Physical Chemistry [3 ed.] 0395918480, 9780395918487

Citation preview

HEMISTRY

3rd edition

Laidler/Meiser

00




o>

§

0)

r-

Ar

Helium 4.002602

2

*

S

Kr

39.948

C

CT

80 Argon

r ^

Krypton

83

3 18

36

Lu

^

ND 00

in

Lr

174.967

Lutetium

(262)

ium

Lawrenc-

103

71

i>

r~< "I?

-r

2.2

CI

-

r

2.4

T

o |

~>

o

o

S

in

E

S

2.5

Se

Sulfur 32.066

u

Z

6iJ

rsC

£

8

o

ri

—

^

n£

&

s

K J

c

5

*

S

* -

E §

on

1 —3
'

*

i

o 3

"fill q

E

3

O -

«-i

00 r;

o

c

-
1

b'

>

X

o^

2

oo

-

g

Cerium

58

Th

140.115

Thorium

90

Os

o

In order to determine the value of the constant C",

hypothesis proposed in 1811 by the Italian physicist

1856) that_a given volume pf any gas rmiiginjji e

(at a fixed

same nnmher nf independent

we can utilize an important Amedeo Avogadro (1776-

temp e rature and pror ur e must .r.

l

partic les. Furthermore, he specified that

16

Chapter

1

The Nature

Chemistry and the Kinetic Theory of Gases

of Physical

atoms or combinations of atoms, coining the word

the particles of gas could be

Avogadro's Hypothesis

molecule for the

V oc where n Definition of the

Mole

is

One may

latter case.

n

V = C"n

or

amount of substance,

the

state

Avogadro's hypothe sis as

(valid at constant

the SI unit for

which

P

is

and T)

(1.26)

the mole.

One_jnflleJsjhe_amo unt of a ny substance_c ontaining the same number-of

me ntary

entitie s (atoms,

ele-

molecules, ions, and so forth) as_lhere_are_in^exactly 0.012

kg of carbo md2_isee also Appendix A). The number of elementary

entities is re-

amount of substance by the Avogadro constant; this is given the symbol L [after Joseph Loschmidt (1821-1895) who first measured its magnitude] and has -1 23 the value 6.022 137 X 10 mol The numerical value of the Avogadro constant, 23 6.022 137 X 10 is the number of elementary entities in a mole. Since Eqs. 1.20, 1.24, and 1.26 show that the volume of a gas depends on T, MP, and /?, respectively, these three expressions may be combined as lated to the

.

,

nRT

lim(PV)

we keep in mind the limitations imposed by may be approximated by

If

Equation of State of an Ideal

(1.27)

the lim P ^

requirement, the equation

PV = nRT

Gas

(1.28)

where R is the universal gas constant. The value of R will depend on the units of measurements of P and V. In terms of the variables involved, R must be of the form

PV _

R =

(force/area)

nT

mol

= Various units of

R

used values of

R

energy

•

K

volume

•

=

K

•

_1 •

mol

force

•

length

is

are required for different purposes,

and some of the most often

R

X

1.987 19 cal

is

one of the most impor-

states is that in

any sample of gas

maintained.

10

R

mor (SI unit) dm 3 K" mol" commonly

in

Various Units

1

1

1

,

7

K

ergK "'

*In SI the volume

expressed

is 1

dm 3

0.082 057

L atm K"

1

'

mol"'*

mol"'

dm 3

defined as

listed as

1

mol

'

0.083 145 10 bar

now

is

Numerical Values of

1.1

8.314 51 JK"'

is

it

allowed to change, the values of the other three variables will always be

0.082 057 atm 8.314 51

What

one of the four variables (amount, pressure, volume, or temper-

ideally, if

such that a constant value of

TABLE

mol"

'

Equation 1.28, the equation of state of an ideal gas,

ature)

K"

are given in Table 1.1.

tant expressions in physical chemistry.

behaving

•

K"

.

1

moP in

1

cubic decimeters (dm

3 ).

The more

familiar unit, the

liter,

The Kinetic-Molecular Theory

1.9

A useful relation can be

of Ideal

17

Gases

obtained by rearranging Eq. 1.28 into the form

m/M RT = V

n

P = —RT = V

p —RT

(1.29)

M

or

RT

M= where

mW

n, the

is

amount of substance, p of

the density

EXAMPLE 1.2 the density of air

Solution

Using Eq.

the

is

mass

divided by the molar mass

At sea

1.29 kg

m~ 3

0°C

if

.

level the pressure

pRT

may

m~ 3 X

1.29 kg

P

be taken equal to

atm or 101 325 Pa.

1

m~ 3 X

8.3145 kg

m

2

101 325 kg

-

0.0289 kg mol

-1

-

K" mol

J

N m"

2

m

s~

2

'

273.15

K

(or Pa)

s~- K~'

28.9 g mol

X

1

8.3145

101 325

1.29 kg

_

mol"

m~

1

X

273.15

K

2

-1

THE KINETIC-MOLECULAR THEORY OF IDEAL GASES An

M, and

the gas.

Calculate the average molar mass of air at sea level and is

4

1.29,

M=

1.9

m

I

experimental study of the behavior of a gas, such as that carried out by Boyle,

cannot determine the nature of the gas or der to understand gases, one could

first

why

the gas obeys particular laws. In or-

propose some hypothesis about the nature

of gases. Such hypotheses are often referred to as constituting a "model" for a gas.

The

properties of the gas that are deduced from this

model

are then

experimental properties of the gas. The validity of the model ity to predict the

We how

compared

reflected in

model leads

fit

to the

its

abil-

behavior of gases.

will simply state three postulates of the kinetic-molecular model,

this

seen to

is

to the ideal gas laws. This

the behavior of

many

and show

model of an idealized gas

will be

real gases.

The gas is assumed to be composed of individual particles (atoms or molecules) whose actual dimensions are small in comparison to the distances be-

1.

tween them.

4

The molar mass

M

r

is

is

mass divided by

the

amount of substance. The

usually called the molecular weight and

is

dimensionless.

relative

molecular mass of a substance

18

Chapter

1

The Nature

of Physical Chemistry

and the Kinetic Theory

of

Gases

motion and therefore have kinetic energy.

2.

These

3.

Neither attractive nor repulsive forces exist between the particles.

particles are in constant

how

In order to see

we must

this

model predicts

the observed behavior quantitatively,

portant characteristics of the gas,

namely

the

mass, and the velocity with which they move.

mass

single molecule of

The

and volume of a gas to the imnumber of particles present, their

arrive at an equation relating the pressure

m

confined

in

We

will first focus our attention

particle traverses the container with velocity

u

(a vector quantity indicating

both speed and direction). Because of the particle's X-component of velocity the

component ux

as

shown

in

on a

an otherwise empty container of volume V.

(i.e.,

Figure 1.7) the molecule will traverse the container

of length x in the X-direction, collide with the wall YZ, and then rebound. In the

Fw on the wall. This force is exactly counteron the molecule by the wall. The force F is equal

impact the molecule exerts a force balanced by the force to the in

change of

F exerted

momentum p

of the molecule in the given direction per unit time,

agreement with Newton's second law of motion,

F=

— = ma = dp

d(mu)

dt

dt

du

m—

(1.30)

dt

The molecule's momentum in the X-direction when it strikes the wall is mu x Since we assume that the collision is perfectly elastic, the molecule bounces off the wall .

with a velocity

— ux

The change

in the opposite direction.

in velocity of the

Au x = [—ux

= — 2u

FIGURE

molecule on each collision

(after collision)]

—

is

[ux (before collision)]

(1.31)

r

Vaxis

1.7

Container showing coordinates

and

velocity

particle of

components

for

gas

mass m.

Molecule colliding

with wall

in

YZ plane (area = A)

Zaxis

Xaxis

u

The Kinetic-Molecular Theory

1.9

The corresponding change of momentum

—2mux

is

collisions a molecule

makes

number of collisions

by

2x.

The

time

in unit

(i.e.,

may be

in unit time

the distance u x the molecule travels in unit time

19

Gases

Since each collision with

.

wall occurs only after the molecule travels a distance of 2x

number of

of Ideal

=

one round

this

trip), the

calculated by dividing

result is

— ux

(1.32)

2x

The change of momentum per

F= The

unit time

(

,

x \

x

Fw exerted on the wall by the particle with opposite sign:

Fw Since the pressure

is

=-F =

Px

,

= the

The Pressure far

we have

mu

x

volume of

of a

(1.34)

x

F — = A

—

mul -

xyz

= V is

^ A

is yz,

we may

write the pres-

as

p =

j>ince xyz

(1.33)

-f

exactly equal in magnitude to

is

force per unit area and the area

sure in the X-direction,

mul

du „ = (-2u m— m)[^j=

force

this but

So

thus

is

Fw yz

mul

= -rf V

d-35)

the container.

Gas Derived from

Kinetic Theory

considered only one molecule that has been assumed to travel

constant velocity. For an assembly of

N

at

molecules there will be a distribution of

if the molecules all began with the same velocity, would occur altering the original velocity. If we define wf as the square of velocity component in the X-direction of molecule i and take the average of this

molecular velocities, since even collisions

the

2

over molecules rather than the sum over u h o

"2

where u x

is

X-direction.

Pressure of a OneDimensional Gas

o

=

we have

i

o

w-i yv

=

n

o

^r

(1

the mean of the squares of the normal component of velocity in The pressure expressed in Eq. 1.35 should therefore be written as

Nmux p _ [

V

-

36)

the

(1.37) \

ThisJ * \\)p pguntinn fnr thp pressure of a one -dimensional gas For the components of velocity in the K-direction and in the Z-direction, we would obtain expressions .

20

Chapter

1

The Nature

of Physical

Chemistry and the Kinetic Theory of Gases

now

similar to Eq. 1.37 but

involving u

2

and u

2

respectively.

,

T he word

terms of the squares of the velocity components.

magnitude of the velocity; speed

|//|

we

average over

2

we

molecules,

all

U Since there

mean of the

is

no reason

;

that

for

2

sum of

is

=

V u\ + u

2

UX

2

+

Uy

2

+

2

and

is

2

uz

(1.38)

as the

means

2

UZ

4*

one direction

the

in

used for th e

obtain

same

u x values will be the

the u\ values. Hence, the

t«

2

speed

the Pythagorean theorem:

Vu =

=

more convenient

defined as the positive square root of u

components by

related to the velocity

If

is

It is

magnitude of the velocity u rather than

to write these expressions in terms of the

to

(1.39)

be favored over the others, the

mean of the uy 2

values and the

equal to u and each

is

mean

is

mean of equal to

is,

Ux

=

=

Uy

UZ

=

..2

1

(1.40)

Substituting Eq. 1.40 into Eq. 1.37 gives the final expression for the pressure on

any wall:

Pressure of a Gas of

in Terms Mean-Square Speed

Nmu' P =

PV = -Nmu 2

or

3V

This

is

the fundamental equation as derived

We

see that Eq. 1.41

law

if

mu"

is

is

in the

(1.41)

from the simple

form of Boyle's law and

is

kinetic theory of gases.

consistent with Charles's

directly proportional to the absolute temperature. In order to

relation exact,

we

use Eq. 1.28 and substitute

nRT for PVin

make

this

Eq. 1.41:

nRT = \Nmu 2

(1.42)

or 1

T

-^Lmu

2

„T = RT

2

— jRT

equal to the Avogadro constant L, and

since N/n

is

Kinetic

Energy and Temperature

We

u

or

have seen

how

mL = M,

the kinetic molecular theory of gases

experimental form of two gas laws.

It

(1.43)

M the molar mass.

may be used

to explain the

can also shed lighten the nature of kinetic

energy. In order to determine the exact relation between u variable u of Eq. 1.41 must be related to the temperature, variable.

and 7, the mechanical which is not a mechanical

For our purpose of determining the relation between kinetic energy and

temperature, Eq. 1.41 that the average kinetic

may be

converted into another useful form by recognizing

energy e k per molecule 1

-jinu

2

is

(1.44)

The Kinetic-Molecular Theory

1.9

of Ideal

Gases

21

Substitution of this expression into Eq. 1.41 gives

PV = ±N-2ek = %N6k

(1.45)

at constant pressure., t h e volume of a gas is proportion al t o the numb erof gas molec ules apd the a verage Unptir pnprgy of the molecules. S ince

Thus,

substitution into Eq.

Since Le k

is

1

N=nL

(1.46)

PV = \nLlk

(1.47)

.45 yields

the total kinetic energy

Ek per mole of gas,

PV = ^nEk The connection between Ek and ideal gas law,

we

PV =

the temperature

is

provided by the empirical

nRT. By equating the right-hand sides of the

last

two equations

obtain

\nEk = nRT

Ek = jRT

or

The average kinetic energy per molecule the Avogadro constant L:

£k

where k B

Boltzmann Constant

(1.48)

=

R/L.

Named

is

obtained by dividing both sides by

2^bT

(1.50)

ek

after the Austrian physicist 5

(1844-1906), the Boltzmann constant k B jhf^jwerage. kjnetic energ y nf the molecules ture

.

Since e k in Eq. 1.50

all at

the

is

Ludwig Edward Boltzmann Thus

the gas constant per molecule.

prop o rtional to the absolute tempera -

independent of the kind of substance, the average

is

all

substances

interesting aspect of this fact

ent gases

is

same at a fixed temperature. when we consider a number of differsame temperature and pressure. Then Eq. 1.41 may be written as

molecular kinetic energy of

An

(1.49)

N^n^Uy

is

Nm 2

is

the

seen

2 U2

Njtnjiij

(1.51)

3V?

3 V,

3V,

or

N2 (1.52)

3Vy

5 /t

B

=

1.380 622

X

10"

IK" ] 1

3V2

3V,

22

Chapter

1

The Nature

Chemistry and the Kinetic Theory of Gases

of Physical

Thus yv, gases is

= N2 =

at the

•

= Nj when the volumes are equal. In other words, equal volumes of

•

same pressure and temperature contain equal numbers of molecules. This

just a statement of Avogadro's hypothesis already seen in Eq.

Law

Dalton's The

1

of Partial Pressures

studies of the English chemist John Dalton (1766-1844)

the total pressure observed for a mixture of gases

virp^jhat pqfh individual container

at the

.26.

mmpnnp nt

pas would exert had

same temperature. This

is

known

as,

showed in 1801 that sum of the p res-

equal to the

is

it

alone occupied the

Dalton's law of partial pres-

Of course, in order for it to be obeyedi no chemical reactions between component gases may occur, and the component gases must behave ideally?) The term partial pressure is used to express t he pressure exerted by one co mponent of the gas mixtu re. Thus sures.

Partial

Pressure

=

P,

where

P,

is

+

Px

n

x

RT -

+

P,

n^RT

(1.53) 1

n

-+••• +

-

V

:

.28,

we may

write

RT

V

+

(«]

+

form of Eq.

the total pressure. Then, using a

P,=

where the

+

Pi

n2

P,'s are the partial pressures

+

+

and the

RT «,-)

(1.54)

V /2,'s

ual gases. Dalton's law can then be predicted

are the

amounts of the individ-

by the simple kinetic molecular

theory (Eq. 1.41) by writing expressions of the form

P

yV,m,w,/3V for each gas.

t

Thus

Pt

=

£ Pi =

Nm

NimiU]

2

3V

2u2

3V

+

+

NinijUj

(1.55)

3V

Dalton's law immediately follows from the kinetic theory of gases since the aver-

age kinetic energy from Eq.

1

.49

is

^RT

and

is

the

same

for all gases at a fixed

temperature.

Application of Dalton's law

The pressure

of

^water in Torr, is

equal to

in

temperature.

water vapor,

approximately

°C near room

is

particularly useful

subsequently collected over water. The

total

when

a gas

the water vapor present in addition to the pressure of the gas that

vapor pressure of water

is in

the order of

is

generated and

gas pressure consists of the pressure of is

generated.

The

20 Torr near room temperature and can be

a significant correction to the total pressure of the gas.

Graham's Law

of Effusion

Another confirmation of the kinetic theory of gases was provided by the work of the Scottish physical chemist

movemen plates

t

Thomas Graham (1805-1869). He measured

the

of gases through plaster of Paris plugs, fine tubes, and small orifices in

where the passages~for the gas are small as compared with the average

dis-

The Kinetic-Molecular Theory

1.9

of Ideal

23

Gases

ance that the gas m olecules travel between collisio ns (see the following section on Molecular Collisions). Such movement is known as effusion In 1831 Graham showed that the _gz£f of fffr j *' n " nf a g aressurp Henraasp.s

wjth an

to the lejigtjj_gjjhe_colu mn,

and since

increase in heig ht.

general, the density depends on the pressure. For liquids, however, the density

tin is

practically independent of pressure, that

of the liquid

is

is,

the compressibility,

k(— — 1/VdV/dP),

very small compared to that of a gas. For liquids, Eq. 1.72 can be inte-

grated at once:

f

where

P

height

z-

is

dP =

P-P

the reference pressure at the base of the

This quantity

P—P

is

dz

-pg

(1.73)

column and P

is

the pressure at

the familiar hydrostatic pressure in liquids.

In a vessel of usual laboratory size, the effect of gravity on the pressure of a

gas is

a

is

negligibly small.

marked

On

a larger scale, however, such as in our atmosphere, there

variation in pressure,

on the density of the

gas.

and we must now consider the

For an ideal gas, from Eq.

1.29,

p

is

effect of pressure

equal to

PM/RT, and

substitution into Eq. 1.72 gives

FIGURE

1.9

Distribution of

a gas

in

a gravity

P

field.

(1.74)

\RTj

Integration of this expression, with the boundary condition that

P = P when

z

=

0,

gives

P

Mgz

In

(1.75)

RT

or

P=P

^ M8Z/KT

(1.76)

This expression describes the distribution of gas mnlp^njpsjn the

at mosphere as

a

fu nction of the ir molar mass, height, temperature, and the a cceleration due togra^v-

Barometric Distribution

Law

the

ity.Jt,

The

barometric distribution law.

distribution function in Eq. 1.74,

formative

when

the sign

is

p Here —dP/P represents a tiplied

by the

which

transposed to the

dP

is in

differential form, is

more

in-

term:

Mgdz RT

(1.77)

relative decrease in pressure;

differential increase in height. This

it is

means

a constant, that

it

Mg/RT, mul-

does not matter

1.11

The Maxwell

Distribution of Molecular

Speeds and Translational Energies

29

where the origin

is chosen; the function will decrease the same amount over each inrrpmpnt of heigh^

-

The that, for

fact that the relative decrease in pressure is proportional to

a given gas, a smaller relative pressure change

expected

is

at

Mg/RT shows

)

high tempera-/

tures than at low temperatures. In a similar manner, at a given temperature agasA ha^ingahigher molar mass s expected to have a larger relative decrease in j>res-\ i

surf fharLq

p-aV u/ith a

Equation

1

nents in a gas. the_e arth's

Inwpr rnnlaj

'

rP r"" 5

.76 applies equally to the partial pressures of the individual Tt

follows from thp previous trpatmpnt that in

j.tmosphprp

compo-

npppf re aches o f

trip

th^ partial prpssnrp will h e relatively higher for a very light

gas such as helium This fact explains .

why helium must be

extracted from a few

helium-producing natural-gas wells in the United States and lium occurs underground.

It is

in

Russia where he-

impractical to extract helium from the air since

tends to concentrate in the upper atmosphere.

The

it

distribution function accounts

satisfactorily for the gross details of the atmosphere, although

winds and tempera-

ture variations lead to nonequilibrium conditions.

Since

Mgz

is

the gravitational potential energy,

P= P Since the density p

is

,

Eq.

=

Pot-

1

.76 can be written as

b '/Kl

(1.78)

directly proportional to the pressure,

p

Boltzmann Distribution Law

c'

Ep

we

also

may

write

Er /RT

(1.79)

where p represents the density at the reference state height of z — 0. These equations, in which the property varies ex pnnpntially with —F.p /RT, are s pecial cases of the Boltzma nn distri bution law We deal with this law in more detail in

Chapter

15,

where we

will encounter a

tions in physical chemistry. Before

we

number of

proceed, note that

its

important applica-

we have

only considered

the average velocity of gas particles without investigating the range of their veloci-

To treat this problem we will now deal with another special case of Boltzmann distribution law, the Maxwell distribution of molecular speeds. ties.

1.11

THE MAXWELL DISTRIBUTION OF MOLECULAR SPEEDS AND TRANSLATIONAL ENERGIES

^HH

Maxwell's famous equation for the distribution of molecular speeds

oped

in the early 1860s, inspired

that dealt with energies of

will derive

The

any kind.

Boltzmann's equation

Distribution of

Boltzmann

We

later in

will

to

in gases, devel-

produce a more general equation

now

Chapter

the

obtain Maxwell's equation, and

15, Section 15.2.

Speeds

In his derivation of the distribution of molecular speeds,

Maxwell paid no

attention

to the mechanics of collisions between molecules. Instead he based his arguments on

probability theory, such as the theory of errors that had been given by Pierre-Simon

30

Chapter

1

The Nature

Chemistry and the Kinetic Theory of Gases

of Physical

Laplace (1749-1827). important

result,

The reason

It is

interesting to note that although

Maxwell had obtained an

he himself was not convinced that his work was of real significance.

was

for his doubt

he had not been able to explain the specific heats of

that

we now

gases on the basis of his kinetic theory;

realize that those results require the

quantum theory for their explanation. The treatment we will describe is essentially that given in 1860 by Maxwell. The speed u of a molecule can be resolved into its three components along the X, Y, and Z axes. If ux is the component along the X axis, the corresponding kinetic 2 energy is -^mu x where m is the mass of the molecule. We can then ask: what is the probability dPx that the molecule has a speed component along the X axis between ux and ux + dup. We must, of course, consider a range of speeds, since otherwise the probability is zero. This probability is proportional to the range du x and _/3< Maxwell deduced on the basis of probability theory that it is proportional to e \ : ^ /2 where /3 is a constant. We can therefore write or e~'"" ,

,

,

dPx = Bt-' nu^ n dux where

B

is

We

the proportionality constant.

(1.80)

also have similar expressions for the

other components:

dP y = Bc-"m The product of these

>

m duy

three expressions

dPx dP y dPz = 5 Since u

2

=

2

ux

+

u

2

+

u\,

3

-m„^/2

e

where u

is

e

+ dux

are really interested in

,

is

uy and u y

-,™^/2

^ ^ .

(1.82)

the speed, mttlfi/2

du r du„ du.

+ du y and ,

(1.83)

components of speed have values and u z + du z However, what we

u.

.

the probability that the actual speed of the molecule lies

+ du, and this can be achieved by a variety of combinations of the components of speed. We construct a speed diagram, shown in Figure 1.10, with the components ux uy and u z as the three axes. We then construct a spherical between u and u

Volume 4ku 2 du

three

,

shell,

,

they

What we want is the probability dPx dPy dP. has only given us the

of radius u and thickness du.

within the shell, but the product

shown

within the small cube

lie

du y du z and ,

FIGURE 1.10 A diagram with axes

£

the probability that the three

is

between ux and u x

(1.81)

is

-„,„;/3/2

dP r dP„dP=B3 e~ This expression

dP z = Be" m"^/2 du z

and

the

probability

dP

dux du y du.

in Eq.

volume of

that the 1

.83

in the

lies

systems

this

cube

,

Equation 1.33 gives the probability speed is represented by the cube of volume du„ duy duz We are interested in the speed corresponding to the volume 2 Attu du that lies between the that the

.

concentric spherical surfaces.

is

dux

Attu" du. In order to obtain the

is

+

between u and u

du,

we must

therefore replace

2

by Attu du:

dP = ATTB\~'m^ n u 2 du u x u y and uz

lie

probability that

diagram. The volume of

the spherical shell

speed

that

(1.84)

.

The is

fraction

dN/N of

given by the ratio of

since that

is

the

this

N molecules

expression to

the range of possible speeds.

dN

that

its

have speeds between u and u

value integrated from u

=

to u

+ du = °°,

Thus

4ttB\-'"

u2p/2

2

u du (1.85)

N

3

4-n-fl

f

-mu-fil2

u"

du

— The Maxwell

1.11

Distribution of Molecular

Speeds and Translational Energies

31

The integral in the denominator is a standard one, 6 and when we evaluate it, we obtain (1.86)

N We now

\2tt

)

consider what the mean-square speed

The point of doing

we already know we have

this is that

from Eq. 1.43

theory. Thus,

is

on the basis of

this treatment.

the mean-square speed

from

kinetic

3RT (1.87)

Lm 4

8

12

20

16

u/ 10 2 ms-

24

This quantity

1

is

obtained from Eq. 1.86 by multiplying u~ by the fraction

dN/N and

integrating:

FIGURE

—

1.11

The Maxwell

distribution of

molec-

speeds for oxygen at 300 K and 1500 K. Note that the areas below the two curves are identical

=

u

since the total

cules

is

number

of

mole-

ndN

f°°

u

]

The

integral

evaluating

is

it

-mu MB/2 4 u du 2

i

e

477i

N

JQ

ular

(1.88)

'Qn

1277/

-'

again a standard one (see the appendix to this chapter,

p. 44),

and

leads to

fixed.

"2 u

Comparison of the expressions

=

— 3

(1.89)

mp

in Eqs. 1.87

and 1.89 then gives the important

result

that

Boltzmann Constant

H where R/L has been written 1.85

may

as k B

N

This

is

which

is

(1.90)

kB T called the

Boltzmann constant. Equation

thus be written as

dN = Maxwell Distribution Law

,

RT

usually

known

.

I

m

3/2

47T \ 27rk B

as the

„ —nur/2k Ba T

e

T

Maxwell distribution

u

2

j

du

(1.91)

law.

Figure 1.11 shows a plot of (l/N)(dN/du) for oxygen gas

at

two temperatures.

Near the origin the curves are parabolic because of the dominance of the u term in the equation, but at higher speeds the exponential term is more important. Note that the curve for

300

becomes much

K are

flatter as the

raised. Indicated

on the curve

the average speed w,

and the most

temperature

the root-mean-square speed

vu

2 ,

is

probable speed u mp The mean-square speed is given in Eq. 1.89, and insertion of l/k B T for j8 gives the expression in Table 1.3; the root-mean-square speed is its .

square root.

The mean speed or average speed

is

given by

dN

r6

The appendix

problems.

to this chapter (p. 44) gives a

N

number of

(1.92)

the integrals that are useful in distribution

32

Chapter

1

The Nature

of Physical

TABLE

Chemistry and the Kinetic Theory of Gases

Quantities Relating to the Maxwell Distribution

1.3

of Molecular

Speeds

~ u =

Mean-square speed

3k » T

Square root of the mean-square speed

Most probable speed

is

the speed at the

resulting expression for this quantity

The We

2k » T

to the expression in Table 1.3.

maximum

entiating (\/N)(dN/du) with respect to u

l

= —k B T

e

and integration leads

Insertion of Eq. 1.91

/

=

u mp

Average translational energy

probable speed

Sk* T

- = u

Average speed

of the curve and

and setting the

is

result equal to zero.

The

also given in Table 1.3.

is

Distribution of Translational

Energy

can convert Eq. 1.91 into an equation for the energy distribution.

e

The most

obtained by differ-

We

start

= \mu 2

with

(1.93)

and differentiation gives m\

du

\

We can therefore replace du in Eq.

—m

/?e\ U2

de

^- = mu =

1.91

\

=

{2em)

m

by de/(2em)

U2

and u~ by

^ = ^[^-T *-^ —^ T

(TTk B T)

Maxwell-Boltzmann Distrubution

This

Law

is

often

An

known

as the

me

(1.94)

J

e

2e/ra,

and we obtain

de

(1.95)

de

(1.96)

Maxwell-Boltzmann distribution

expression for the average translational energy

is

law.

obtained by evaluating

the integral

e=\\^J

The

result is -^k B T,

o

which we already found,

(1.97)

N

in another

way, in Section

1

.9.

J

1.12

33

Real Gases

REAL GASES

1.12

The Compression Factor The gas laws

that

we have

treated in the preceding sections hold fairly well for

most gases over a limited range of pressures and temperatures. However, when the range and the accuracy of experimental measurements were extended and improved, real gases were found to deviate from the expected behavior of an ideal

PVm product does not have the same value for all gases nor is dependence the same for different gases. (V,„ represents the molar volume, the volume occupied by 1 mol of gas.) Figure 1.12 shows the deviations of N 2 For instance, the

gas.

the pressure

and Ar from the expected behavior of an ideal gas under particular isothermal (constant temperature) conditions. It is difficult, however, to determine the relative deviation from ideal conditions from a graph of this sort or even from a

P

against

A more

convenient technique o ften used to show the devi ation from ide al b ehavior involv e s the llgp "f g ra ph? or t abl es r>f t h e c ^rapr? «sion ( o r c o pirssihility) facto r Z. defined b v

\IV

plot.

m

V/dm

FIGURE

nRT

3

1.12

volume for nitrogen and argon at 300 K. Nitrogen follows the ideal gas law very closely, but argon shows Plots of pressure against

significant deviations.

¥or the

id pal gas,

Z=

1

Therefore, departures from the value of unity indicate

nonideal behavior. Since each gas will have different interactions between

molecules, the behavior of Z can be expected to be quite varied. In Figure

of Z versus

P for several initial

slop es tial

(Z>

among

the curves for

all

Z>

1

Irish physical

chemist

(C0 2 can be

Z=

stant.

P/10

FIGURE

80

5

100

Pa

To

his Surpris e

at 31.1 °C. state.

\W

at

about

1,

at

compression

factor, Z,

against pressure for several gases at

273

K.

Z=

T he ini-

0.25. All gases at sufficiently

P —> 0,

studied the behavior of

varying temperatures. Using a sample of liquid

room temperature using

sufficiently high

temperature while maintaining the pressure conhnn nHary hetween the gat and liquid regions H kappe areH its

Further increase in pressure could not bring about a return to the liquid

This experiment led Andrews to suggest that a ajticaLtem per a tu reJZ^ex-

each gas. Above thisJemperat ure, pressure alone could not liquefy he gas. Jnnther wnrrfc T is the hi g hest tem per a ture at which a liquid can exist as a d istinct phag e nr region He introduced the word vapor for the gas that is in equilibrium ,

Plot of the

can be related to

although with different slopes.

isted for

1.13

1)

Critical Point

liquefied at

pressure), he gradually raised

60

some of these curves (Z
-(b + ZfjvZ +

I

At Tc the volume has three

j;Vm

-f =

real roots that are all identical. This

(V,„-Vc

3

\

may be

(1.103)

expressed as

=

)

(1.104)

or

V*

- 3VC V 2 + 3VC2 V„, ~ K 3 =

Since Eqs. 1.103 and 1.105 describe the same condition

by

Pc and Tc

in Eq. 1.103,

we may

3Vc = b +

and

finally,

in

Vm

when P and Tare

equate coefficients of like powers of

replaced

Vm From .

we have

the coefficients of V~

From terms

(1.105)

RT —

(1.106)

,

3VC2 =

—

K3 =

—p

(1.107)

from the constant terms,

1

we

these last three equations,

a

/

= 3PC V 2 C

(1.108)

c

obtain

&

,

SPC VC R=

= y,

(1.109)

3TC

Although the van der Waals constants may be evaluated from these equations, the

method of choice

is

to

Alternatively, the 1

.99.

d etermine a and b empirically from experimental

same

results

may be determined

PVT data

.

using the expressions in Eq.

Application of the mathematical conditions in these equations to the van der

Waals equation

will eventually yield Eq. 1.109.

See Problem 1.57 for a similar ap-

plication. If the

expressions obtained in Eq. 1.109 are inserted into the van der Waals

equation for

1

mol of gas, we obtain

L+ P..

iYl\(^-L\ = 3L Vl

\V,

3

37\

(U10)

40

Chapter

1

The Nature

and the Kinetic Theory

of Physical Chemistry

It is

convenient

Vn and Tn

Gases

of

each of the ratios P/Pc V/Vc and T/Tc by

at this point to replace

,

,

respectively; these represent the reduced_2ressurs_ P r reduced

volume

,

a nd red uce d temperature

Tr and

Pn Vn

are dimensionless variables. Equation 1.110 then

takes the form

/

r It is

thus seen that

all

+

*\ /

?)

r

\

8

-?)

3

i

(1.111)

r

gases obey the same equation of state within the accuracy of

r the van der Waals relation when there are no arbitrary constants specific Law

of

Corresponding

States

to the indi

law of corresponding states As an illustration, two_gases having the same reduced temperature andjgduced pressure- are n co rresponding-Slates and shouULQceupy the s ame_reduced vol ume. 5 Thus, if 1 mol of helium is held at 3.43 X 10 kPa and 15.75 K, and 1 mol of car5 bon dioxide is held at 1 10.95 X 10 kPa and 912 K, they are in corresponding states (in both cases, P/P c = 1.5 and T/Tc = 3) and hence should occupy the same reduced volumeXThe law's usefulness lies particularly in engineering where its vidual gas. This

is

a statement of the

i

range of validity

is

sufficient for

experi mental behavior is

is

nicely

many

T he

applications.

shown

in

Figure

1

.

1

ability

of the law to pre dict

where the red uced pressure

ft T

plotted against the compression factor for 10 different gases at various reduced

temperatures

"1

FIGURE

1.18

Compression

factor versus

reduced pressure

for ten

gases.

(Reprinted with permission from Industrial istry,

and Engineering Chem-

Vol. 38. Copyright

American Chemical

1946

Society.)

x Methane o Ethylene

5 © 6 © o

A Ethane © Propane n-Butane

Iso-Pentane

n-Heptane Nitrogen

Carbon Dioxide Water

Average curve based on data on hydrocarbons 2.5

3.0

3.5

4.0

I

I

I

4.5

5.0

5.5

Reduced pressure, Pr

I

7.0

1.14

The

Virial

Equation

41

Other Equations of State There are two other major equations of

common

state in

use.

P.

A. Daniel Berthelot

(1865-1927) developed the equation

P +

Berthelot Equation

which

is

the van der

attractive term.

It

~

Waals equation modified

may also be

slightly

it

provides high accuracy

at

The

dependence of the

terms of the reduced variables as

27 (1.113)

low pressures and temperatures.

state is the

one introduced

in

1899 by C. Dieterici.

Dieterici equation involves the transcendental number, e (the base of the nat-

ural logarithms);

however,

near the critical point.

Dieterici

(1.112)

647/

1287;

V,„

Another major equation of

in

+

1

= RT

for the temperature

modified

RT

where

B)

vItF"

It

gives a better representation than the other expressions

it

may

Equation

be written as

(Pe

where a and b

a/VJK-

)(Vm

-b) = RT

(1.114)

are constants not necessarily equal to the

van der Waals constants. In

reduced form, Eq. 1.114 becomes

P = r

2Vr -

exp

I

2

(1.115)

Tr Vr

1

Several other equations have been proposed. In 1949 Otto Redlich and Joseph

N.

Redlich and Equation

S.

Kwong

introduced the equation

Kwong

n a

P + T

U2

V(V + nb)

(V-

bn)

= nRT

(1.116)

which provides a simple and accurate two-parameter expression applicable over a wide range of temperature and pressures.

Another expression that has gained some popularity is the Benedict-WebbRubin equation of state, which relates pressure, molar density, and temperature. It has been used to predict fairly accurately the thermodynamic properties of complex hydrocarbon systems.

1.14

THE VIRIAL EQUATION The advantage of the equations discussed kept to a is

minimum and

to use a large

in the last sections is that the

relate to theoretically defined parameters.

number of constants

to

fit

constants are

Another technique

the behavior of a gas almost exactly, but the

42

Chapter

1

The Nature

Chemistry and the Kinetic Theory of Gases

of Physical

resulting equation

is

then less practical for general use and particularly for thermody-

namic applications. Furthermore, as the number of constants

more H. Kamerlingh

Onnes

the Nobel Prize 1

in

received

physics

91 3 for liquefying helium

and

for the

in (1

908)

discovery of

becomes

The Dutch physicist Heike Kamerlingh Onnes (1853-1926) suggested in 1901 power series, called a virial equation, be used to account for the deviations linearity shown by real gases. The general form of the power series for Z as a from that a

P

is

PVm RT 1

Virial

it

expressions are of such general usefulness that they are discussed here.

function of

superconductivity.

increased,

is

with physical parameters. However, two such

difficult to correlate the constants

Equation

Z(P, T)

However,

this

r

1

+

B'(T)P

+ C(T)P 2 + D'(T)P 3 +

does not represent the data as well as a series in

(1.117)

where the odd

1/V„,

powers greater than unity are omitted. Thus the form of the equation of gases presented by Kamerlingh Onnes

PV =

B(T)n

+

1

third,

coefficients B'(T),

and fourth

the right

is

C(T)n

+

2

D(T)n

(1.118)

|

C(T), D'(T) and B{T), C(T), D(T) are called the second, and the notation indicates that they

When Eq.

R; sometimes, therefore,

1

.

1 1

8

is

multiplied through by R, the

R is called theirs?

tures the coefficients are functions of both temperature

how this is done, Eq.

1

.

8

1 1

is

virial coefficient.

term

For mix-

and composition, and they are

rewritten in terms of the molar density p,„

=

1

first

PVT data by a graphical procedure. To illus-

found experimentally from low-pressure trate

4

|

virial coefficients, respectively,

are functions of temperature.

on

of real

V

nRT where the

state

is

B(T)

+

= n/V.

C(T)p„

(1.119)

P,„RT

The 180 270 360 450 540 7/ K

FIGURE

1.19 Plot of the second virial coefficient, B{T), against T for several

gases.

left-hand side of Eq.

value of

B

at the intercept

Another way of as

P —>

plotted against

is

—

0.

p =

C

B on

T,

yielding a

=

C

stants A, b,

B(T)

is

made showing

Evaluate the Boyle temperature in terms of the

1.5

and

R

the dependence

B(T) =

0.

(b

T = TB b

con-

A/RT m )P

The Boyle temperature occurs when This condition

condition

known

for a gas having the equation of state

PVm = RT+ Solution

0.

for that particular temperature. In Figure

temperature for several gases.

EXAMPLE

this

for fixed

becomes zero derivatives.) The slope of

for a discussion of partial

gives the value of

1.19 a plot of the second virial coefficient

of

p

At the Boyle temperature, B(T)

stating this is that the partial derivative [d(PV)ldP] T

(See Appendix

0.

the curve at

1.119

where p

.

is fulfilled

Solving for

= A/RTB X

,

when

the second virial coefficient,

the quantity (b

A/RT-") =

TB

Tf =

A/bR;

TB = (A/bR) 3

0.

Under

The

1.14

The importance of ods of

statistical

Virial

the virial coefficients lies in the fact that, through the meth-

mechanics, an equation of

of a real gas

state

may be developed

The empirically derived coefficients can thus be related to counterparts, which (it turns out) are the intermolecular potential

the virial form. oretical

In this interpretation, the second virial coefficients, for instance, are lar pair interactions; the other coefficients are

The

equation

virial

43

Equation

is

due

due

in

their the-

to

energies.

molecu-

to higher order interactions.

not particularly useful at high pressures or near the

criti-

because the power series does not rapidly converge under conditions of higher order interactions. Furthermore, if one is to proceed on a theoretical basis cal point

rather than an empirical one, the calculation of the constants

chanics

is

difficult

evaluation of the multiple integrals involved

The

final

from

statistical

me-

because the potential functions are not well known and the is

very

expression to be considered here

difficult.

1927—

the equation proposed in

is

1928 by the American chemists James Alexander Beattie and Oscar C. Bridgeman. 7

RT[l P =

Beattie-Bridgeman Equation

(c/Vm T

3 )]

+

(V„

B)

(1.120)

~i

where

A = An

B=

1

B, v,„

and

a, b,

man

A

,

B

,

and c are empirically determined constants. The Beattie-Bridge-

R and

equation uses five constants in addition to

work, especially

Table

in the high-pressure range.

1

.6

is

well suited for precise

gives the Beattie-Bridgeman

8

constants for 10 gases.

TABLE

1.6

Constants for Use

Gas Pa

A> mol

m6

the Beattie- Bridgeman Equation with

in

a 2

10

6

m3

R

--

=

8.3145 J

"1

10

6

m3

1

mol

b

So

mol

K

mol

-1

10~ 6

m3

1

c

mol

1

10m 3 K 3

mol

He Ne

0.00219

59.84

14.00

0.0

0.0040

0.02153

21.96

20.60

0.0

0.101

Ar

0.13078

23.28

39.31

0.0

H2

0.02001

-5.06

20.96

-43.59

N o

2

0.1362

26.17

50.46

-6.91

2

0.1511

25.62

46.24

Air

0.13184

19.31

46.11

-11.01

4.34

CO,

0.50728

71.32

104.76

72.35

66.00

CH 4 (C H

0.23071

18.55

55.87

-15.87

12.83

124.26

454.46

119.54

33.33

2

5 )2

3.1692

7

Am. Chem. Soc,

4.208

A. Beattie and O. C. Bridgeman,

J.

A. Beattie and O. C. Bridgeman, Proc. Am. Acad. Arts ScL, 63, 229(1928).

J.

4.20

4.80

49, 1665(1927); 50, 3133, 3151(1928).

J. 8

5.99

0.0504

1

44

Chapter

APPENDIX:

1

The Nature

SOME

of Physical

Chemistry and the Kinetic Theory of Gases

DEFINITE INTEGRALS

OFTEN USED

IN

PHYSICAL CHEMISTRY

r^-Kif 2a

-Vi

r t~

axi

Vjr} m

2

x dx

•VI

r e~ &

ax2

3

x dx

=

1

2a

-^x4

e~ f o

ax

l

dx=^ dx

=

J

1/2

1

a

.-,»*/;

""

'

2a

\

a

= —z 1

e

jc

dx

a

r\~ ax x dx= i

2

in

KEY EQUATIONS Equation of state of an ideal gas:

Definition of kinetic energy:

PV = nRT

Ek = —mu

Pressure Potential energy for a body obeying Hooke's law:

of a gas

derived

from the kinetic -molecular

theory:

~2

EP =

Nmu P = 3V

-^k,x-

2

where

Boyle 's law:

P«

— V

or

PV =

u~

is

the

mean-square speed.

Relation of kinetic energy to temperature: constant

(at

constant T)

Gay-Lussac's (Charles's) law:

V

oc

T

or

—V = constant

where k B (at

constant P)

is

the

Boltzmann constant.

)

45

Problems

Mean

Dalton 's law of partial pressures: P,

=

(«,

+n +

RT

+ ,

2

«,)

's

Mean-square

/p(gas 2 ) _

rate (gas 2 ) is

V 3M

law of effusion

rate (gas.)

where p

%RT

V

'

Graham

VP

(g as

the density and

1

/M(gas 2 )

—

M (gas

u

V

)

M

molar mass.

the

is

s

„

Collision density (SI unit:

m~ 3

):

2V

+

TrB )

y2

PV,„

RT

Van der Waals equation:

and

2

PV nRT

_1 s

\f2TTd\uiN\

2

M

Compression factor:

2V

2

3RT

p = Poe -WT

N

ird AB (u A

velocity:

Barometric distribution law:

):

\flTrd A uA A A

=

i

_1

Collision frequency (SI unit: 2

velocity:

P+

NA NB

an —)(V

= nRT

nb)

where a and b are the van der Waals constants.

Mean free path:

V

A =

2

N

V2-rrd A A

PROBLEMS 9 (Problems marked with an asterisk are more demanding.)

Classical Mechanics Thermal Equilibrium

required for a 1600-kg car (typical of a Ford Taurus) under

same

1.2.

that a rod of

copper

is

used to determine the

temperature of some system. The rod's length

What

is

0.

10

X

at the

the

temperature of the system temperature

expansion of copper

is

of

the

at

it

system?

0°C

is

is

27.5

28.1 cm.

Atoms can

transfer kinetic energy in a collision. If an

X

1

,

10~ 24 g and travels with a

what

is

the

maximum

kinetic

head-on elastic collision 10" 23 g? 1 X

to the stationary

in a

atom of mass

The

linear

1.4. Power is defined as the rate at which work is done. _1 The unit of power is the watt (W = 1 J s ). What is the power that a man can expend if all his food consumption of 8000 kJ a day (~ 2000 kcal) is his only source of energy and it is used entirely for work?

given by an equation of the form

problems in this book, temperatures and other quantities given 3 as whole numbers (e.g., 25°C, 300 K, 2 g, 5 dm ) may be assumed to be exact to two decimal places. Other quantities are to be considered exact to the number of decimal places specified. all

,

energy that can be transferred from the moving atom

1.5.

In

K

,

atom has a mass of velocity of 500 m s

or 9

+ at + )3r) where a = 0.160 X 10" 4 B = 10" 7 K" 2 / is the length at 0°C, and /, is the length

(1

conditions.

Assume

cm, and

/

1.3.

work required to accelerate a 1000-kg car (typical of a Honda Civic) to 88 km hr"' (55 -1 miles hr ). Compare this value to the amount of work the

=

at t°C.

and

Calculate the amount of

1.1.

/,

State

whether the following properties are intensive

extensive:

(a)

mass;

(d) gravitational field.

(b)

density;

(c)

temperature;

46

Chapter

The Nature

1

of Physical

Chemistry and the Kinetic Theory of Gases

Gas Laws and Temperature The mercury

1.6.

tube

level

1.4a

Fig.

in

containing bulb. The

arm

cm

34.71

is

arm

left

arm of

the left

in

attached

is

cm

10.83

is

the J-shaped

thermostatted

a

to

gas-

changes

Assume

temperature-induced

that

reading of the barometer and

in the

J

mercury spills over the on the trapped air?

the

right

above the bottom of the manometer. If what is the

of the gas?

pressure

When

funnel into the open end.

and the

barometric pressure reads 738.4 Torr,

the

A J-shaped tube is filled with air at 760 Torr and 22°C. The long arm is closed off at the top and is 100.0 cm long; the short arm is 40.00 cm high. Mercury is poured through a

*1.14.

tube are

top of the short arm, what

Let h be the length of mercury in the long arm.

A Dumas

1.15.

conducted

Vacuum technology has become increasingly more many scientific and industrial applications. The unit torr, defined as 1/760 atm, is commonly used in

experiment to determine molar mass is which a gas sample's P, 6, and V are 3 a 1.08-g sample is held in 0.250 dm at 303 K

in

determined.

small enough to neglect.

the pressure

is

If

and 101.3 kPa:

1.7.

important in

a.

mmHg

Find the relation between the older unit

and 3

The density of mercury is 13.5951 g cm 0.0°C. The standard acceleration of gravity is defined the torr.

m

9.806 65

s"

b. Calculate

m

in 1.00

3

the

X 10~ 6

Torr and at 1.00

vacuum

X

10" 15 Torr

obtainable).

gravitational acceleration

1.9.

1

atm

Dibutyl phthalate

is

cm

3

is

is

9.806 65

m

s

",

calculate the

in kPa.

1.047 g

often used as a

What

is

manometer

the relationship

1.10.

The volume of a vacuum manifold used

gases

is

calibrated using Boyle's law.

at a pressure

pumpdown,

of 697 Torr

the manifold

between the manifold and

is

is

at

A

isothermal conditions, what

is

3

flask

opened and the system

the

volume of

the mani-

3

volume of 0.300 dm at a pressure of 1.80 X 10 Pa. What is the new volume of the gas maintained at the same temperature if the pressure is 5

X

5

10 Pa?

-3

mol dm of an ideal 101.325 kPa (1 atm), and (b)

1.13. Calculate the concentration in at

X

298.15 K and at (a) 10" 4 Pa (= 10~ 9 atm). In each case, determine the

number of molecules

in 1.00

2 is

collected over water at a

What

1.19.

are the

mole

L

and

fractions

partial pressures

of

container into which 100.00 g of

is

the total pressure?

The decomposition of KC10 3 produces 27.8 cm of The vapor pressure of

2

collected over water at 27.5°C.

water

at this

temperature

is

27.5 Torr. If the barometer reads

751.4 Torr, find the volume the dry gas would occupy

25.0°Cand 1.00 1.21. Balloons

at

bar.

now

are used to

move huge

trees

from

their

cutting place on mountain slopes to conventional transportation. Calculate the

volume of

a balloon needed

if it is

1000 kg when the temperature is 290 K at 0.940 atm. The balloon is to be filled with helium. Assume that air is 80 mol % N 2 and 20 mol % Ignore the mass of the superstructure and 2 have

to

a

lifting

force

of

.

propulsion engines of the balloon.

1.12. If the gas in Problem 1.11 were initially at 330 K, what will be the final volume if the temperature were raised to 550 K at constant pressure?

1.00

H

K

3

The stopcock

ideal gas occupies a

reduced to 1.15

sample of

and at a pressure of 99.99 kPa. What is the pressure of hydrogen in the dry state at 298.15 K? The vapor pressure of water at 298. 15 K is 3. 17 kPa.

desired

An

3

temperature of 298.15

1.20.

to transfer

fold?

gas

A 0.200-dm

What

reaches an equilibrium pressure of 287 Torr. Assuming

1.11.

molar mass.

its

nitrogen and 100.00 g of carbon dioxide are added at 25°C?

attached, and after system

flask is

mass of

fluid. Its

0.251 -dm

10.4 mTorr.

the molar

is

between

mm height of this fluid and the pressure in torr?

.000

at

.

calculate

each gas in a 2.50

.

273.15 K,

The density of air at 101.325 kPa and 298.15 K is dm 3 Assuming that air behaves as an ideal gas,

1.159 g

1.18.

The standard atmosphere of pressure is the force per unit area exerted by a 760-mm column of mercury, the 3 density of which is 13.595 11 g cm" at 0°C. If the

at

molar mass of the sample?

150 kPa and 298 K. What

the sample?

1.8.

density

at

as

number of molecules present

volume be

gas that behaves ideally has a density of 1.92

3

dm

g

1.17.

K

298.15

(approximately the best

1

A

1.16.

the

is

at

.

at 1.00

pressure of

What

2

at

the sample's

constant pressure? b.

measurement of low pressures.

the

What would

a.

dm 3

.

A gas

*1.22.

% to

mixture containing 5 mol

argon (such as

is

used

% butane and 95 mol

in Geiger-Miiller

counter tubes)

be prepared by allowing gaseous butane to

atm

evacuated

cylinder

cylinder

then weighed. Calculate the

gives

is

the

maintained final

desired at

mixture.

at

1

composition

fill

3

The 40.0-dm mass of argon that

pressure.

the

if

25.0°C. Calculate the

total

The molar mass of argon

is

an

is

temperature

is

pressure of the

39.9 g mol

.

47

Problems

The

1.23. -

km

of

'

gravitational constant g decreases

Modify

a.

account.

by 0.010

ms

2

the barometric equation to take this variation into

Assume that the temperature remains constant.

assuming

that sea-level pressure is exactly

that the temperature of

Suppose

1.24.

ammonia

is

that

1

atm and

K is constant.

298.15

on another planet where the atmosphere on the surface at h = 0, is 400

that the pressure

at

the earth.

aware

that, in the

lower part of the

atmosphere, the temperature decreases linearly with altitude.

This dependency

may be written

proportionality constant, z

temperatures

is

as

ground level and

at

T— T —

the altitude, and

az,

where a

is

a

T and Tare the

at altitude z

,

respectively.

Derive an expression for the barometric equation that takes this into account.

energy

kinetic

total translational kinetic

By what

changed

s"

if

The

*1.32.

of

each

factor

a gas

is

Work to a form involving In P/P

and 101.325 kPa.

ZA

thermometer and a mercury thermoat 0°C and at 100°C. The thermal expansion coefficient for mercury is

N2

mean

3.74

is

number of

unit time for

N2

10" 10

X

average speed

Its

ZM

m

at

474.6

m

is

number of

free path, the average

collisions

speeds

K to 400 K?

experienced by one molecule

the average

in unit time,

and

per unit volume per

.

mean

free path of a gas in terms of the

variables pressure and temperature, which are

more

easily

measured than the volume. 1.34. Calculate

ZA

ZM

and

for argon at 25°C and a ,0 d = 3.84 X 10" m obtained

pressure of 1.00 bar using

from X-ray crystallographic measurements. 1.35. Calculate the

.

The

ideal gas

meter are calibrated

root-mean-square

the

heated from 300

Calculate the

.

are

energy in the system.

collision diameter of

K

collisions

bar.

An

translational

*1.33. Express the

1.25. Pilots are well

1.26.

average

298.15

250 K. Calculate the pressure of ammonia at a height of 8000 metres. The planet has the same g value as Torr

The

e.

1.31.

b. Calculate the pressure of nitrogen at an altitude of 100

km

The

d.

molecule.

altitude.

1.36.

mean

Hydrogen gas has

K and (a)

1.00

a molecular collision diameter of

0.258 nm. Calculate the

298.15

Ar at 20°C and 10 d = 3.84 X 10" m.

free path of

collision diameter

mean

free path of

hydrogen

133.32 Pa, (b) 101.325 k Pa, and

(c) 1.0

at

X

8

a =

10 Pa.

-^-{dV/dT) P

=

1.817

X 10" 4 +

5.90

X

1O~

9

+

3.45

10" 10 6> 2

X

space

interstellar

exists at a concentration of

is

it

meter. If the collision diameter

where 6

is

V

=

6

at

the value of the Celsius temperature and 0.

mercury scale

V =

What temperature would appear on when the ideal gas scale reads 50°C?

the

estimated

1.37. In

hydrogen the is

mean

free path A.

is

that

atomic

one

particle per cubic

X

10" I0 m, calculate

2.5

The temperature of

interstellar

space

2.7 K.

*1.38. Calculate the value of Avogadro's constant from a

made by Perrin [Ann. Chem. Phys., 18, 1(1909)] in which he measured as a function of height the distribution study

Graham's Law, Molecular Collisions, and Kinetic Theory 1.27.

It

takes gas

orifice as the

A

of bright yellow colloidal gamboge (a

2.3 times as long to effuse through an

same amount of

nitrogen.

What

the molar

is

mass of gas A? 1.28. Exactly

1

dm 3

of nitrogen, under a pressure of

it

1.29. ideal

take for helium to effuse under the

What

is

monatomic gas confined

to 8.0

dm

its

molar mass

is

at

kPa

in a

2.00-dm

28.0134 g mol"

The amount of N 2 present. The number of molecules present. The root-mean-square speed of the molecules.

relative

3 1

Some

data

at

6

number of gamboge

gum

resin) particles

15°C are 5

35

100

47

particles at height z

Pgamboge

=

1.206

gCm" 3 3

=

mol of an 200 kPa?

calculate:

c.

long

same conditions?

3

1.30. Nitrogen gas is maintained at 152

b.

bar,

the total kinetic energy of 0.50

vessel at 298.15 K. If

a.

How

1

in water.

height, z/10" TV,

takes 5.80 minutes to effuse through an orifice. will

suspended

0.999 g Cm Pwater radius of gamboge particles, r

=

0.212

X

10

m.

(Hint: Consider the particles to be gas molecules in a

column of

air

and

that the

tional to the pressure.)

number of

particles

is

propor-

.

48

Chapter

The Nature

1

Chemistry and the Kinetic Theory of Gases

of Physical

Real Gases

Speeds

Distributions of

and Energies 10 2

v u

values for (a) the ratio

Note

and

/u,

and

(b) the ratio u/u mp

1.40.

The speed

body of any mass must have to 4 1.07 X 10 m s _1 At what

that a

would molecule, and (b) an temperature

is

.

speed

average

the

of

molecule be equal

2

(a)

to this

H2

an

H2

gas

25°C, calculate the

at

fraction of molecules that that

have a speed 2w

have the average speed

the van der

V as

using the values in Table

Compare

1.49.

weighing

w.

How

does

depend

depend on the mass?

at

1

isotherm over the same

K and 450 K for Cl 2

350

.4.

the pressures predicted for 0.8

17.5

g

K

273.15

at

using

(C 2 H4 ) has a

of Cl 2

ideal

gas

Pc = fc = 308.6 K. T — 97.2°C and

pressure of

critical

temperature of

critical

dm

the

(a)

Calculate the molar volume of the gas at

90.0 atm using Figure 1.18. Compare the value so found

Assuming

1.51.

ideal gases are heated to different

methane

that

a

is

perfectly

spherical

molecule, find the radius of one methane molecule using the value of b listed in Table 1.5.

Determine

1.52.

two

that

.

with that calculated from the ideal gas equation. that

have the average speed «i o°c at 100°C to the fraction that have the average speed u 25 °c at 25°C. How does this ratio

Suppose

1

1

61.659 atm and a

to the fraction

this ratio

PV

Waals

Figure

in

of the

ratio

Waals

Justify this relationship for a

equation and (b) the van der Waals equation.

on the mass of the molecules and the temperature? b. Calculate the ratio of the fraction of the molecules

1.42.

Draw

1.48.

1.50. Ethylene

For

stated that the van der

escape

speed? 1.41. a.

was

gas composed of spherical molecules.

range of P and

on these quantities?

escape from the earth

it

approximately four times the volume occupied

is

by the molecules themselves.

independent of the mass and the do the differences between them depend

How

constant b

.

that these ratios are

temperature.

In Section 1.13

1.47.

1.39. Refer to Table 1.3 (p. 32) and write expressions

Boyle

the

temperature

terms

in

of

constants for the equation of state:

temperatures such that their pressures and vapor densities

same.

the

are

What

is

the

relationship

between

PVm = RT{\ +

their

-

%I51{P/PC )(TC /T)[\

4(7,/T)

2

]}

average molecular speeds? 1.43. a. If U25°c

is

the average speed of the molecules in a

25°C, calculate the ratio of the fraction that will have the speed w 25 c at 100° to the fraction that will have the gas

R,

Pc and Tc ,

are constants.

at

1.53. Establish the relationships

same speed b.

Repeat

1.44.

On

at

25 °C. speed of 10 m 2 5°c-

the basis of Eq. 1.80 with for

the

fraction

of

/3

=

a.

l/k B T, derive

molecules

in

one-

a

is

the

+ dux

PV RT

.

most probable speed?

*1.45. Derive an expression for the fraction of molecules in a

one-dimensional gas having energies between ex and ex dex Also, obtain an expression for the average energy ex .

+

*1.46. Derive an expression for the fraction of molecules in

+

du. (Hint: Proceed

by analogy with the derivation of Eq.

Then obtain

the expression for the fraction having

1.91.)

energies between and e

+

de.

What

fraction will have

energies in excess of e*?

b.

2

1

+ x + x +

to the integrals

given in the appendix to this chapter,

••,

b)

=

(1

expand

44.

RT

V„

- b/VJ -i (1 - b/V„,y

and

',

(1

-

x)

'

=

l

to the quadratic (a).

Group terms containing the same power of Vm and compare to Eq. 1 1 8 for the case n = 1 d. What is the expression for the Boyle temperature in terms of van der Waals parameters? .

1

*1.54. Determine the Boyle temperature of a van der Waals

gas in terms of the constants

p.

1

c.

1.55.

Refer

-

a

V„

v in

Since V„/(Vm

The

critical

36.5°C, and l0

that

term and substitute into the result of part

.

a two-dimensional gas having speeds between u and u

show

Starting with Eq. 1.101,

an '

dimensional gas having speeds between u x and u x

What

between van der Waals B and C of

virial coefficients

Eq. 1.118 by performing the following steps:

this calculation for a

expression

parameters a and b and the

a, b,

temperature

its critical

pressure

and R.

Tc of P c is

nitrous oxide

7

1

.7

(N 2 0)

is

atm. Suppose that

'

Suggested Reading

mol of N 2

compressed

atm at 356 K. Calculate and use Figure 1.18, interpolating as necessary, to estimate the volume occupied by 1 mol of the gas at 54.0 atm and 356 K. 1

is

to 54.0

the reduced temperature and pressure,

Compare

*1.61.

mol

van der Waals,

equations. For

will H 2 be in a CH 4 at 500.0 K and 2.00 bar pressure? Given Tc = 33.2 K for H 2 190.6 K for CH 4 P, = 13.0 bar for H 2 46.0 bar for CH 4

1.56.

,

;

*1.57. For the Dieterici equation, derive the relationship of

a and b

Remember

that at the

2

P/dV

the critical

to

)

=

T

volume and temperature.

[Hint:

=

and

point {dP/dV) T

critical

0.]

the

find

volume of a van der Waals

gas.

reasonably accurate estimates of volumes can be deriving an expression for the compression factor

of

P from

the result of the previous problem.

substitutes for the terms

However, made by

expression and use

Beattie-Bridgeman

C0 2 the Dieterici equation constants are

m6 mol" 2 10 m3 mol -l

a

=

0.462 Pa

b

=

4.63

3.040

X

5

Z in

terms

gas obeys the van der Waals equation with

X

10

value of the

Vm on the right-hand side in terms of V,„ = RT/P. Derive this

Pa (= 30 atm) and Tc = 473 K. Calculate the van der Waals constant b for this gas.

powers of V~ in order to cast it into the virial form. Find the second and third virial coefficients. Then show that at low densities the Dieterici and van der Waals equations give essentially the same result for P.

Expand

]

the Dieterici equation in

volume of CC1 2 F 2

Essay Questions 1.64. In light of the

and 5.00 bar pressure. What will be the molar volume

liquefaction of gases.

to find the

Pc =

6

30.0°C

it

bulb using the ideal

One simply

gas law expression

ideal

the

A

*1.62.

*1.63.

1.58. In Eq. 1.103 a cubic equation has to be solved in order to

3

and

.

,

2

Dieterici,

At what temperature and pressure

corresponding state with

(d

the values obtained for the pressure of 3.00

CO z at 298.15 K held in a 8.25-dm

gas,

49

at

van der Waals equation, explain the

computed using the ideal gas law under the same conditions? *1.59.

gases

A is

1.65. State the postulates of the kinetic molecular theory of

general requirement of

low pressures. Show

that this

is

true for

The van der Waals constants

literature are

a

=

found

5.49 atm

to

1.66. Eq. 1.23 defines the ideal-gas thermometer. Describe

how

van der Waals equation.

1.60.

gases.

that they reduce to the ideal gas equation (Eq.

1.28) in the limit of the

equations of state for

all

for

C 2H6

an actual measurement would be

thermometer in the older

starting

made

using such a

with a fixed quantity of gas

at

a

pressure of 150 Torr.

be

2

L mol"

Express these constants

and

b

in SI units

=

(L

0.0638

=

L mol"

= dm).

liter

SUGGESTED READING The references

at the end of each chapter are to specialized books or papers where more information can be obtained on the particular subjects of the chapter.

For an authoritative account of the discovery of Boyle's law, see I. Bernard Cohen, "Newton, Hooke, and 'Boyle's Law' (discovered by Power and Towneley)," Nature, 204, 1964, pp. 618-21. For interesting accounts of the H. A. Boorse and L.

Atom,

New

lives of early chemists, see

Motz

(Eds.),

The World of the

York: Basic Books, 1966.

For a discussion of pressure gauges, see R. cal

Chemistry:

Methods,

J.

Techniques,

Sime, PhysiExperiments,

Philadelphia: Founder's College Publishing, 1990, pp.

314-35.

For problems and worked solutions covering many of the types of problems encounted in this book, see B. Murphy, C. Murphy, and B. J. Hathaway, A Working

Method Approach for Introductory Physical Chemistry Calculations,

New

York: Springer- Verlag (Royal

Society of Chemistry Paperback), 1997.

For an in-depth account of virial coefficients, see J. H. Dymond and E. B. Smith, The Virial Coefficients of Gases, A Critical Compilation, Oxford: Clarendon Press, 1969.

For much more on the use of equations of state, see O. A. Hougen, K. M. Watson, and R. A. Ragatz, Chemical Process Principles: Part II, Thermodynamics, (2nd ed.),

New

York: Wiley, 1959, Chapter

14.

50

Chapter

1

The Nature

of Physical

Chemistry and the Kinetic Theory of Gases

For more on supercritical fluid use in chromatography, see M. J. Schoenmaker, L. G. M. Uunk, and H-G JanJournal of Chromatography, 563-78. seen,

506,

1990,

pp.

N. G. Parsonage, The Gaseous State, Oxford: Pergamon Press, 1966.

M

J.

tific,

For accounts of the kinetic theory of gases, J.

see:

Cambridge: University Press, 1959. L. B. Loeb, Kinetic

1961. This

is

applications.

Theory of Gases,

New

York: Dover,

a particularly useful account, with

many

1987.

Kinetic Theory of Gases, New York: McGraw-Hill, 1960. For an account of the development of physical chemistry, with biographies of many of the scientists mentioned in this book, see K. J. Laidler, The World of Physical Chemistry, Oxford University Press, 1993.

R

H. Jeans, Introduction to the Kinetic Theory of Gases,

Modern Gas Kinetics, Theory, Experiand Applications, New York: Black well Scien-

Pilling (Ed.),

ments,

D.

Present,

The First Law of Thermodynamics

PREVIEW According

to the first

law of thermodynamics, energy

cannot be created or destroyed but can only be converted into other forms.

Heat and work are forms of energy, and

the law can be expressed by stating that the change in the internal energy of a system

q supplied

to the

AU is the

system and the work

AU = In this chapter the nature of

q

w

sum of the done on

heat

is

will be introduced

It

will be

shown

The concept of standard these conditions.

law

is

that

An

we can

states

is

introduced to define

important consequence of the

first

write balanced equations for

we

will

show how they can be

manipulated algebraically so as to obtain enthalpy considered, and the

and explained.

that the internal energy

important property for processes that occur

U is

at

internal energy plus the product of the pressure

are

formed from

which we

call

states.

number of processes involving considered. It is shown that for an ideal

In Section 2.6 a

the

and the

when compounds

the elements in certain defined states

standard is

A vast

amount of summarized in the form thermochemical information is which are the of standard enthalpies offormation, changes for other reactions.

enthalpy changes

an

constant

volume. For p rocesses at_constant pressure the fundamental property is the^ejithalrjy^//, which

ideal gases are

gas the internal energy and the enthalpy depend only on

volume:

the temperature and not

H = U + PV

is

This chapter

concerned primarily with heat changes when

changes, and

important concept of thermodynamically reversible

work

is

chemical reactions, together with their enthalpy

it:

+ w

work

law and

a process occurs under precisely defined conditions.

is

field that is

based on the

not true for real gases, with which Section 2.7

is

concerned.

also concerned with thermo-

chemistry (Section 2.5), a

on the pressure or volume. This

first

51

v^hapter

1

was mainly concerned with macroscopic properties such as pressure, We have seen how some of the relationships between

volume, and temperature.

these properties of ideal and real gases can be interpreted in terms of the behavior

of molecules, that

of the microscopic properties. This kinetic-molecular theory

is,

of great value, but

it

is

many of

possible to interpret

macroscopic properties without any reference without assuming that molecules exist. This

to the is

the relationships

behavior of molecules, or even

what

treatments of the science of thermodynamics, which

done

is

between the various forms of energy, including

shall present

thermodynamics

clarify

some of

the basic ideas

most formal

in the

concerned with the general

is

relationships

in a less formal

is

between

heat. In this

way, and from time to time

by showing how they

relate to the

book we

we

shall

molecular

behavior.

At

might appear

first it

true.

It

thermodynamics, with no

that the formal study of

gard to molecular behavior, would not lead us very

far.

The

re-

contrary, however,

is

has proved possible to develop some very far-reaching conclusions on the

basis of purely

thermodynamic arguments, and these conclusions are

convincing because they do not depend on the truth or

atomic and molecular behavior. Pure thermodynamics

falsity

starts

all

the

more

of any theories of

with a small number of

assumptions that are based on very well-established experimental results and makes logical deductions

from them,

finally giving a set

be true provided that the original premises are

2.1

of relationships that are bound to

true.

ORIGINS OF THE FIRST LAW There are three laws of thermodynamics (aside from the zeroth law, which was

mentioned

The Nature

of Heat

in

Section

of energy Before the .

1.4). first

of heat to be understood. long time for

this to

The

first

law

is

essentially the principle o f conservati on

law could be formulated

We know

today that heat

it

is

a

was necessary

for the nature

form of energy, but

it

took a

be realized and become generally accepted. In the seventeenth

century Robert Boyle, Isaac Newton, and others proceeded on the correct assumption that heat is a manifestation of the

motions of the particles of which matter

is

composed, but the evidence was not then compelling. In the next century investigators

such as Joseph Black and Antoine Lavoisier, both of

experiments on heat, were convinced that heat

is

whom

did important

a substance, and Lavoisier even

it as one of the chemical elements. Firm evidence that heat is related to motion and

listed

is,

therefore, a

form of energy

came only toward the end of the eighteenth century. Experiments were carried out which showed that expenditure of work causes the production of heat. The first quantitative experiments along these lines were performed

born Benjamin Thompson (1753-1814),

who had

by the MassachusettsEurope and

a colorful career in

was created Count of Rumford while serving in the Bavarian army. During his supervision of the boring of cannon at the Munich Arsenal, he became interested in the generation of heat during the process.

He

suggested in 1798 that the heat arose

from the work expended, and he obtained a numerical value generated by a given amount of work.

52

for the

amount of heat

A

53

States and State Functions

2.2

more persuasive contribution was made by

German

the

physician Julius

Robert Mayer (1814—1878). His medical observations led him to conclude

work performed by humans

derived from the food they eat, and in 1842 he

is

the important suggestion that the total energy

conserved. At the same time, and

is

independently, precise experiments on the interconversion of

by the English

a variety of conditions, were carried out

First

Law

under

heat,

scientist

(J).

The experiments of Joule

The

work and

James Prescott of energy, work, and heat is

Joule (1818-1887), and in his honor the modern unit called the joule

that

made

in particular led to the conclusion that t he

the universe remains constan t, which

a compact statement of the

is

thermodynamics. Both work and heat are quantities

energyjjf

first

law of

that describe the transfer of en-

ergy from one system to another. If

two systems are

temperatures, heat can pass from one to the

at different

other directly, and there can also be transfer of matter from one to the other. Energy

can also be transferred from one place

which tal

2.2

considered later (Section

is

form of work, the nature of

to another in the

2.4).

No

matter

how

these transfers occur, the to-

energy of the universe remains the same.

STATES AND STATE FUNCTIONS Section 1.3 emphasized the important distinction between a system and roundings.

We

also

explained the differences between open

systems, and isolated systems. ings

particularly

is

The

important

distinction

in

between a system and

thermodynamics,

concerned with transfer of heat between a system and

concerned with the work done by the system on roundings on the system. In

all

its

we

since

its

its

closed

surroundconstantly

are

surroundings.

sur-

its

systems,

We

are also

surroundings or by the sur-

cases the system must be carefully defined.

Certain of the macroscopic properties have fixed values for a particular state of the system,

whereas others do

ter in a vessel at

to

1

cm 3 These .

25°C and quantities,

of the system. Whenever

same

state,

and

this

not.

Suppose, for example, that

a pressure of 10

g of

1

we

means

Pa

H 2 0, 25°C,

(1 bar);

it

will

5

10 Pa, and

1

satisfy these four conditions,

that the total

same. As long as the system

5

These macroscopic properties

that

cm 3

will

,

all

we have

amount of energy

is in this state, it

we maintain 1 g of wahave a volume of close

in the

specify the state the water in the

molecules

is

the

have these particular specifications.

we have mentioned

(mass, pressure, temperature,

and volume) are known as state functions or state variables.

One

very important characteristic of a state function

ifieHjtie st ate of a sy stem

valu es nf

we have

all

is

that

once

we have

other ct-ne fnnrtinny are fixed Thus, in the example just given, once

energy

in the

Another important

is

The

pressure and temperature, in fact,

in the system.

rjiar arterktir

chang ed thp change

volume

molecules that make up the system, and en-

therefore another state function.

depend on the molecular motion s ystem is

sp ec-

state functions , the

specified the mass, temperature, and pressure of the water, the

fixed. So, too, is the total

ergy

is

b y giving the values of some of the

in

any

of

a state

st ate

function

is

that

when

the jstateof a

function depends only on the

initial

an d

54

The

Chapter 2

First

Law

of

Thermodynamics

final states

example,

of the

if

we

s ystem, a nd

not on the path followed in

heat the water from

equal to the difference between the

25°C

initial

AT =

and

T,fi nal

making

26°C, the change

to

in

the _change. For

temperature

is

final temperatures:

1°C

initial

(2.1)

The way in which the temperature change is brought about has no effect on this result. This example may seem trivial, but it is to be emphasized that not all functions have this characteristic. For example, raising the temperature of water from 25°C to 26°C can be done in various ways; the simplest is to add heat. Alternatively, we could

stir

the water vigorously with a paddle until the desired temperature rise had

this means that we are doing work on the system. We could also add some heat and do some work in addition. This shows that heat and work are n ot

been achieved;

function s.

sta |e

In the

point

A

meantime,

on the

4000

that is

m

it

is

useful to consider an analogy. Suppose that there

earth's surface that

above sea

level.

is

The

1000

m

above sea

difference,

3000 m,

level is

is

a

and another point

B

the height of

B

with re-

spect to A. In other words, the difference in height can be expressed as

Ah =

hR

where h A and h B are the heights of A and is

thus a state function, the difference

chosen. However, the distance

pendent on the path; Distance traveled

2.3

is

we can go by

h.

(2.2)

B above

Ah

we have

-

sea level. Height above sea level

being in no

way dependent on

to travel in order to

go from A

to

the path

B

is

de-

the shortest route or take a longer route.

therefore not a state function.

EQUILIBRIUM STATES AND REVERSIBILITY Thermodynamics Force,

state functions

F= PA Area cross

piston

section

directly

concerned only with equilibrium It

states, in

which the

provides us with in-

formation about the circumstances under which nonequilibrium states will

of

Frictionless

is

have constant values throughout the system.

toward equilibrium, but by =A

itself

Suppose, for example, that

movable piston (Figure

tionless

it

tells

we have

a gas confined in a cylinder having a fric-

2.1). If the piston is motionless, the state

of the gas

can be specified by giving the values of pressure, volume, and temperature. ever, if the gas

is

compressed very

move

us nothing about the nonequilibrium states.

rapidly,

it

How-

passes through states for which

pressure and temperature cannot be specified, there being a variation of these properties

throughout the gas; the gas near to the piston

heated than the gas

FIGURE A gas at

in a

2.1

pressure

P

maintained at

equilibrium by an external force, F, equal to PA, where A is the area of cross section of the pis-

The force applied depends on mass of the piston, and any masses that are placed on

ton.

the

it.

nonequilibrium

though

it

could

tell

at the far state.

is at first

more compressed and

end of the cylinder. The gas then would be said

Pure thermodynamics could not deal with such a

us what kind of a change

would spontaneously occur

to

be

state, al-

in

order

for equilibrium to be attained.

The

criteria for

equilibrium are very important. The mechanical properties, the

chemical properties, and the temperature must be uniform throughout the system

and constant

in time.

The

force acting on the system must be exactly balanced by

the force exerted by the system as otherwise the

volume would be changing.

If

we

Energy, Heat, and

2.4

55

Work

we see that for the system to be at must exactly balance the pressure P of

consider the system illustrated in Figure 2.1, equilibrium the force the gas; if A

is

F exerted on

the piston

the area of the piston,

PA = If

we

if

Suppose sure that

we

we

that

decrease

we

dP as compression we are

by removing mass, the gas

on the gas

(i.e., it

can make

it,

F by

increase the force

are exerting

pressure of the gas

Reversible Processes

(2.3)

increase the force, for example, by adding mass to the piston, the gas will be

compressed;

We

F

will be

small as

we

now be

will

P +

dP).

like,

and

will expand.

an infinitesimal amount dF. The presinfinitesimally greater than the

The gas

compressed.

will therefore be

at all stages

during the infinitely slow

therefore maintaining the gas in a state of equilibrium.

We

refer to a process of this kind as a reversible process. If

we

reduce the pressure to

P —

is,

reversibly.

dP, the gas will expand infinitely slowly, that

Reversible

processes play very important roles in thermodynamic arguments. All processes that actually

slowly, there

2.4

ENERGY, HEAT, AND

occur is

however, irreversible; since they do not occur

are,

necessarily

some departure from

infinitely

equilibrium.

WORKHHHHaiHHHHHHHHHiHHI^H^^

We come now which the

total

to a statement of the first

amount of energy

law of thermodynamics, according to

in the universe is conserved.

Suppose

that

we add

heat q to a system such as a gas confined in a cylinder (Figure 2.1). If nothing else is

done

energy

to the system, the internal

U increases

by an amount

that is exactly

equal to the heat supplied:

Internal

MJ =

Energy

(with no

q

This increase in internal energy

is

work done)

(2.4)

the increase in the energy of the molecules

which

comprise the system.

Suppose instead

that

no heat

is

transferred to the system but that by the addi-

amount of work w is performed on it; the details of this are considered later (Eqs. 2.7-2.14). The system then gains internal energy by an amount equal to the work done

tion of

mass

to the piston an

:

At/ In general,

if

heat q

is

= w

(with no transfer of heat)

supplied to the system, and an amount of

formed on the system, the increase

AU = This

is

'The

that

many

is

to use the

(work done on the system)

(2.6)

We

on absorbing some

by

can understand the law heat,

can store some of

w for the work done on the system. The w for the work done by the system.

symbol

older treatments use the symbol

w

also per-

law of thermodynamics.

that a collection of molecules,

IUPAC recommendation

warned

first

+ w

work

is

in internal energy is given

q (heat absorbed by the system)

a statement of the

by noting

(2.5)

reader

is

56

Chapter 2

The

First

Law

of

it

Thermodynamics

and can do some work on the surroundings. According

internally,

convention the work done by the system

is

—w,

q (heat absorbed by the system)

which

=

At/

— w

(2.6a)

equivalent to Eq. 2.6.

is

In applying Eq. 2.6 q,

it is,

of course, necessary to employ the same units for U,

and w. The SI unit of energy

is

the joule (J

=

sponding to a force of one newton (N metre. In this

book we

=

kg

m s~

kg

)

m

2

s~

); it

is

the energy corre-

operating over a distance of one

although

shall use joules entirely,

2

many thermodynamic 2

ues have been reported in calories, one thermochemical calorie being

We

IUPAC

to the

so that

4.

1

84

val-

J.

should note that Eq. 2.6 leaves the absolute value of the internal energy

indefinite, in that

we

practical purposes this

is

U

most

are dealing with only the energy change At/, and for

adequate. Absolute values can in principle be calculated,

although they must always be referred to some arbitrary zero of energy. Thermody-

namics

concerned almost entirely with energy changes.

is

The

State Functions

internal_energy (Vis a state fn nrtinn of the system- that

the state of the system and not on

we saw

that a

change from one

how

the system achieved

its

is, it

depends only on

particular state. Earlier

such as from 25°C to 26°C, can be

state to another,

achieved by adding heat, by doing work, or by a combination of the two. experimentally that however

we

bring about the temperature

always the same. In other words, for a particular change

+

equal to q

behavior

is

and work

w,

is

way

independent of the

w

It is

found

sum q + w

is

in state, the quantity At/,

which the change is brought about. This example demonstrates that heat q

in

characteristic of a state function. This

w

change can be brought about by various

are not state functions since the

divisions of the energy between heat and work; only the

The

rise, the

distinction

between

state functions

that are not state functions

may be

such as

sum q + w

U and

is fixed.

quantities such as q

and

considered from another point of view.

Whether or not a property is a state function is related to the mathematical concept of exact and inexact differentials. The definite integral of a state function such as U,

I is

a quantity

occurs.

On

work; that

U2 —

U\

=

At/,

which

independent of the path by which the process

the other hand, the integral of an inexact differential such as heat and

is,

\

is

is

dU

dq

dw

or

a quantity that does not have a fixed value but depends on the process by which the

change from

state

the quantities

1

to state 2 occurs;

we have used

would therefore be wrong Aq and Aw have no meaning.

act differential.

It

"There are other calories: the "15° calorie"

is

~

the

symbol dio indicate an inex-

q 2 — q\ = Ag or w 2 - w = Aw; and worJcjaate-memsefves-evrdent

to write

H eat

4.1855

J;

the "international calorie"

{

is

~

4.1868

J.

2.4

Work

Energy, Heat, and

57

o nly during a change from one state to another and have no significance when a svs remain^ jn a pa rrimlar stal e: they are properties of the path and not of the state A

jgm

.

such as the internal energy U, on the other hand, has a significance in

state function

relation to a particular state. If

U were not a state function, we could have violations of the principle of conser-

To see how a violation could two states A and B, and suppose that there are two alternative paths fromA and B. Suppose that for one of these paths U is 10 J; and for the other, 30 J: vation of energy, something that has never been observed.

occur, consider

=

At/,

We

10 J

At/,

=

30

J

A to B by the first path and expend 10 J of heat. If we then reA by the second path, we would gain 30 J. We would then have

could go from

turned from

B

the system in

to

and would have a net gain of 20

original state,

its

therefore have been created from nothing. nitely,

with a gain of energy

been made to create energy

machines of the

at

all

have ended

would only work

make

inability to

Many

indefi-

attempts have

way, by the construction of perpetual motion

in this

stantly rejecting devices that

were violated! The

each completion of the cycle.

but

first kind,

Energy would

J.

The process could be continued

in failure if

the

—

first

patent offices are con-

law of thermodynamics

perpetual motion machines provides convinc-

ing evidence that energy cannot be created or destroyed.

thermodynamic studies

In purely

Nature of Internal Energy

energy really consists

of;

it

is

not necessary to consider what internal

however, most of us like to have some answer to

this

question, in terms of molecular energies. There are contributions to the iniemal-ene rgy of a su bstance

from motion of the individual molecules,

1

the kinetic energy of

2.

the potential energy that arises

3.

the kinetic

from interactions between molecules,

and potential energy of the nuclei and electrons within the individual

molecules.

The

somewhat complicated, and it is a great we can make use of the concept of internal energy

precise treatment of these factors

strength of thermodynamics that

without having to deal with

The Nature

of

it

is

on a detailed molecular

basis.

Work

in which a system may do work, or by which work may be done on a system. For example, if we pass a current through a solution and electrolyze it, we are performing one form of work glecta cal wor k. Conversely, an

There are various ways Electrical

Work

—

electrochemical cell

Chemical Work

osmotic work

,

may perform work. Other forms

and -mec hanical

ways, involved when

chemical substances. verse osmosis

concentration

(p. is

Chemical work

of work are chemical work, is

usually, but not quite al-

larger molecules are synthesized from smaller ones, as in

Osmotic work

living organisms.

Mechanical Work

work.

It is

the

involved, for

223) and

much

is

work required

to transport

example, when seawater

and concentrate

is

in the formation of the gastric juice,

purified

by

higher than that of the surroundings. Mechanical work

performed, for example,

when

a weight

is lifted.

re-

where the acid is

58

Chapter 2

The

First

Law

of

Thermodynamics

Reversible work done on

wtev

the system

= PAl = - PAl/

\< Gas

Volume decrease,

Pressure

at ,

constant pressure

FIGURE

applied ,

P

=

-AV = Al

i

P+dP

2.2

work done by a P moving a piston. A simple way for a gas to be at constant pressure is to have a

The

reversible

constant pressure

vapor

in

equilibrium with

Area

its liquid.

One simple way

in

of cross section

which work

is

done

=A

when an

is

external force brings

about a compression of a system. Suppose, for example, that

ment

which a gas or

in

liquid

is

against a

movable piston (Figure

we must

apply a force

P

maintained 2.2). In

we have

an arrange-

constant pressure P, which

at

order for the system to be

to the piston, the force

at

it

exerts

equilibrium

being related to the pressure by

the relationship

F = PA

(2.7)

where A is the area of the piston. Suppose that the force is increased by an infinitesimal amount dF, so that the piston moves infinitely slowly, the process being reversible. If the piston

on the system

moves

to the left a distance

w However, Al

/,

the reversible

work

w rev

done

is

is

the

r

,

Fl

=

PAl

(2.8)

volume swept out by the movement of the piston, that is — AV. The work done on the system

crease in volume of the gas, which

is,

the de-

thus

is

= -PAV In. _our is

we have compressed the gas and AV would have been positive and the w ork system wnnlH have hepn ripgafivp- fhat is the gas would have don e a

example

this is a positive quantity since

negative. Tf the gas harl e xp anded,

Hnne_on fpe positive

amount of work on

EXAMPLE 2.1 which The tube to

(2.9)

is is

AV

the surroundings.

Suppose

that a

chemical reaction

is

caused to occur

in a

attached a capillary tube having a cross-sectional area of 2.50

open

to the

atmosphere (pressure

course of the reaction the rise in the capillary

by the reaction system.

is

=

bulb

mm

2 .

101.325 kPa), and during the

2.4 cm. Calculate the

work done

Reversible work done on the system, wrev =

-

r

P dV

The volume

Solution

2.50

X

increase

10~ 6

m

2

=

6.08

X

2.40

1.01325

10" 3

X

59

is

X

10" 2

m=

The work done by the system, which following written as -w, is PAV:

-w = PAV =

Work

Energy, Heat, and

2.4

X

5

= N

[Pa

J

the

X

10 Pa

6.00

X

2 ;

m

10

IUPAC

6.00

m

X

convention must be

10" 8

Nm=

m

3

J]

S-H Initial

i_ (._„>

-

=

V,

pressure

If the

volume

by a process of

P moves volume dV is sure

Final

P

volume change, we must obtain the work done The work done on the system while an external presso that the volume of the gas changes by an infinitesimal

varies during a

integration.

the piston

volume*

Jw rPV = -PdV If,

as illustrated in Figure 2.3, the

the reversible

FIGURE

work done on

V

x

volume changes from a value

the system

V

to a value

x

V2

,

is

2.3

The reversible work performed when there is a volume decrease from

(2.10)

to

V2

-f

w„

PdV

(2.11)

.

In the

example shown in Figure

2.3,

V > V2 (i.e., we have compressed the gas) and this x

work is positive. Only if P is constant is it permissible to integrate this directly to give rV2

w rev = -p[

P(V2 -

„

Vi)

= -PAV

(2.12)

'V,

(compare Eq.

2.9). If

P

is

not constant,

we must

express

it

as a function of

V before

performing the integration.

We

have already noted

work done

that

not a state function, and this

is

further stressed with reference to the mechanical

derivation has

shown

that the

work

is

related to the process carried out rather than to

the initial and final states. We-caft-considei the

volume

V

x

to

-

volume

V2

.

and can

may be

work of expansion. The previous

als o-consider

l

e veisible

expansion of a gas

an irreversible proc e ss

,

in

from

which ras e

work will be done bv the system. This is illustrated in Figure 2.4. The diagram to shows the expansion of a gas, in which the pressure is falling as the volume increases. The reversible work done by the system is given by the integral le ss

the left

"Wrev

which

is

PdV

(2.13)

represented by the shaded area in Figure 2.4a. Suppose instead that

pressure to

P2

we

by instantaneously dropping the external the final pressure P 2 The work done by the system is now against this throughout the whole expansion, and is given by

performed the process pressure

=

irreversibly, .

-w m = P

2

(V2

-

V,)

(2.14)

60

Chapter 2

FIGURE The

The

First

Law

of

Thermodynamics

2.4

left-hand diagram (a)

trates the reversible

work

Reversible work done by the gas

illus-

of

expansion from I/, to V2 The right-hand diagram (b) shows the irreversible work that would be performed by the system if the external pressure were suddenly dropped to the final value P2

Irreversible

work done by the gas = P2 (Vj>-«/

Vo

.

f

^

P dV

1 )

V

:

.

V

V

a.

b.

This work done by the system is

less than the reversible

is

represented by the shaded area in Figure 2.4b and

work. Thus, although

system has changed from A

to B, the

work done

is

in

both processes the state of the

different.

This argument leads us to another important point.

Maximum Work

tem

in a reversible

A

The work done by

B represents changing from A to B.

expansion from

the system can perform in

to

the

the sys-

maximum work

that

EXAMPLE 2.2

Suppose that water at its boiling point is maintained in a cylinder For equilibrium to be established, the pressure that must be applied to the piston is 1 atm (101.325 kPa). Suppose that we now reduce the external pressure by an infinitesimal amount in order to have a reversible expansion. 3 If the piston sweeps out a volume of 2.00 dm what is the work done by the system? that has a frictionless piston.

,

Solution

The

external pressure remains constant at 101.325 kPa, and, there-

fore, the reversible

work done by

-w rcv = PAV = Since Pa

= kg m _1

work done by

s~

2

(see

the system

For many purposes

it

is

is

the system

101 325 Pa

X

Appendix A), 202.65

is

2.00

dm 3 =

the units are

202.65 Pa

kg

m

2

s~

m

2

3

=

J;

thus the

J.

convenient to express the

first

with respect to an infinitesimal change. In place of Eq. 2.6,

dU =

dq + dw

law of thermodynamics

we have (2.15)

Energy, Heat, and

2.4

Work

61

d denotes an inexact differential. However, if only PV dw may be written as — P dV, where dV is the infinitesimal in-

where again the symbol

work

is

involved,

crease in volume; thus,

dU = dq-PdV Processes It

Constant Volume

at

follows from this equation that

PV work is

ume, and only

(2.16)

an infinitesimal process occurs

if

dU = where the subscript

V indicates

is

(2.17)

supplied

is

at

constant volume. (Note

an exact differential, so that the d has lost

AU=U

2

- U

x

=q v

its

(2.18)

The increase of internal energy of a system in a rigid container ume) is thus equal to the heat q v that is supplied to it.

In

constant vol-

dq v

that the heat

under these circumstances dq v bar.) This equation integrates to that

Processes

at

involved,

(i.e., at

constant vol-

Constant Pressure: Enthalpy

at

most chemical systems we are concerned with processes occurring

in

open ves-

which means that they occur at constant pressure rather than at constant volume. The relationships valid for constant-pressure processes may readily be desels,

duced from Eq. 2.16. For an infinitesimal process absorbed dq P

is

f fiflt

change from

P is

state

1

=

]

(

u

process involves a

=

\

dU+\ PdV

hi

(2.20)

Jv.

dV

+ P(V2 -

This relationship suggests that

PV, and

it

(2.21)

-V,

t

y)

is

known

it

In the older scientific literature

it

is

V,)

= (U2 + PV2 )

would be convenient

as the enthalpy,

H 3

If the

v2

dU + P\

I

= (U2 - U

U+

mpd

to state 2, this equation integrates as follows:

rU2

tity

iv-fH>rfnr

(2.19)

constant,

qP

Definition of Enthalpy

PV wnrk

nn wnrk nthpr than

qP Since

constant pressure the heat

= dU + PdV

dq P prnviHpfJ

at

given by

known

3

+

PV,)

to give a

name

(£/,

(2.22) to the

quan-

with the symbol H:

=U +P y

(2.23)

as the heat content, but this term can be misleading.

62

Chapter 2

The

First

Law

Thermodynamics

of

We

thus have

=

qP This-MptHUct- n

i?

val i d

nnhiJf

at

H

= AH

x

(2.24)

wnj±Js_ all PV wo rk. Under

these circumstances

equal to the heat q P that

constant pressure. Since U, P, and

V

are

is

functions,

all state

is

it

supplied to

follows from

also a state function.

at constant pressure for which q P and AH are which a positive amount of heat is absorbed by the system. Such processes are known as endothermic processes (Greek endo, inside; therme, heat). Conversely, processes in which heat is evolved (q P and AH are negative) are known

A

chemical process occurring

positive

Exothermic Process

-

is

Eq. 2.23 that enthalpy

Endothermic Process

2

system

the increase in enthalpy it

ihp

AH of a

H

as

is

one

in

exothermic processes (Greek

exo, outside).

Heat Capacity The amount of heat required to raise the temperature xtf-aoy-sHbsfrmf-f-~hy~t K (which of course is the same as 1°C) is known as its h eat capacit y, and is given the symbol C; its SI unit is J K~'(jrhe word specific before the name of any extensive physical quantity refers to the quantity per unit mass. The term specific h eat capa city is thus the amou nt of heat required to raise th eje mperature of unit mass of a material by 1 K; if the unit mass is 1 kg, the unit is J K^ kg which is the SI unit for specific heat capacity. The word molar before the name or a quantity refers to the quantity divided by the amount of substance. The SI unit for the molar heat ca1

1

pacity

is J

K

_1

mol~\

Since heat

is

not a state function, neither

ways necessary, whep te mperature is raised

Isochoric Process

1.

by

The heat capacity choric process);

stating 1

K.

is

the heat capacity.

It is

a he a t-capacity, to sp ecify the_ptocer r ,

Two

.

therefore al-

hv which t he

heat capacities are of special importan ce:

related to a process occurring at constan t_volume-( an iso-

this is

denoted by

C v and

Cywhere j&/J s_ the heat supplied

at

is

defined by

dqy_ (2.25)

dT

constant

volum e. Since q v

is

AU,

equal to

it

follows that

Cv

Heat Capacity at Constant Volume If

we

are

working with

1

mol of

(2.26)

the substance, this heat capacity

heat capacity at constant volume, and

4

The

subscript

m may

be omitted

when

there

is

is

represented by the symbol

no danger of ambiguity.

is

the

C Vm

molar 4

.

Energy, Heat, and

2.4

Isobaric Process

2.

The heat capacity isobaric process)

CP and is defined by _ (dH\ ~df ' [~df) p dq P

_ "

quantity

The heat required constant volume

is

represented by the symbol

to raise the temperature of

(2.27)

CPm

.

mol of material from

1

C Vm is

7, to

T2

at

is

c If

63

related to a process occurring at constant pressure (an

is

Heat Capacity at Constant Pressure The molar

Work

independent of temperature,

-r Cp m dT

(2.30)

This integrates to 1p, m if

CP

is

,

respectively, per

For liquids and quently,

CV m and

solids,

CPm

C Vm

and

CP m

.

AUm

A//„,

(2.31)

mole of material.

and

AHm

are essentially the

however, the A(PV) term

tween

—

T{)

independent of temperature. The expressions in Eqs. 2.29 and 2.3 1 repre-

AUm and A//m

sent

— CP

H 2 0(/)

Combustion processes also frequently occur

When

stoichiometry.

carbon the

an organic compound

is

present as

usually

to

completion

with

simple

burnt in excess of oxygen, the

C0 and the hydrogen into H 2 0, while N 2 in the final products. Often such

practically all converted into

is

nitrogen

is

2

combustions of organic compounds occur cleanly, and much thermochemical information has been obtained by burning organic compounds in calorimeters. 2.

fnrfirfrf Cnlr\rimotK\i

flrf

nf

f-f^ss's

J

Few

nu>

reactions occur in a simple

manner, following a simple chemical equation, with the result that the enthalpy

changes corresponding

Hess's

Law

to a

simple chemical equation often cannot be measured

For many of these, the enthalpy changes can be calculated from the

directly.

values for other reactions by making use of Hess's law,

named

Henri Hess (1802-1850). According to

permissible to write

this law,

it

is

after

Germain

stoichiometric equations, together with the enthalpy changes, and to treat

them

as mathematical equations, thereby obtaining a fhermochemically valid result.

For example, suppose 1.

A + B-»X Suppose

that

that a substance

A//,

X

A reacts with B according to the equation

= -lOkJmol"

1

reacts with an additional molecule of

A

to give another

product Y: 2.

A+

X^Y

According

A//2

= -20kJmol"'

to Hess's law,

it

is

permissible to add these two equations and

obtain: 3.

2A +

B^Y

A//3

=

A//i

+ A//2 = -30kJmor'

The law follows at once from the principle of conservation of energy and from the fact that enthalpy is a state function. Thus, if reactions

evolution of 30 kJ into 2

A+ B

different

when

1

mol of Y is produced.

by the reverse of reaction

from 30

kJ,

we

3. If

1

and 2 occur, there

In principle

is

a net

we could reconvert Y

the heat required to

do

this

were

should have obtained the starting materials with a net

gain or loss of heat, and this would violate the principle of conservation of energy.

68

Chapter 2

The

First

Law

of

Thermodynamics

EXAMPLE

The enthalpy changes

2.4

talline

a-D-glucose and maltose

liquid

H 2 0,

at

complete combustion of crys-

in the

C0 2 and

298 K, with the formation of gaseous

are:

A,//7kJmor' a-D-Glucose,

C 6 H, 2

C 2 H 22

Maltose,

- 5645 .5

(c)

, ,

,

-2809.1

6 (c)

Calculate the enthalpy change accompanying the conversion of talline

1

mol of

crys-

glucose into crystalline maltose.

The enthalpy changes given

Solution 1.

C 6 H 12

relate to the processes

+ 60 2 (£) -» 6C0 2 (£) + 6H,0(/) AC H° = -2809.1 kJmor C 12 H 22 O u (c) + 120 2 (g)-> 12C0 2 (£) + 11H 2 0(/) \H° = -5645.5 kJ mol -1 6 (c)

1

2.

We

are asked to convert

mol of glucose

1

C 6 H 12 Reaction 2 can be written 2'.

6 (c)

we add

reactions 1 and

2:

+ 6Q 2 (g)

2822.8 kJ mol"

2',

we

obtain the required equation, with

AH° = -2809.1 +

3.

+ ±H20(Z)

and divided by

-» yC, 2 H 22 0,,(c)

2822.8

=

13.7 kJ

mol"

Vari ation of Equilibrium Constant with Temperature,

of measuring

AH

will only be

the second law of

method

is

is

5645.5

AH° = If

-» jC, 2 H 22 O n (0

in reverse

6C0 2 (g) + -jH 2 0(/)

into maltose; the reaction

mentioned here very

thermodynamics and

is

1

A third

general method

briefly, since

it is

based on

considered in Section 4.8. This

based on the equation for the variation of the equilibrium constant

K

with the temperature:

If,

therefore,

1/7, the slope

d\nK

A//°

d{\IT)

R

we measure of the line

at

A//7J mol (2.40)

8.3145

A' at a series

of temperatures and plot In

any temperature will be

— A//78.3145

J

K

mol

against -1 ,

and

hence A//° can be calculated. Whenever an equilibrium constant for a reaction can be measured satisfactorily provides a very useful

way of

at

various temperatures, this method thus

obtaining AH°.

reactions that go essentially to completion, in

The method cannot be used

which case a

reliable

obtained, or for reactions that are complicated by side reactions.

for

K cannot be

Thermometer

Ignition

leads

pxygen

inlet

Calorimetry The he n ry

r volvfri in r-nmhiistinn processes ar e

,

—Stirrer

two types of which burnt

are

shown

Figure 2.5.

in

determined

A weighed

bomb

signed to withstand high pressures. The heavy-walled steel

tric

is filled

Water

with oxygen

tion wire,

Oxygen

inlet

Stirrers

is

is

be

to

is

de-

of about 400

mL

more than

by passing an elec-

initiated

evolved rapidly

in the

combustion

two different ways in the two types of calorimeter. In the type of calorimeter shown in Figure 2.5a, the bomb is surrounded by a water jacket which is insulated as much as possible from the surroundings. The water in the jacket is stirred, and the rise in temperature brought about by the combustion is measured. From the thermal characteristics of the apparatus the heat is

determined

in

evolved can be calculated.

Ignition leads

calorimete rs,

a pressure of perhaps 25 atm, this being

cause complete combustion. The reaction

to

a.

Thermometers

at

current through the ignition wire. Heat

process and

Sample

bomb

in

sample of the material

placed in the cup supported in the reaction vessel, or bomb, which

is

volume enough

Air jacket

69

Thermochemistry

2.5

a

known

and

it

A

correction

is

made

for the heat produced in the igni-

customary to calibrate the apparatus by burning a sample having

is

heat of combustion.

The type of calorimeter illustrated in Figure 2.5b is known as an adiabatic calorimeter. The word adiabatic comes from the Greek word adiabatos, meaning impassable, which in turn is derived from the Greek prefix a-, not, and the words dia, through, and bainen, to go. An adiabatic process is thus one in which there is no flow of heat. In the adiabatic calorimeter this

is

achieved by surrounding the inner water

means of a heating coil is maintained at the When this is done the amount of heat supplied

jacket by an outer water jacket which by

same temperature

as the inner jacket.

to the outer jacket just cancels the heat loss to the surroundings. This allows a simpler

determination of the temperature rise due to the combustion; the measured

— Sample Water

(Tfinai

By

initial) is

directly related to the

the use of calorimeters of these types, heats of

in

combustion are

large,

Many Schematic diagrams of

bomb

compounds than

AT

combustion.

is

necessary, since heats

and sometimes we are more interested

ferences between the values for two 2.5

in the

combustion can be measured

with an accuracy of better than 0.01%. Such high precision

evolved

FIGURE

amount of heat evolved

in the dif-

in the absolute values.

other experimental techniques have been developed for the measurement

of heats of reactions. Sometimes the heat changes occurring in chemical reactions of

two types

calorimeter: (a)

A

con-

ventional calorimeter, (b) an adiabatic calorimeter.

are exceedingly small, and

it

is

then necessary to employ very sensitive calorime-

Such instruments are known as microcalorimeters. The prefix micro refers to the amount of heat and not to the physical dimensions of the instrument; some mi-

ters.

crocalorimeters are extremely large.

Another type of microcalorimeter

is

the continuous flow calorimeter,

which

Adiabatic Calorimeter

permits two reactant solutions to be thermally equilibrated during passage through

Microcalorimeter

change

separate platinum tubes and then brought together in a mixing chamber; the heat in the reaction

is

measured.

Relationship between

Bomb

AU and AH

calorimeters and other calorimeters in which the volume

internal energy

therefore give

change AU. Other calorimeters operate

AH

values.

Whether

AU or AH

is

at

is

constant give the

constant pressure and

determined, the other quantity

is

70

Chapter 2

The

First

Law

Thermodynamics

of

easily calculated

we

see that

from the stoichiometric equation

AH = AU + If all reactants

and products are solids or

A(PV)

has a volume of less than

ways be

less than

dm 3

1%

dm 3

1

(i.e.,

liquids, the

,

1

dm 3

At

).

in a reaction will al-

bar pressure, with

1

AV —

,

quite negligible

is

in volume if a reacmol of a solid or liquid

change

and the volume change

less than 0.01

A(PV) = 100 000 Pa X 10 This

2.23

(2.41)

tion occurs at constant pressure is quite small. Usually

0.01

From Eq.

for the reaction.

AH and AU are related by

order of kilojoules, and

-5

m

3

mol

'

=

1.000

compared with most heats of is

much

less than the

J

mol

reaction,

-l

which are of the

experimental error of most determi-

nations. If

gases are involved in the reaction,

AU and AH may

EXAMPLE 2.5

amount of heat produced,

moF

1

at

25°C. Calculate

Solution

— 1364.47

eithe^ as reactants or product s.

thefollowing example.

For the complete combustion of ethanol,

C 2 H 5 OH(/) + 30 2 (g) the

how ever

differ significantly, as illustrated in

Since the -

AC H

bomb

as

2C0 2 (g) + 3H 2 0(/)

->

measured

in

a

bomb

calorimeter,

is

1364.47 kJ

for the reaction.

calorimeter operates at constant volume,

AC U —

The product of reaction contains 2 mol of gas and the reactants, 3 mol; the change An is therefore - 1 mol. Assuming the ideal gas law to apply, A(PV) is equal to AnRT, and therefore to kJ mol

.

(-l)RT = -8.3145 X 298.15

J

-2.48 kJ mol"

mol"

Therefore,

A C H = -1364.47 + The difference between

1366.95 kJ mol

(-2.48)

AU and AH is now

large

enough

to

be experimen-

tally significant.

Temperature Dependence of Enthalpies of Reaction Enthalpy changes are commonly tabulated to

have the values

at

at

25°C, and

it

is

frequently necessary

other temperatures. These can be calculated from the heat

capacities of the reactants and products.

The enthalpy change

in a reaction

can be

written as

AH = H (products) - H (reactants)

(2.42)

2.5

Thermochemistry

Partial differentiation with respect to temperature at constant pressure gives

aan fdAH\ /

ion )H :

\

(products) (products;

\\

/

71

6

d//( reactants)

(2.43)

**

= CP

(products)

- CP

(reactants)

lpJ

= ACP

(2.44)

Similarly,

For small changes

may be

in

temperature the heat capacities, and hence

peratures

T and T2

AH -

If there is a large difference

not satisfactory, and

temperature. This

is

2

a

good approximation

to

1500

CPm a

=

c=

29.07 J

X

20.1

28.99 J

Alternatively,

(2.46)

7/,)

x

T2

,

this

procedure

CP with CPm as a power series:

a

+ bT+ cT 2 +

(2.47)

the values can be calculated over a

first

wide temperature

three terms of this expansion. For hydrogen, for

0.5% over

If

K" mol" '

10~

7

b

'

= -0.836 X

10" 3

J

K" 2 mol"

JK" 3 mor'

1

1

at

273

and somewhat more

e,

each of the

K

and

Q>.,„

satisfactorily,

and/are given

CP

=

32.34 J

we can

K" mol" 1

1

at

„,

in

ACP m for the

reaction will have the

(2.48)

is

AHm (T - AHm {T,) = 2)

T2

leads to

I'

ACP dT

first

deduced

in

J

written in the

form of

same form:

= Ad+ AeT + A/T" 2

These relationships (Eqs. 2.43, 2.44) were

K

Table 2.1.

Integration of Eq. 2.44 between the limits T, and

Kirchhoff (1824-1887).

1500

use an equation of the form

values for products and reactants

AC>.„,

6

K

to

K" mol"

values of d,

Eq. 2.48, the

exam-

from 273

the temperature range

CPjn = d + eT + fT~ 2 Some

,

following constants are used:

These values lead

CPjn =

= ACP (T2 -

between the temperatures T and

values are fitted to within

K if the

A//,

often done by expressing the molar value

range by using only the ple, the

AC V

necessary to take into account the variation of

it is

CPjn = To

and

to give

x

A(AH) =

is

ACP

taken as constant. In that case Eq. 2.44 can be integrated between two tem-

(2.49)

(2.50)

T,

1858 by the German physicist Gustav Robert

72

Chapter 2

The

First

Law

of

Thermodynamics

TABLE

Substance

d

State

He, Ne, Ar, Kr,

Xe

H2 o2 N2

Gas

20.79

Gas

27.28

Gas

H2 H2 C (graphite) NaCl

known

J

2

K

3.26 4.18

Gas

28.58

3.76

Gas

28.41

4.10

44.22

Vapor

30.54

Liquid

75.48

Solid

16.86

8.79 10.29

4.77

45.94

Solid

AHm (T

1

29.96

Gas

2

fT

e

1

K~ mol

J

CO co

CPm = d + eT +

Parameters for the Equation

2.1

16.32

f 1

mol

X X X X X X X X

J

10" 3

5.0

10~ 3

-1.67

10" 3

-5.0

10^ 3

-4.6

10" 3

-8.62

x

)

A//,„(r2 )

is

for

10" 3

x

= AHJTi) + =

AHm {T + x

EXAMPLE 2.6

)

(Ad + AeT + Af T~)

Ad(T2 - T

bomb-calorimetric study of

mol"'. Calculate

Solution

From

Ad =

= Ae =

AH°

+ Q 2 (g)

the values in Table 2.

X

-

(2

=

5.29

X

8.79

X

—

10" 3

(2

- Afly -

7f)

T2

is

25 °C leads 2000 K.

28.41) - 29.96

=

)

-

X

(2

4.10

to

AH° = -565.98

1.66 J

K "'

mol"

X 10~ 3 )

- 4.18

X

1

10" 3

]

/(reactants)

5

10

5 )]

+

(2

JKmol"'

X

0.46

X

10

5

thus

2C0 2 (g)

we obtain

1

JK" 2 mor

10

->

4

10

y

e(reactants)

10

X (-8.62 X

= -14.65 X

X

-3

X

Af = /(products) — [2

10

(2.51)

d(reactants)

-

44.22)

e(products)

=

=

10

5

4

-8.54 X 10 5

dT

this reaction at

for this reaction at

(/(products) (2

+ ^Ae(T22 -

)

4

10

Consider the gas-phase reaction

2CO(g)

A

x

10

10" 3

AH

T = 25°C,

x X X X X

10" 3

the m (T2 ) at any temperature given by this equation, and substitution of Eq. 2.49 leads to If

K mol" 1

5 )

+

1.67

X

10

5

kJ

73

Thermochemistry

2.5

Then, from Eq. 2.52,

A//°(2000K)/Jmor'

565 980 1

-

+

+

-3/ 10'"(2000-

X

5.20

14.65

-

1.66(2000

X

10

5

-

298")

I

298/

,2000

A//°(2000 K)

298)

-565 980

+ 2825 +

-557 169

J

mol"

1

10169

- 4183

= -557.17

mol

kJ

Note that when numerical values are given, it is permissible to drop the subscript from A//°, since the unit kJ mol" avoids ambiguity. Remember that the mole referred to always relates to the reaction as written.

m

1

Enthalpies of Formation The

number of known chemical

total

inconvenient

if

one had

reactions

enormous, and

is

it

would be very

We

to tabulate enthalpies of reaction for all of them.

can

by tabulating molar enthalpies of formation of chemi cal Co mpound s Which arp thp p nthfl lpy nhangps astnri^pH u/ith thp forma t on of m ol nf t he snhstanre from the elements in their standard state s. From these enthalpies of avoid having to do

this

i

,

formation

We

it

is

possible to calculate enthalpy changes in chemical reactions.

have seen

that the standard state of

be the most stable form that

1

we form methane,

which

in at

1

occurs

it

each element and compound at

1

bar pressure and

at

bar and 25°C, from C(graphite) and

is

taken to

25°C. Suppose

H 2 (g),

which are

the standard states; the stoichiometric equation is

+ 2H 2 (g)

C(graphite) It

we

does not matter that

that

we cannot

directly

cannot make

measure

its

methods can be used. In such ways kJ mol"

Standard Enthalpy of Formation

1

CH 4 (g)

-^

occur cleanly and, therefore,

this reaction

enthalpy change; as seen previously, indirect

it

known

is

found

that

A//° for

this reaction is

-74.81

molar enthalpy of formation k f H° of methane at 25°C (298.15 K).ffrfe"term standard enthalpy of formation refers to the enthalpy change when the~compound in its standard state is formed ,

and

this quantity is

from the elements

as the standard

in their standard states;

it

must not be used

Obviously, the standard enthalpy of formation of any element in

in

any other sense.

its

standard state

is

zero. ''A

Efftnalpies of formation of organic their enthalpies of 1

mol of methane

we can 1.

compounds

are

commonly

obtained from

combustion, by application of Hess's law. When, for example, is

burned

in

excess of oxygen, 802.37 kJ of heat

is

evolved, and

therefore write

CH 4 (g) + 20

2 (g)

->

C0 2 {g) + 2H 2 0(g)

A,//°

- -802.37

kJ

mol

74

Chapter 2

The

First

Law

of

Thermodynamics

we have

In addition,

2.

C(graphite)

3.

2H 2 (g) +

If

we add

the following data:

+

2 (g)

AH° = -393.50

C0 2 (g)

AH =

2H 2 0(g)

->

2 (g)

->

kJ

mol"

1

2(-241.83)kJ mol"

reactions 2 and 3 and subtract reaction

1

,

the result

1

is

+ 2H 2 (g) -> CH 4 (g)

C(graphite)

Af H°(CH 4 =

2(-241.84)

)

-

-

393.50

(-802.37)

= -74.80

kJ

mol"

1

many other compounds can be deduced in a similar way. 7 Appendix D gives some enthalpies of formation. The values, of course, depend on the state in which the substance occurs, and this is indicated in the table; the value for liquid ethanol, for example, is a little different from that for ethanol in Enthalpies of formation of

aqueous solution.

(included

in

D

Appendix

are enthalpies of formation of individual ionsAThere

is

an arbitrariness about these values, because thermodynamic quantities can never be

determined experimentally for individual ions;

it

is

always necessary

to

work with

assemblies of positive and negative ions having a net charge of zero. For example,

Af H°

HC1

for

aqueous solution

in

is

—167.15 kJ mol"', but

one can make experimental determinations on the individual

A H°

convention

is

to take

be zero;

it

then follows

to

— 167.15

t

1

on

,

this basis, the

a

Af H°

A//°(NaCl, aq)

= way

H+

is

no way

and Cl~

in its standard state (1

value of

value for the

Af H°

NaCl

in

that

The mol kg" ) ions.

1

for the CI

ion

is

aqueous solution

is

we have

A / //°(Na + ) =

In this

that,

aqueous ion

kJ mol"'. Then, since the

-407.27 kJ mol"

Conventional Standard Enthalpies of Formation

for the

there

H+

whole

set

- Af H°{C\~) = -40121 +

-240.12 kJ mol"

167.15

1

of values can be built up. Such values are often

known

as

conventional standard enthalpies offormation; the word conventional refers to the value of zero for the aqueous proton. In spite of the arbitrariness of the procedure, correct values are always obtained

making calculations

when one

for reactions; this follows

uses these conventional values in

from the

always a

fact that there is

balancing of charges in a chemical reaction. Enthalpies of formation allow us to calculate enthalpies of any reaction, pro-

vided that

we know

any reaction ucts and the

is

the

sum

of the A,

AH° =

7

The

table in

A f H°

the difference

Appendix

D

H°

values for

the reactants

all

between the sum of the values for

all

and products. The

Af H°

values for

all

AH°

for

the prod-

the reactants:

]T kfH° (products)

-

^

Af H°

(reactants)

(2.53)

also includes, for convenience, values of Gibbs energies of formation; these

are considered in Chapter 3.

EXAMPLE 2.7 sis

Calculate, from the data in

H NCONH 2 (a

75

Thermochemistry

2.5

3 (a