1 72nRT 2 5 7

2R 5 29.1 J/mol ? K (21.34)

x

z

y

yx

z

Translational motion of the center of mass

Rotational motion about the various axes

Vibrational motion along the molecular axis

a

b

c

Figure 21.5 Possible motions of a diatomic molecule.

(Classical) Equipartition Prediction for DiatomicGases

In total:

Eint = N

(7

2kBT

)We can also write the internal energy:

Eint =7

2nRT

And so,

CV =7

2R

This is what we would expect for a diatomic gas based on theequipartition theorem.

It is not quite what is observed, however.

What Actually Happens in Diatomic Gases

Prediction:

CV =7

2R

For most diatomic gases, such as H2 and N2,

CV =5

2R

at moderate temperatures (around room temperature).

And at low temperatures for these gases CV = 32R.

It is almost as if degrees of freedom become “activated” oncethere is enough energy...

What Actually Happens in Diatomic Gases

Hydrogen gas:

52519-10 A H I NT OF QUANTU M TH EORYPART 2

HALLIDAY REVISEDHALLIDAY

19-10 A Hint of Quantum TheoryWe can improve the agreement of kinetic theory with experiment by includingthe oscillations of the atoms in a gas of diatomic or polyatomic molecules. Forexample, the two atoms in the O2 molecule of Fig. 19-13b can oscillate towardand away from each other, with the interconnecting bond acting like a spring.However, experiment shows that such oscillations occur only at relatively hightemperatures of the gas—the motion is “turned on” only when the gas moleculeshave relatively large energies. Rotational motion is also subject to such “turningon,” but at a lower temperature.

Figure 19-14 is of help in seeing this turning on of rotational motion andoscillatory motion. The ratio CV/R for diatomic hydrogen gas (H2) is plotted thereagainst temperature, with the temperature scale logarithmic to cover several ordersof magnitude. Below about 80 K, we find that CV/R ! 1.5. This result implies thatonly the three translational degrees of freedom of hydrogen are involved in the spe-cific heat.

As the temperature increases, the value of CV /R gradually increases to 2.5,implying that two additional degrees of freedom have become involved.Quantum theory shows that these two degrees of freedom are associated with therotational motion of the hydrogen molecules and that this motion requires acertain minimum amount of energy. At very low temperatures (below 80 K), themolecules do not have enough energy to rotate. As the temperature increasesfrom 80 K, first a few molecules and then more and more of them obtain enoughenergy to rotate, and the value of CV /R increases, until all of the molecules are ro-tating and CV /R ! 2.5.

Similarly, quantum theory shows that oscillatory motion of the moleculesrequires a certain (higher) minimum amount of energy. This minimum amount isnot met until the molecules reach a temperature of about 1000 K, as shown inFig. 19-14. As the temperature increases beyond 1000 K, more and more mole-cules have enough energy to oscillate and the value of CV /R increases, until all ofthe molecules are oscillating and CV /R ! 3.5. (In Fig. 19-14, the plotted curvestops at 3200 K because there the atoms of a hydrogen molecule oscillate somuch that they overwhelm their bond, and the molecule then dissociates into twoseparate atoms.)

20 50 100 200 500 1000 2000 5000 10,000Temperature (K)

0

1

2

3

4

C V/R

7/2

5/2

3/2

Translation

Rotation

Oscillation

Fig. 19-14 CV/R versus temperature for (diatomic) hydrogengas. Because rotational and oscillatory motions begin at certainenergies, only translation is possible at very low temperatures.Asthe temperature increases, rotational motion can begin.At stillhigher temperatures,oscillatory motion can begin.


At 3200K the molecules begin todissociate.

21.4 Adiabatic Processes for an Ideal Gas 637

and vibrational quantum states of a diatomic molecule. The lowest allowed state is called the ground state. The black lines show the energies allowed for the mol-ecule. Notice that allowed vibrational states are separated by larger energy gaps than are rotational states. At low temperatures, the energy a molecule gains in collisions with its neighbors is generally not large enough to raise it to the first excited state of either rotation or vibration. Therefore, even though rotation and vibration are allowed according to classical physics, they do not occur in reality at low temperatures. All molecules are in the ground state for rotation and vibration. The only contribution to the mol-ecules’ average energy is from translation, and the specific heat is that predicted by Equation 21.29. As the temperature is raised, the average energy of the molecules increases. In some collisions, a molecule may have enough energy transferred to it from another molecule to excite the first rotational state. As the temperature is raised further, more molecules can be excited to this state. The result is that rotation begins to contribute to the internal energy, and the molar specific heat rises. At about room temperature in Figure 21.6, the second plateau has been reached and rotation con-tributes fully to the molar specific heat. The molar specific heat is now equal to the value predicted by Equation 21.33. There is no contribution at room temperature from vibration because the mole-cules are still in the ground vibrational state. The temperature must be raised even further to excite the first vibrational state, which happens in Figure 21.6 between 1 000 K and 10 000 K. At 10 000 K on the right side of the figure, vibration is con-tributing fully to the internal energy and the molar specific heat has the value pre-dicted by Equation 21.34. The predictions of this model are supportive of the theorem of equipartition of energy. In addition, the inclusion in the model of energy quantization from quan-tum physics allows a full understanding of Figure 21.6.

Q uick Quiz 21.3 The molar specific heat of a diatomic gas is measured at constant volume and found to be 29.1 J/mol ? K. What are the types of energy that are con-tributing to the molar specific heat? (a) translation only (b) translation and rota-tion only (c) translation and vibration only (d) translation, rotation, and vibration

Q uick Quiz 21.4 The molar specific heat of a gas is measured at constant volume and found to be 11R/2. Is the gas most likely to be (a) monatomic, (b) diatomic, or (c) polyatomic?

21.4 Adiabatic Processes for an Ideal GasAs noted in Section 20.6, an adiabatic process is one in which no energy is trans-ferred by heat between a system and its surroundings. For example, if a gas is com-pressed (or expanded) rapidly, very little energy is transferred out of (or into) the system by heat, so the process is nearly adiabatic. Such processes occur in the cycle of a gasoline engine, which is discussed in detail in Chapter 22. Another example of an adiabatic process is the slow expansion of a gas that is thermally insulated from its surroundings. All three variables in the ideal gas law—P, V, and T—change during an adiabatic process. Let’s imagine an adiabatic gas process involving an infinitesimal change in volume dV and an accompanying infinitesimal change in temperature dT. The work done on the gas is 2P dV. Because the internal energy of an ideal gas depends only on temperature, the change in the internal energy in an adiabatic process is the same as that for an isovolumetric process between the same temperatures, dE int 5 nCV dT (Eq. 21.27). Hence, the first law of thermodynamics, DE int 5 Q 1 W, with Q 5 0, becomes the infinitesimal form

dE int 5 nCV dT 5 2P dV (21.35)

Rotationalstates

Rotationalstates

Vibrationalstates

ENER

GY

The rotational states lie closer together in energy than do thevibrational states.

Figure 21.7 An energy-level dia-gram for vibrational and rotational states of a diatomic molecule.

1Left diagram, Halliday, Resnick, Walker; right diagram Serway & Jewett

Question

Quick Quiz 21.43 The molar specific heat of a gas is measured atconstant volume and found to be 11R/2. Is the gas most likely tobe

(A) monatomic,

(B) diatomic, or

(C) polyatomic?


A Useful Ratio

The quantity γ is defined as:

γ =CP

CV

For a monatomic gas:

γ =5

3

What is γ for a diatomic gas near room temperature?

γ =7

5

Adiabatic Process in Ideal Gases

For an adiabatic process (Q = 0):

PVγ = const.

and:

TVγ−1 = const.

(Given the first one is true, the second follows immediately fromthe ideal gas equation, P = nRT

V .)

Adiabatic Process in Ideal GasesWhere this relation comes from:

∆Eint = W

Considering a small volume change in time that produces a smallchange in temperature:

dEint

dt=

dW

dt

nCVdT

dt= −P

dV

dt(1)

The ideal gas law derivative:

PdV

dt+V

dP

dt= nR

dT

dt

ndT

dt=

1

R

(P

dV

dt+V

dP

dt

)Substitute n dT

dt into our energy equation (1).


CV

(n

dT

dt

)= −P

dV

dt

CV

R

(P

dV

dt+V

dP

dt

)= −P

dV

dt

VdP

dt= −

(1 +

R

CV

)P

dV

dt

Notice: γ = 1 +R

CV

then, dividing by PV :

1

P

dP

dt= −

γ

V

dV

dt

Integrating both sides:

lnP = −γ lnV + c


lnP = −γ lnV + c

Implies:PVγ = const.

This equation characterizes an adiabatic process in an ideal gas,along with this one, which follows from PV = nRT :

TVγ−1 = const.

Example

Based on problem 28, Chapter 21.

How much work is required to compress 5.00 mol of air at 20.0◦Cand 1.00 atm to one-tenth of the original volume in an adiabaticprocess? Assume air behaves as an ideal diatomic-type gas.


Weather and Adiabatic Process in a Gas

On the eastern side of the Rocky Mountains there is aphenomenon called chinooks.

These eastward moving wind patterns cause distinctive cloudpatterns (chinook arches) and sudden increases in temperature.

Weather and Adiabatic Process in a GasAs the air rises from the ocean it expands in the lower pressure ataltitude and cools. The water vapor condenses out of the air andfalls as precipitation.

As the air passes over the mountain it absorbs the latent heat fromthe water condensation, then it stops cooling. As it descends, it iscompressed (nearly) adiabatically as the ambient pressureincreases. The air temperature rises!

Temperature and the Distribution of Particles’Energies

In a gas at temperature T , we know the average translational KEof the molecules.

However, not all of the molecules have the same energy, that’s justthe average.

How is the total energy of the gas distributed amongst themolecules?

Temperature and the Distribution of Particles’Energies

Ludwig Boltzmann first found the distribution of the number ofparticles at a given energy given a thermodynamic system at afixed temperature.

Assuming that energy takes continuous values we can say that thenumber of molecules per unit volume with energies in the range Eto E + dE is:

N[E ,E+dE] =

∫E+dE

EnV (E ) dE

WherenV (E ) = n0e

−E/kBT

and n0 is a constant setting the scale: nV (0) = n0.

The Boltzmann DistributionThis particular frequency distribution:

nV (E ) ∝ e−E/kBT

is called the Boltzmann distribution or sometimes the Gibbsdistribution (after Josiah Willard Gibbs, who studied the behaviorof this distribution in-depth).

This distribution is even easier to understand for discrete energylevels.

The probability for a given particle to be found in a state withenergy Ei drawn from a sample at temperature T :

p(Ei ) =1

Ze−Ei/kBT

where Z is simply a normalization constant to allow the totalprobability to be 1. (The partition function.)

The Boltzmann Distribution

p(Ei ) =1

Ze−Ei/kBT

If we know the energies of two states E1 and E2, E2 > E1, we canfind the ratio of the number of particles in each:

nV (E2)

nV (E1)= e−(E2−E1)/kBT

States with lower energies have more particles occupying them.

The Boltzmann Distribution

Lower temperature Higher temperature

1Figure from the website of Dr. Joseph N. Grima, University of Malta.

Summary

• molecular models

• rms speed of molecules

• equipartition of energy

• molar heat capacities

• adiabatic processes

Quiz in class, Monday, May 8.

Collected Homework due Monday, May 8.

Test on Thermodynamics, Monday, May 15.

Homework Serway & Jewett:

• new: Ch 21, CQs: 7; Probs: 15, 17, 21, 23, 25, 29, 31, (41),47, 59, 65

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