2.6 – Livre caminho médio

UFABC – Fenômenos Térmicos – Prof. Germán Lugones MÓDULO 2 – TEMPERATURA E A TEORIA CINÉTICA DOS GASES

A Figura mostra a trajetória de uma molécula típica quando ela se move através do gás.

Entre colisões, a molécula se move em linha reta com velocidade constante mudando abruptamente tanto o módulo quanto o sentido da velocidade quando ela col ide elasticamente com outras moléculas.

Embora a figura mostre as outras moléculas como se estivessem paradas, elas também estão se movendo.

Caminho Livre Médio

Um parâmetro úti l para descrever este movimento aleatório é o caminho livre médio ! das moléculas. Como o seu nome indica, ! é a distância média percorrida por uma molécula entre colisões.

Cálculo do Caminho Livre Médio

Consideremos uma molécula que está viajando com uma velocidade constante e que todas as demais moléculas estão em repouso (mais tarde, vamos abrir mão desta condição).

v

Supomos que as moléculas são esferas de diâmetro . Uma colisão ocorrerá se os centros de duas moléculas se aproximarem dentro de uma distância d um do outro.

d

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

HALLIDAY REVISED

At a given temperature T, all ideal gas molecules—no matter what their mass—have the same average translational kinetic energy—namely, . When we measurethe temperature of a gas, we are also measuring the average translational kinetic energy of its molecules.

32kT

CHECKPOINT 2

A gas mixture consists of molecules of types1, 2, and 3, with molecular masses m1 !m2 ! m3. Rank the three types according to(a) average kinetic energy and (b) rmsspeed, greatest first.

Fig. 19-5 A molecule traveling througha gas, colliding with other gas molecules inits path.Although the other molecules areshown as stationary, they are also moving in a similar fashion.

Fig. 19-6 (a) A collision occurs whenthe centers of two molecules come within adistance d of each other, d being the molec-ular diameter. (b) An equivalent but moreconvenient representation is to think of themoving molecule as having a radius d andall other molecules as being points.Thecondition for a collision is unchanged.

dd

2d

m

(a)

(b)

d

m

m m

Using Eq. 19-7 (k " R/NA), we can then write

(19-24)

This equation tells us something unexpected:

Kavg " 32kT.

19-6 Mean Free PathWe continue to examine the motion of molecules in an ideal gas. Figure 19-5shows the path of a typical molecule as it moves through the gas, changing bothspeed and direction abruptly as it collides elastically with other molecules.Between collisions, the molecule moves in a straight line at constant speed.Although the figure shows the other molecules as stationary, they are (ofcourse) also moving.

One useful parameter to describe this random motion is the mean free pathl of the molecules. As its name implies, l is the average distance traversed by amolecule between collisions. We expect l to vary inversely with N/V, the numberof molecules per unit volume (or density of molecules). The larger N/V is, themore collisions there should be and the smaller the mean free path. We alsoexpect l to vary inversely with the size of the molecules—with their diameter d,say. (If the molecules were points, as we have assumed them to be, they wouldnever collide and the mean free path would be infinite.) Thus, the larger the mole-cules are, the smaller the mean free path. We can even predict that l should vary(inversely) as the square of the molecular diameter because the cross section ofa molecule—not its diameter—determines its effective target area.

The expression for the mean free path does, in fact, turn out to be

(mean free path). (19-25)

To justify Eq. 19-25, we focus attention on a single molecule and assume—asFig. 19-5 suggests—that our molecule is traveling with a constant speed v andthat all the other molecules are at rest. Later, we shall relax this assumption.

We assume further that the molecules are spheres of diameter d. A collisionwill then take place if the centers of two molecules come within a distance d ofeach other, as in Fig. 19-6a. Another, more helpful way to look at the situation isto consider our single molecule to have a radius of d and all the other moleculesto be points, as in Fig. 19-6b.This does not change our criterion for a collision.

As our single molecule zigzags through the gas, it sweeps out a short cylinderof cross-sectional area pd 2 between successive collisions. If we watch this mol-ecule for a time interval #t, it moves a distance v #t, where v is its assumed speed.Thus, if we align all the short cylinders swept out in interval #t, we form acomposite cylinder (Fig. 19-7) of length v #t and volume (pd 2)(v #t).The numberof collisions that occur in time #t is then equal to the number of (point) mole-cules that lie within this cylinder.

Since N/V is the number of molecules per unit volume, the number of moleculesin the cylinder is N/V times the volume of the cylinder, or (N/V )(pd2v #t).This is alsothe number of collisions in time #t.The mean free path is the length of the path (and

$ "1

√2%d2 N/V

halliday_c19_507-535v2.qxd 28-10-2009 15:56 Page 514

Uma maneira equivalente de se olhar para a situação é considerar nossa molécula como tendo um raio e todas as outras moléculas como sendo puntiformes.

d

Quando a molécula ziguezagueia através do gás, ela varre um pequeno cilindro de área de seção transversal entre colisões sucessivas.

Se observarmos esta molécula por um intervalo de tempo , ela se desloca por uma distância , onde é sua velocidade.

πd2

ΔtvΔt

v

Assim, se alinharmos todos os pequenos cilindros varridos no intervalo , formaremos um cilindro composto de comprimento e volume

.

ΔtvΔt

(πd2)(vΔt)

51519-6 M EAN FR E E PATHPART 2

HALLIDAY REVISED

of the cylinder) divided by this number:

(19-26)

This equation is only approximate because it is based on the assumption thatall the molecules except one are at rest. In fact, all the molecules are moving;when this is taken properly into account, Eq. 19-25 results. Note that it differsfrom the (approximate) Eq. 19-26 only by a factor of .

The approximation in Eq. 19-26 involves the two v symbols we canceled. The vin the numerator is vavg, the mean speed of the molecules relative to the container.The v in the denominator is vrel, the mean speed of our single molecule relative to theother molecules, which are moving. It is this latter average speed that determines thenumber of collisions.A detailed calculation, taking into account the actual speed dis-tribution of the molecules, gives and thus the factor .

The mean free path of air molecules at sea level is about 0.1 mm.At an altitudeof 100 km, the density of air has dropped to such an extent that the mean free pathrises to about 16 cm.At 300 km, the mean free path is about 20 km.A problem facedby those who would study the physics and chemistry of the upper atmosphere in thelaboratory is the unavailability of containers large enough to hold gas samples (ofFreon, carbon dioxide, and ozone) that simulate upper atmospheric conditions.

12vrel ! 12vavg

1/12

!1

"d2 N/V.

# !length of path during $t

number of collisions in $t !

v $t"d2v $t N/V

Fig. 19-7 In time $t the moving mole-cule effectively sweeps out a cylinder oflength v $t and radius d.

2d

v ∆t

CHECKPOINT 3

One mole of gas A, with molecular diam-eter 2d0 and average molecular speed v0,is placed inside a certain container. Onemole of gas B, with molecular diameterd0 and average molecular speed 2v0 (themolecules of B are smaller but faster), isplaced in an identical container. Whichgas has the greater average collision ratewithin its container?

Sample Problem

collisions for any given molecule? At what rate does the mol-ecule collide; that is, what is the frequency f of its collisions?

(1) Between collisions, the molecule travels, on average, themean free path l at speed v. (2) The average rate or fre-quency at which the collisions occur is the inverse of thetime t between collisions.

Calculations: From the first key idea, the average time be-tween collisions is

! 2.44 % 10&10 s ! 0.24 ns. (Answer)

This tells us that, on average, any given oxygen molecule hasless than a nanosecond between collisions.

From the second key idea, the collision frequency is

(Answer)

This tells us that, on average, any given oxygen moleculemakes about 4 billion collisions per second.

f !1t

!1

2.44 % 10&10 s! 4.1 % 109 s&1.

t !distance

speed!

#

v!

1.1 % 10&7 m450 m/s

Mean free path, average speed, collision frequency

(a) What is the mean free path l for oxygen molecules at tem-perature T ! 300 K and pressure p ! 1.0 atm? Assume thatthe molecular diameter is d ! 290 pm and the gas is ideal.

Each oxygen molecule moves among other moving oxygenmolecules in a zigzag path due to the resulting collisions. Thus,we use Eq.19-25 for the mean free path.

Calculation: We first need the number of molecules per unitvolume, N/V. Because we assume the gas is ideal, we can usethe ideal gas law of Eq. 19-9 (pV ! NkT) to write N/V ! p/kT.Substituting this into Eq. 19-25, we find

! 1.1 % 10&7 m. (Answer)This is about 380 molecular diameters.

(b) Assume the average speed of the oxygen molecules is v ! 450 m/s. What is the average time t between successive

!(1.38 % 10&23 J/K)(300 K)12"(2.9 % 10&10 m)2(1.01 % 105 Pa)

# !112"d2 N/V

!kT12"d2p

KEY I DEA

KEY I DEAS

Additional examples, video, and practice available at WileyPLUS

halliday_c19_507-535v2.qxd 28-10-2009 15:56 Page 515

O número de colisões que ocorrem em um tempo é então igual ao número de moléculas (puntiformes) que estão dentro deste cilindro.

Como é o número de moléculas por unidade de volume, o número de moléculas no cilindro é: multiplicado pelo volume do cilindro, ou seja:

.

Este é também o número de colisões que ocorrem no intervalo . O caminho livre médio é o comprimento da trajetória (e do cilindro) dividido por este número:

Δt

N/VN/V

(N/V )(πd2)(vΔt)

Δt

λ =  comprimento do caminho durante Δt número de colisões em Δt = vΔt

πd2 . vΔt . N/V = 1πd2 ⋅ N/V

Esta equação é apenas aproximada porque está baseada na suposição de que todas as moléculas, exceto uma, estão em repouso.

Porém, todas as moléculas estão se movendo. Quando isto é levado em conta adequadamente, chega-se à seguinte expressão para o caminho livre medio:

λ = 12πd2 ⋅ N

V

Exemplos:

O caminho livre médio de moléculas de ar no nível do mar é cerca de 0,1 μm.

Em uma altitude de 100 km, a densidade cai para níveis tão baixos que o caminho livre médio aumenta para cerca de 16 cm. A 300 km, o caminho livre médio é cerca de 20 km.

!

1. ! varia inversamente com a densidade de moléculas. Quanto maior for N/V, maior deve ser o número de colisões e menor a distancia percorrida pelas moléculas entre duas colisões sucessivas.

2. ! varia inversamente com o tamanho das moléculas. • Se as moléculas fossem puntiformes, elas nunca colidiriam e o caminho

livre médio seria infinito. • Alias, varia (inversamente) com o quadrado do diâmetro molecular

porque a seção de choque de uma molécula determina sua área efetiva como alvo.

λ = 12πd2 ⋅ N

V

Análise da equação do livre caminho médio: