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Unformatted text preview: 4 z a H" ﬂaw wpuwm.“ Description of an ideal gas Figure 21 —2 A cubical box of side
1, containing an ideal gas. A molecule
is shown moving toward A1. 21 3 Ideal Gas—A
Microscopic Description From the microscopic point of view we deﬁne an ideal gas by making the following
assumptions; it will then be our task to apply the laws of classical mechanics sta
tistically to the gas atoms and to show that our microscopic deﬁnition is consistent
with the macroscopic deﬁnition of the preceding section: 1. A gas consists of particles, called molecules. Depending on the gas, each
molecule consists of one atom or a group of atoms. If the gas is an element or a
compound and is in a stable state, we consider all its molecules to be identical. 2. The. molecules are in random motion and obey Newton’s laws of motion.
The molecules move in all directions and with various speeds. In computing the
properties of the motion, we assume that Newtonian mechanics works at the
microscopic level. As for all our assumptions, this one will stand or fall depending
on whether or not the experimental facts it predicts are correct. 3. The total number of molecules is large. The direction and speed of motion of
any one molecule may change abruptly upon collision with the wall or another
molecule. Any particular molecule will follow a zigzag path because of these colli
sions. However, because there are so many molecules we assume that the re
sulting large number of collisions maintains the overall distribution of molecular
velocities and the randomness of the motion. The volume of the molecules is a negligibly small fraction of the volume 0c—
cup‘ied by the gas. Even though there are many molecules, they are extremely
small. We know that the volume occupied by a gas can be changed through a large
range of values with little difﬁculty, and that when a gas condenses the volume oc
cupied by the liquid may be thousands of times smaller than that of the gas.
Hence, our assumption is plausible. 5. No appreciable forces act on the molecules except during a collision. To the
extent that this is true a molecule moves with uniform velocity between collisions.
Because we have assumed the molecules to be so small, the average distance
between molecules is large compared to the size of a molecule. Hence, we assume
that the range of molecular forces is comparable to the molecular size. 6. Collisions are elastic and are of negligible duration. Collisions between mol
ecules and with the walls of the container conserve momentum and (we assume)
kinetic energy. Because the collision time is negligible compared to the time spent
by a molecule between collisions, the kinetic energy that is converted to potential
energy during the collision is available again as kinetic energy after such a brief
time that we can ignore this exchange entirely. 21 —4 Kinetic
Calculation of Pressure Let us now calculate the pressure of an ideal gas from kinetic theory. To simplify
matters, we consider a gas in a cubical vessel whose walls are perfectly elastic.
Let each edge be of length 1. Call the faces normal to the xaxis (Fig. 21—2) A1 and
A2, each of area [2. Consider a molecule that has a velocity v. We can resolve v
into components or, by, and v, in the directions of the edges. Since each wall is
perfectly elastic, this particle will rebound, after collision with A1, with its x— i
E
g Pressure and molecular speed component of velocity exactly reversed. There will be no effect on v” or 122 , so that:
the change Ap in the particle’s momentum will be " AP = p:  pa = (mvx) — (map) = —2mv,, normal to A1. Hence, the momentum imparted to A1 will be 2mvx, since the total
momentum is conserved. Suppose that this particle reaches A2 without striking any other particle on
the way. The time required to cross the cube will be 1/ 12,. At A2 it will again have
its xcomponent of velocity reversed and will return to Al'. Assuming no collisions
in between, the round trip will take a time 21/ 1),. Hence, the number of collisions
per unit time this particle makes with A1 is um /21, so that the rate at which it
transfers momentum to Al is ' 2
Pi = ﬂ
2va 21 1
To obtain the total force on A1 , that is, the rate at which momentum is imparted to
A1 by all the gas molecules, we must sum up mvxz/l for all the particles. Then, to
ﬁnd the pressure, we divide this force by the area of A1, namely, 12. .If m is the mass of each molecule, we have ,
m _.
P=7§(Ux12+9122 + ' ’ '), where v,r1 is the xcomponent of the velocity of particle 1, 12,2 is that of particle 2,
and so on. If N is the total number of particles in the container, we can write this
as \
But mN is simply the total mass of the gas and mN / V is the mass per unit volume.
that is, the density p. The quantity (1);“2 + @122 +   )/Ij__is the average value of
v,2 for all the particles in the container. Let us call this vac”. Then p = Putz For any particle 122 = v32, + 12,,2 + 0,}. Because we have many particles and
because they are moving entirely at random, the average values of 01,301,”, and 1);2
are equal and the value of each is exactly onethird the average value of 122. There
is no pref_erence among the molecules for motion along any one of the three axes.
Hence, of = 5?, so that p = b??? = ipﬁ. (213) Although we derived this result by neglecting collisions between particles,
the result is true even when we consider collisions. Because of the exchange of ve
locities in an elastic collision between identical particles, there will always be
some one molecule that will collide with A2 with momentum mvx corresponding to
the one that left A1 with this same momentum. Also, the time spent during colli
sions is negligible compared to the time spent between collisions. Hence, our
neglect of collisions is merely a convenient device for calculation. Likewise, we
could have chosen a container of any shape —the cube merely simpliﬁes the cal
culation. Although we have calculated the pressure exerted only on the side A1 , if
we neglect the weight of the gas, it follows from Pascal’s law that the pressure is
the same on all sides and everywhere in the interior. The square root of 725 is called the rootmeansquare speed of the molecules Speed of molecules and is a kind of average molecular speed (see Section 21—9). Using Eq. 21—3, we
can calculate this root—meansquare speed from measured values of the pressure and density of the gas. Thus,
um, = ﬁ = ‘1—353. 1 (21—40) In Eq. 21—3 we relate a macroscopic quantity (the pressure p) to an average
value of a microscopic quantity (that is, to F or vmf). However, averages can be
taken over short times or over long times, over small regions of space or large
regions of space. The average computed in a small region for a short time might
depend on the time or region chosen, so that the values obtained in this way may
ﬂuctuate. This could happen in a gas of very low density, for example. We can ig
nore fluctUations, however, when the number of particles in the system is large enough. Example 3 Calculate the rootmeansquare speed of hy 3p 3 x 1.01 x 105 N /m2
drogen molecules at 000°C and 1.00 atm pressure, assuming hy vrms = ‘p— = = 1340 m/S ‘ drogen to be an ideal gas. Under these conditions hydrogen has a density p of 8.99 X 10‘2 kg/m". Then, since p = 1.00 atm = This is of the order of a mile per second, or 3600 mi/h. 1.01 x 105 N/m”, 1
\ \Table 21—1 gives the results of similar calculations for some gases at 0°C.
These molecular speeds are roughly of the same order as the speed of sound at the
same temperature. For example, in air at 0°C, um, = 485 m/s and the speed of
sound is 331 m/s, and in hydrogen vrms = 1838 m/s and sound travels at 1286 m /s.
These results are to be expected in terms of our model of a gas; see Problem 37.
We must distinguish between the speeds of individual molecules, described by
vrms , and the very much lower speed at which one gas diffuses into another. Thus,
if we open a bottle of ammonia in one comer of a room, we smell ammonia in the
opposite corner only after an easily measurable time lag. The diffusion speed is
slow because large numbers of collisions with air molecules greatly .reduce the
tendency of ammonia molecules to spread themselves uniformly throughout the room. Table 21 —1 Some molecular speeds Translational
Molecular kinetic energy
weight," um.s (at 0°C), per mole (at 0°C),
Gas g/mol m/s inmsz, J /mol
H2 2.02 1838 3370
He 4.0 1311 3430
H20 18 615 3400
Ne 20.1 584 3420
N2 28 493 3390
CO 28 493 3390
Air 28.8 485 3280
02 32 461 3400
C02 44 393 3400 " The molecular weight and the mole are deﬁned on p. 375. We
, will discuss the last column in this table in the next section. 21 —5 Kinetic
Interpretation of Temperature If we multiply each side of Eq. 21—3 by the volume V, we obtain
[7V = éhpvvrz'ms, ‘ where pV is simply the total mass of gas, p being the density. We can also write
the. mass of gas as nM, in which n is the number of moles and M is the molecular
weight. Making this substitution yields 1 Kinetic energy per mole Kinetic energy per molecule pV = énMvﬁms. The quantity. énMvEms is twothirds the tolal kinetic energy of translation of the
molecules, that is, iGnvams). We can ther\ write pV = glGnMvzz'ms) The equation of state of an ideal gas is
pV = nRT. Combining these two expressions, we obtain worm. = m. ' (21—5) That is, the total translational kinetic energy per mole of the molecules of an ideal
gas is proportional to the temperature. We may say that this result, Eq. 21—5, is ‘
necessary to ﬁt the kinetic theory to the equation of state of an ideal gas, or we
may consider Eq. 21—5 as a deﬁnition of gas temperature on a kinetic theory or
microscopic basis. In either case, we gain some insight into the meaning of tem
perature for gases. The temperature of a gas is related to the total translational kinetic energy
measured with respect to the center of mass of the gas. The kinetic energy asso
ciated with the motion of the center of mass of the gas has no bearing on the gas
temperature. The temperature of a gas in a container does not increase when we
put the container on a moving train. Let us now divide each side of Eq. 21—5 by the Avogadro constant. NA,
which (see p. 375, footnote) is the number of molecules per mole of a gas. Thus M/NA (=m) is the mass of a single molecule and we have 1.4L 2 l _3 R
2(NA)vrms_2mv%ms_E(E)T Now imvfms is the average translational kinetic energy per molecule. The ratio
R/NA ——which we call k, the Boltzmann constant —plays the role of the gas con
stant per molecule. We have émvﬁms = ‘EkT (216)
in which _ R _ ‘ 8.314 J/mol K _ _23 k — ————NA — —————————————6.022 x 1023 molecules/mo] — 1.381 x 10 J/molecule K.
We shall return to the Boltzmann constant in. Section 21—9. In the last column of Table 21—1 we list calculated values of % vams. As Eq.
21—5 predicts, this quantity (the translational kinetic energy per mole) has
(closely) the same value for all gases at the same temperatures, 0°C in this case.
From Eq. 21—6 we conclude that at the same temperature T the ratio of the
rootmeansquare speeds of molecules of two different gases is equal to the square root of the inverse ratio of their masses. That is, from 2
_ _2_ mlvlrms _ l m2v§rms T ‘ 3k 2 ” 3k 2
we obtain
1 “m = ﬂ. (21—7) \ UZrms m1 \ We can apply Eq. 21—7 to the diffusion of two different gases in a‘container
wi‘h porous walls placed in an evacuated space. The lighter gas, whose molecules
mo e more rapidly on the average, will escape faster than the heavier one. The
ratio of the numbers of molecules of the two gases that ﬁnd their way through the
porous walls for a short time interval is equal to the square root of the inverse ratio
of their masses, \/m2/m1. This diffusion process is one method of separating
(readily ﬁssionable) 235U (0.7% abundance) from a normal sample of uranium con taining mostly 238U (99.3% abundance). ...
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 Fall '10
 Feebeck
 Kinetic Energy, Momentum, Kinetic theory

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