Download 15-1 Note 15 Properties of Bulk Matter

Note 15
Properties of Bulk Matter
Sections Covered in the Text: Chapter 16
In this note we survey certain concepts that comprise
the macroscopic description of matter, that is to say,
matter in bulk. These include temperature, pressure
and changes in phase. We shall touch on the various
states of matter, namely solids, liquids and gases, but
we shall focus primarily on gases, the simplest macroscopic systems. We have already discussed certain
macroscopic aspects of fluids in Notes 13 and 14.
These ideas were organized by scientists who are seen
today as the founders of modern chemistry.
Matter in Bulk
By matter in bulk we mean matter in a quantity to be
seen, touched or weighed. At a given temperature and
pressure a substance is either a solid, liquid or gas.
Water, the best-known substance, is a solid (ice) at a
pressure of 1 atm and temperature of 0 ˚C and below.
Between 0 ˚C and 100 ˚C water is a liquid, and above
100 ˚C it is a gas (in the form of steam or water vapor).
The solid, liquid and gaseous states of a substance are
called phases. When a substance melts (or freezes) and
boils (or condenses) it is said to undergo a phase
change. We shall discuss phase changes in more detail
later in this note. The three phases of matter can be
described succinctly as follows.
Solid
A solid is a rigid macroscopic system with a definite
shape and volume. It consists of particle-like atoms
connected together by molecular bonds. Each atom
vibrates about an equilibrium position, but an atom is
not free to move around inside the solid. A solid is
nearly incompressible, meaning that the atoms are
about as close together as they can get.
Liquid
Like a solid, a liquid is nearly incompressible too. A
liquid flows and deforms to fit the shape of its
container. The molecules are held together by weak
molecular bonds and are therefore free to move
around.
Gas
Each molecule in a gas moves through space as a free,
noninteracting particle until, on occasion, it collides
with another molecule or with the wall of the container. A gas, like a liquid, is a fluid. It is also highly
compressible, meaning that a lot of space exists
between molecules.
State Variables
Matter in bulk is described by so-called state variables.
These variables include volume V, pressure p, mass M,
mass density ρ, thermal energy Eth and temperature T. If
any one state variable is changed then the state of the
system as a whole is changed.
All Matter
The fundamental building block of all matter is the
atom. An example is atomic hydrogen. However, some
states of matter, many gases for example, are best
described in terms of a grouping of atoms called a
molecule. To form a molecule two or more atoms bind
together via a combination of electrostatic and
quantum mechanical processes that we shall not go
into here. The complexity of molecules varies widely
from the simplest, hydrogen gas, which consists of
only two atoms of hydrogen, to the complex, like
DNA, that typically consists of many thousands of
atoms.
Atoms and Molecules
The idea that matter consists of indivisible entities
called “atoms” dates back 2500 years. The Greek
philosopher Democritus (460-370 BC) speculated that
repeated subdivision of matter would eventually
yield a smallest unit which could not be further
subdivided. The earliest quantitative ideas were
formulated at the beginning of the 19th century by
Dalton (1800) and others.
A Few Definitions
From experiments too numerous to go into here we
know today that an atom consists of a nucleus made
up of protons and neutrons, and electrons moving in
stable orbits about the nucleus. A proton has a
positive charge +e, an electron a negative charge –e
and a neutron no charge. The best measurements of e
yield the value 1.609 x 10 –19 Coulombs (C) (to 4
significant digits). An atom is electrically neutral; the
number of electrons surrounding the nucleus equals
the number of protons in the nucleus. 1
The atomic number Z is the number of protons in the
nucleus (or the number of electrons circling the nucleus). The atomic mass number A is the number of protons plus neutrons in the nucleus. The number A is
1
We shall be discussing electric charge in detail starting in Note
20.
15-1
Note 15
conventionally written as a leading superscript on the
symbol for the atom, for example 1H for hydrogen.
Isotopes are atoms that differ only in the number of
neutrons in the nucleus. Atomic hydrogen has 3
isotopes, hydrogen 1H, deuterium 2H and tritium 3H.
A selection of elements and their A numbers is listed
in Table 15-1.
Table 15-1. A selection of elements and their atomic mass
numbers
Element
A
1
H Hydrogen
1
7
He Helium
4
12
C Carbon
12
14
N Nitrogen
14
16
O Oxygen
16
the atomic and molecular masses are nearly integers.
The fourth column in Table 15-2 contains the atomic
or molecular mass rounded to an integer; this number
is sufficiently accurate to use in most calculations.
NOTE: An element’s atomic mass (e.g., 1.0078 for 1H) is
not the same as the atomic number (1 for 1H). The
atomic number, which is the element’s position in the
periodic table, is the number of protons in the
element’s nucleus.
Mole and Molar Mass
In chemistry where substances in everyday amounts
are weighed on a balance and mixed it is convenient
to use what is called a molar mass. (Note the use of the
lower case mole, not Mole.) By definition,
one mole of a substance (abbreviated 1 mol), whether
solid, liquid or gas, is the amount of substance that
contains as many basic particles as there are atoms in 12
g of 12C. 3
Atomic and Molecular Mass
The mass of an atom or molecule is expressed most
fundamentally in the absolute unit kg. The mass of an
atom of atomic hydrogen is 1.674 x 10 –27 kg (expressed
to 4 significant digits). This is a very small number.
It is convenient to express an atomic (and molecular)
mass not in this absolute unit but in a relative unit.
The relative unit is based on a scale in which the mass
of the isotope of carbon, 12C, is taken to be exactly 12
units—atomic mass units (abbreviated u). That is,
m(12C) = 12 u. The atomic mass of any other atom is its
mass relative to 12C. For example, careful experiments
with hydrogen have established that the mass ratio
m(1H)/m(12C) is 1.0078/12. Thus the atomic mass of
hydrogen is m(1H) = 1.0078 u. Some examples of
atomic masses are listed in Table 15-2. 2
Table 15-2. Examples of atomic and molecular masses.
The approximate values can be used in calculations.
Type
atomic
molecular
atomic
atomic
molecular
molecular
Symbol
1
H
H2
12
C
N
N2
NH3
At. mass (u)
1.0078
2.016
12.000
14.007
28.014
17.031
Approx (u)
1
2
12
14
28
17
This number, denoted NA , is called Avogadro’s number.
NA has units; they are number of basic particles per mole.
Thus€the number of moles n in a substance containing
N basic particles is
n=
N
.
NA
…[15-1]
€
12(g)
m=
= 1.993 × 10−23 g = 1.993 × 10−26 kg .
6.02 × 10 23
€
The author of our textbook uses the lower-case letter m to denote
the mass of an atom or molecule, and we shall do likewise.
15-2
N A = 6.02 ×10 23 mol–1.
Since NA 12C atoms have a mass of 12 g, the mass of a
single 12C atom is
A molecule’s molecular mass is the sum of the atomic
masses of the atoms forming the molecule. Notice that
2
A basic particle may be an atom or a molecule. For
example, the basic particle of Helium, which is a
monatomic gas, is the helium atom. The basic particle
of oxygen gas, which is a diatomic molecule, is the
oxygen molecule. The number of basic particles in a
mole has been carefully measured. It has the value
The conversion factor between atomic mass units and
kilograms is
3
The definition of the mole given here may differ from its definition in your chemistry text or in the physics text you used in high
school. This is the definition given in our textbook, and to avoid
confusion the definition we shall use.
Note 15
12
1u =
€
m( C) 1.993 × 10
=
12
12
−26
(kg)
24
= 1.661 × 10−27 kg.
n=
Some examples of molar masses are listed in Table 153. Thus the molar mass of a substance is the mass in
grams of 1 mol of the substance. We shall denote the
molar mass by M mol. The number of moles n in a
system of mass M (g) consisting of atoms or molecules
with molar mass Mmol is given by:
n=
M(g)
.
M mol
Example Problem 15-1
Calculating the Number of Moles in a Quantity of
Oxygen
How many moles are contained in 100 g of oxygen
gas?
Solution:
The basic particle in oxygen gas is the oxygen molecule (O2). The molecular mass of the O2 molecule is m
= 32 u. Converting this to kg, we get the mass of one
molecule to be
1.661 × 10−27 (kg)
= 5.31× 10−20 kg.
1u
The number of molecules in 100 g = 0.100 kg is
N=
€
€
It is important to note that M and Mmol in eq[15-2] are
expressed in grams not kilograms. Let us consider an
example using moles.
€
Alternatively, we can use eq[15-2] to find n more
directly
n=
M(g)
100(g)
=
= 3.13 mol.
−1
M mol (g.mol ) 32(g.mol−1 )
…[15-2]
Table 15-3. A selection of molar masses
Substance
€
molar mass (g)
molecular hydrogen (H2)
2.016
atomic nitrogen (N)
14.007
ammonia (NH3)
17.031
m = 32(u) ×
N 1.88 ×10
=
= 3.13 mol.
N A 6.02 ×10 23
M
0.100(kg)
=
= 1.88 ×10 24 .
−20
m 5.31×10 (kg)
The number of moles is therefore this number divided
by Avogadro’s number:
The Periodic Table
Electronic properties of elements are summarized in
the periodic table of the elements. The table was
invented by Mendeleev in 1869. He found that by
arranging the elements in order of increasing atomic
mass, regularly-recurring properties were revealed.
By so arranging the 62 elements then known, he
predicted that unknown elements would soon be
discovered to fill the gaps. With the guidance of his
predictions of their chemical and physical properties,
three new elements were discovered in the next few
years, providing striking confirmation of the general
correctness of his ideas. However, a full understanding of the periodic table had to await the development
of the quantum mechanics of many electron atoms.
We shall leave this aspect of matter to a course in
chemistry.
Another topic concerning matter in bulk is thermodynamics. Thermodynamics is the study of the flow of
heat. The laws of thermodynamics relate heat flow,
work, and thermal energy. In order to study the flow
of heat, it is necessary to introduce a new physical
quantity called temperature. But before doing so we
introduce two related concepts, thermal contact and
thermal equilibrium.
Two objects are said to be in thermal contact if heat can
be exchanged between them without work being
done.
If two objects are placed in thermal contact with each
other, and no net heat exchange occurs between them,
then they are said to be in thermal equilibrium. 4
We are now in a position to discuss temperature in
qualitative terms.
€
4
We don’t yet know precis ely what heat is. For the moment we
shall have to be content with our intuitive sense of the concept. We
shall be discussing heat in detail in Notes 16, 17 and 18.
15-3
Note 15
Temperature
We live with the idea of temperature every day. We
know that temperature has something to do with
“hotness” or “coldness”. We know that we can
“measure” the hotness or coldness with a device
called a thermometer.
The idea of temperature can be understood by considering a number of everyday situations. Consider
two objects that are not in thermal equilibrium. If they
are placed in thermal contact, then heat flows between
them until they reach thermal equilibrium.
If two objects A and B are not in thermal contact,
then how can we tell whether they are in thermal
equilibrium with each other? We can tell by doing the
following:
solid responds to a change in temperature. In other
words, we need to know something about the thermal
expansion/contraction of the substance.
Thermal expansion
Thermal expansion is a macroscopic property. It is a
consequence of the temperature-induced change in
the average separation between atoms. There are two
kinds of thermal expansion: linear expansion and
volume expansion.
Linear expansion is commonly described by an
expression of the form:
ΔL
= αΔT ,
L0
1 Place a thermometer C in thermal contact with A,
and wait for thermal equilibrium to occur (i.e. wait
until the reading of the thermometer is steady).
2 Now place thermometer C in thermal contact with
B, and wait for thermal equilibrium to occur. If the
two readings of the thermometer are the same, then
A and B are in thermal equilibrium with each other.
where α is the coefficient of linear expansion and L 0 is the
length of the solid at STP. 5 Linear expansion only
applies to €
solids. For solids α ≅ 10 –5 T –1.
Volume expansion is commonly described by an expression of the form
These observations are expressed by the zeroth law of
thermodynamics.
ΔV
= βΔT
V0
Zeroth law of thermodynamics
The zeroth law of thermodynamics can be stated in
these words:
where β is the coefficient of volume expansion and V0 is
the volume at STP. Volume expansion applies to
solids, liquids
gases. For liquids β ≅ 10 –4 T–1 while
€ and
–4
–1
for gases β ≅ 10 T .
If two objects A and B are separately in thermal equilibrium with object C, then they are in thermal equilibrium
with each other.
Temperature Scales
In order to establish a quantitative scale of temperature, it is necessary to adopt some particular kind of
thermometer.
Thermometers
A thermometer measures the temperature of a system
by displaying the change in some physical property,
such as
a)
b)
c)
d)
e)
f)
length of a solid
volume of a liquid
pressure of a gas at constant volume
volume of a gas at constant pressure
electrical resistance of a conductor
colour of a very hot object
To understand the working of a thermometer based
on a) or b) we need to know something about how a
15-4
The most commonly-used scale of temperature in the
world today is the Celsius scale. On this scale the
freezing point of water at 1 atm pressure is set at 0,
and the boiling point of water is set at 100. Between
these two points are 100 Celsius degrees. A
temperature of 10 degrees Celsius is written 10 ˚C.
The Fahrenheit scale (that actually predates the Celsius
scale) is still in use in the U.S. On the Fahrenheit scale
the freezing point of water is set at 32 and the boiling
point of water is set at 212, making for 180 Fahrenheit
degrees between the two points. Thus a temperature
of TC ˚C and TF ˚F are related by
9
TF = TC + 32 .
5
5
…[15-2]
STP stands
€ for standard temperature and pressure, or 0 ˚C and 1
atm. STP was introduced in Note 13.
Note 15
Both scales have disadvantages, making it desireable
to define an absolute scale of temperature. This we do
in the next section.
2 Whatever gas is used in the bulb the corresponding
straight-line graph extrapolates to the same temperature at zero pressure, namely T0 = –273 ˚C.
Absolute Zero and Absolute Temperature
These facts form the basis of the absolute temperature
scale. The temperature at zero pressure is taken to be
the zero of the absolute temperature scale. (Zero
pressure implies zero movement of gas molecules and
therefore zero thermal energy and zero temperature.)
On this scale the temperature of –273 ˚C is taken to be
zero—zero Kelvin, denoted 0 K. The freezing point of
water (0 ˚C) is thus 273 K. This means that one Kelvin
unit and one Celsius unit are of the same size. On this
scale there is no negative temperature. The absolute
temperature scale is the SI scale of temperature.
The conversion between the Celsius scale and the
Kelvin scale is
One of the most important scientific thermometers is
the constant volume gas thermometer (Figure 15-1). The
working of the thermometer is based on the fact that
the absolute pressure of a gas in a sealed bulb container (whose volume cannot change) varies linearly
with temperature. A pressure gauge is provided for
measuring the pressure. The bulb is immersed in the
system whose temperature is to be measured.
TK = TC + 273.
…[15-3]
Phase Changes
€ knowledge that water can exist in three
It is common
Figure 15-1. A constant volume gas thermometer (a) and
typical results for three gases (b).
Before using the thermometer it is first calibrated by
recording the pressure at two reference temperatures,
the boiling and freezing points of water. The points
are plotted on a graph (Figure 15-1b) and a line drawn
between them. The bulb is then brought into contact
with the system whose temperature is to be measured.
The pressure is measured then the corresponding
temperature is read off the graph.
With this thermometer you would observe two facts:
1 The gas pressure depends linearly on temperature
phases, solid (ice), liquid (water) and gas (water vapor
or steam). An idealized apparatus for the study of the
phase changes of water is sketched in Figure 15-2a. A
quantity of ice at an initial temperature of –20 ˚C is
placed in a sealed container equipped with a thermometer. The container is then heated (an open flame
will do nicely). The heating is done slowly enough to
ensure that the thermometer registers a temperature
that applies to the whole container.
The kind of graph of temperature versus time that
would be obtained with this apparatus is shown in
Figure 15-2b. The temperature rises from the initial
–20 ˚C to 0 ˚C where it remains for some time. During
this time the ice melts and the temperature of the icewater mixture remains at 0 ˚C; the solid-liquid
mixture is said to be in a phase equilibrium.
Once all the ice has melted the liquid water begins to
warm. Warming takes place uniformly until the temperature of 100 ˚C is reached, at which temperature
boiling occurs. During this time the water is transformed from the liquid to the gas phase. As the water
boils the temperature remains at a constant 100 ˚C; the
liquid-gas mixture is now in a phase equilibrium.
Once all the water has boiled off the system is completely gaseous. Subsequent warming just raises the
temperature of the gas (steam).
Unhappily, the results in Figure 15-2b apply only if
the apparatus is at a pressure of 1 atm. Thus the graph
only applies to a location at sea level. To avoid this
disadvantage many experiments have been carried
15-5
Note 15
out at various pressures yielding the so-called phase
diagrams shown in Figures 15-3. Figure 15-3a for water
shows three regions corresponding to the solid, liquid
and gas phases. The boundary lines separating the
regions delineate the phase transitions.
Figure 15-2. An idealized apparatus for the study of phase
changes in water (a) and the temperature-time graph
obtained with the apparatus (b).
A point of special interest on the diagram for water is
the triple point where the phase boundaries meet. The
triple point is the one value of temperature and
pressure for which all three phases can coexist in
phase equilibrium. For water, the triple point occurs
at T3 = 0.01 ˚C and p3 = 0.006 atm.
The triple point of water is actually used to complete
the definition of the Kelvin temperature scale. Recall
that the Celsius scale required the boiling and melting
points of water as two reference points. These
temperatures are not satisfactory because their values
depend on pressure. The Kelvin scale requires only
one reference point since the low end, zero temperature at zero pressure, is fixed. The triple point of
water is taken as the second reference point because it
is a fixed, identifiable point (same pressure and
temperature) regardless of the location of study.
The Kelvin temperature scale is therefore defined to
be a linear temperature scale starting from 0 K at
absolute zero and passing through the temperature
15-6
273.16 K at the triple point of water.
Figure 15-3. Phase diagrams (not to scale) for water (a) and
carbon dioxide (b).
Because the temperature at the triple point of water is
T3 = 0.01 ˚C on the Celsius scale, absolute zero on the
Celsius scale is T 0 = –273.15 ˚C. This completes the
definition of the absolute temperature scale.
A gas is the simplest macroscopic system. A real gas is
known to consist of small, hard atoms or mole-cules
moving randomly at high speeds which, on occasion,
collide with each other and with the walls of the
container. No matter how simple a real gas is, it is still
useful to model it as a so-called ideal gas.
The Idea of an Ideal Gas
The potential energy curve of two atoms or molecules in an ideal gas can be represented as in Figure
15-4. This curve represents the interaction of two hard
spheres that have no interaction at all until they come
into actual contact, at separation r contact, and then
bounce. It is found experimentally that the ideal gas
model is a good model for real gases if the following
conditions are met:
Note 15
constant. This means we can write eq[15-4] as
pV = nRT .
…[15-5]
Eq[15-5] is called the ideal gas law. Remarkably, all
gases yield the same graph and the same value of R.
For a gas€in a sealed container the number of moles n
is constant. Thus we can write from eq[15-5]:
pV
= nR = const .
T
Figure 15-4. The “idealized” potential energy diagram for
two atoms or molecules in an ideal gas.
This means that for any two states i and f of the gas,
p f V f piVi
=
.
Tf
Ti
€
1 the density is low
2 the temperature is well above the condensation
point
…[15-7]
The ideal gas law written in terms of the number N of
molecules in the gas is
The Ideal Gas Law
Recall that the macroscopic state of a system is described by the state variables—volume V, number of
moles n, temperature T and pressure p.
Numerous experiments have shown that for any gas,
whether monatomic, diatomic or polyatomic, a plot of
pV vs nT yields the kind of graph shown in Figure 155. The pressure p must be in units of Pa and temperature T must be in units of Kelvin. The graph is a
straight line, with pV and nT being related by
pV = (const)nT .
…[15-6]
€
N
R
pV = nRT =
RT = N
T.
NA
NA
…[15-8]
The factor R/NA is known as Boltzmann’s constant and
is denoted kB :
€
kB =
…[15-4]
R
= 1.38 ×10−23 J.K –1.
NA
Thus the ideal gas law can be written in terms of kB :
pV = NkB T .
€
€
…[15-9]
Eqs[15-5] and [15-9] are equivalent.
€
Figure 15-5. A graph of pV vs nT for an ideal gas.
The slope of the graph, found experimentally, is
const = 8.31 J.mol –1K–1.
The constant, denoted R, is called the universal gas
€
Ideal Gas Processes
The state of an ideal gas changes in a process in which
p changes, V changes or T changes. These changes are
most easily represented on a so-called pV diagram
(Figures 15-6). Each state of the gas is represented as a
point on the diagram (Figure 15-6a).
A process in which a gas changes state is
represented by a path or trajectory on the pV diagram
(Figure 15-6b). Different processes take a gas from one
state to another via different trajectories (Figure 156c).
The ideal gas law applies only to gases in thermal
equilibrium. Thus the kind of processes we shall
consider here are those which keep the gas in thermal
equilibrium as it is taking place. These are processes
15-7
Note 15
that take place very slowly and are therefore called
quasistatic processes. A quasistatic process is also a
reversible process.
We now consider three of the simplest processes in
turn, a constant volume process, a constant pressure
process, and a constant temperature process.
process appears on a pV diagram as a vertical line.
Figure 15-7. Illustration of an isochoric process.
Example Problem 15-2
The Constant Volume Gas Thermometer:
An Isochoric Device
Figure 15-6. A pV diagram of an ideal gas
Constant Volume Process
A constant volume process is called an isochoric process. In such a process
V f = Vi .
…[15-10]
An idealized representation of an isochoric process
involving a gas in a container with an unchangeable
volume is shown
in Figures 15-7. If the gas is heated
€
its pressure rises but its volume remains the same.
The path between any two states of the gas is shown
in Figure 15-7b. If the gas were cooled a similar
process would occur in reverse. Thus an isochoric
15-8
A constant volume gas thermometer is placed in contact with a reference cell containing water at the triple
point. After reaching equilibrium, the gas pressure is
recorded as 55.78 kPa. The thermometer is then placed
in contact with a sample of unknown temperature.
After the thermometer reaches a new equilibrium, the
gas pressure is 65.12 kPa. What is the temperature of
the sample?
Solution:
This is an example of an isochoric process. The
temperature of the triple point of water is T1 = 0.01 ˚C
= 273.16 K. The ideal gas law gives
p2V2 p1V1
=
,
T2
T1
which, for V1 = V2 yields for the new temperature
€
Note 15
T2 = T1
p2
65.12(kPa)
= (273.16K)
p1
55.78(kPa)
The gas pressure inside the cylinder is
p = patmos +
= 318.90 K = 45.75 ˚C.
€ Keep in mind here that pressures are expressed in Pa,
temperatures in K.
Constant Pressure Process
A constant pressure process is called an isobaric process. In such a process
p f = pi .
…[15-11]
An idealized apparatus for producing an isobaric process is shown in Figure 15-8. A cylinder of gas has a
tight-fitting€piston of cross sectional area A on which
is placed a mass m. As the temperature is raised the
volume increases but the pressure remains
unchanged. If the temperature is lowered the volume
decreases but still the pressure remains unchanged.
mg
.
A
…[15-12]
This pressure is constant so long as m remains constant. Thus an isobaric process appears on a pV
diagram
€ as a horizontal line.
Example Problem 15-3
An Isobaric Process
A gas occupying 50 cm 3 at 50 ˚C is cooled at constant
pressure until the temperature is 10 ˚C. What is its
final volume?
Solution:
This is a constant pressure, isobaric process. Using the
ideal gas law as in Example Problem 15-2 but this
time putting p 2 = p1 we have for the final volume:
V2 = V1
T2
(10 + 273)K
= (50cm 3 )
= 43.8 cm3
T1
(50 + 273)K
The volume decreases as it must if the temperature
decreases.
€
Constant Temperature Process
A constant temperature process is called an isothermal
process. In such a process
Tf = Ti .
…[15-13]
Using the ideal gas law we can put eq[15-13] into the
form:
€
or
Figure 15-8. Illustration of an isobaric process.
pfVf
pV
= Tf = Ti = i i
nR
nR
p f V f = piVi .
…[15-14]
€
An idealized
apparatus for studying an isothermal
process is shown in Figure 15-9.
The container
of gas equipped with a movable close€
fitting piston is held in thermal equilibrium with a
constant temperature bath. If the piston is moved
downwards (slowly) or upwards (slowly) then the gas
and the bath remain at thermal equilibrium and therefore at the same temperature.
15-9
Note 15
Example Problem 15-4
An Isothermal Process
A gas cylinder of the type shown in Figure 15-9a
containing 200 cm 3 of air at 1.0 atm pressure initially
floats on the surface of a swimming pool in which the
water is at 15 ˚C. The cylinder is then slowly pulled
underwater to a depth of 3.0 m. What is the volume of
the gas cylinder at this depth?
Solution:
The water in the swimming pool acts as a constant
temperature bath. This process is an isothermal one.
The pressure of the gas, initially 1.0 atm, increases to a
value
pwater = p0 + ρgd
The ideal gas law gives, for T2 = T1,
€
V2 = V1
p1
p0
= V1
p2
p0 + ρgd
Plugging in numbers we obtain
€
Figure 15-9. Illustration of an isothermal process.
In an isothermal process both p and V change. Since T
is constant we have
p=
nRT const
=
.
V
V
…[15-15]
V2 = 155 cm3.
As a result of the pressure under water the volume of
the cylinder is reduced from 200 cm 3 to 155 cm3.
Isochoric, isobaric and isothermal processes are
special cases. An arbitrary process may consist of a
number of these cases followed one after the other.
This is called a multistep process. Such a multistep
process is illustrated in Figure 15-10. We shall treat it
as an example problem.
This means that an isothermal process at some temperature T follows an hyperbolic path on a pV diagram
(as shown
€ in Figures 15-9b and c). Each curve of the
family of hyperbolas (one for each temperature) is
called an isotherm.
Figure 15-10. An example of a multistep process.
15-10
Note 15
Example Problem 15-5
An Example of a Multistep Process
A gas at 2.0 atm pressure and a temperature of 200 ˚C
is first expanded isothermally until its volume has
doubled. It then undergoes an isobaric compression
until it returns to its original volume. First show this
process on a pV diagram. Then find the final temperature and pressure.
Solution:
This process is shown on the pV diagram of Figure 1510. The gas starts in state 1 at pressure p 1 = 2.0 atm
and volume V1. As the gas expands isothermally it
moves downward along an isotherm until it reaches
volume V2 = 2V 1. The pressure decreases during this
process to a lower value p 2. The gas is then compressed at constant pressure p2 until its final volume
V3 equals its original volume V 1. State 3 is on an
isotherm closer to the origin, so we expect to find T 3 <
T1.
Since from eq[15-15], p1 = nRT1/V1 and p2 = nRT2/V2
we have
p2  nRT2  V1  V1 1
=
= .

=
p1  V2  nRT1  V2 2
p2 =
Thus
€
1
p1 = 1.0 atm.
2
During the isobaric process we have V3 = V 1 = V 2/2
and so
€
T3 =
V3
V /2 1
T2 = 2 = T2
V2
V2
2
= 236.5 K = –36.5 ˚C
Thus,
€ indeed, T3 < T1.
15-11
Note 15
To Be Mastered
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Definitions: atomic mass unit, mole, Avogadro’s number, molar mass
Definitions: Thermal contact, thermal equilibrium
Statement: zeroth law of thermodynamics
Definitions: Celsius temperature scale, Fahrenheit temperature scale, kelvin temperature scale
Definitions: coefficient of linear expansion, coefficient of volume expansion
Physics of: Ideal gas, the ideal gas law
Physics of: Ideal gas processes: isochoric, isobaric, isothermal
Physics of: pV diagram of an ideal gas
Typical Quiz/Test/Exam Questions
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15-12