CHAPTER 22
•••••••.•••
r•••••
••••
/(
7...
·
• M'MlYnnnz ••
TEMPERATURE
Up to now we have dealt in turn with the mechanics ofsingle particles,
systems 0/ particles, rigid bodies, andfluids. In each case, we tsed one form or
another of Newton's laws to analyze the dynamics a/the system and [a/allow ill time the
motions a/individual particles or elements of the system.
Beginning with this chapter, we broaden our perspective to deal with systems that are too
complex io treat in terms a/the motion a/individual particles, These systems usually
appear disordered because a/the large number of particles involved and the many differeni
ways they can share the energy available to the system, To analyze such systems Ive use the
principles a/thermodynamics, In our study of thermodynamics lVedefine a nl'lVset of
physical variables to describe the state 0/ a system, and we deduce a nelv set of laws that
govern the behavio; q( systems. We also show how it is possible to understand these Ilell'
laws on the basis oj our previous laws 0/ mechanics.
A central concept a/thermodynamics is temperature. In this chapter we define temperature
and discuss its measurement.
22-1
MACROSCOPIC
IVnCROSCOPIC
DESCRIPTIONS
AND
A liter of gas contains about 3 X 1022 molecules. Let us
take the simplest possible case and treat the gas molecules
as point particles that collide elastically with one another
and with the walls of the container. If we specify the initial
vosition and velocity of every particle, we could then
apply Newton's laws and deduce the position and velocity
of each particle at any future: time. Given that information, W~ could calculate certain measurable properties of
tlie system, such as the net impulsive force exerted on an
clement of area of the container. We call this the microscopi: description of the system. Because the number of
particles is so large, it is advantageous to treat the system
using average values of the microscopic quantities. T11is
approach is called statistical mechanics and is discussed
in ('hapter 24.
.'\ different approach is based on the following question: Can we describe the system, including its interacti(~0:; with its environment, in terms ofa small number of
overall properties that are measurable by relatively easily
performed laboratory operations? In the ("l~.e of a gas confined to a container, we cau indeed obtain such a descriptiou in terms of the macroscopic quautitics-vpressurc,
volume, temperature, quantity of matter, and internal
energy, among thers. Fur systems other than a gas, we
can d fine and measure different macroscopic variables.
For instance, j 11 a ferromagnet such as iron, the particles
interact not by impulsive for;; in collisions but by magnetic forces; in the macroscopic description of a fClTOmagnet, the magnetization must be includ ed among the
macroscopic quantities.
Macroscopic properties can usually he measured directly in the laboratory, for example, the pressure of a
confined gas or the magnetization of a piece of iron. We
can also easil y measure the variation of any such property
with the temperature and derive an equation ofstate that
describes the dependence of the macroscopic variables 'on
one another.
For any system the macroscopic and the micr 'scopic
quantities must be rela eel because they are simply differcnt way" of describing the same situation. ln particular,
we' should be able to express one in terms of the other. The
pressure of a gas, a macroscopic quantity, is measured
operationally
using a manometer.
Microscopically,
pres-
493
494
Chapter 22
Temperature
sure is related to the average rate per unit area at which the
molecules of the gas deliver momentum to the manometer fluid as they strike its surface. In Section 23-3 we quantify this microscopic definition of pressure. Similarly (see
Section 23-4), the temperature of a gas (also a macroscopic quantity) is related to the average kinetic energy of
translation of the molecules.
If the macroscopic quantities can be expressed in terms
of the microscopic quantities, then the laws of'thermodynamics can be quantitatively expressed in terms of stat istical mechanics. This accomplishment
is one of the landmark achievements in the development of physics. As we
proceed through our study of thermodynamics,
this
theme of the relationship between macroscopic and microscopic variables will arise frequently.
22-2 TEMPERATURE AND
THERMAL EQUILIBRI~M
Consider the two systems A and B illustrated in Fig. I a.
They are "isolated" from one another and from the environment. By "isolated" we. mean that neither energy nor
matter can enter or leave either system. For example, the
systems might be surrounded by walls made ofthick slabs
of Styrofoam, presumed 10 be both rigid and impermeable. The walls in this case are said to be adiabatic. (The
word "adiabatic" comes from the Greek for "cannot be
crossed." You can think of "adiabatic" as meaning "insulating.") Changes in the properties of one system have
no effect on the other system.
We can replace the adiabatic wall separating A and B
with one that permits the flow of energy (Fig. Ib) in a form
that we shall come to know as heat. A thin but rigid sheet
of copper might be an example. Such a wall is called
diathermic. (The word "diathermic,"
loosely translated
from the Greek, means ','heat passing through." You can
think of "diathermic" as meaning "conducting.")
When two systems are placed into contact through a
diathermic wall, the exchange of energy causes the macroscopic properties of the two systems to change. If the
systems are confined gases, for example, the pressure
might be one of the macroscopic quantities that change.
The changes are relatively rapid at first, but become
slower and slower as time goes on, until finally the macroscopic properties approach constant values. When this
occurs, we say that the two systems are in thermal equilibrium with each other.
One way of testing whether bodies are in thermal equilibrium is to bring them into contact through a diathermic
wall and to observe whether the macroscopic properties of
the systems change with time after they are brought into
contact. If no changes in the macroscopic properties arc
observed with time, the systems were originally in thermal
equilibrium. It might, however, be inconvenient or even
impossible to move two systems so that they would be in
contact with one another. (The systems might be too
bulky to move easily, or they might be separated by a very
large distance.) We therefore generalize the concept of
thermal equilibrium so that the systems need not necessarily be brought into contact with each other. The separated
bodies can be said to be in thermal equilibrium if they are
in states such that, if they were connected, they would be
in thermal equilibrium.
The way to test whether such separated systems are in
thermal equilibrium is to use a third system C. By placing
C into contact with A and then with B, we could discover
whether A and B are in thermal equilibrium without ever
bringing A and B into direct contact. This is summarized
as a postulate called the zeroth law of thermodynamics,
which is often stated as follows:
If systems A and B are each in thermal equilibrium
with a third system C, then A and B are ill thermal
equilibrium with each other.
(a)
(b)
Figure 1 (a) Systems A and narc separated by an adiabatic
wall. The systems have different temperatures 1'.1 and T,1'
(b) Systems A and B are separated by a diathermic wall. The
systems, having come to thermal equilibrium, have the same
temperature 1".
This Jaw may seem simple, but it is not at all obvious. If
A, B, and Cwere people, it might be truethatA and Crnay
each know B but not know each other. If A and Care
unmagnetized pieces of iron and B is a magnet, then A
and C are each attracted to B without being attracted to
each other.
The zeroth Jaw has been called a logical afterthought. It
came to light in the 1930s, long after the first and second
laws of thermodynamics
had been proposed and accepted. As we discuss later, the zeroth law in effect defines
the concept of temperature, which is fundamental to the
first and second Jaws of thermodynamics,
The law that
e~tablishes the temperature should have a lower number,
so it i~'called the zeroth Jaw.
Section 22-3
:mperature
ten two systems are in thermal equilibrium, we say that
:y have the same temperature. Conversely, temperature
.hat property of a system which equals that of another
item when the two systems are in thermal equilibrium.
r example, suppose the systems are two gases that inil\y have different temperatures,
pressures, and volaes. After we place them into contact and wait a suffi:ntly long time for them to reach thermal equilibrium,
eir pressures will in general not be equal, nor will their
.lurnes; their temperatures,
however, will always be
ual in thermal equilibrium. It is only through this argu-
Measuring Temperature
495
There exists a scalar quant ity called temperat ur '
whic I IS a prope;:tyO/ all thermo 'ynamic systems in
eiiiii71 num.
\VO systems are 111 ierma equt I rium
if and only ijthelr temperatures are equal.
The zeroth law thus defines the concept of temperature
and specifies it as the one macroscopic property of a system that will be equal to that of another system when they
are in thermal equilibrium. The zeroth law permits us to
build and use thermometers to measure the temperature
of a system, for we now know that a thermometer
in
thermal contact with a system will reach a common temperature with the system.
ent based 011 thermal equilibrium that the notion of tern-rature call be introduced into thermodynamics.
Although temperature in its everyday use is familiar to
Iof us, it is necessary to give it a precise meaning ifit is to
! of value as a scientific measure. Our subjective notion
[temperature is not at all reliable. For example, suppose
ou are sitting indoors in' a chair that is made partly of
oth, wood, and metal. Touch the various parts of the
nair and decide which is "coldest," that is, which is at the
iwest temperature.
You will probably decide that the
ietal parts are coldest. However, we expect that all parts
f the chair have been in the room long enough to come
rto thermal equilibrium with the air and should all thereore be at the same temperature as the air. What you are.
esting when you touch the metal. is not only its temperaure but also its ability to conduct heat away from your
presumably warmer) hand. In this case your hand is givng a ·subjective and incorrect measure of temperature.
-urthermore, that subjective judgment will change with
.imc if you hold your hand on the metal, as your hand and
.he metal approach thermal equilibrium
with one an-
other.
You can also test your subjectivity by soaking one hand
in cold water and another in warm water. When you then
grasp an object of intermediate temperature, you will find
that the first hand senses a higher temperature than the
second. You can be somewhat more objective in comparing two different samples of the same material at different
temperatures by touching each with the same hand,
which may distinguish the "hotter" from the "cooler:"
This procedure might reveal which object is at the higher
temperature, but it is hardly sufficient to be quantitative
about the difference. It is therefore necessary that we carefully specify an objective way of measuring temperature,
which is our goal in this chapter.
In practical use of the zeroth law, we wish to identify
system C as a thermometer.
If the thermometer
comes
separately into thermal equilibrium with systems A and B
and indicates the same temperature, then we may conclude that A and B are in thermal equilibrium and thus do
indeed have the same temperature.
Another statement of the zeroth law, more formal and
more fundamental, is the following:
. 22-3 MEASURING TEMPERATURE
In Chapter I we described a two-step procedure for establishing a measuring standard for a physical quantity: we
define a base unit, and we then specify a procedure for
making comparisons with the base unit. For instance, in
the case of time, we defined the base unit in terms of the
frequency of light of a certain wavelength emitted by cesium atoms. To make I second takes 9,192,631.770 of
those vibrations. We can (at least in principle) use tills
scale to measure a human lifetime or even the age ot the
universe by counting the corresponding number of vi brations.
Temperature is one of the seven base units (see Table I
of Chapter I), and we might therefore attempt to treat
temperature as we treated other base units in the SI system: establish a standard and relate all other scales to the
standard. However, temperature has a nature different
from that of other SI base units, and so this scheme will
not work in quite that simple a form. For instance, if we
define one period of vibration of the light emitted by a
cesium atom as a standard of time, then two such vibrations last for twice the time, and any arbitrary time interval can in effect be measured in terms of the number of
vibrations. But even if we define a standard of tempera. ture, such as that of water boiling under certain conditions, we have no procedure to determine a temperature
twice as large. Two pots of boiling water, after all, hilve;h •••
.same temperature as one pot. There is no apparent way
using only this standard that we can relate the temperature of boiling water to that, for example, of boiling oil; no
amount of boiling water will ever be in thermal equilibrium with boiling oil.
To establish a measuring scale for temperature we
adopt the following procedure, which differs from the
usual procedure for the SI base units: we find a substance
that has a property that varies with temperature, and we
measure that property. The substance we choose is called
the thermomelric substance, and the propert.y that depcnds on tempera lire IS ca cd the thcrmo/lwlne nroprrtv,
I
.1
'J.
496
Chapter 22
Temperature
,
,
Examples might be the volume ofa liquid (as in the common glass-bulb mercury thermometer), the pressure Of a
gas kept at constant volu me,' the electrical'resistance 'Of a
wire, the length of a strip or metal,' or the 'cdlor' of a'larrtp
filament, all of which vary with temperature. The choice. ,
0/ one a/these substances leads to an individualte'mpe~~ture scale that is defined only/or that substance and that
does 1101 necessarily agree with other independently defilled temperature scales, Removing this disagreement req-uires the adoption of standards for the choice ofa particular thermometric substance, a particular thermometric
property, and a particular relationship between that property and a universally accepted temperature scale. Each
individual temperature
scale can thus be calibrated
against the universal scale. We describe the accepted universal scale in Sections 22-4 and 26-5.
Let us assume that our particular thermometer is based
on a system in which we measure the value of the thermometric property X. The temperature Tis some function of
X, T(X), We choose the simplest possible relationship
between T and X, the linear function given by
,
I
I
;1
T(X) = aX+
b,
(I)
where the constants a and b must be determined. This
linear scale means that every interval of temperature llT
corresponds to the same change S X in the value of the
thermometric property. To determine a temperature on
this scale, we choose two calibration- points, arbitrarily
detine the temperatures T, and T2 at those points, and
measure the corresponding values XI and X2 of the thermometric property.
The most familiar examples of this type of scale are the
Celsius and Fahrenheit scales used in common thermometers, in which the thermometric substance may be mercury and the thermometric property may be its volume,
observed by means of the length of the mercury column in
a thin glass tube. The linear behavior in this case means,
that the intervals between degree markings on the glass
tube of a thermometer are of uniform size.
The Celsius and Fahrenheit Scales*
In nearly all countries of the world, the Celsius scale (formerly called the centigrade scale) is used for all popular
and com mercial and most scientific measurements. The
Celsius scale was originally based on two calibration
,
;.
• Anders Celsius (1701-1744) was a Swedish astronomer who,
in addition to developing the temperature scale named for him,
made measurements of the length of the arc of a meridian that
verified Newton's theory of the flattening of the Earth at the
poles. Daniel Fahrenheit (1686 - 1736), a contemporary of Celsius, was a German physicist who invented both the alcohol and
mercury. liquid thermometers and used them to study the boiling
and freezing points of liquids.
points: the:;horma: freezing point of water, defined to be
O°C, and the normal boiling point of water, defined to be
100°0, These two points were used to calibrate thermometers,' and other 'temperatures were then deduced by interpolation or extrapolation. Note that the degree symbol (0)
is used to express temperatures on the Celsius scale.
The Fahrenheit scale, used in the United States, employs a smaller degree than the Celsius scale, and its zero is
set to a different temperature.Tt was also originally based
on two fixed points, the interval between which was set to
100 degrees: the freezing point ofa mixture of ice and salt,
and the normal human body temperature. On this scale,
the normal freezing and boiling points of water turn out to
be, respectively, 32 OF and 212 "F. The relation between
the Celsius and the Fahrenheit scales is
(2)
The degree symbol is used in expressing temperatures on
the Fahrenheit scale, as, for example, 98.6°F (normal
human body temperature).
Transferring between the Fahrenheit and Celsius scales
is easily done by remembering
a few corresponding
points, such as the normal freezing point (O°C = 32°F)
and boiling point (100°C = 212 OF) of water, and by making use of the equality between an interval of9 degrees on
the Fahrenheit scale and an interval of 5 degrees on the
Celsius scale, which we express as
9FO = 5C.
(3)
Note that these intervals are expressed as FOand CO, /lVI as
OFand "C. Readings on the temperature scale are given in
OF or °C (degrees Fahrenheit or degrees Celsius); differences in readings are given in FO or Co (Fahrenheit degrees or Celsius degrees).
The Kelvin Scale*
On the Kelvin scale, one of the calibration points is defined to be at a temperature of zero, where the thermometric property also has a value of zero; in effect, the
constant b in Eq. 1 is set to zero, in which case
T(X) = aX.
(4)
To determine a temperature on this scale, we need only
one calibration point P. At that point, the temperature is
defined to be T" and the thermometric property has the
measured value Xp. In this case
• Lord Kelvin (William Thomson, 1824- 1907) was a Scottish
physicist and engineer who made fundamental contributions to
a wide variety of subjects, including not only thermodynamics
but the law of conservation of energy, electricity and magnetism,
acoustics, and hydrodynamics, His-scientificcontributions were
so highly regarded that he was accorded the honor of burial in
Westminster Abbey in London.
-,
;'alet .••
a"",
Vial
(
320·F
J
li
t,·
1hcrmomeler wen
d
Vacu\lm
lIa~k
t". ~1
t I
c
!ffi!k>;::"·:.TI1I'\:-:i~§=!tl;:;f"':oTlt;;:ig::-;u:":,o;-:n::-:"~::-rE::-'-:-:>"l~'~1K
r----~-ll
Abl.Olute
[,.ro ••' CI K
U~ -. r1
- 2n) 5:C
Figure:.1 The Kelvin, CdSlU , .\IId I ahrenhch
scales compared.
-
(5)
(6)
.11)'general agreement, we choose for eur calibration the
temperature at which ice, liquid water, and water vapor
C01;, h! Hi equilibrium.
I'his point, whi h is very close to
the normal Ireezrng point of water. is called the triple f aim
01 war 'I (Fig, 2) 1 he temperature
t the triple point has
been set b) intcmatlonal agreement 10 be
1~, - 213.16 K.
where . (- kelvin) is the Sf base unit of temperature on
the absolute seale, ~hi h is identical with the ideal gas
: n per' rure scale di cussed in the next section. The kel••m IS thus d 'tmed as 1/273.16 of the temperature of the
tl ph: p in 01 water. \Vith this choice of calibration point.
Eq 6 becomes
1(X) = (273.16 K) ~
•
(7)
Al!
where A" IS the value of the thermometric property at the
triple point.
;\ temperature determined from Eq. 7 is valid only for
IhJ! particular
thermometric
PI perty; other thermometric properties and thermometric substances might
bi e ditlerent temperature readings (see Sample Problem
I) 10 elnninate this confusion between the readings of
<hffetl:nl thermometers, we choose as an accepted stand .rd I PC t) pe )1'thermometer in \\ hich the temperature
~_-3'-<-'·"1""'C""""''''''
• __ 0","-",,-",,-'
temperature
can be determined indcpendcruly
ot the a,lIlt\"
f the
thermometric
substance, I hi" choice h di 'U~' 1 III Ih
next section.
The size of the degree is the ••ame on the CdMUS and the
Kelvin scales. but the zero of the Celsius scale IS sluftc d to
a more convenient value Today we no longer use tv.\..!
fixed pcietsto deftne the Cclsius scele; instead, the Kelvin
scale is defined, and the relationship between the ( elsius
temperature
1~: and the Kelvin temperature
'I I \lOW
set as
7~"""T-273.15.
(~,
The freezing and boiling ponus of water are I ()
measured on the Kelvin scale arid then converted t ) (.
sius using Eq. 8. The e perimental values arc r pc
tively.O.OO· and 99.975·('. Figure ~ compares th
enheu. Celsius, and Kelvin scales.
Sample Problem 1 Tilt: resistance 01 a reruun plaunu 11 wire
increases by a factor of I. 92 between the trip!' poim of \>. arer
and the normal boiling point of water. Find the platinum resrstance temperature 01 boiling water,
Solution
We use Eq. 7. with the resistance R 'IS the thermometric property X We life not given the value of R,,. but w do
know that at the boiling POint of water, R 4. 1 392R" , Fhus
T(R) = T"
f.t
= (273.16 K)(I.J92)'~
J80 2 K.
This. value gives the "platinum
resistance temperature"
01
boiling water, Other thermometers will give dulcreru value for
example, the temperature
ix)lhn£ water 'l';d)Hli,. tt) a
or
-,
copper -cOn taman thermocouple is 440 K. LiCI'! of these readillS 15a rernpcrctur
•determined on a "private" seale, valid only
for that dc\ Ice [he accepted temperature of the normal boiling
1 'lilt ot \ ater IS )7.\ I 5 . which is dcterrmned USIOg the constant-volumc ga', thcrm nnctcr described 10 the next section.
Scale
h
22··. THE IDEAL ,AS
_ TU\lf>ERATURE SCALI~
The temperature of a system should have a well-defined
value. mdependcnt of the particular means used to
measure 11. According to Eq. 7, different thermometric
substances all give the same temperature at the triple
pouu. hut (as we have seen in Sample Problem 1) their
readings at other pomts may differ, We might imagine
doing n series of measurements
in which we simultaTIl'\1Ihl:
use different thermoraetric
properties to deternunc the t ~mperatur . of a system. Results of such a test
\\ \iuld ho« that the thermometers <Illgive different readIn' \\ c ought conunuc h .choosing a particular thermometric prop .n •. uch :1" the resistance of n wire. and
mcusurm the temperature of the System using different
~inds or \ ires In Hi of differing materials.
gain we
waul I hml a Wid" '. anauon in the measurements.
1 Obi,1I0 J lefimte temperature scale, we must select
on particulnr kind of thermometer as the standard The
choi e Will be made. not all the basis of experimental
convcmcn 'C, ut by inqumng whether the temperature
,11 defined D\ .1 particular
thermometer
proves to be
u ctul !D Iorrnutating the laws 01 phYSICS. The smallest
vanauon in readings h. found amongmnstalU·yolwncgas
thermometers using different gases, which suggests that
we choose a gas as the standard thermometric substance.
It turns out that as we reduce the amount of gas and
t!1\ rd,He it pressure, the vanation in readings between
~as lb.' mornetcrs 1l~1!1I'. different kinds of &U~ is reduced
I ) lien
there ecrns to he something fundamental
~'\{ut IIC bCI1..l\ ••• r of. constant volume thermometer
c ta nmr a ga at i \\pre ure 1 ct u therefore consider
the pro} rue 01 the constant-volume ga~ thermometer.
It the v lurne of a gas is kept constant. its pressure
dep nd on the temperature and increases linearlv with
11 • -ig temperature.
I he COt~ taut-volume gaS thermometer \l c~ th: pre' sur ot a gas at constant volume as the
>
tho r mornctru
PiOJ ell
I igure .:1 shows a diagram of the thermometer. II cooI IS of a bulb of gla . porcelain. quartz, platinum,
or
platmurn-ciridium
(depending on the temperature range
over which It is to be used), connected by a capillary tube
to a mercury manometer. The bulb JJ containing some gas
is put Into the bath or environment 'whose temperature T
is to he measured; by raising or lowering the mercury
reservoir R, the mercury in the left branch of the U-tube
R
'/}{
FigUf~ 4 A constant-s olume -as thermometer. 'I he bulb 81\
immersed in a l ath \\ hose temperature
is to be measured
The difference between the pressure of the &:\S III tht' bull 111d
atmospheric pressure is determined b~ the he: hI 11of Ill' col-
r
umn of mercury.
can be made to coi ncidc \\ ith J fixed reference mark, thu
keeping the confined gas at a constant volume. The dillerence between the pressure fl of the con ti ncd gas on the Iell
branch of the tube and the pressure Po of the atmosphere
on the right branch of the tube is indicated hy the height II
of the column of mercury. and thus
p = Po - pgh,
(9)
where II is the density of the mercury in the manometer.
in practice the apparatus is very elaborate, and we must
make many corrections. for example, ( I) (0 allow rOl the
small volume change owing to slight contraction or 1:\pansion of the bulb and (2) to allow for the fact that not 1'1
the confined g,IS (such as rhat in the caoillar
11.1\ "'11
immersed in the bath 1 e\ u assume that all con c II I
have been ma Ic. and that p is the correct d \ alue 1 th
absolute pressure at the temperature of the oath 1I11.:n the
temperature is given pro VI tonally by
1'([1)- (273.16 K)£
P.,
(constant
11.
(10)
Let a certain amount of gas, mtrogen. tor mstauce.
put into the bulb so that WIlNl the bulb is surrounded U)
water at the triple POlOt the pressure ri, is equal to a def
nite value, say, 0 em Hg. Now we immerse the bulb In
the system whose temperature Twe wish to measure, and
with Ih: volume kept constant at Its previous value. we
measure the gas pressure /1, as in Eq. 9, and calculate the
provisional temperature Tofthe system using Eq I( . rile
-,- ----.-PI;),$m in fu ion lest rea lor
10'
I(II
'enter of Sun
urface of un
Mehina point of tungsten
freezing point of wat 'r
ormal boiling point of
(. X 101
(, X 10)
~ 7" 10]
He
~Ie::'$p'_n_l_·t)(_)_h_ng,;;.,...~
20
40
60
,~ 2
'l '
~
Aniabaric derna nelLl.lllon 01 p:lT;tnllgul:ll( salt
Hz
o
17
'1
Normal boiling point 01 He
[ean temperature of univer-,e
lIIe- "He diluuon rcmvcrutor
eo
1(
IU
J
101
~ Ill'
~__
100
Per (em Hg)
Fi ure 5 As the pressure of the nitr gen gas in a
con iant-volurne gas thermometer is reduced from 0 em Hg
to 40 nd then to 20, the temperature deduced for the system
appro
hes a limit correspondin
to a pressure
Other
gase
pproach the: me limit, which IS the ideal gas temperature 7' of the system. The full range of the vertical sc le is
about I K fur I. pical conditions.
oro.
result of this measurern nt is plotted as a point in ig, 5.
• '0 v we I ·turn the thermometer
lO the triple-point
cell
and remove some ot'the gas, so that Pir has a smaller v: lue,
say, 40 .rn Hg. Then we return the thermometer to the
unknown system, measure the new value of P. and calculate another provisional temperature T, also plotted in
FIg. 5. We continue
this same procedure, reducing the
amount of gas 10 the bulb and at each new lower value of
JlI(ca\culaling the temperature T. If we plot the values ofT
a rainst p". we can extrapol te the resulting curve to the
Inter cction with the a is where p" - O. The data points
lor 2 nd the resulting straight-line extrapolation are
shown in Fig. 5,
\ e r peat this procedure with gases other than nitrogen
In the thermometer, obtaining results shown in Fig. 5.
The lines show that the temperature readings of a onst nt- . lurne gas thermometer depend on the gas used at
ordinary values of the r ference pressure. However, as the
retercnce pre ure is decreased, the temperature readings
of constant- olume gas thermometers
using different
rases appr ch the same
lue T, which we can regard as
the temperature of the system. The extrapolated v lue of
itic temperature depends only on the general properties of
gases and not 011 any particular gas. We therefore define
the ideal gas temperature scale:
·r-(273.16K) lim..E.
p,.-o PI,(
(constant
V).
(11)
OUf stand rd thermometer
is chosen to be a constantvolume gas thermometer
using a temperature scale defined by Eq. II.
If rernpcrnture IS to be a truly fundanrental ph ~iC;ll
quantity, one an hich the laws otihcnno I> nauucs IOU,
be expressed, It is absolutely ne essarv that us definuion
hi: independent 01 the ropcrtics 01 i ccrlic material
It
wo lid 110l do. for e. ample, to have such a OJ IC quanuty
as temperature dt!J~I\(.J un the c pan I it oln crcur .thc
electrical re isiivity [platinum.
r any othcr su h "handbook" property. \\ e chose the gas thermomcrer a o r
standard instrument
precisely because no su h sp .cilic
properties of marenals art' involved in its 0ll'l auon. '1011
can use any gal, and you alwnvs ~d the same :tIlS\\CI .
Although our l mpcrature scale j., indep ·Il.! nt ,II the
pr pertics or anyone particular gas. 11 10t:\ dep 'Hd on I he
properties of gases in general (that is, on the propc IIII:) ora
o-called ideal gas). The lowest temperature Ih,lt can be
measured with a gas thermometer IS about I K. 1\) obtain
this temperature \\C rnu: t lISC low-pressure helium. \ hich
remains a gas at lower temp rarures than an~ lither gas.
\Ve cannot give e: I crim mal 1lH.':l111ng h) ll'IIlIl\!I.I!\l!\'\
belov about I
b}' means of <1 gas therm unctcr.
We would like to define :i tem] .ruturc calc in 'I W:1~
that is independ nt of the propcrues 01 11) paruc ular
subst n e. \
show in Section 26- - that the ah ilute
temper turc . 'ale. called the Kelvm
uch a scale. Wc als show 110 I the IIk,11 IS ~ OlIC
thermody n mi
scale. IS
and the Kelvin ale arc identi :\1in the ran 'f.! ut tempe 1·
tures in which a gas th rmorneter may be u ed. lor tlus
r son we can use units of kclvinsf r thcideal gu temper3tuTC, as we ha 'C already done in 1 c II. J able 1 1t~1 the
temperatures in kelvin oi'variou I xlic ani pro I:'~~.
'I e also show in Section ::b-5 that the Kelv III :.< ;I~ Ins
an absolute zero of 0 K and that it IS impossible to (001
system below 0 K. I'he absolute (em 01 temperature has
defied all attempt 10 reach it e. pcrimcntally, hut icmj vr
atures within a sm II range (I O'~ K) 01 absolute zero have
been achie ed.
..•.
Although there is a dir t onnection, as we how 10
Chapter 23. between the microscopic mouon of m If,-f
.
"
n
cutes and the ma -roscopi temperature,
II l~ not true 1I1:11
3\1molecular monon cease at the absolute It..: ro ofternper-
500
Chapter 12
Temp rature
ature. The connection between temperature and molecular kinetic energy is based on classical concepts, while the
quantum theory tells us that there i a nonzero lower limit
t the molecular kinetic energy, even at absolute zero.
This zero-point energy cannot be inferred from classical
calculations.
The International Temperature Scale
Precise measurement
of a temperature with a gas thermometer is a difficult task, requiring many months of
painstaking laboratory work. In practice, the gas thermometer is used only to establish certain fixed points that
can then be used to calibrate other more convenient secondary thermometers.
For practical use, as in the-calibration of industrial or
scientific thermometers,
the International Temperature
calc has been adopted. This scale consists of a set of
procedures' for providing in practice the be t possible approximations to the Kelvin scale. The adopted scale consi <, of a set offi cd point ••, along with the Instrument
to
be used for interpolating between these fixed points and
for c. trapolating beyond the highc t fixed point A new
scale h s been adopted by the International Committee of
Wci hts and Measures abo t every 20 years; the fixed
points of the most recent one (1990) arc shown in Table 2.
1,
m,
2
PRIMARY FIXED POINTS
1990 Il\.TTER ATIONAL
TEMPERATURE
SCALE"
ON THE
Substance
State
lelium
I !}drogen
Hydrogen
II ydrogen
, 'con
O"'Y en
Argon
Mercury
Water
Gallium
Boiling point
Triple point
Boiling pointb
. Bolli ng point
Triple point
Triple point
Triple point
Triple point
Triple point
Melting point
Indium
Freezin
Tin
Zinc
Aluminum
Silver
Freezing point
Fre zi g point
Fre zin point
Freezin point
Freezing point
692.677
933473
1234.9
) 337.33
Freezing point
1357.77
Gold
Copper
Temperature (K)
point
3-5r
13.8033
17.025-17.04Sc
20.26- 20.28c
24.5561
54.3584
83.8058
234.3156
273.16
302.9146
429.7485
505.078
• See "The lnternauont I Temper ture Scale of 1990 (ITs... 90)," by
H. Preston-Thomas, M~(ro!ollla. 27 (1990), p. 3.
~ Thi bollin point is for a pressure or t atm. AU other boiling
points. melting points, or freezing points are for a pressure of I atm.
• The: temperature of the boiling pomt varies scmewhat with tb
pressure of the gas above the liquid. The temperature scale gives
the relationship between T and p that can be used \0 calculate: T
ior a ilven p.
You can often loosen a tight metal Jar hd bi holding it
under a stream of hot water. As its temperature rises, the
metal lid expands slightly r lative to the glass) r. Thermal
expansion is not alway desirable, as Fig. 6 suggcs . \: e
have all seen e pansion lots in the roadways of bridge.
Pipes at refineries often include n cxpan Ion 10 p, so that
the pipe will not buckle as the ternpcratur
nses. Materials
used for dental fillings have expan ion properne SImilar
to those of tooth enamel. In aircraft manufacture. rivet
and other fasteners are often d igned so that they arc to
be cooled in dry ice before in ertion and then allowed to
expand to a tight fit. Thermometers and thermostats 111:1\
be based on the differences in expansion between the cornponents ofa bimetallic strip; see FIg. 7.1n a thermometer
ofa familiar type, the bimetallic strip is coiled into J heli
that winds and unwinds as the temperature change:
ce
fig. 8. The familiar liquid-in-gla s thermometers arc
based on the fact that liquids uch as mercury or al ohol
expand 10 a different (greater) e tent than do their lus\
containers.
We can understand thi expansion hy considcrin
simple model of the structure of a crystalline solid. Ihe
atoms arc held together in a regular arr . by ele In .1
force) which are like those that would beexerted by a t
of springs connecting the atoms. We can thus visualiz he
solid body a a microscopic bedsprin
(Fig. ). 1 he
"springs" are quite tift and not at all ideal ( • Problem 3
"
1' •• To
T> To
Figur 7 A bimetallic stop, consisting of a strip of brass and
3t temperature To. At temperatures higher than To. the strip bends as shown; at lower ternperaiures 11 bends the other way. Many thermo tats operate
on this principle, usin the motion of the end of the strip to
ma e or break an electrical contact.
a strip of steel welded together,
proportional
8e.flng
POinter '
to the temperature
change 6.'1 and
10 the
original length L. l Ience we C.1nwrite
AL=uLtl/,
called the wet It/,'II( ') line..•r C'\P IISiO/j h s
.alues for dif erent r I te rials Re nun this lor
mula, we obtain
-where
Q,
different
(13)
Helical bimetal
..-
element
II ure
thermometer based on a bimetallic strip, The
sinp is formed into helix, which coils or uncoils as the temper ture i changed.
o Chapter 15), and there are about IOn of them per cubic
centimeter. At any temperature the atoms of the solid are
vibraung. The mplitude of vibration is about 10-9 em,
bout one-tenth of an atomic diameter, and the frequency
is bout lOll Hz
\ h n the temperature IS iner ased, the tom vibrate at
larger amplitude, and the average distance be ween aioms .
increases. ( ee the discussion of the microscopic basis of
thermal expansion at the end of this ection.) This leads to
an expansion of the whole solid body. The change in any
linear dimension of the solid, such as its length, width, or
thickness, is called a. linear expansion. If the length of this
lmear dimension is L. the change in temperature flT
causes a change in length flL. We find from experiment
that, if AT is small enough, this change in length AL is
so that a has the meaning. of a fractional eh nge in length
per degree temperature change.
Stri t1y speaking, the \ alue of u depends on the a tu I
temperature and the referen e 1 mperarure ho n t determine L (see Problem 2}). llo ' ver. its van non I~
usually negligible compared to the ac .ur C) \~ ith which
measurern nts need t be made. It is Il n sufficie I to
choose an average alue that CuI) e treated as a on \.,.ot
over a ertain temperature range. In f able
e li')t Ihi:
TABLE 3
SOME AVERAQ!.: COEFFlCIE;
L1
AR EXPAl SI
r 01:
r
ubstance
--------------------------~~--
502 .Chanter 22
.•
1cmperature
experimental values for the average coefficient of linear
cxpan ion of several common solids. For all the subtanccs listed. the change in Si7C consists of an c parisien
as the temperature rises, because ex is positive. The order
of magnitude of th expansion i about 1 millimeter per
meter length per 100 Celsius degrees. (Note the use of C',
not -C, to expre s temperature changes here.)
(0)
m ITTlT]
9
0
Fis:urc 10
• mple Probl m 2 A steel metne scale is to be ruled so that the
millimeter intervals are ac urate to within about 5 X lO-s mm
at n certain temperature. What is the maximum temperature
variation allowable during the ruling?
Solution
A steel rule at two different temperatures
I he exin proporuon in all dimensions the scale,
the numbers, the hole. and the thickne s arc all in rea! d bv
the same factor. (The expansion hewn I greatly exaggerated;
(0 obtain such an cxpan ion would require a temperature
in-
pansron increase
crease of about 20,000
'!)
From EQ. 12. we have
4£
4T-
(Xl. ••• (II
5XI~smm.
X 1Q-4/C'XI.0 mm) -4.5 C ,
\\ here we have used the value of a for steel from Table 3. The
• temperature during the luling must be kept constant to within
about 5 C'. and the scale OIUSl be used within that same interval
of temperature at which It wa made.
tore that If the alloy in. r were used instead of steel, we could
achieve the same precision over a temperature interval of about
75 "; or. cquiv lently, if'we could maintain the same temperature variation (5 CO). we.could achieve an accuracy due to temperature chang of about 3 X 1O-~ mm.
For many solid, called I otropic. the percent change in
length for a given temperature change is the same for all
lines in the solid. The expansion is quite analogous to a
photographic
enlargement. except that a solid is threedimensional. Thus, if you have a flat plate with a hole
punched in it, tlLIL(-ex
T) for a given tlTis the same
for the length, the thickness, the face diagonal, the body
diagonal, and thc hole diameter. Every line, whether
straight or curved, lengthens in the ratio ex per degree
temperature
rise. If you scratch your name on the plate.
the line representing your name has the same fractional
change in ten th a any other line. The analogy to a photo-graphic enlargement i hown in Fig. to.
With these ideas to mind, you should be able to show
(see Problems 30 and 31) that to a high degree of accuracy
the fractional change in area A per degree temperature
change for an isotr pic solid is 2a, that is,
~A - 2aA AT.
(14)
anti the fraction Ichang in volume V per degree temperature change for n isotropic solid is 3a, that is,
V-3
tl T.
( 15)
Because the shape f a fluid i not definite, only the
change in volume with temperature is signifrcant, Gases
respond strongly to temperature
or pressure changes.
••••
hereas the change in volume of liquids with changes in
temperature or pressure is very much smaller. If we kl /1
represent the coefficient )f volume expansion for ,}liquid
so that
\ 1'/1'
( 161
fJ --;:[ ,
find that fJ IS rclauvely independent of the tempe r.
turc. Liquid ••typically expand with incrcasi ng tempe .1lure, th If volume expansion being generally about III
times reaicr than that of -ohds.
However. the most common liquid. water, d( cs not
behave like most other liquids. In FIg. 11 \C sho« Ih
volume expansion curve for water. otc that above -l'
water expands as the temperature rises. although not linearly. (That is, P i not constant o\'er these large tern .ralure intcrvals.) As the temperature is lowered from 'C hI
O·C, however, water e pands instead of contractin ,
which is the reason that lakes freeze fir t at their upper
surface. Such an c pension with decreasing temperature
is not observed in any other common liquid; it is obscr, ed
in rubberlike substances and in certain crystalline solids
over limited tcmr raturc intervals. The densit of water is
a maximum at 3.')g-C. where its value is 999.971 'gjm1
(The standard kilogram and meter were originally supposed to correspond to a maximum den. ity for water 0
1000 kg/ml or I g/cml. Accurate measurements sh iw,
howey r, that II. ' international standards do not correspond exactly to this \ alu )
W'
Microscopic
On the micro
Basis of Thermal E pan ion
(Optional)
orne level. thermal e nansion of a solid su csts
an increase in the average sepa uon betw en the atoms In the
solid. The potential energy curve for t ••••o adjacent atoms in :l
crystalline solid as a function of their internuclear separauon I~
an asymmetric cur .•c like that of Fig. 12. As the atoms move
close together. their separation decreasing from the equilibrium
value ro, strong repulsive forces come into pby. and the potential energy nSC5 ~l~'epl} (F.,. - dUldr); as the atoms move: Ianhcr
apart, their S~.1 : ••uion mcreasm from the equilibn um \ 1Iu('.
somewhat weal-a attractive forces' take over and the potential
energy nses more .lowly, t a given VIbrational energy the scparation or the atoms changes periodically from a minimum to a
QllnlftltH
01--1----,....
100
503
-
Figure 12 Potential energy curve i~}r IWO adjacent .•tOIl1' m .I
solid as :l Iuncuon (it their uuemuctear scparauon ,h~l:Jnl'c.
The equilibrium separation l~ ro. Because the curve h :1\) mmetnc, the average separation tr" r~) InCrC;lS(.,'S
;IS tlw temperiure 1'1' 1;) , nd the vibrauonal cu.:rgy (fl' f.' ) increase
creases as the temperature
rises, the ,tH'T<lg.: separation between
atoms increases wuh ternperarure.un i the entire solid e I ;Jl1d,
I ote that rt IhC·POLClltl.!1 cller!?,} curve were 'i) nuueu h': about
the equifibnum
separ.mon. in .n me average ~p.H.t!!On would
equal the equilibrium s<:p,jl:'1tion:n
mailer \10\\ klf!;C Ih~ .unpluude of the \ ibration. Hence thermal c,\pan\lon " ,\ -tiled
Temperature l'C}
11 (.;) I'he specinc volume (the volume occupied b
:.Iflku!:n mass) of water a i1 function of us temperature.
I ne sp.:-:lIi ••' volume IS the inverse of the density (th mass per
UOIt volume). (b) An enlargement
of the region near·\ "C,
shuv,lf\g :1 minimum specific volume (Of a maximum density).
Jot'LOH'
:t
maximum value, the average separatlon being greater than the
separation because orthe asymmetric nature of the
potential energy curve. At $(iII higher vibrational energy the
average separation is even greater. The effect is enhanced because. as suggested by Fig. 12. the kinetic energy is smaller at
larger eparauons; thus the parti Its move slower and spend
more lime :11 large separauons, ~ hich then contribute a l:lrg r
sh. re to the time a erage, Because the vibrational energy in-
equilibnum
consequence
01 the J.::, l ••uon
trom "ytnmdl)-
ol'thc I:hara: lc'fI,
lie potcnua! enCfr) curv 01 'vilLI.,
Some crj SMl1IIH: ~oluh, H\ •..\~II,jH\ 1<:lflik! HUh' ! ••..
gl\lIl~ "1.\)
contract us the tcmperatur e rises the .tb,"" anal SlS TCm.un\"
valid if one assumes 111:\\ (lilly compresvioual (longl\udmal}
modes ofvibraticn exist or that these modes predommate } low-
ever, solids may vibrate If! shetlrlo\-.:e (transverse) mode» a:; \\cU
and these modes of vibration allow the solid 10 CMtr3Ct as the
temperature rises, the average separation of the planes 01 atoms
decreasing, For certain typ¢5 of crystalline Structure and In cerlain temserature regions. these transversemodes
of vibrauon
may predominate over the longuudinal ones.l.4ivlO :I net ncp
,the coefficient of thermal expansion.
It should
be emphasized
that
the
microscopic
mllJeh
presented here are ovcr-amplincauunv 0/ ;J complc pl.cnoml'
non thnt C' fl be treated \\ ith gre:lIet II1sigjll usin ' st: mil" l me
chanics and qu 111 m theory. •
Q ESTIONS
I. Is temperature a microscopic or macroscopic concept?
2. Can \\e define temperature as a derived quantity, in terms of
length, mass, and time? Think of a pendulum, for example.
J. bsolute zero is a minimum temperature.Is there l\ maximum temperature? ,
4. Can one object be houer than anotherifthey are at the sarne
temperature? Explain.
5. Are there ph}sical quantities other than temperature th:u
tend to equalize if tWO different systems ate joined?
6. A piece of ice and a warmer thermometer Me suspended In
an Insulated evacuated en losure so that the)' are not in
contact. Why does the thermometer
reading decrease II••a
time'!
7. What qualities make 1I particular thermomeuie
suitable for use in a practicel rhermerneeer?
pr PI.'I\\
B. What diftkulties would arise if you defined iempcnuu •.•..1!1
terms of the dem,llY of water?
9. Let III be the pressure lrt the bulb of a. constant-volume p~
thermometer when the bulb ts ••t the inplc-pouu \emp,'I.I·
ture an73.ln K and lelfl be lile pressure when the bulb h at
room temperature Given three constant-votumc g:h Ihermometers. forJ the g:t ISO h:C!l and p~ ~Ol)l Ili~:1M /I
50... Chapter 12
Tcmperatvre
the gas is also oxygen but Pj - 40 em Hg; lor C the gas is
hydrogen and p) ~ 30 em Hg, The measured values of p for
the three thermometers are I'A' PII> :mdpc. (tl) An approximate value of'the room temperature Tean be obtained with
each of the thermometers using
TA - (273.16 K)(P••/20 ern Hg),
TII •••
(27 3.16 K)(p.J40 ern Hg),
Tc "" (21 .16 K.){PcJ30 em Hg).
Mark each of the folio'>' ing statcments.true or false: (I) With
the method described, alt three thermometers wm give the
same value of T. (2) The two oxygen thermometers will
agree with each other but not with the hydrogen thermometer. (3) Each of the three wtll give a different value ofT. (b) In
the everu that there is a disagreement among the three thermomcters. explain how you would change the method of
using them to cause all three to give the same value of T.
10. The editor-in-chief
of a well-known business magazine, discussing po sible warming effects associated with the increasing concentration of carbon 'dioxide in the Earth's atmosphere (greenhouse effect), wrote: "The polar regions mIght
be three rimes warmer than now: ....
" What .do you
suppose he meant, and what did be say litemlly? (See
"Warmth and Temperature: A Comedy of Errors," by
Albert A. Bartlett. The Physics T(,(II:Ju!r: November 1084,
p.517.)
I!. Although
the absolute zero of tempera ure seems 10 be
c pcnrncntally
unattainable.
temperatures
as low as
0.00000002 K have been achieved in the laboratorv. Why
would physicists strive, as indeed titey do, 10 obtain still
lower temperatures? Isn't this low enough for all practical
purposes?
12. You put two uncovered pails of water, one contnming hot
water and one containing cold water, outside in belowfreezing weather. The patl with the hot water will usually
begin to freeze first. Why? What would happen ifYOll covered the pails?
U. Can a temperature be' assigned to a vacuum?
14. Does our "temperature sense' hav e a built-in sense ordireelion: that iv, docs hotter necessarilv mean 111~hertemperalure. or is Ihls just an arbit racy convention? Celsius, by the
way, originally cho: e the steam point as O'C and the ice
point as IOO·C.
15. Many medicine labels inform the user to store below 86"F.
v hy !l6? (lllni. Change (0 Celsius.) (See The Science )1/.
mana(". 1985-1986. p, 430.)
16. How would
you suggest measuring the temperature
of
upper atmosphere, (c) an insect,
(a) the Sun, (0) the Earth's
(d) the Moon, (c) the ocean floor, and (f) liquid helium?
11. Considenng the clSIUS, Fahrenheit, and Kelvin scales. docs
nny one stand outas "nature', scale"? Discuss.
]8. Is one gas any better than another for purposes of 1\ standard
coostant ••volume
gas thermometer?
WhJ( properties
UI\,'
oesirable In a gal>for such purposes".
19. S ate some objections to usmg \ ater-in-glass as a therrnorne-
ter. Is mercury-in-glass an improx emcnt? If'so, explain why
20. Explain why the column of mercury fir:.1 descends and then
rises when a mercury-in-glass thermornct r IS put m a flame
21. What are the dimensions of 0:, the cocllkielll ot linear expansion? Does the value of 0: depend on the unu ofknf,lh
used? When Fahrenheit
degrees are used mstcad of CdSlUS
degrees as the unn of'temperature change, docs the nurncncal value of 0: change? If so, how? If not. prove It.
'll. A metal ball can pass through a metal ring. Wht'n the hall is
heated, however, it gels stuck in the nng. What would hap.
pen if the ring. rather than the ball. were heated?
23. A bimetallic strip, consisting of 1WO different metal &trips
riveted together. is used as a control dement m the common
thermostat Explain how it works.
24. Two strips, one of iron and one 01 line. arc riveted together
side by side to form a stratzht bar that ClWT when heated
Why 1$ the Iron on the inside of the curve?
25. Explain how the period of a pendulum clod, can be ept
constant with temperature b} attaching vertical tubes 01'
mercury to the bottom of the pendulum.
26. Why should a chimne, be frecstandmg, lhJI is, 1'10\ part of
the structural $UPDOI t of the house'!
27. Water expands when it freeze . Can we define: a coctlicient
olume •...xpansioc for the freezing process?
or ....
'lB. ExpJllin why the apparent expansion of a liquid in a glass
bulb does not give the true expansion 0 the liquid
29. Does the change in volume of an object when its temperature is raised depend on whether (he Object has cavuies
inside. other things beina equal?
3(). Why is it much more difficult to make a precise determination of the eoefficienr of expansion of a liquid than of a
solid?
Jl. A common mode! of a solid a sumec the alOlll'; to re POInts
executing imnle harmonic rnouon about mean lattice POSInons. \\'hltl would be the coefficient of linear expansion of
such a lattice.
:'1, Fxplam the fact that the temperature of the ocean at ueat
depths is vcry censtant the year round, at a temperature 01
about 4'('.
33. Explain why lakes freeze first at the surface
J4, What causes water pipes to burst in the winter?
35, What can you conclude about how the mclung pomt otice
depends on pressure from the fact that Ice floats on water?
-----.------.------- --------
PROBLEMS
Section n~JMeasuring Temperature
1. A resistance thermometer is a thermometer
in which the
electrical resistance changes with temperature. We are free
to define temperatures measured b1 such n thermometer 10
kdvim (K) \0 be directly PlOp0t110tl31 0 the resistance R
"111M 'II! I
measured in ohms (n). A zrtain resistance thermometer is
found to have resistance R of 90.3 n when its bulb is
pia ed In water 31 the triple-point temperature (27 J.t K).
Wha temperature is indic led b ' the thermometer if ih
bulb i plac d in n environment su h lh t Its resistance i
other 4.0 CO? Assume that the outside temperature
change and rhat
pre ious problem.
therm
ouple, with one june-
12. A particular ga!>thermometer is constructed of two g:! -cuntalfllng but . each of which i rut 101\) a water t> ih, ;1$
sho n in PI'. 13. I he press ! di er nee t .tween the two
hul
is mea ure b: a mercury man m ier ~ h \\1\ in the
iion held t O·C. the output voltage vanes linearlyfrom 0 to
2!10 IIIV as the ternperaiure of the other junction is raised
from 0 to 510' • Find ine temperature of'the variable juncnon when the therm ouple output is 10.2 mY.
tigure, Appr
3. Th arn] 11 icauon or am of a transi tor amplifier m y depend in the temperature. The
in for a certain amplifier at
n »n temperature (200'C) is 300. whereas at 5 .0'
it I>
3 2 \: 'hat woul the gam be ar 28.0·C if the gain depends
linearly on temperature OH~r this limned range?
4. Absolute zero is -273.15·c' Find absolute
Fahrenheit scale.
zeroon
does not
pplies: see the
t rn
i Iti 373.1 K.v.h.tti.
the limiting value f the u tio 0 he pressures 1.1
at the
team point and at Ih triple point 01 v ter v.Len the g.. IS
kept fit constant volume]
ay that a small voltage j produced
are at different temperatures. In :1
p rticular iron-iconstantan
law 01 C oltn'
n. Ifthc!!;3stempe aiure auhe
is termed rom two different metals Joined
at 1"'(1 points in SU h :
wh n the t vo junctions
ewion's
Src/ion n-4 Tht Id('ul IIU 1emperature Seat«
:>6.2 n!
2. A thermocouple
50S
priate reservotrs,
maintum constant
to
Ih' diagram,
vol me in the two bulb
n
1h re 1 no
I
hV"1I
difleren e in pt suo _ vhen both baths re at the triple point
of water. The pre sure difference IS 120 mrn H when one
bath is at the tnpl polnt and the th r i at th 'lx)I\tng point
of water, Finallv, the PII:S ure ditleren <: IS <)0 mm Hg
when one bath h. ut the trrple POIllI alld the other IS \II ,\0
unknown tcmncrature to 1)1:measured hod the lll\knov.n
tempera; 1\'.
the
5. If)
ur doctor tells you lhat your temperature is 31Q kelvins
above <l olute zero, should you worry? Explain your anwcr,
6. (n) f he temperature
of the surf ce of the Sun I:> about
()
'. Exprc
thb on the Fahrenheit scale, (b) L press
n I m I human body temperature,
986 'F, on the Celsius
..tie. ( ) In the continental United 'tateS. the lowe t Oftl~
I \I re
rdcd temperature I -70'r at Rogers Pas . MOlltnna 1 xpre s this on the CelsIUS scale, (d) ExpreSS the nor01••1 t Illllg point of oxygen, - 183 ,on the Fahrenheit
sc le, (e) At what
elsius temperature
would you hnd a
room to be uncomfortably
warm?
( J
Fil!Uft' J 3
Problem
1_.
7. At what iemperat ire, ifan)" do the following pairs of scales
t-i .•.. the
me readin : (al Fahrenheit and Celsius (b) Fahrenheit and Ke! -in, and (c)
lsius and Kelvin?
mperarure i the Fahrenhelt scale re ding equal to
(allv.ice that of the Celsius nd (b) h If that of the Celsius?
.'" t\ hatt
9. It is an ev rvday ob nation that hOI and cold objects cool
do n or warm up to the temperature of their surroundings,
If the temper' ture difference
J between an obje t ' nd its
surroundings (6T- T<illJ - T••••
,) is not tOOgreat, the rate of
cling or \\ rrning of the obie
maiely, to this temperature
t is proportional,
dilfercnc
JAT --A(~:t)
dl
approxi-
. Tv o consram- olum gas thermometers arc as n I d.one
U ing nitrogen
th WO! king
nd Ihe other usina hehum. BOth ont In enou 1 b'I
that Pit
1
m Hg,
What is the difi rence between the pee ure in the IWO
thermometers II both are Inserted InIO a '" ter b.tlh .II the
boiling point? Vhich pre. sure IS ihe higher 01 the two? See
Ft
; that is,
'
where A IS a constant, The minus ign appears bee use AT
decrea ~ with time if AT is positive and increases if t:.T is
neg tive This i known
'ell/on's law of cooling. (0) On
what Iaciors does A depend'? What are its dimen ions? (b) If
at
me instant 1-0
the temperature
difference is A1'o.
show that it i
t a ti me tIler.
10 ..
arl} In th.:momingtheheatcrofahousebrea
down. The
outside iemperatur IS -7 O·C. ( a result, the inside temperature drops (rom 22 to 1 • , \0 45 Olin. !Iov.; much longer \\111 ': t
e t r the insrde rernperat .. :,'lO fall b)
n-
1 '
5,
Chupter i: Temperature
506'
18. A gJ
window is 200 em hy 300 em at 10·C. Oy how much
ha'i its area increased when its temperature is40' ? A ume
that the glass IS free t expand.
19.
l r.t s cube ha an edge length of 33.1 em at 20.0'C. Find
(0) the increase 10 surface area and (b) the increase In volume when it is heated to 75.0·C.
if its volume at
20. What is the volume of a lead ball. t -12"C
160· is 530 em)?
21.
how that when the temperature of a liquid in a barometer'
change b) AT, and the pres urc is const nt, the height I:
ch: ngcs by 6h ~ Ph I. whcrefJis the coefficient of volume
e pansicn of the liquid. 'eglccr the expansion of the glass
tube.
:U. In a ertam c .pcnment, It WJ necessary lobe able to mOve a
srnal! radroacnve source at selected, extremely slow speeds.
cicnt of volume expansion of mercury and cr I Ih ellcicnt 0 linear expansion of tass.
27. (a) how that if the lengths OIIWO rods of dill r nt \ ,lid ur
inverseh proportional to th rr rc I '<;II\':C 'lliCl~nls I hI!
car expansion at the same initial temperature, th dilleren •.
m length between them w III be constant 31 1\ tern rature
(b) What should be the lengths of a steel and a brass rod 1\
O'C SO that at alltemperatures
their difference in lcnath I
0.30 m?
28. As a re ultofatemperat\lrcn~orJ::!·
its center bu Ides upward. a shown
.a bar wirb a erack at
m h~. 15. If ihc n: ed
3.,17 m and the coefficient of lincar c pansion
distance /-0 10 6/(', find .v, the UI\t;lO('C to which the: center
is 25
rises.
This was accomplished by fastening the source 10 one end 0
an aluminum rod and heating the centralsection of the rod
I.n a controlled way.If the effective heated section of the rod
In Fig, J 4 is 1.8 m, at what onstant rate must the temperaturc of tbc rod be made to change if the source is to move at ~
constant speed of 96 nm/s?
Figure 1
29. A steel rod IS 3.0(') em to diameter at 25· . \ I ras nn has
an intcrior diarne cr of •. 9. 2 em at 2S' . AI what common
temperature vill ihe nng just slide onto the rod?
Figure 14
2:\.
Problem 22.
how that it (1' is dcpend
I.
1'1
nt on the temperature
[,1 I
J~
n(T)dT]
T, then
1
to the temperature,
l
,
• \\ here Lo is the kngth :H u reference temperature rr>'
24. Soon after the Farth rorm« . heat rele •.ised h the dccav of
radioactive elements raised the average internal temperalure (rom 300 to :'1000 K. at about which value it remains
today. ssumin . an average coefficient of volume cxpansion 0 3;; X IO-s K-I. by how much has the radius of the
Earth increased since it formation?
15. Arod i me sun;dto\>e20.05
m long u ing a steel rulcr at a
room temperature of 20·C. Both th rod and the ruler are
placed In an oven at 270'
where the rod now measures
20 II em using the arne ruler. Calculate the coefficient of
thermal expansion for the material of which the rod i made.
26. Con ider a rnercury-in-gla s thermometer. Assume that the
cross section of the capillary is con tant at A. and that JI is
the volume of the bulb of mercury at O.OO·C. Suppa e that
the mercury JUSt fill the bulb at O.OO'C Show that the
length L of the mcrcurycolurnn
in the capillary at a temperalure T. in 'C, i
that IS. proportional
30. The area .tI of a rc rangular plate l. lib, Its coefficient 01
line r e. pansion is <t,. Iter a temperature rise A J • SlJ illS
longer by Aa and side "i Ion er by Sb. Show that If we
neglect the small qll,1tI1;t} Au blab (see fig. I l, then
-20'..161', •• 01)""Eq.14.
wherejJls thc coeffi-
Figur.: 16
31.
Problem
a
;0.
th I if w neglect extremely small qu nuue . the
hange in volume of'u solid u n expan ion thr ugh a It mperature rise Tislti\cnby'~r'-3
I ~T.v.hcre((1
the
c f1icient of linear expansion. See Eq, 15.
I'I\lVe
32. When the temperature ora copper penny (which I not pure
copper) is raised by 100 CO, its diarnet er mer .ases by 0.1
1111 ! the percent increase 10 «(I) the area of a face, (/1) III
thle!
55, (c) the volume, and (ff) the muss of the penny.
(c) C'll ulate us coefficient of linear expansion.
33. Dens.: J 1<, mass divided by volume. If the volume I' I~ ternpcrau.. , Iepcndcnt, so is the density p Show that the chanae
in dcnsuy Ap With change in temperature
'l is given b;
l'rubian«
up
-Pld.T.
H'/> is the coeificient of velum
mmus sign.
wh
507
to the lime given by the clock IS necessary It ihc end of JU
days'}
expansion, Explain the
Vhen the temperature of a metal cylinder is raised from 00
to 100·C. il5 length Increases by 0.092%. {a} Find the perceru change in d"n~iIY_ (b) Identify the metal,
35. A I 100-C ,J gia~ flas is c srnplerely tilled by 89! g of mer.
cury \\ hat mass of mercury is needed to fill the flask at
- 35'C7 (The coefficient of linear expansion of ~as$ b
9.0 X 10 "IC"; the rodficicnt of volume expansion of mercury is 1.8 x to-t/C·.)
,t-t
.'6. Figure 17 shows rhc variation of the coefficient of votu me
expansion of water be! ee1l4·C and 20'C. The density of
.•••ter at 4o'C IS 1000 kgjm'. .alculale the densnyofwater a;
41. A pendulum clock wuh a pendulum made \)f bru ~ I de.
signed to keep accurare IITnc :H 20 CHow much \~111the
error be, in secunds per h ur, if the dock opera!\'\ .Il 0'( V!
42. A n aluminum cup of I 10 ml L.tlxjcit~ is filled \\ ith gl}(l;'nn
at 2Z·C. How much glycerin. Ifany, wi1!plll
out of the cup
if the tempel" turc olrhe cup and glycerin JSfaj~tl to 28'C?
(The coefficient of volume expansion
H14/C·.)
43. A 1.28.m·!OrIg vertical glass tube is half-filled w ith a liquid
at 20.0·C, 110 w much will the height of the liquid column
change when the IUb~ b heated to 310·C? Assume lh,\\
(rll••••••Ll
44.
20
of g.!yn:rin IS 5.1
.
'10-$/("
:lflclPI",,,,J -4.2
X 10
»c-,
steel rod at 24·C I) boiled securely at holt. ends and then
cooled. 1\1 what tcmperuiurc will u begrn to )Idd? See Table
1, 'haptcr 14.
1\
45. Three equal-Iengrh straight r
of aluminum. invar, and
Sled. aU at 20'C, forrn an cquilaterul tnangle with hinge PIl1~
at the vertices, At ~\h,1t tcmperutur« will the angle ')PJ){) ire
the mvar rod be 5'.1:).' Sc\' Appendi« II {l'f ncnbt tng.lI1o,
rncmc formulas
40.
I WUrt: 17
PI oblcm 36.
heated.so that their temperature increases by.\ 1 (a) Sho»
that the rod Interface IS dtsplaccd upon hc;.,lIn· bi all
amount given by
J7. l\compositebarofkngthL-/1
+ L1ismntiefromnbarof
materml I and length 1., attached to a bar of malena12 and
length 1'2, ;4 shown in Fig. lit (a) Show that the effective
coetheient of linear expansion 0: for Ihis bar is given hy
,
1 I:J.T
u 1~=(OIEj: -+ C'<lE!)
F'
,
•.t
+ O:zL11/L. (fJ) Usin
steel and brass,design such
OJcomposae bar whose length is 52.4 em and whose effective
cocthcrent of linear expansiou is. 13 X 1O-6/C·.
(t·- (Ul!.t
l i~ur J 8
1 W{}rods o(dlilercH( mate rals but having tilt' arne It Ilf,lh)
L and cross-sectional arca-, ,1 .1ft' arranged end tu end 1>(
tween nxed, rigid SUPJ)vHS, as shown In rig. 19u fhe trmperature IS T and there IS no uuual )trt'~S the rods arc
,~!
where 0:, ,fr! arc Ihe coefficieurs of linear expansion and 1:',.
[~1are Young's moduli of Ihe m.uerials Ignore dlOJnl:c~ In
cross-sectronal urea ; see rIg 19b. (b) r ind the strevs al till'
mterface after he.Hint'
T
Problem J 7_
{a}
.'~. (a) Prove that the chan
e in rotational inertia Iwith temperof OJsolid object is given by tl. [,.. 2aI liT. (b) A thin
brass
puuun
aboui
\"< ( en icular to it ;.II Its center, is heated without mccharuc~t ,'oOI"CI until Its temperature increases by 170 C·. CalcuJalt' the change III an ular velocity.
L1IUfC
unif
39
rm
rod,
free!
uI230
rev/s
n
T
.:>1
avi
cvlinder placed 10 frictionless bearings is: set rotating
ab0~( its axis '} he cylinder 1\ then heated, without mechanical contact, until ns radius is increased by O.! 8%. What is
the percent change in the cylinder's (a) angular momentum,
velocity, and (c) rotational energy?
40. (II) Prove that the change in period Pofa physical pendulum
with temperature j~ given bv L\ P"'" jnP 61' (b) A clock
re:ndululll mad", t)j invar has a period of 0500 s and is
ccurate "I 2(}- __II the clock IS used in a climate where the
temperature averages 30'e, what approximate correction
(b)
Figure 19
Problem 4(,.
-47. An aluminum cube 20 ern on all edge floats Qf! mercury,
I low much farther will the block Sink when the temperature
rises from 270 to 320 K7 (The coetliciem of volume expnn-
(b) angular
sion of mercury IS I.~ X I O-"/C" ,)
48,
A glass tube nearly
filled wuh mercury is attached in tandem
rod 100 ern long How
high must the mercury be in Ihe glass tube so thru the center
ot mass of this pendulum will not nse or fall \\ IIh changes III
to the bottom of an 11011 p.:ndulullt
temperature?
('I he t'fll\\·set·1I0nJ}
area or the tube 1\ l'(11.1:11
O~
(1IIf1/"
[CTIlp
r tur
to that of the iron rrx' Neglect the ma S orlh glass. Iron ha
!OJ ylt'm' and
coeff icnt af hncnr
expansion equal to 12 1O-6/C·. I he co lficicnt of v {_
ume e pansion of mercurv i~ I X I0 ~/ ~",}
49. The distance between the towers r the rnam span of the
Golden Gale Brid C OC:1f 'an Fmnciscn is 4200 n (Fig. 20t
The sag of the cable half\\ay between the towers at 50' F is
470 ft. Take 0: 6.5 iO J-/P for the cabl and c mpute
(0) the change in length of the cable and (b) th chanKc 1Il ~1
f r temperature ch: ngc from 10 to 90T. As urne no
bendin or separation of (he tow r . no a pa.abolic sha
for III cable,
a def1~IIY of 7. 7
------------------------------------------------------Fif,;ur('20 Pro I
49.
CHAPTER 23
T
KINETIC THEORY AND
THE IDEAL GAS
.,
1
PI
TI.t' basic laws oftkermodynam! s deal with llle fetationslllJu between
m croscomc propt!((iCS, such as the pressure, temperatur», volum», and tnterna!
energv of an Ideal g.h The laws 'ay nmiling about the.farr mot nuuter is made IIJI uJ
particles (ouun or molecules). 0:l'i1l8 to .11,· large numbe« of panicle: involved It 1.\1/01
"'<ldIC,11 to tlPf!~ Ihi:' {<i"S qf lIIe hantcs (0 jiud ill.> nI( tton of t'l ct I'paruc}; ut : ;:,! .
ln vt, d 1\ use asrraging tedmiou, ta t: press th, {h,T/flll,!I,/1lI
( ":OI'-'II.I,;r pr ·pcfll<s.irrlll!
mm:!' 1 (l{p niclcs 11 \£'I\!."
/"t'PUI/t'\,
,;, ,'f"';"J
u I, ./I.rll~,' hid
n,
f!~
d 'Jm" quantu it '
In this chapter, Wi' consider an approacllw averaging C. llrd kineri theory, In It/lit it ",'
folk» (he monon of represenuuive particles if I tl }.It s and (hell ovcrae« this behavior OWf u!l
particles. Kinetic th.:ory !~J;l$ dcvt?ioped in the J Zth to J 9th (ell III' ies by Royle. J) lh·rnoulli.
JOllie. Kronig, Clousius, and Afa.'(I\eit. among others. Another appr(){lcit {(J averaging' IS
stati rical mechanics. in willen laws of probabilllY ale applied 10 suuistic I dtstrtbutiens of
mol -cular properties. This approach ts considered In Chapter 24.
• "a.'lfft
23·1
$
$
err
1.11
II
f\,1ACROSCOPIC
PROPERTIES 01" A GAS
AND 'I'n
IDEAL (~AS LAW
Figure 1 shows a gas confined to a cylinder fitted w ith a
movable piston. We wish to arry out a series of mea uremen!'. of tne macros: spic properties of the gas: the type
and amount 01 gas and it' nressure, volume, and absolute
IKelvin) temper. tur . We assume mat we have au <.:h~
tv the c Iin er suitable devices for me suring these prop
rtie . W also, um III l \' 11 ve at our dlSPO I th '
me us to chang' any 0 these propenie
For example, w _
s ppo
the gas to be in contact with an ide; lized devic
called a thermal reservou, which we can regard 4\S a bod,
maintained at a temperature T.such t11 t the t rnperature
01 the reserv if does not hange when our gas cylinder
comes into therrn 1equilibrium
viih it. \Ve assume that
I"e can easily ch nge the temperature of the reservoir,
fht>reby changing the temperature of the gas. Ifwe wish to
change the pressure p, We add or subtract weight on the
piston, (The "pace above the Piston is assumed to be evaeuated, so that there is no air pressure pushing down on the
pi' Ion) The volume V can be changed simply by chang.
Cas ~lIllllft
FiK Ire I
ras is couf ned to a cylinder th I is ill contact with
a thermal reservoir I th (adjustablel temperature 7~ The \.)1Ston exerts :l total downward force Mg on the gas. whi h in
equilibrium is balanced by the upward. force due 10 the ~
pressure. The volume of the gas can be determined from a
measurement of the height h of the piston above the bottom
of the cylinder. ant! the temperature of the gas IS measured
with a suitable thermometer. A gas supply permits additional
gas to be added to the cylinder; we assume th.l.! a mech"lU~m
IS also provided for tell10\,Log gas and fOf ch!ltlgmg the supply
to admrt different kmds of gas
5U9
5J 0
Chartt'r 23
A'/fICW
'TIIN·ry mu! the {den/ Gns
tog the posuion Of the \11"[\'1), and the amount of gas mirht
be d\ang\~d by allowing gas to enter the chamber, thereby
changing the number of molecules N. After each change.
we all ( w enough time- for the gas to reach thermal equilibnum and \0 acquire a new set of rnacrosconic therrnodynanuc van bles.
Let us now conduct the following experiments
On the
g,)<;,
C'
P :" I'
(IV. T constant).
Here C represents another constant, which would have ,\
different value if we had chosen different values of Nand
T. Equation
2 i called Bo)'I(".\ law and like Fq. I is a
As we discuss in Section
23·S, real gases deviate somewhat from tillS ideal be
somewhat ideal generalization.
havior.
1. j)e[lClIrlcIICI' of I' Ol! N. Keeping the temperature and
pressure constant (that is. the g.,1S is in contact with the
thermal reservoir at a particular temperature T, and the
weight on the pisron is constant), we allow gas to enter or
leave the chamber, and we measure the resulting volume
rl by observing the height oruic piston. (We a some that
we know the mass of each molecule and the total mass of
gas that is present in the cylinder. Thus we (';\0 determine
N, the total number of molecules.) Figure 2 shows typical
results of such experiments. The data points appear to'
follow a straigh; hue, and we conclude that, to a sl,)ffi~
ciently good approximation,
there is a direct proportion
between Vand N; that is. tile volume increases linearly
with the number of particles, Furthermore, by replacing
the gas in the cylinder with an equal number of molecules
of a different gas at the same pressure and temperature, we
find that the new 1;a5occupie the same volume. Thus we
would con .lude that the volume occupied by a gas at a
parucular pressure and temperature is indcpen lent of the
type of gas or of the she or mass of its molecules; the
volu me depends only on the IIlimber of molecules. Mathew
mattcally, f'u; N. Of
r
CN
(p,Tconstant).
(I)
Here C is a constant, equal to the slope of the line in Fig. 2
and determined b . the v alucs of fl Hod T. If we repeated
this experiment with different constant values of p and 'T:
wc v-ould still fwd Eq. I to hold, hut with a different value
3, Depcndenco of V (JII T Keeping II and N constant.
vary the temperature T (by changing the temperature
vc
of
the thermal reservoir), and we measure the resulting volume V. We find (Fig. 4) a direct relationship: the volume
increases as the temperature increases; thus "'Y 1'. Of
V= CUT
(p.
constant).
{31
where C" is yet another constant. Lqu: lion 3 I'; called
Charles' law or Gay-Lussac'. jaw. Like Eq5. I and 2, u rs
an idealization of the bchavi r of real gases
• Gas 1
o Cas 2
" G<l53
v
Figure 2 The volume !'occupied by (he gas In fig J depends on the number of molecules N. At 0 given temperature
and pressure, different gases follow the same lmear relationship.
of' the constant C.
Equation 1 is s. mctimcs known as Avogadro's law. It
holds to a very good approximati
n for all gases, espedally at low density, where the molecules are vcry fur
apart and the volume occupied by the molecules themselves is indeed a negligibl small fraction of the volume
of the container to which the gas is confined. We can
generalize from the beha vior of these rea! gases to that of
all idea!
th,lt docs follow Fq. 1 exactly. in the next
section we 'on idel the microscopic
properties of an
'f
v
L....-l.._L-.l......._L_-1_t_ ',__
(el)
p
\
1-
idea gas,
2. Depond: !I( (II r on p, Keeping the number of partides S and the temperature
r constant, we change the
pressure (h} changing the weight on the prston) and
mea urc the resulting \OIOIrH' The result i~shown in Fig.
3d. w hich ~ug.(.\~!itS an mVt'!'Sc relationship: as the pressure
p increases, the volume !' decreases. To check 101<:,we
instead plot p \ ersus V I, as in Fig, 311, which confirm a
linear rclauonshrp, We therefore conclude p C( v-! ~or
V )-
.,?/
(lo}
_ •.L
L__ l_1---1__1_
,.
Figure 3
(u) The volume Voccupied by the gas tlpprars In
depend inversely on the pressure 1'. With the temperature and
the number otparticlcs held constant. (h> PlotllTlg J \ agains:
p shows that the relationship i!. indeed an inverse linear one,
Section 23·2
Figure 4 The volume V occupied by the gas varies linearly
with the temperature T, when the pressure and the number of
molecules are held constant.
Eq uation of State
Equations I. 2. and 3 summarize experimental results
strictly valid only for our hypothetical ideal gas but to a
high degree approximately valid for most real gases. We
can combine the three equations into a single equation
that includes all three of the observed relationships, as
follows:
pV
Sample Problem 1 An insulated cylinder fnted with u piston
(Fig. 1) contains oxygen at a temperature of20'Cand a pressure
of 15 aim in a volume of 22 liters. The piston is lowered, decreasing the volume of the gas to 16 luers, and imultaneou I)
the temperature is rai ed to 25' . Assuming oxygen to behave
like an ideal gas under these condition . what i the hnal pressure
of the gas?
Solution
From Eq, 7, since the quanuiy of ga rernams unchanged, w.: have
(4)
NT-k,
or
here k is a constant. By rewriting Eq. 4 we can show that
it is consistent with Eqs, I - 3:
(~)N=CN
(kNT) C'
p""'_
••••
-
V
V
V
T constant),
(N. T constant),
(Sa)
(P.
PT- P,
_
Pr'" ( I:> atrn)
(273 + 25 K) (22 L)
.
273 + 20 K
16 L •••21 atm.
(5c)
The constant k in Eq. 4 is called the Boltzmann C01lconstant with a value determined by
c perirnent to be
k"" 1.38066 X 10-2) 11K.
It is more common to write Eq. 4 in a slightly different
We e press the quantity of gas not in terms of the
number of molecules N but in terms of the number oj
moles n The mole was defined in Section 1-5. In terms of
the Avogadro constant NAJ the number of moles is
1"0011.
(6)
and we can rewrite Eq. 4 as
pV =kN
n'T
".
pV=
'
r
into Sf units, but we mUSII!X{lft':>S T in absolute (Kl'hill) temperature units Thus
(5&)
(k;) T= CUT (p,Nconstant).
(Tf)T, (v.)V
Because this is in the form era ratio. e need not con vert p and V
Wllt.1t is a universal
or
511
Equation 7 is called the ideal gas law ot Ideal gas equation
oj tate. An equation of state oT a system gives a fundamental mathematical relationship among macroscopic
thermodynamic quantities. Experiments reveal that. at
low enough densities, all real gases approach the Ideal gas
abstraction described in Eq. 7.1 hi IS the same limit that we discussed in connection with the ideal a temperature
scale in Section 22-4. The constant R has the same value
for all gases and is called .111 univer (II gas constant.
v
V
The ldeal Gas: A stod«!
(7)
!lRT,
where
R=N"k
= 8.3145 J/mol·K.
(8)
'IODEL
When phy ideas want to under tand a complcx s, tern,
they often inv ent a model A model i a simplified ver ion
of the system that permits calculation to be made but sull
yields physical insight. A model might begin with a t or
simplifying assumptions that permit the system to be analyzed using an existing set of laws. for c ample. Iewtonian mechani s. The analysis might then lead to an equation or set of equations describing the original phy i al
system. Because the model IS asrrnphficauon of nature,
the final result is generally not a true or complete description of nature. but if we have been clever at forming the
model. the final result may prove to be •.I very good approximation of the behavior of the system. Whatis more
important, the final result may give us a way of studyin ' ~
the system in the laboratory and gaining still mort: msrght.
Previously in this text, we have u e 1 a model (wuhout
calling it one) to describe the motion of a complicated
object as a point particle under certain circumstances. \\ e
. •..
c
512
(I; in r 23
Ktnetic Theory and the Ideal Gas
have also sometimes modeled the force between atoms in
a molecule. or between atoms in tI solid, in terms of the
pnng force. F - kr which is itself based on a kind of
model thnt simplifies (under certain clastic conditions)
the complicated internal processes in a solid under stress.
A gas confined ill a container is an example of a complex system that is difficult to analyze using Newton's
ln vs. The molecules can C llide inelastically, and the energy (If the collision can be absorbed by the molecules as
internal cner y m n vaucty ofway . Kecpin track of these
PIO\.C~ s foi III the J1)o!rCIJ!"s would be a pr [ect of hope-
less cornplc
11_.
model Ih.1l I
W <;Ilnplif' this problem by inventing a
'I It cs
th
microscopic
properties
of the
real ~.lS. r hi III del, vhich we call the ideal ga model
prove to be entirely \ onsistcnt with the concept of the
ideal gn' that w developed experimentally in Section
2 J-l. In tha section we saw that, especially at low density,
the macroscopic pro erucs tif real gases approximately
follow n g ncral I csult, the ideal gas law of Eq, 7.
Prom the rnrcroscorxc
point of view our model of an
ideal as includcs th followin a surnptions. Based on
these assumptton
. w use I cwton's law to analyze the
m chani s of the Ideal :as; this proc dure forms the basis
of kind ;('[11cory. Lat r we relate t hi micro copic descriplion to a macroscopic one.
I. A I:fl.1 consists 0/ panlcles, called molecules. Dependingon the gas, each molecule may consist 0 one atom ora
group of atoms, If tit ,;-t. is an clement or a compound
and is in a table tate, we con. id r all its molecules to be
Identical.
2. The molecules are ill random motion and obey Newton's laws ofmotion. The molecules move in all directions
and with a range of velo ities. In describing the motion,
we assume that cwtonian mechanics is valid at the rnicroscopic level.
3. The total number of molecules is larg '. The velocity
(rnagnitud
and direction) of anyone
molecule may
change abru ptly on collision with the wall or another molecule. An, particular mole ulc will follow a zigzag path
bccau c of these colli ions. How ver. because there are S()
ma Ii' molccul
W'
a. ume that the resulting large nurn-
1 r r collisions maintains the overall distribution of molccular vclo iti s and the randomness of the motion.
4. 7 he volume of (he 1I1011'CII/(>.I' is a negligibly small fracfirm (Illite volumo on IIp/('cl hy the gas. Even though there
are many molecules, they arc extremely mall. We know
that the volume occupied by a gas can he changed through
a large range.of'valucs with lillie difficulty, and that when
a gas cond .nscs the volume occupied by the liquid may be
thou .nds 01 times "mailer than that of the gas. H .ncc our
assurnpuon ls piau iblc. Later we hall invcstigat the actual sile of rnolecul c; and see whether we need to modify
this assumption,
.•5. So <Jl'lmnab/r torccs act 011 (he melee ules except dur-
ing a calli: ion. That is, we assu me that the range of molecular forces is comparable to the molecular size and much
smaller than the typical distance between molecules. To
the extent that this is true a molecule moves with constant
velocity between collisions. Therefore the motion of a
particular molecule is a zigzag path consisting mostly of
segments with constant velocity changed by impulsive
forces.
6. Collisions are elastic and of negligible duration. Collisions of on molecule \ ith another or with the walls of th
container conserve momentum and (we assume) kin lie
energy. Molecules arc not true point panicle and do hav
internal structure: thus some kinetic energy may be coovetted into internal energy during the collision We assume that the molecule docs not retain this internal energy, which is then available again as kinetic energy after
such a brief time (th time between colli ion!'» that we can
ignore this exchange entirely.
23-3
Let us now calculate the pressure of an ideal gas from
kinetic theory. For simplicity, we consider a g...'1S in a cubical container of edge length L whose walls are perfectly
elastic. Call the fac normal to the x axis (Fig, 5) 41 and
A1, each of area V. Consider a molecule of IIIa III with
velocity v, which we resolve into components OX' r,.• and
l' •• When this particle collides with A I' it rebounds with 115
x component of velocity reversed. that is. 1\ - ( "
Th re is no effect on Vy or VI' 0 that the change in
the particle's momentum has only an X component.
given by
final momentum
- initial momentum - t:1r;~-
BecauSe the total momentum
sion, the momentum imparted
j
(mvx)
= - 2mv(.
(9)
conserved in the collito Al is + 2n1V~.
y
Figure 5 A cubical box or edge L containing an ideal a. A
molecule of the gas is shown moving with v locity l' toward
side A r-
Section 23·3
Suppose that this panicle reaches Al without striking
any other panicle on the way. The time required to cross
the cube is Liv.: (If the molecule strikes one or the other
faces of the box on the way to Al, the x component of its
velocity does not change, nor does the transit time.) AtA1
it again has its x component of velocity reversed and
turns to AI' Assuming no collisions with other molecules,
the round trip takes a time 2L/vx which is the time between collisions with AI' The average impulsive force
exerted by this molecule on AI is the transferred momentum divided by the time interval between transfers,
or
F ••.•2mv .• _ m~
(10)
x
2L/v..
L'
re-
To obtain the total force on AI. that is. the rate at which
momentum is imparted to AI by all the gas molecules, we
must sum the quantity mv;/L for all the particles, Then,
u find the pressure, we divide this force by the area of A I'
n rmely, U. Tl1( pressure is therefore
Kinetic Calculation ojillt! PI{,H!lfC
negligible compared to the time spent between collisions.
Hence our neglect of collision) is merely a convenient
device for calculation. Similarly, we could have chosen ,\
container of any shape: the cube merely simplities the
calculation. Although we have calculated the prcssur c· ~,,erred only on the side A \I it follows from Pascal's law thai
the pressure is the same on all sides and everywhere in the
interior. (This is true only if the density of the gas is \l ai.
form. In a large sample of gas, gravitational effects might
be significant, and we should take into account the varying density. See Section 17·3 and Problem 6 of uus chapter.)
The square root of Vi is called the root-tnean-souorc
speed of the molecules and is a kind of average molecular
speed. (We consid r this avera~e in mote detail i \Section
24~3.) Using Eq. 14, we can alculatc the root-meansquare speed from measured values of the pressure and
density of the gas, Thus
R
'Ill'
L
== -L (1)'2xl + lP",1+ ... ),
111
In Eq 14 we relate
( 11)
v.here /) I is the x component of the velocity of particle 1.
1112 is that of particle 2, and so on. If N isthe total number
01' panicles in the container, then Nm is the total mass and
Nrn/U is the density p. Thus mlU ••••pIN, and
p=p
(V;l + v~ + ... ) .
(12)
The quantity in parentheses in Eq. 12 is the average value
of for all the particles in the container, which we represent by i? Then
(13)
p-p~.
513
13p
I
-
'p
rnacroscopi
•
quanuty {the pressure
p) to an average value ora microscopic quamitv (llhlt is,
to i} or v~,). However, averages can be taken over suo •.t
time or over long times, over small regions 01 $P ce or
large regions of space. The average computed in a sm II
region for a short time might depend on the time or n..'Slon
chosen, so that the valu 's obtained in this way may fluctuate. This could happen in a gas of very low densil)" for
example. We can ignore Iluctuntions, however, when the
number of particles in the system is large enough.
v;
v
v; v;,
v;, u;. V;
For any panicle, 2 •••• ~ + +
Because we have
man, particles and because they are moving entirely at
random. the average values of
and are equal, and
the value of each is exactly one-third the average value of
v2• There is no preference among the molecules for motion along anyone of the three axes. Hence ~= 1;;' so
that Eq. 13 becomes
P = lpiJi.
'2 Calculate the root-mean-square
speed of
hydrogen molecules lit O.OO·C and 1.00 atm pressure, assumin
hydrogen to be an ideal gas. Under these conditions hydrogen
has a density p or8.99 X lO-l kg/ml.
Sample Problem
Solution
Since p - 1.00 aIm'"
1.0 I
IO~ Pa,
I 3(1.01 Ie-Pal'
Ii' """'I 8.99 X 10 kg.lm
f3P
I)rm • .., "
1
1 •..•
!840 mho
This is of the order of a mile per ~cond, or 1600 £111/h
(14)
Although we derived this result by neglecting collisions
between particles, the result is true even when we consider
collisions. Because of the exchange of velocities in an
elastic collision between identical particles, there will
always be a molecule that collides with Al with momentum lnlJ ~ corresponding to the molecule that left A 1 with .
this same momentum. Equation 14 holds even Utile box
contains a mixture of molecules of different masses, because momentum is conserved in collisions, and the wall
must receive the same impulse regardless of which molecules strike it. Also, the time spent during collisions is
Table I gives the results ofuimilar calrulations for or Ie
gases at room temperature. These molecular speeds .irc
roughly of the same order as the speed of sound at the
same temperature. For example, in nit at we, '1)"••• - 485
m/s and the speed of sound is 331 m/s; in hydrl.'l n
tlftI"l"" 1838 m/s and sound travels at 1286 m/s. These
results are to be expected in terms of our model of a has;
see Problem 38. The energy of the sound wa ve is car ried :\'.
kinetic energy from one molecule 10 the next one with
which it collides. We might therefore expect sound W:l\ t,
to propagate with a speed that is roughty the same ,IS tIll'
514
Chapser 23
Kinetic Theory and the ideal Gas
~r\mI: 1 SOME: MOLECULAR
_~
SPEEDS
AT
__ R._O_O_fv_1_T~1Pr:RATURE {30e K)
Translational
Kinetic Energy
.'(orar mass A/Vrm .•
per Mole
(g/mol)
. Hydrogen
Helium
Water
(m/s)
(lImo!)
1920
2.0
VUI')()f
Nitrogen
. Oxygen
Carbon dioxide
Sulfur dioxide
4.0
, 1370
18.0
645
3720
3750
3740
28.0
32.0
517
3740
483
Our initial assumption .• hal the speed of sound In a gas 1\ the
same as the root-mean-square
speed or the molecules. I, onl\
crudely correct. In reality, (he speed of sound i~proport iona I 10
Ii"".,> Does this change the conclusions
of this sam plc problem
regarding the dependence of'the speed of sound on the temperature? See Sample Problem 6 for a derivation of the speed of
sound in a gas .
.$
3730
44 ••
0
412
3730
64.t
342
3750
'1hc molar mass, sometimes also known as the molecular weight, is
given here for convenience in g/mnl: its S1unit is kg/mol.
----
characteristic speed of molecular motion, which is in fact
what we observe. The molecules themselves, In spite of
their high speeds, do not move very far during a period-of
the sound vibration; they are confined to a rather small
space by the effects of a large number of collisions. This
explains why there is a time lag between opening an ammonia bottle at one end of a room and smelling it at the other end. Although molecular speeds arc high. the large
number of collisions restrains the advance of the ammonia molecules. They diffuse through the air at speeds that
are very much less than molecular speeds.
23-4 KINETIC INTERPRETATION
OF THE TEl\tlPERA T RE
or
If we multiply each side
Eq. 14 by the volt! me I'. we
obtain
pV'= 1PVt)l,
where pVis the total mass ofgas,p being the density. We
can also write the mass
gas as IIA1, in which n is the
number of moles and M is the molar mass. Making this
or
substitution
yields
pV- -jJlM(;l.
The total translational
(17)
kinetic energy of the gas is
!m(vf+vi+
...
v;")=lm(NVZ),
where Nis the total number of molecules. The total mass
of the gas can be written as mN ••• ntH. The right side of
Eq, 17 is therefore two-thirds
the total translational
kinetic energy. We can write Eq. 17 as
or
Sample J'roblcm3
Assuming that the speed cf'sccnd inages is
the same as the root-rnean-square speed of the molecules, snow
how the speed of sound for an ideal gas would depend on the
temperature.
Solution
"'" 3BnMi)2).
Combining this with the equation of state of an Ideal g.a
(p V"" nRT), we obtain
The density of a gas is
iMVl = iRT.
nM
That is, the average ttunslational
mole of an ideal gas is proportional
V'
P-
in which M is the molar mass (the mass of I mole) and n is the
number of moles. Combining this with the ideal gas law pVnRTyields
p
RT
-...p
M'
We obtain from Eq IS
t'r ••
,-It-~3~T,
(16)
so that the speed of sound 1'( at a temperature 71 is related lO the
speed of sound 1': In the same gas at a temperature Tl by
E.t ••• IT;
tll.
1'01 example.
If the speed
'(7;'
of sound at 273 K is 331 mls in air,
us speed in arr at 30(1 K is
••
p
.
(.331 m/s)
Note tha: the absolute
y[3OOl<.
fir K - 347 m/s.
(Kelvin) temperature
is used here. Why?'
I
(I )
kinetic energy per
temperature
10 the
This result connects the kinetic theory with the equation
of state ofan ideal gas. Equivalently, we may consider Eq.
l8 as a connection between a macroscopic property, temperature, and a microscopic property, the kinetic energy
of a molecule. Either way, we gain some insight into the
meaning of temperature for gases.
The temperature of a gas is related to the average translational kinetic energy measured with respect to the
center of mass of the gas. Th kinetic energy a~"()<:HII('d
with the motion of the center of mass of the gas has no
bearing on the gas temperature. In ection 23-2 we assumed random motion as part of OUf statistical d 'bnitioll
of an ideal gas and in Section 23- we calculated v10n(hi"
basis. For a distribution of molecular vclociucs having
random directions, the center of mass would be ill rest
Thus, to calculate V2,we must use a reference frame 1[\
which the center of mass of the gas is at rest. In 'lli other
frames the molecules each have velocities greater by u (the
velocity of the center of mass in tha: frame) than 111 the
Seclljm ] J·5
center-of-mass frame; hence the motions will no longer be
random, and we obtain different values fer Vi.The temperawn: of'a gas in a ontainer does not increase when we put
the container in a moving cart
Let us now divide each side ofEq. 18 by the Avogadro
constant N ", which is the number of molecules per mole
of a gU3. Thus .\f/S = m, the mass of a single molecule,
anu we ha -e
.'
HM/N;.)u1= !miJi-HR/NA)T.
(19)
10v.
m~ is the average translational kinetic energy per
molecule. The ratio RjNA is, from Eq. 8, the Boltzmann
con rant k whi h plays the Tole of the gas constant per
molecule. We then h ve
!mi)l- fkT.
W(lrk
Dtnu) Oil (111 Ith'{l/ (;01
515
The molar mass M er=ur, b 0.349 hymo] and th..tl
ofl3'UF6 is 0.3$2 kg/mol. Thus after passage through a POfIJUS
barrier ihe gas will be en riched in m U by Ihe sepal a Iion ract or a,
given by Eq, 22:
Solution
ct _
n:r; _
y M7
Each successive
0.352 kg/mol _ \ GO.n
0.349 kg/mol
.
.
passage through
iI porous
WJll mcreases
the
relative fraction of Wu by a factor of o. After 11such paS!j;IAe~,
the relativeconcentralion oP}~U will increase by 0-". To mrrease
the concentration of mU from 0.7%, characteristic of natural
uranium. to 3%, an enrichment commonly used III 1)(1\ er reactors, the number
determined from
II of
a
(20)
rous barners
that must be pa
cd l~
"(0.007)
. (o.oJ)
0.993. ••. 0.97 .
Lquation 20 is the molecular analogue of Eq. 18, which
Solving, we obtain n w 350. In practice, this 1$ accomplished
dealt with molar quantities. Here we see that the average
translational kinetic energy of 3. molecule is determined
through successive stages. in which II portion 01 the g:J\ th,l{
passes most easily through a b mer (and thus h ~ltgh\l) ellridj~'J
in lJ)U) advances \0 the ne: t stage. vlule the TI!IH. indcr Crhl\\
slightly depleted OPHU) is returned to Iced the previous \()\'CI
stage. To obtai n nearly pure
such U~!~ requued IN nuc 1e.1!
.weapons, might require seve ul thousand }h:I)S
by the temperature.
.
In the last column of Table 1 we list calculated values of
!-\h~n,.. As Eq. 18 predicts for an ideal gas, this quantity
(the tt anslational kinetic energy per mole) has nearly the
same value for real gases at a given temperature (300 K in
this caser From Eq. 20 we conclude that at a particular
temperature Tthe ratio of the root-mean-square speeds of
molecules of two different gases is equal to the square root
of the inverse ratio of their masses. That is, from
23~5 WORK DONE ON Al
IDEAL GAS
1'=2. m,V1 ••..•2. m i1
1
3k
2
3k
2
we obtain
(21)
We an apply EQ. 21 to the diffusion of two different
gases in a container with porous walls placed in an evacuated space. The lighter gas, whose molecules move more
rapidly on the average, will escape faster than the heavier
one The ratio of the number of molecules of the two gases
that pass through the porous walls in a short time interval,
which is called the separationfactor a, is equal to the ratio
of their rms speeds, and thus according to Eq. 21 to the
square root of the inverse ratio of their molecular masses
or. equivalently. their molarmasses:
ex'" Jm1/m1
-
-IM2/M1•
=u.
(22)
The diffusion process through porous walls is one method
used to separate the atoms of an element by mass into its
different isotopes.
,
Sample Problem 4 Natural uranium consists primarily of two
isotopes, mU (07% abundance) and =u (99.3% abundance).
Only m
is easIly fissionable. In a sample of the gas UF 6 (uraruurn hexafluoride), it is desired to increase the abundance of
• If'we raise the temperature ofthe gas in the cylinder off l'
1, the gas expands and raise the weight against gravity;
the gas does (positive) work on the ••••
eight. The upward
force exerted by the gas due to its pressure p is given by p.),
where A is the area of the piston. By Newton's third law.
the force exerted by the piston 011 the gas is equal and
opposite to the force exerted by the gas Oft the piston.
Using EQ. 7 of Chapter 7, we can therefore write the work.
W done on the gas at.
w- J F'dx= J (-pA)dA.
(23)
Here dx represents the displacement of the piston. and the
minus sign enters because the force exerted by the piston
on the gas is in a direction opposite to the displacement or
the piston. If we reduce the temperature
of the gas, it
contracts instead or expanding; the work done on the gas
in that case is positive. We assume that the process described by Eq, 23 is carried out slowly. so {hot the gas can
be considered to be in equilibrium at all intermediate
stages. Otherwise, the pressure would not be dearly defined during the process, and the mtegral in Eq. 23 could
not easily be evaluated.
We can write Eq. 23 in a more general form that turns
,"u from 0.7% to 3% by forcing the gas n ti rnesthrough a porous
out to be very useful. If the piston moves through <I dis-
barrier. Find fl.
tance
dx, then .the volume of the gas changes by an
5 6
Chaptcr Li
K!I1tilC
Theon and the Ideal Gas
p
P
••1 __
P,
V,
V
Figur 6 The magnitude of the work IV done on a gas by a
proce of rbitranlv varying pressure is equal to the area
under the pressure curve on a p V diagram between the initial
volume'P, and the final volume VI'
• __
-.
2
A,
C
E'
IF
v
-----------.--Figure 7 A gas is taken from (lie pressure and volume a(
point A to the pressure and volume at point D along two dlf·
ferent paths. ABD and A CD. Along path 1 (ABD) the v.ork I~
equal to the area of the rectangle BDFE. while along path 2
(ACD) the work is equal to the area of the rectangle ACFE
amount aV - A ax. Thus the work done on the gas can be
written
W--
J
pdV ..
(24)
The integral is carried out between the initial volume VI
and the final volume Vr•
•
Equation 24 is the most general result for the work done
o a g s. It make: no reference to the outside agent that
d s the work; It states simply that the work done on the
en -an be calculated from the pre sure and volume of the
".1 i elf. 'otethat
he algebraic sign of the work. is irnplicitly c nrained in EQ. 24: if the gas expands, dVis positive
and HI i<; negative, P being a scalar quantity having only
p sitive values. Conversely, if the gas contracts, dV is
negative and the work done on the gas is positive.
Equation 24 is analogous to the general result for the
work done on a system l:;y a varir ble force F. You will
recall from Fig. 7 of Chapter 7 that if we plot F'against x,
the vork done by F is just the area under the curve between XI and Xr. Figure 6 shows the similar situation for
the work done on the gas. A graph in the form of Fig. 6 is
called a pV diagram, with P plotted on the vertical axis
(like F) and V plotted on the horizontal axis (like x). The
magnitude oithe work done on the gas is equal to the area
WIder the pressure curve on a P V diagram. The sign of Wis
d terrnined according to whether Vr> Vi (in which case
IV is ne ative, as in Fig. 6), or Vr < VI (in which case Wis
positive), Once again, the work done on the gas is negative
I th
pr
ss increa s the volume of the gas and positive
the process reduces the volume of the gas.
The pre sure force i clearly nonconservative, as FIg. 7
d monstrates. Let us suppose we wish to take our ideal gas
om the initial conditions VI and PI (point A') to the final
c nditions Vrand pr<poinfD). There are many different
paths •.••
e can take between A and D, of which two are
own in Fig. 7. Along path 1 (ABD), we first increase the
ure from p, to Pr at constant volume. (We might
a
plish this by turning up the control knob on the
th • 1 I reservoir, increasing the temperature of the gas,
'1 we sirnultar -ously add just the right amount of
iuonal weight .0 the piston to keep it from moving.)
'~eri tollcw ~a.th BD by increasing the temperature
but adding no additional weight to the piston, so that the
pressure remains constant at the value Pc while the volume increases from Vi to Vr. The work done in this entire
procedure is the area of the rectangle BDrE (the area
below the line BD).
We can find W" the work done on the gas along path 1,
by considering the \ ork done along th two segments ,.IE
and BD:
~ WI = WAB+ WDD·
Because the volume is constant along AB, it follows from
Eq, 24 that WAD ••••O. Along BD, the pressure is constant
(at the valuep-) and comes out of the integral. The result is
J.+\ -
WAS
""0-
+ WaD
J pdV--pt
J:"dV=-pr(Vr-
V,),
To follow path 2 (ACD), we first increase the temperature while holding the pressure constant at PI (that is,
adding no additional weight to tbe piston), so that the
volume increases from Vi to Vr' We then increase the
pressure from PI to Pr at the constant volume Vr by increasing tbe temperature and adding weight to the piston
to keep it from moving. The work done in this case is the
area under the line AC or the rectangle ACFE. We can
compute this as
W1-
WAC
+ WeD
- - J P dV+ 0 -
-PI
J:" dV-
-p,<J'r-
V,).
Clearly WI .;. W1, and the work depends on the path.
We can perform a variety of operations on the gas and
evaluate the work done in each case.
Work Done at Constant Volume
The work is zero for any process in which the volume
remains constant (as in segments AB and CD in Fig. 7):
W"" 0
(constant
V).
(25)
•
We deduce directly from Eq. 24 that W- 0 if Vis CQnstant, Note that It is not sufficient that the process start
and end with the same volume; the volume must be constant throughout the process for the work to vanish. For
example, consider process ACDB in Fig. 7. The volume
starts and ends at V" but the work is certainly not zero.
The work is zero only for vertical paths such as AB. represennng a process at constant volume.
Work Done at Constant Pressure
I lere we can easily apply Eq. '24, because the constant p
comes out of the integral:
W=-p
f
<IV
= - p (Vf -
(constant
Vi)
p).
(26)
Examples are the segments •.tCand SD in Fig. 7. Note that
the work done on the gas- is negative for both of these
segments, because the volume increases in both processes.
Work Done at Constant Temperature
If the gas e pands or contracts at constant temperature,
the relationship between p and V, given by the ideal gas
law. is
.
pV = constant.
On a pI' diagram, the plot of the equation pV"" constant
is exactly like a plot of the equation xy'" constant on an
Xl' coord I nate system:
it is a hyperbola, as shown in Fig. 8.
A process done at constant temperature is called an
isothruua! process, and the orresponding hyperbolic
curve on the p V diagram is called an isotherm. To find the
work done on a gas during ,in i othermal process, we use
Fq. 24, but we must find a way of carrying out the integral
v. hen I' varies. I'o do this we use the ideal gas equation of
state to write p = uRT/ V, and thus
W=-
I
Vt
v,
II" nwr
pdV=-
"tTdV
v,
-IIUl'
J-" -.€IV
i ,
I
where the last step can be made because we are la\..in' 7 to
bea constant. Carrying out the integral. v,e fwd
H' "'" -
nRT In Vr
(constant
V,
1').
Note that this is also negative whenever Vr> I t (In .\ is
positive for x > I) and positive whenever VI < l'"
Work Done in Thermal Isolation
Let us remove 'the gas cyll nder in ria. 1 from contact with
the thermal reservoir and rest it on a slab ot insulating
material. The gas will then be H\ complete thermal isolalion from its surroundings: if we do work on u, its temperature will change, in contrast to itsbehavior when il wa in
contact with tbt~ thermal reservoir, A process arried out
in thermal isolation is called .10 adiabutic process.
lt'we allow the gas to expand with no other constraint •
the path it \\ ill follow is represented by the hyperbola-like
- curve
(28)
P !l1 "'" constant,
••s hewn in Fig. 9. The parameter y, called the- rau» (if
specifu: heats. must be determined empirically Ior any
particular gas. Its values arc tYPI(ally III the range 1.1
1.8. (In Section ::~·4 We discuss the specific heals or g,I'>C),
and we derive Eq. 2(}in Secuon 25·6.) Because "I is greater
than I, the curve p~/Y - constant is a bit steeper than tl e
curve [1 !I ••• constant, and hence the work done In this
process will be somewhat smaller in magnitude than the
work done in expanding from Vi to VI at constant I. as
can be seen from Fig, 9.
The con-stunt in Eq. 28 is determined trom the pressure
I"
I~
1
"
pV
COIll,IMt!
I
'"
1I
" ....
--. "v1' •• (Qn~t..(11
p
1
I
W
,.->'''<",
~
P,---I-------~
J
w
v
Figure 8 A process do lie-at constant temperature (isothermal
process) IS repr esemed by a hyperbola on a pV diagram. The
work done in changlng the volume is equal to the area under
the curve between Vj and VI'
-",,-I
~ __~ __--------L--V,
------~----~--.---~
VI
Figure 9 An adiabatic process IS represented on u {II <.Iia
gram by the hyperbola-like curve pVt •••constant '1 he "()r~
done in changIng the volume IS equal to the area under the
curve between V, and VI' Because t> 1. the adiabatrc curve
has a Sleeper negative $Iupe than the isothermal curve pi' •••
constant.
518
Chatuer 23
Kinetic Theory and the Ideal Gas
and volume at any particular point on the curve. Let us
choose the initial point PI. V, in Fig. 9, and so
The work done at constant volume is zero (see Eq. 25), so the
total work for path I is
WI ,.
P r_ J)IV/
or
r
P-~
(29)
p'
Path 2 represents an isothermal process. alon whi h T
constant. Thus Pi V, - PrV, = nRT. the work done during the
isothermal process can be found using Eq. 27. substituting p.}',
for nRT, which gives
.
We can now find.the adiabatic work:
"1'"
w- -
p dV
-
Il"PJni+ dV- -PiVt' IVf V1
v,
P,
r
- _l?L.L (V1-Y
)1_
I
r
I
batic work as
:~'l[(~~y~'
-I]
1
- (Pr (~p
}'- I
V.)
(adiabatic).
V
-III VI In ..:..!V I
.
10m)
-(10 PaX4.O mJ) In -4'0
.
.
m
1-
5S J.
Path 3 consists of a process at constant volume. for which the
work is again zero. followed by a process at constant pressure.
and so the total work for path 3 is
VI-,).
-
First by bringing a factor of V{-l inside the parentheses,
and second by using PI V{ - PrVl, we can write the adia-
w-
W1 -
dv
".
30 J + 0 •••30 J.
(30)
1
If the gas expands. then VdVr< 1. and since a number
less than I raised to any positiv power remains less than
I.-the work again is negative.
W,- 0 -
p,{Vr- V,) - -(40 Pa)(1.0 m' - 4.0 m')'" 120 J.
Note that the work is positive for all three processes, and that the
magnitudes increase according to the area under each path on
the p V diagram.
Sample Problem 6 (a) Find the bulk modulus B for an adl batic process involving an ideal gas. (b) Use the adiabauc bulk
modulus to Clod the speed of sound in the gas a a funcuon of
temperature. Evaluate for air lit room temperature (20'0
Solution (a) In the differenual limit, the bulk modulu (see leq
5 of Chapter 17) can be written
1J ••• _ I' tip
ampl
Problem 5
dV'
A sample or gas consisting of 0.11 mol is
compressed from a volume of 4:0 mJ to 1.0 in) while its pressure
increases from 10 to 40 Pa. Compare the work done along the
three different paths hown in Fig. 10.
elution
Path I consists of two processes, one at constant pressure followed by another at constant volume. The work done at
constant pressure is found from Eq. 26.
W - - p( ~ r- VI)" -( 10 PaX 1.0 m' - 4.0 Inl) - 30 J.
For an adiabatic proce
Eq. 28 (pV' - constant) give.
the derivative with respect to j'.
1)_(d,,)
d(PV
dV
II'
takin
p(I'I,,-I) =0.
dll
or
dp
v---'1P
dV
.
Thus
B - yp
for an adiabatic process involving an ideal gas.
(b) In Section 20·1. we determined that the speed of'sound In a
gas can be written
(V,. PI)
40
v3
30
~•..
20
10
where B is the bulk modulus andpis the densuy of the gas. log
the result of pan (o)and the ideal gas equati n of state( Eq. 7). we
obtain
v'----:-----~
(V,, p~
234
fifiP.
'1P
-.... ~y(nRTIV)
~ p
p
c::.
The Quantity p V is the total mass of the gas. which can also be
written 11M. where /I is the number of mole and M I~the molar
mass. Making this subsutution, we have
V(m~
--- --- - ---ligure 10 Sample Problem S. A gas taken from initial
IS
point i to !lnnl point f along three different paths. Path 2 is an
isotherm.
Thus the speed of sound in II gas depends on the SQuare roo! of
the temperature. as we inferred in Sample Problem 3.
Secuon 23·6
For air, the average molar mass is about 0.0290 kg/mol. and
the parameter '1is about 1.4. Thus for T- 20·C - 293 K.
I
(1.4X8.31 .t/mol· KX293 K) ••• 343
0,0290 kg/mol
m s,
The Internal
the internal energy the contributions of rotational krn uc
energy as well as translational kinetic energy.
The rotational kinetic energy of a diatomic molecule,
illustrated in Fig, 11)can be written
KlOI
23-6 THE INTERNAL ENERGY
OF AN IDEAL GAS
j
f
OUf model of the ideal gas is based on molecules that are
considered to be point particles. The temperature, as we
have seen. depends on the translational kinetic energy of
the molecules. For point particles, there lS no other form
for the internal energy Eint to take. There is no molecular
potential energy. nor is-there any internal energy associated with the rotation or the vibration of'the molecule.
For all ideal gas, the internal energy can only be translational kinetic energy. If we' have n moles of an ideal gas at
temperature T, then
E,m = nUAJVl) "'" jnRT
(31)-
using Eq. 18. The interna] energy of aft ideal gas depends
cnly on the temperature. It does not depend. for example,
on the pressure or the volume of the gas.
One way to change the internal energy of an ideal gas is
to do work on il (or to allow the gas to do work on its
environment). Suppose the gas in the cylinder shown in
Fig. 1 is isolated from the thermal reservoir. Let the environment do work tV on the gas. The generalized law of
conservation of energy (see Eq, 28 of Chapter 8) then gives
(32)
519
Eller ')' ()j (111 Idea! {,(J5
= 11x-w;. + J)'ow;.
where 1 is the rotational inertia of the molecule for\otauons about a. particular axis. The x'y'z' eoordinatesystern is fixed to the center of mass of the molecule. For
point masses, there is no kinetic energy associated with
rotation about the :' axis, because J:, - O. The total kinetic energy of the molecule is the sum of the translational
and rotational parts:
K - lmv~ + mv; + 1m!}; + 11.•..w;.. + !I)-.w;
«,
(34)
Because kinetic energy is the only type of energy the molecule can have. Eq. 34 also represents the contribution of
one molecule to the internal energy of the gas. To find the
total internal energy of the gas, we must find the sum of
expressions such as Eq. 34 over all N molecules. A simpler
way is to evaluate the average energy per molecule. and
then multiply by the number of molecules, N.
Suppose we do work Won the gas, increasing its internalenergy. How much of this increase appears as translationa! kinetic energy and how much as rotational kinetic
energy? This determination is very important for understanding the macroscopic proQerties of the gas, because
only the average translational kinetic energy of a gas CIJI/tributes to its temperature. That i . two gases with the
same average translational kinetic energy have the same
temperature, even if one has greater rotational energy and
thus greater internal energy.
, To determine the relative contributions of translational
and rotational kinetic energy (and possibly other forms as
because Internal energy is the only way the gas can store
energy, and the work gives the only contribution to the
change in internal energy of the gas.
Suppose the environment does work on the gas, so that
Wis positive in Eq. 32.lt then follows that t:.E101 must be
positive, and using EQ. 31 we can write
)I'
(33)
so that the temperature change is also positive.
If the piston moves upward, the envircnrnentdoes negative work on the gas, and by Eq. 32 the change in internal
energy is negative. According to Bq. 33 the change in
temperature is also negative.
Let us now modify one of the basic assumptions in our
model of the ideal gas. Instead of considering a molecule
to be represented as a point particle, let it be considered as
two point particles separated by a given distance. This
.odel gives a better description of diatomic gases, those
with two atoms in each molecule, including such common gases as O2• N1• or CO (carbon monoxide). Such a
molecule can acquire kinetic energy by rotating about its
center of mass. and it is therefore necessary to consider in
r.
--·---.L~x
V .
i'
-------~------ - --Figure 1) A diatomic molecule.consisting of two atoms
to be point particles. is shown with its axis along
the z! axis of a coordinate system. In this orientation. the roo
tat{onal inertia for rotations abour lhe z' axls is zero, and thu!>
there is 110 term in the kinetic energy corresponding to such
rotations. The rotational inertias for rotations about the x'
and y' axes are not Zero. and thus there are kinetic energy
terms for such rotaticns.
considered
520
homer 21
Kineuc Theory and the Ideal Gas
well) to th internal energy, it i necessary to consider the
avcrag valu of each different term in the expression for
the internal energy of a gas, such as the five terms in Eq.
34, which is based on the a urnption of a rigid diatomic
molecule. For other gasc , we might need to include a
third rotational term. and for nonrigid molecules it 'is
necessary to include terms in the energy corresponding to
the Vibrational motion (se Section 15-10), From classical
statistical mechanics, which we consider in Chapter 24,
we can how that, when the number of particles is large
and Newtc nian mechanic holds, each o] these independent terms has the same average energy of !kT, In other
words, the available cnerg, depends only on the temperature and is 01. tributed in equal shares to each of the independent ways that a molecule can store energy. This
theorem, deduced by Maxwell, is called the equipartition .
a/energy.
Each independent form that a system's energy can take,
as, for example. the five terms of Eq, 4 is called a degree
of freedom. A monatomic gas has only three degrees of
freedom
r molecule, since it has only translational kinetic energy (H,"t· !nll'; + .inTI'; !mv;),
diatomic
g ••hasfire degrees of freedom per molecule, if the molecule IS rigid.
1 ct us use the equipartition of energy theor m to write
an cxpr Ion for the internal energy of a monatomic
ideal gas. The average internal energy per molecule is lkT
( degre s of freedom X !k'f per degree of freedom), and
the total int mal energy of the N molecules is
l:.',ftI -
'{ k'Tv=
nRT
(monatomic
gas).
(35)
where we have used Eq .6 and 8. Equation 35 is identical
with EQ. 31.
For a diatomic gas, with S degrees of freedom, the result is
E'nl - N(!kT} - inRT
(diatomic gas).
(36)
A polyatornic gas (more than two atoms per molecule)
generally has three possible axes of rotation (unless the
three atoms Ii in a straight line, as i'i\ CO2), The internal
kinetic energy per molecule could then have a sixth term,
Vl'W~'. For 6 degree of fseedorn. the internal energy is
E'n! - i '( kT) - 3nRT
(polyatornic
gas).
(37)
So far we have considered only the contributions
of the
overall translational or rotauonal kinetic energy to the
internal energy of a gas. Other kinds of energy may also
contribute. For c ample. a diatomic molecule that is free
to vibr te (imagine the two atoms to be connected by a
spring) has two additional comribuuons to the energy: the
potential energy of the pring and the vibrational kinetic
encrg of the atom, Thu a diatomic molecule free to
translate, rotate, and vibrate would ha e 7 (- 3 + 2 + 2)
degrees of Irecdom. For polyatornic mole ules, the number of vibrational terms in the energy can be greater t ha n
two. The vibrational modes in the internal energy arc
usually apparent
nly at high temperature.
\.\ here the
more violent collisions can cause the molecule I vibrate.
In Section 25-4. we how that the results defied in this
section give a very good description of the relauonship
between the internal energy and the temperature of real
gases. We also see that. as the temperature
of a gas is
lowered, the vibrational and rotational m ti IlS can he
"frozen," 0 that at a low enough temperature only the 3
translational degrees of freedom are present, The 010$1
senous shortcoming of this ideal gas model is Its failure to
account for (he quantum effects inherent In atomic and
molecular structure, E pcrimcnts with ga collisions provided earl) evidence that the internal energy of an a (1111 I'>
quantized. We can thus say that the seeds of quantum
theory lay in the kinetic theory of gases .•
'ample Problem 7
cnsidcr once agam the situauon otSam-
pic Probl m 5, III \.\ hich the ~a begin at the initial pouu \ ith
\olumc~',-40m)andpf
ssurcp,-IOP,I
I t ihe cvlin crb
removed from the thermal fee; rvorr, and let \I omprcs the a
adiabatically
until its olurnc IS I',
internal energy of the gas.
<,sumlllg
I.() O1l h. d the ch II' 10
it to t • helium (a 1)( n
as with r= 1.61\).
atomic
Solution
To find the change
In internal
energy, we can u •••.t q
n ifwc know the change in temperature.
We an nnd the rru I 1
temperature USIng the ideal ga 1\\\\ ( ince p, and 1,;\1' 110 \ III
and we can find the final temperature If we now rh PI ure
and volume of the final porru. Th final pres, urc can be found
using the adiabatic relauonship of I'q. 29:
flr-
p V1
;,(;-
(10 Pa)(4.0 m1)'
(I.Orn')''''
-1(lOP
On the pr' diagram of Fig, 10, the Iinal pomt reached in Ih~
adiabatic process lies vertically Iar above the final POlOt reached
in the Isothermal process (40 Pal, This
consisteut With the
adiabatic curves being steeper than the rsotherrnat CIlf\C~. 'IS
shown in FiS. 9,
We can now proceed to fin 1the initial and final temperature
and then the change In Internal energy'
I,
fI,",
T --"'"
,
T
tlR
(lOPa)(4.0rn1)
(0.11 mol)(8.31 J/mol'K)
Prl'(
(100 PaX 1.0 m
--=
nR
(0.11 0101)( .31 J/mol· K)
I)
r
K
-44
--I}
.
o
£,ftI-~nRtJ.1
.3IJ/mol·
-HO.llmol)(
')(IOQK-
-lK
• J
The change in internal
nel'%Y i po IIi\(', c nsistcnt \ ith I q '2
for this adiabatic proccs , because the wor done In cornnrcs In
th gas is similarly posiuvc
I
•. See "On Teaching Quantum Phe-nomena," by rr ·.!.:-"1 0\\.
COrl/C'IIlP{)rtJr)
Physln, ugust 1964. p, 401.
F
L
-,
:n~7 I~TEHMOL£
I·ORCc~
CLAR
(Optional)
}'OC\;C$ between molecules are-of electromagnetic
origin. All molecules com In electric charges in motion. These molecules are
electncally neutral in the sense that the negative charge ef the
electrons is equal and opposite to the positive charge of the
nuclei. This does riot mean. however. that molecules do not
interact electrically. For example, when two moteeuleaapproach
each other. the charges on each are disturbed and depart slightly
from their usual positions in such a wny that the averagedistance
between opposite charges in the two molecules is a little smaller
than that between like charges. Hen e an attractive intermolecular force results. This internal rearrangement takes place only
when molecules are fairly close together, so that these forces act
only over short distances; the)' Are short-range forces.If'the molecules come very close together, so that their outer charges begin
10 overlap, the intermolecular
force becomes repulsive. The molecules repel each other because there is no way tor a molecule to
rearrange itself internally to prevent repulsion of the adjacent
external electrons. It is this repulsion on contact that accounts
or the billiard-ball character of molecular collisions In gases. If
It were nor for this repulsion, molecules would move right
through each other instead of rebounding 00 collision.
Lei us assume that molecules are approximately spherically
symmetrical. Then we can describe inrermclecutarforcesgraphic311yb) plotting the mutual potential energy of two molecules.
C, as a function of distance r between the.r centers. The force F
JCling on each molecule is related \0 the potential ene~~y U 'I
F= -dV/dr.
In Fig. 12a we plot a typical VCr). Here we can
imagine one molecule to be fixed at O. Then the other molecule
ISrepelled (rom 0 when the slope of Uis negative and is attracted
100 when the slope is positive, At '0 no force acts between the
molecules; the slope is zero there. In Fig. lib we plo: the mutual
force F(r) corresponding to this potential energy function. The
lme Em Fi . 12a represents the mechanical energy of'rhe colliding molecules. The intersection of U(r} with this line is a "turn109 point" of the motion (see Section &-4),The separation of the
centers of two molecules at the turning point is the distance of
closest approach. The separation distance ar which the mutual
potential energy is zero may be taken as the approximate disranee of closest approach in a low-energy collision and hence as
th, diarneter of the molecule. Fersimple molecules the diameter
IS about 2.5 X 10-10 m. The distance '03t which the potential is a
minimum (the equilibrium point) is about J.5 X IO-H) m fot
simple molecules. and the force and potential energy approach
zero as r Increases to about 10-" m, or about <\ diameters. The
molecular force thus has a very short range. or course, differcillt
molecules have different sizes and internal arrangement of
charges so that intermolecular forces vary from one molecule to
another. However. they alway show the qualitative behavior
Indicated In Fig. 12.
ln J sotid, molecules vibrate about the equilibrium position
'er Their total energy E is negative. that is,lying below the horizontal axis in Fig. 12(/. The molecule do not have enough energy to escape from the potential valley (that is, from the attractive binding force), The centers of vibration 0 are mote or less
frxcd in a lid. In a liquid the molecules have greater vibrational
energy about centers that are free to move but that remain about
the same distance from one another. Molecules have theirgre.at-
R~ u!~,ve
Attr(je\,~f---'t-----~-
r
(b)
--~.
------ -Figure 12 «(I) The mutual potential energy U of \\';0 molecules as a function of their separation distance r. The mcchunical energy E. is indllatrd by the horizontal hne. (/1) The raJI.!1
Jort'e between the molecules. given b)
(/(ildr. correspondmg
to this potential energy. '111C potenual energy is u minimum at
the equilibrium separation 'v. at which POlO! the force IS zero.
est kinetic energy in the ~{)US
state. In a gas the average cisranee between the molecules is onsiderably greater than the
eflecuve range of intermolecular Iorces, and the molecules move
in straight lines between collisions. Maxwell discusses the relation between the kinetic theory model of a gas and the uuermolecular forces as follows: "Instead of~} 109 that the parucles arc
hard. spherical. and clastic, we may ir we please say th!ll the
particles are centers of force. of which the action I:' insensible
except 31 a certain small distance, When it sudden! . appears it) a
repulsive force of very great mlen"ily. It IS evident that either
assumption will lead to the Slime results."
It is interesting to compare the measured intermolecular
forces with the. gravitational force of attraction between molecutes, If we choose a separailon distance of.j X 10 IU 10. for
example, the ferce between tWO helium atom is about 6 X 10-'1
N. The gravitational force III that separauon IS about 7 IO-~:
N. smaller than rhc imermolecolar force by a factor or lO.N! I his
IS a t -pieal result flIHJ shows that gravnauon 1\ ncghgihle 10
iutermole ular forces.
Although the miermolccular forces appear to be small h)
ordrnary standards, we must remember Ih,1\ the mass 01 a mulecule is scsmall (about 10 ~o k.\l) that these rorces 'In unpart
instantaneous accelerations of the order of IO" mrs' ( 10'· Ri.
These ac elerations may laM for only a very short ume. of course.
because one molecule can very QUickly mO\1: out ot the ran~t' f
influence of the other, _
522
Chapter 23
23-8
THE
A DER \
AI.S EQ ATIO
OF TATE (Optional)
Kinetic Theory and the Ideal Gas
Solving for p, we obtain
Kinetic theory provides the microscopic description of the behavior of an ideal gas, but certain of the assumptions of our
model of the ideal gas are not valid when applied to real gases.
Many modifications to the equation of tale oftbe ideal gas have
been suggested to correct for these deficiencies, In the previous
section. we showed that n realistic way of'looking at the intermolecular force lead us 10 conclude that molecules have a small but
certainly nonzero diameter(which may contradict assumption 4
of the ideal gas model) and that the range ofthe force may extend
beyond the "collision diameter" (which contradicts assumption
5). In this section we devefOp a modified equation of state that
takes these factors into c....,..t:
To consider the effect of\ilae finite size of the molecules, let us
regard each molecule as a hard Qilere or diameter d. Two molecules are not permitted to apprcacn 0WiC. anot her so close that the
distance between their centers woul~
than d(Fig. 13). The
"f e volume" available (or one n
IS therefore decreased
by the volu me of a hemisphere ot
d centered on the other
molecule. Let b represent the decrease in the available volume
due to the molecules in I mole of a gas. The total volume available to the entire collection of molecules in /I moles is thus the
volume V of the container less an amount nb that represents the
volume occupied by the molecules. If we take the estimate from
the previou section of d - 2.5 X 10-10 m, then we estimate b as
ro....s
b - ~N ..••
H1tcf3) - 2 X 10-' m Imol
- 2 X 10-) LImo!.
.
(The factor of comes about because. as two molecules approach one another, the volume within which they interact is not
a full sphere but the hemisphere facing the direction of approach.) Under normal conditions, I mole of a gas has a volume
of 22.4 L, and thus the correction b is normally small (0.010.1%), but it can become much more significant if we study a gas
at high density.
'.
The "free" volume available to the gas is thus V - nb, and we'
can modify the equation of state accordingly:
p(V - lib) - nRT.
nRT
P-V-nb'
Equation 39 indicates that the pressure of a real gas is incre d
relative to that of an ideal gas under the same conduions. In
effect. the reduced volume available to the molecules mean that
they make more collisions with the walls and thereby rncrca c
the pressure.
To account for the effect of tlie range of the force between
molecules, let us consider a region of the gas within a distance d
of one of the walls of the container (Fig. 14). \ e choo d to
correspond to the range of the force between molecules. unci we
focus our attention on a particular molecule C that 1\ about to
strike the wall. When it strikes th wall, the impulsemomentum theorem, 6p - J F dt. can be used to relate the
change in momentum of'the molecule to the impulse of the II t
force F that acts on it during the collision. In the ideal gas mod I,
the molecules exert forces on one another only during collisions:
thus the only fore that acts on a molecule colliding with the wall
is exerted by the wall. This force, by Newton's third Jaw. is equal
to the force exerted on the wall by the molecule and thus is
responsible for the pressure that the gas exerts on the wall of the
container, as we discussed in Section 2 - .
Now suppose that molecule C also experien es force from
the attraction of other nearb molecules (tho: 'I~\O within a
hemisphere of radius d, the range of the force) For a mol ule
near the wall, the sum of all the intermolecular orce give a
resultant that acts away from the wall. ( 101ecules near the surface of a liquid experience a similar inward force, which I rcsponsible for surface tension; e Section \ 7-6.) 1 nus during the
collision the component of the force acting away from the \'"311
-- ", . ,
,I
.... ...•
(38)
.:
"
•
I •
\ 1
II
•I
C
I
I
I
-II
~
d
• I
IC --,.,.-".....-
I
I
1-(
;'
I
I
I
I'
• 1
I
,, .
I
I.
Ii
Figure 13 If molecules of a ga arc considered to behave like
hard spheres, then the center of molecule B is not permitted
to move within the hemisphere of radius d centered on molecule A. Here d is the diameter of a molecule. The free volume
available for molecule B is reduced by the volume of such a
hemisphere centered on each molecule of the gas.
A gas molecule C (here considered to he a point)
near the wall of the container experiences a net force away
from the wall due to the attraction pf the surrounding molecules within the range d of the force between molecules. The
net pressure on the walls of the container is reduced by all
such molecules within a distance d
the walls,
Fiaure 14
or
Section 2]·8
has two corunburlons; one from the wall and another from the
surrounding molecules. For a given change in momentum from
a collision with the wall, the force exerted by the wall during the
collision is therefore smaller, the reaction force exerted by the
molecule is maller, and the pressure exerted by the gas is likewise smaller
Tim reduction In pressure owing to the collision of molecule
".lIh the wall is proportional to the number of mole ules in the
hemisphere of radius d surrounding molecule C and thus to the
number
f molecules per unit volume of the gas. N/V. The uet
effect due to all the molecules like C in the surface layer of
Hue ness d I~ proporuonal to the number of molecules in that
layvr , which I'> also proportional to the number of molecules,
per unit volume of the s. The towl reduction in pressure resuiting from the force between molecules is thus proportional to
(N/I-y
That is" if we triple the number of molecules but keep the
volume of the container constant, OUf imaginary hemisphere
will have three times as man), molecules and hence molecule C
will suffer three times the force pulling it away from the wall. In
the entire gas there will be three times as many of these molecules, each suffering the same effect The overall effect thus
Increases ninefold.
..•.
The net effect of the intermolecular force is to introduce a
correction to the pressure, proportional to (N/V)l. Instead of
wnring thi$ correction in rerms of the number of molecules N,
we write it in terms of the number of moles It, so that the
corrected pressure becomes
nRT
(h)l-V'
p-----a
V-nb
where a is a proportionality
state can be written
The
an der WUUI.I Equation t!rSIMI'
(I'
a ;~)
(V -
11b)- !TRi.
constant. The modified equation of
4
!I
!
3
2
---
~
•...
o
I
I
!
I
I
I
I
I
1
T c 264 K
I
I
I
I
I
-rf~-r--~~~
_
IB
o
(oJ)
I
C
V (10-4 ml..
J
3
I
I
4
0
(b)
523
(4 I)
This expression. first deduced by J. D. van der Waals (18371923), is called the van der Wa«/Jl:q/IGIIOtl ofstate. Note that Eq.
4! reduces to the ideal gas equation of state (Eq. 7) when the gas
occupies a large volume (t1l31 is, the molecules are very far apart
and the gas density is small) ..
The values, of the onsiarus
and b must be deiernuned by
experiment, which makes the equanon empirical in this respect.
Like the ideal gas equaticn of state, it IS also based on ,I model
with oversimplifying assumptions. No simple formula can be
applied 10 all gases under all conditions, and only through exPerimenis can we learn w hether one equation rs supenor 10 another
in its de ription of reality over a certam set of condmons.
Figure 15 compares isotherms for an ideal. gas with those
calculated for CO2 USing the van der Waals equation or stale.
Note that Ihe deviation from ideal behavior occurs primarily at
h:igh pressure and at low temperature. For CO) at temperatures
below 304 K, the isotherms begin to co rve down ward. indiea ti ng
that as we decrease the volume, the pressure likewise decreases.
Since this behavior is contrary to expectauons for a gas, it suggests that some _he CO, is condensing uuo allqllid,leaving less
of lt in the gascO\lS state. The van der Waals model thus suggests
the existence of mixtures of different phases, which the ideal gas
model cannot do. If we were 10 compress a sample of CO:, we
would find that tire actual T- 264 K isotherm would not dip
downward as th van der Waals equation predicts, but instead
would follow the horizontal segment A1:J in Fig. 15, as the gas
condenses into a liquid llt constant pressure. The van der Waal
model gives an improvement over the ideal gas model. but no
simple mode! is able: to account for the behavior ofthe gas under
all possible cireurnstunces,
(40)
,......_T •• 304 K
(OptlOl/al)
b
1 Vcr
2
3
•
V(lO-'m)
l
"
-- -- - --;:---;---:--:--:-:--:-~:-:-:-------Figure 15 (0) Isotherms for I mole of an idea! gas. (b) Isotherms for I mole of eo. d-et-e-rm-·.-in-ed-rrom-
the van der Waals equation, Note that at large volume, tne ideal and van der Waals·isotherms behave
sunilarly. As the temperature is raised, the van del' Waals isotherms behave more like those of tbe-ideal
a) IOle also thai, as the pressure becomes very large, the volume approaches the value of b. as EQ. 40
requires, rather than the value of sere, a the ideal gas equation of state would predict. The dashed line
1/l shows a more re hsuc representation of the T- :!6·1 K isotherm. As the as is compressed along this
isotherm, some of the gas ondenses into a liquid, and the pressure remains constant
524
Chapter ']J . Kinetic Theory and rhe Meal Gas
\ e also find that om ether results for the ideal gas are only
approximately correct in their application to real gases. For example. the internal energy ofa real gas depends on the volume as
well 3<; on the temperature. If tbcre are attractive forces between
the molecules. then the internal potential energy increases as we
increase the average distance between the molecules. We therefore expect the Internal energy ofa gas to increase slightly with
volume. and thi expectation IS consrstcnt with experiment in
most gases. If the state of the gas L sue; th: I repulsive forces are
more significant than attractive forces, then increasing. the disranee between the mole ulcs decreases the pctential energy. For
some gases (hydrogen and helium at ordinary temperatures. for
instance) the internal energy i observed to decrease as the volume increases. In either case, III internal energy is not simply a
function of temperature but depends on the volume as well.
In taking both derivatives we assume constant 7. us I appropnate for an isotherm.
Setting both derivatives equal to zero and SOh·1O these cquutions simultaneously for (1 nod b. we fino
(J ~
27R2T2
Ct
64p.,
II"" RT., .
aptt
Reading PH = 0.15 X 10' Pa from Fig. J Sb, we then calculate
a-OJ64J'm)/moI2
Sample Problem 8 The isotherm drawn in Fig. ISh for CO2 at
a temperature of T- 304 K is called the critical isotherrn.Jt is
distinguished by having a minimum and a point of inflection
(the point where the curvature changes from downward to upward) that coincide at a single point.
sing this information
along with the value of the critical pressure p,•• estimate the
values of the van der Waals constants a and b for COl'
Solution
The minimum of a curve on a pI" diagram is determined by the point at which the lope dp/dV is zero, and in
calculu: we learn that at a point of inflection the second deri •.ative IS zero, We can find the derivative when the van der Wallis
equation of slate is written in the form of Eq. 40:
and h"'4._7XlO~$lU}/mol.
Although the van der Waals model gives a much more r'::III,ti.:
description than me ideal gas model of (he bchav ior of a real gas
such as CO2, it still represents only an approximation or the
actual behavior. In the case of CO~. for 1OSI;1I1(£. the above
calculation gi ..es I'" = 311/1 ••• 1.28 x 10" m) for .hc volume ,,1
i molt at the critical pomr. 1 he measured value. howc 1:1, h
0.96 X 10-' m ', Nevcrthele s. it is .1successful fIrst step III irnproving the ideal cas model in t::lSCS ill which the molecule arc
sufficienti) close together that the ba ic assurnpuons of the ideal
gas model do nOI hold. and it even suggests condensation due to
the force between molecules. which the ideal gas model i, cornpletely unable to do. •
QUESTIONS
1. In discussing the fact that it is impossible to llpply the laws of
rncchamcs individually to atoms in a macroscopic system,
Mayer and Mayer state: "The very complexity of the problem (that is, the fact that the number of atoms is large) is the
secret of its rolution:' Discuss this sentence.
2. In kinetic theory we assume that there is a lar&c number of
molecules in a gas. Real gases behave like an ideal gas at low
densities. Arc these statements contradictory? If not, what
conclusion can you draw from them?
3. We have assumed that the walls of the container arc clastic
for molecular colli ions. Actually, the walls may be.inelastic,
Why docs this make no diff~"'encc as long as the walls are at
the same temperature 3\ the !'\s?
4. On a humid day. some say that the air is "heavy," Hew does
the densuy of humid air compare with tltat of dry cir at the
same temperature and pressure?
S. Where does the root-mean-square speed of molecules in still
air at room temperature fit into this sequence: 0: 2 mts
(wal 109speed); 30 't(lts (fast car); 500 m/s (supersonic airplane); 1.1 X 10' mts (escape speed from Earth): 3 X 10'
mts (speed of light)'!
6. '1wo equal-size rooms communicate through-an open door- wav. However, the average temperatures in the two rooms
are maintained at different values. In which room is there
more air?
7. Molecular motions nrc maintained by no outvide Iorrc. yt:1
continue indefirutely with no sign of dimirustnug
<pc•.'d.
What is the reason that friction does nOl bring these 1lI1]
particles to rest, as it does other movmg panicles?
8. WhatjllstifIC~Hio"
is there In neglecti ng the changes rn gravrtational potential energy of mole ules in a gas'!
9, We have assumed that the force exerted by molecules on the
wall of a container i~steady in time. How is tbis JU~lIficd?
10. [I is fo.und that the weight of an empty Oat thin plastic hag I~
not changed when the bag is filled with air. \vh> nOI"
11: We know tha; a 5t ne Will Iall t the ground If we release It.
We put no constraint on molecules of the air, yet Ihe~ don't
all' fall to the ground. Why not?
12. Justify the fact that the pressure of a gas depends on the
square of the speed of its parucles by explairung the dependence of pressure on the collision frequency and the momentum transfer of the particles.
13. flow IS the peed ot'sound rc!atl'd
kinetic theory model'!
10 the g<l~ ,.1Wlhk,
111 till'
14. Consider a hot, stationarygol ball silting on a tee and a cold
golf ball just moving off the tee alter being hi. The tn !
kinetic energy of the molecules' motion relative 10 the t('\'
can be 'he same in the two cases, Explam how. What is the
ni!Tcrence between the two cases'!
Problems
. F r bove the Earth's surfa e 'he gas kinetic temperature is
reponed to be on the order of 1000 K. However, a person
placed in such an envir nment would freeze to death rather
than vaporize. Explain.
16. Why doe n't the Earth's atmo phere leak away'? At the lOP
of the atmosphere
toms will occasionally be he ded out
with speed e.•.ceeding the e pe speed. Iso't it just a matter
of time?
. Titan, one of'Sarurn's many moons, has an atmosphere, but
our own Moon does nor. What is the explanation?
18. How, if'at all, would you expect the composition ofthe air to
change with altitude?
19 Expl ••in why the temperature decreases with height in th
tower atmosphere.
20. In large-scale inel stic colli ions mechanical energy is lost
through internal friction resulting in a rise of temperature
OWlOg to increased internal molecular agitation. Is there a
] s of mechanical energy to heat in an inelastic collision
between molecules?
.
11. By c nsidcrina quantities that rnu t be conserved in an el
tic colli ion, show th tin general molecule of a gas cannot
have (he same speeds alter colli ion a they had befor .Is It
sible, then, for a gas to consist of molecules th I all have
tl e
me speed?
22. We OHCH )' tha: we see the steam em rgin from the SPOUI
ola ' eule in w hi h Wilier is boiling. However, steam itself is
a c lorless
What is it that you really see?
2). Wh} docs smoke rise, rather than fall, from a lighted candle?
2.t. Would a gas v-hose molecules were true geometric points
obey the ideal gas law?
25. Wh} do molecules not travel in perfectly straight line between collisions and what effect, easily observable in the
l.lbof..ttof) occurs ;IS a result?
'lb. \\'hy must the lime allowed for diffusion separation be relall\d, short?
27. Sup ose we want to obtain mU instead oflUU as the end
product of a diffusion process. Would we use the same pro-
525
cess'? If not, e plain how the separation pr e s waul have
to be modified
28. Considering the diffusion of gases into each other, can you
draw an an logy to a large jostling crowd with many "collisions" on a large Inclined plane WIth a stope of a fewdegree ?
29. Can you describe a cenuifu al de 'ice for gaseous separ .
tion? Is a entrifuge better than a diffusion chamber for
separation of gases?
30. Do the pressure and volume of air in a house change when
the furnace raises the temper ture si nificantly? If'not, i$ the
ideal gas law violated'!
31. Would you expect real mole utes 10 be spherical! '~ymmel·
rieal? If not, how would the potential energy function of Fig.
120 change?
32. Explain why the temperature of a gas drops in an adiabatic
expansion.
33. lfhot air rises. why is It cooler at the lOP of mountain than
near sea level?
34. Comment on tlu taterneru: "There are two W:lY to carry
out an adiabatic proce s. One is to do it quickly and the
other is to do it In an Insulated box."
35. A s led rubber balloon contams d very light gas The balloon IS released and it rises high into the atmosphere, DeS ribe and explain the thermal
nu mechanical beh vior of
the b lloon.
36. Although real gases can
hquified, an ideal as cannot be.
Explain.
37. Show that as the volume: per mole of a t' ~increases, the van
der W als equation tends to the equation of state of an ideal
gas.
38. Extensive quantities have- values that depend on what the
system's boundaries are, whereas intensive quantities are
independent of the choice of boundaries. That is, extensive
quantities arc necessarily defined for a .••.holt' system,
whereas intensi ·c quanuues apply uruforrnly t uny small
part of the system. Of the 101l0w lng quantities, determine
which are extensive an which are intensive: pressure, volume, temperature, density, mass, internal energy.
PROBLEMS
Section 13-) Macroscopic Properties of a Gas and the Ideal
Gas Law
1. ( ) Calculate the volume occupied by 1.00 mol of an ideal
gas t standard COI dition , that is, pressure of 1.00 aim
(- 1.01 X 10i P ) and temperature
of O·C ("" 273 K).
(b) Show that the number of molecules per cubic centimeter
(the Loschrnidt number) at stan tard conditions is 2.68 X
1019•
2, The Nost vacuum thai can be attained in the laboratory
corresponds to II pressure of about 10-1$ arm, or 1.01 X
]0-1) Pa. How many molecule are there per cubic centimeter in such a vacuum at 22"0
3. A quantity of ideal gas at 12.0·C and a pressure of 108 kPa
oc upies a. v lurne of2.47 m-. (.1) How many moles of the
gas are plesenl?(b) If the pressure is now raised to 316 kPa
and the temperature is r ised to 31.0·C. how much volume
will the gas now occupy? Assume no leaks,
4. 0 ygen gas having a volume of 1130 ern! at 42.0·C nd a
pressure of 10I kP e. \lands until its volume i 1530 em)
and its pressure is 106 -Pa. md (a) the number 0 milks of
oxygen in (he: syste
::! (b) its nnal temperature.
S. A weather balloon is loosely inO led with helium at a pressure of 1.00 atm (-7'.>.0 em Hg) and a temperature of
22.0·C. The gas volume is 3.47 ml. At an elevation of 6.50
krn, the atmcspheric pressure is down to 36.0cm Hg and the
helium has expanded, being under no restraint from the
confining bag. At thi elev titian the gas temperature IS
-48.0·C. What is ibe gas volume now?
6. The variation in pressure m the Earth's atmosphere,
assumed to be at a uniform temperature, is given by
526
Chapter 2J
Kinetic Theory and tile Ideal Gas
PrIl-Mtyll!.7', where J\{ is the molar mass 01 the air. (See
Section 17-3.) Show that II" - n v(Je-,liv>II<'. where n" is the
number of molecules per unit volume.
7. Consider a given mass of ideal gas. Compare curves representing constant pressure, constant volume. and isothcnnal
(constant temperature) processes on (a) apV diagram, (lJ) a
pTdiagrllm. and (c) a VTdiagrnm. (d) How do these curves
depend on the mass of gas chosen?
8. Estimate the mass of the Earth's atmosphere. Express your
estimate as a fraction of the mass of the Earth. Recall that
atmospheric pressure equals 101 kPa.
9. An automobile tire has a VOlume 0['188 in.! and contains air
at a gauge pressure or24.2 Ib/in.1 when the temperature is
- 2.60'C. Find the gauge pressure or the air in the tire when
its temperature rises to 2S.6·C and its volume increases to
1020 in.'. (11InI: It is not necessary to convert from British
to S1 units. Why? Use P.'m - 14.71b/in.1.)
10. (0) Consider J.OO mol of an ideal gas at 285 K and 1.00 atrn
pressure. Imagine that the molecules are for the most. part
evenly spaced at the centers of identical cubes. Using Avogadro' constant and taking the dlarnctcr-of'a molecule to be
3.00 X 10-scm. find the length of an edge of such a cube and
calculate the ratio of this length to the diameter
a molecule. The edge length is an estimate of the distance between
molecules in the gas. (b) Now consider a mole of water
ha ing a volume of I8 cm1. Again imagine the molecules to
be evenly spaced at the centers of identical cubes and repeat
the calculation in (a)
11. An air bubble of 19.4 em) volume is at the bottom or a lake
41.5 m deep where the temperature is 3.RO·C: The bubble
rises to the surface, which is at a temperature of 22.6"C.
Take the temperature of the bubble to be the same as that of
- the surrounding w':ltcr and find its volume just before It
reaches the surface.
12. An open -closcd pipe of length L •.••
25.0 m contains air at
atmospheric pressure. U is thrust vertically into II freshwater
lake until the water rises h~lrway up in the pipe, as 5110wn in
Fig. 16. Whatis the depth h of the lower end of the pipe?
Assume that the t .rnperaturc is the same everywhere and
docs not change
fJ -
or
•
.••.•
Problem
I::.
I J. Container A contains an ideal gas at a pressure of 5.0 X IO~
Pa and at a temperature or 300 K. It is connected hy II thin
r
it
:..
A
_ "
.• .•• •
.•
••
•
•.
•.•
•
• •. ~ • • •. •
•
••
4t
-------------, ----figure 17 Problem 13.
••
••
•
•
B
,
tube to container B witl: four time the volume of /v. see Fig
17. B contains the same ideal gas at a pressure of 1.0 X !OJ
Pa and at a temperature or 400 K. The onne ling valve l~
opened. and equilibrium is achieved at a common pressure
while the temperature of each container is kepI constant :lI
its initial value. Whal is the final pressure in the system?
14, Two vessels of volumes 1.22 Land 3.18 LCQnt(l1O krypton
gas and are connected by a thin tube.Jmually.the
vessels arc
at the same temperature. 16,0'C. and the same pre sure.
1.44atm. The larger vessel is then healed to 108'(' while the
smaller one remains at 16,0-C. Calculate the fmal pressure
(flint: There (Ire 1'10 leaks.)
IS. Consider a sample of argon gas at 35.0·C and 1.22 atrn
pressure. Suppose that the radius ora (spherical) argon atom
is 0.710 X 10-10 m. Calculate the fraction of the container
volume actually occupied by atoms.
J 6. A rnercury-fllled manometer with two unequal-length arms
of the same cross-sectional area is scaled off with the same
pressure Pn in the two arms, as in Fig. llt WIth the temperature constaru.an additional 10.0 em I of mcrcur '1\ admitted
through the stopcock at the bouorn The level on lht' Itlt
increases 600 em and thai on the nghl increases 4 ()Oem
Find the pressure Pn
Figure 18
J.'igurc 16
,
·41··.. .•• '?..-:~
~ •.•
.."
.
••••••••
• • ••••4Ir'~:i:!:(5~;.pc::!:~
• • • •
..:~".:.J
.
.... ... .
.. ..
.-::-::0-:-,.......,:-.'"
Problem 16.
Section 23-3 Kinetic Calculation ollht Pressure
17. The temperature in interstellar space IS 2.7 K Find the
root-mean-square spec of h -drogcn molecules at this ternperature. (See Table I.)
18. Calculate the root-mean-square
peed or ammUOI:I (1'<II ,i
molecules at 56.0"('. An atom or nitrogen has a mas, or
r
Problems
2.33 X lO-J6 kg and an atom of hydrogen has 13 mass of
1.61 x lO-H kg,
•
19. At O'C and 1.000 atm pressure the densitiesofair, oxygen,
and nitrogen are, respectively, 1.293 ks/ml 1.429 kg/m",
and t .250 kg{m'. Calculate the fraction by m.nssofnltrogen
in the air from these data, assuming only these two gases 10
be present
20. The mass of the H2 molecule is ~.:) X Hr14 g. It 1.6 X 1011
hydrogen molecules per second strike 2.0 cm2 of wall at an
angle of 55" with the normal when moving with a speed of
1.0 X IOs cmrs, what pressure do they exert on the wall?
21. At 44.0·C and 1.23 X 10"'" atm the density of a gas is
t n
10' &/<:m'. (a) Find p,m. for the gas molecules,
(Ii) Find the molar mass of the gas and identify it
22. Duiron', lull' stares that when mixtures of gases having no
Chemical interaction are present together in a vessel, the
pressure exerted by each constituent at a given temperature
t'.i the same as It would exert ifi! alone filled the whole vessel,
and that the total pressure is equal to the SUm of the partial
pressures of each gas. Derive this law from kinetic theory,
U5! ng Eq. PI.
:no A oruainer encloses twO ideal gases. Two motes of the first
gas are present. with molar mass M\. Molecules oflhe second gas ha vc a molar mass .H! - 3M ,_and 0.5 mol of this
g:t:. is present. What fraction of the total pressure 011 the
coruumer W;JI! is attributable to the second gas? (Hint: See
527
and at IOO·C. (b) Find the traMlalional kinetic energy per
mole otan ideal gas at these. temperatures, in joules,
26. At what temperature is the average translational kinetic energy ofa molecule in an ideal gas equal to LOO eV?
27, Oxygen (01) g.'tS at 1S' C and 1.0 atm pressure is confined to
a.cubical box 25 em on a. side. Calculate the ratio of the
change in gravitational potential energy ofa mole of oxygen
molecules falling the hei&ht of the box. to the total tr'lnsla·
tionalldm:tic eners}' of the molecules.
28, Gold has II raolsr (atomic) mass of t 97 S/rool. Consider a
2.50.,& sample of pure goW vapor. (0) Calculate-the nurnbet
of moles of gold present. (b) How many atoms of gold are
presenfl
Find the average translatlena! kinetic energy orindiv!dua.!
nitrogen molecules at 1600 K (a) in joules and (b) in electron-volts.
30. (0) Find the.number of molecules in 1.00 m) ofnital 20.0·C
and at a pressure of tOO atm. Cb) What is the mass of this
volum e of air'? Assume that 75% of the molecules llre nitrogen (Nt) and 25% tire oxygen (01),
31, Ceuslder a gas at temperature T occupying a volume V to
consist O!a mixture of atoms, nRmcly,N. atoms ofma$~ m•
• each having an rms speed v., and N"atoms of'rnass m" each
having an rms speeo VI>' (0) Give an expression for lhe total
pressure exerted by the gas, (b) Suppose now that N. "'" N~
and that the diiferent atoms combine at constnnt volume to
form molecules t1f mass nI, +
Once the temperature
retums to its original value, what would be the ratio of the
pressure after combination to the pressure before?
31. A steel tank CMlains315 g of ammonia g.as (NH) tit an
29.
m".
Problem 22.)
Section 23-4 Kinetic lnterpretatio» cfth~ Tempitrtlttlrc
2.1 f he Sun is a huge ball of hot ideal gas. Tbeglewsurrounding
me !'HlH in the x-ruy photo shown in Fig. i9 is the corona-
the atmosphere of'tne SUIl. Hs temperature and pressure are
:':.0 'X 10· K and 0.030 P:L Calculate the rms speed of free
electrons in the corona.
absoll.lteprcssureofl.35
X l06Paandtempenlture77,0·C.
(a) What is the volume utthe tank? (Ii) The tank is cheeked
later when the tempc1"dture has dropped to 22.0·C and the
absolute pressure has fallen to 8.68 X IO! Pa, How many
grams orgas leaked out of the tank'l
3:;, ((1) Co
temperatures at which the tHIS speed is
equal to
of escape from the surface ofthe Earth for
molecular ltydtogen and f¢f meleenlar oxygen. (oJ Do the
same for the Moon, assuming the grttvhationai uecelcration
-on itsslltfacc to bell! 6g. (c)1ne temperature lugh in the
Earth's upper atmosphere is aboui !000 K. Would you expect to find much hydrogen there? Mueh ox)'gen'l
34. At what iemperature do the atoms of helium gas h::vc the
same nus speoo as the molecules ofhydtogcn gas n 26.0·C?
35. The envelope and basket of a hor-eir balloon hll'"~ a cornbined mass 01' 249 kg, und the envelope has a capacity of
2180 ml. When fully inflated. what Should be the temperatute uftht enclosed air to give the balloon a lifting CUPllcity,
of
.kg (.in u.ddition to its 0. wn mu.',S)1.A$S1.HI1C tJ~~lhe
surrounding air, at liJ)'C, has a denr:lty 011.22 kgfm'.
zn
~
Seetion. 23-5 Work Don« an tIn laMl (J(JJ
Flgur': 19
Problem 24.
25; (0) C a teulare Ihe a verage value in electron-volts of'the translatlOnal kinetic cnergy of the particles of'an ideal gas atO'C
..,
36. A sample ofgllscxpands from tOto 5.0 ml while:its pfC$$ure
., decreases from 15 to S.OFa, How m uch work is done Oil : he
gas if its pressure changes with volume according to each of
th e three processes shown in t.he pI! diug.rttrn in Fig. 20'?
37. Suppose that a sample of gas expands frorn 2.0 to 8.0 !T\ l
alotlg the diagonal path in the pI! diagram .hown in Fig. 2:.
It is then compressed back to 2.0 ml alQng either path 1 Of
528
Chapte) 23
Kinetic Theory and the Ideal Gas
20
Hi
g: 10
'2
<:I.
5
..
3
0
2
J
5
4
6
V(m~
Figure 20
Problem 36.
o
6
,:.4
8
original volume. Find ItS final pres urc and temperature,
(b) The gas if. now cooled hack to 273 Kat constant pressure
Find the final volume. (c) Find the (0l31 "url.; cone nil (he
gas.
45, The gas in a cloud chamber at n temperature of 292 K
undergoes a rapid expansion, Assuming the process i~adl:lbauc, calculate the final temperature if j' "" 1.40 and the
volume expansion entia is 1.28.
46. An air compressor takes air at 1 .0'( and I e){) nun pre••sure
and delivers compressed air al 2 30 aim pressure I'hc cornpressor operates a12_ 0 W of useful PC>W\)I. '\S ume that the
compressor operate adiabati ally. (oj Frnd lllc temperature
of the compressed air. (b) How much cornprc sed arr In
liters, is delivered each second?
47. A thin tube, sealed at both ends. is 1,00 m tong. It lies
honzontal!y. the middle 10,0 em containing mercury and
the two equal ends containing air al standard atmospheric
pressure. If the tube is now turned to a vertical position. hv
what amount w,1I the mercury be displaced? Assume that
the process is (a) isothermal and (In adiabatic (For .111.
}'= I AO.) Which assumption is more reasonable?
Section 23·6 Tit" Internal Enug)' of an Idea! Ga
\ (Ill))
48. Calculate the mtcrnal .:nc1t:' (II' 1 mole
.•.5.0·C.
r In Id I,'
kmcti(' cncrzv ,it ,In 'Ih: t\,·,1 •
cutes in [ mote of air at 25.0' C.
50. t\ cosmic-ray particle with enctg\ I ..~·!·),'\ IS stoPl'l'J Ifl.\
detecting tube that contains 0.1::'0 mol of II (In /::ts ()n'"
this energy Isdl~tributl'J among ,111:11<'
~\hH!1',. bv 1\0\\ mu :
IS the temperature of the neon mrrcase-I?
S1. An ideal gas experiences an adiabatic comprcssi '0 I:, u
p-122kPa.
T'- IO.7m'. T=-2JO
ClOP"'"
1,\ tH!'.!
I' = I 36 rn I, (a) Calculate the value Ill' 'j (i'l Find the: In I
temperature. k) How man moles of gas ~Il' P:';"1.::1t'
49. Calculate the total rotational
path 2. Compute the net work done on the gas for the complete cycle in each cas::.
JR. The "!X'eOof <ound in different gases at the same temperatuft' depend •• Oil the molar mass of the gas, Show that
t'l/!',
';:H d\l~ (constant T). where
is the speed of
sound m the gas of molar mas M 1 and Ii) is the speed of
v;
~ und in the gas 0 molal mass /If .
39. ,\ ir • t O'(X)'C and 1.00 ntrn pressure has a dcnsirv of
: .29 t '/ 10 ' >rJc·m). and the speed or sound is 33 l mts at
that temperature, Compute (El) the value of y of air and
(b) the effective molar mass
air.
40. Air Ih,lt occupies 0,142 m1 at 103 kPa gauge pressure is
expanded isothermally to zero gauge Pl'(:!>SUfC and then
c ooled at constant prccsure until it reaches its initial volume.
Compute the work done on the gas.
41. Calculate th work done by an external agent in compressing I 12 mol of ox w~n from a volume of 22.4 Land 1.32
aim pressure to 15.3 L at the same temperature.
42. U~ the result of Sample Problcm S tc show that the speed of
sound in lllr increases about 0.59 mls for each Celsius degree
nse In temperature near 20·C.
4;1. Gas occupies a volume of 4.33 L at 1 pressure of I. I? aun
and a temperature or;JH) K, It is compressed adiabatically
to a volume of 1.06 L. Determine (0) the final pressure-and
(tJ) the fmal temperature, assuming the gas to be an ideal gas
for which r - 1.40. (c) How much work was done on the
p?
44. (eI) One liter (If gas with f'" 1.32 is at 27) K and 1.00 31m
pres .urc l~ IS suddenly (adiabatically) compressed \0 halfits
or
-
(d) What
IS the
total translational
kinetic coer!')' per mote
before and after the compression? (e} Calculate the rauo "f
the rms speed befere to that after the eompressk 11.
Section 23-8 Th« Van der Waal5 Equatlo« of Sla(1'
52, Van der Waals b for oxygen is J~ em 1/r~()1. Compute 'I\.!
diameter of an 0) molecule.
53, Using the values of a and h for co) found In amnlc Ph I,
lern 8, calculate tbe pressure at 16.0·C of 2.5) mol ot O.
gas occuo mg a volume of 14.2 L. Assume (a) that the \ an
der Waals equation is correct, then (b) that COl behaves a~
an ideal gas.
54. Calculate the work done on» rnoles o a van der Waalvgas In
an isothermal expansion from volume V, \<1 l'r
55. Show that V
CI"'"
3ltb.
56. The constants a and h in the van dcr W;l;\!S ~q\JMI(1n art'
different for different substances. Show, however, that If W{'
toke V.n Pc<' and Tn as the umts ol volume. pressure. and
temperature, the van dcr Waals C'QU:lhOn rn.'I."<!'11 'S lC!cnll' at
for all substances.