Thermal Physics
Thermal Physics
Thermal concepts
3.2
Thermal properties of matter
3
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3.1
TOK introduction
‘In looking at the ways of knowing described in the
heory of knowledge guide, scientists could legitimately
claim that science encompasses all these. Driven
by emotion, using sense perception, enhanced by
technology and combined with reason, it communicates
through language, principally the universal language of
mathematics.’ (© IBO 2007). Science cannot suppose to
be the truth as so many paradigm shits have occurred
over the centuries. It is hoped that the following
historical perspective will help in showing how science
has changed the way of reasoning in thermal physics.
Phlogiston/caloric Theory
T
he concept of heat has been studied for many centuries.
Aristotle (384 –322 B.C.) considered ire one of the
ive basic elements of the Universe. Over 2000 years ago,
Greek philosophers believed that matter was made of
“atomos”, elemental atoms in rapid motion, and that the
result of this rapid motion was heat. It was understood
that heat lowed from hot bodies to colder ones, somewhat
analogous to water or another luid lowing from a higher
to lower elevation. It is not surprising that the early theory
of heat low regarded heat as a type of luid.
Around the time of Galileo Galilei (1564 –1642), this
heat luid was known as phlogiston – the soul of matter.
Phlogiston was believed to have a negative mass, and,
upon heating or cooling, the phlogiston was driven out or
absorbed by an object.
Further reinements of the phlogiston theory were carried
out by Antoine Lavoisier (1743–1794), and it became
known as the caloric theory. Sir Isaac Newton (1642–1727)
and other famous scientists supported the caloric theory.
Calorists believed that a hot object had more caloric than
a cold object. hey explained expansion by saying that the
caloric illed up the spaces between atoms pushing them
apart. he total amount of caloric was unchanged when a
hot and cold body came into contact.
However, the caloric theory did not adequately explain
some phenomena involving heat. It was diicult to
understand how the conservation of caloric luid applied
to friction and the expansion of liquids and gases. Some
calorists’ answer to the friction concept was that the latent
heat was released which implies that a change of state was
involved. Others argued that during friction the material
is “damaged” and that it “bleeds” heat. No satisfactory
answers were forthcoming.
count rumford
Much of the credit for dismantling the idea that heat was
motion rather than substance or caloric goes to Benjamin
hompson (1753 –1814), also known as Count Rumford of
Bavaria.
During the American Revolution, he was a Tory or loyalist
in the disputes between Britain and its American colonies
serving as a major in a company of militia. It is believed
that he invented a cork lotation system for cannons while
being transported by horses across rivers. He also designed
a gun carriage that could be carried by three horses and
could be assembled ready for iring in 75 seconds. He
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CHAPTER 3
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was knighted by King George III of England, and made a
Count in 1791 by heodor in his brief reign as elector of
the Holy Roman Empire.
In 1793, hompson let England ultimately to take up a post
with the before mentioned heodor, elector of Bavaria. He
was appointed a major general in the Bavarian army. He
designed fortiications and worked as an administrator
in munitions. It was here that he observed that a large
amount of heat was generated in the boring of cannons.
He read the following extracts before the Royal Society of
London in 1798.
‘Being engaged, lately, in superintending the boring
of cannon, in the workshops of the military arsenal at
Munich, I was struck with the very considerable degree of
Heat which a brass gun acquires, in a short time, in being
bored; and with the still more intense Heat (much greater
than that of boiling water as I found by experiment) of the
metallic chips separated from it by the borer.’
‘From whence comes the Heat actually produced in the
mechanical operation above mentioned? Is it furnished by
the metallic chips which are separated by the borer from
the solid mass of metal?”… “If this were the case, then,
according to the modern doctrines of latent heat, and of
caloric, the capacity for Heat of the parts of the metal, so
reduced to chips, ought not only to be changed, but the
change undergone by them should be suiciently great to
account for all the Heat produced.’
Count Rumford was saying that the metal chips should
have undergone some alteration in their properties ater
the production of so much thermal energy. He noted that
some cannon shavings were hot enough to glow, but he
continued:
“But no such change had taken place; for I found, upon
taking equal quantities, by weight, of these chips, and of
thin slips of the same block of metal separated by means
of a ine saw, and putting them, at the same temperature
(that of boiling water), into equal quantities of cold water,
the portion of water into which the chips were put was
not, to all appearance, heated either less or more than the
other portion, into which the slips of metal were put. From
whence it is evident that the Heat produced [by boring the
cannon] could not possibly be furnished at the expense of
the latent Heat of the metallic chips.”
Rumford further went on to explain that he had immersed
cannons in water while they were being bored and noted
the rate at which the temperature rose. His results showed
that the cannon would have melted had it not been cooled.
Rumford concluded that heat was not a caloric luid in
74
which caloric is conserved but rather a concept of motion.
He argued that heat is generated when work is done, and
that the heat will continue to be generated as long as work
is done. He estimated a heat to work ratio within the order
of magnitude accepted today.
However, many scientists of the time were not convinced
because Rumford could not give a clear explanation of
exactly what heat was in terms of the accepted model for
matter at that time. It would take another half century
before Joule supplied the accepted answers.
James Prescott Joule
James Prescott Joule (1818-1889) conducted a series of
brilliant experiments between 1842 and 1870 that proved
beyond doubt that heat was a type of energy – internal
energy – of the particles of matter. he caloric theory lost
popularity very quickly.
Joule was the son of a wealthy brewer in Manchester, UK.
Because of his wealth, he never worked for a living. His
experiments were performed in a laboratory that he built
at his own expense while he was in his twenties. He became
interested in ways to develop more eicient engines that
were driving various components of the brewing process.
Encouraged by the work of Count Rumford and others,
he began to investigate whether mechanical work could
produce heat.
Joule performed a variety of experiments and he reined
and elaborated his apparatus and his techniques. In one
of his irst experiments, he used a falling weight to drive
a small electric generator. he current produced heated a
wire that was immersed in a deinite mass of water, and
the change in temperature was noted. He reasoned that
the work done as the weight decreases its gravitational
potential energy should be equivalent to the heat energy
gained by the water. In another experiment he mounted a
large container illed with air into a tub of water. When the
air was compressed, the temperature of the gas increased.
He measured the amount of work needed to compress the
gas and the amount of heat energy given to the water as a
result of compression.
Perhaps Joule’s most famous experiment consisted of a
paddlewheel mounted inside a cylinder of water that was
driven by falling weights as shown in Figure 301. He wanted
to see if one could raise the temperature of the water simply
by turning the paddles. He repeated this experiment many
times continually improving the apparatus and reining
his analysis of the data. For example, he took great care
to insulate the container so that no heat was lost to the
surroundings, and he developed his own thermometer so
Thermal Physics
that he could measure the temperature with a precision of
a fraction of a degree.
Example
flywheel
spindle
to pulley with
weights atached
Calculate the mechanical equivalent of heat for Joule’s
paddlewheel experiment if a mass of 2.0 kg falls through a
height of 100 m, and increases the temperature of 10 g of
water by 46.8 °C.
to pulley with
weights atached
moving vanes
Solution
calorimeter
(containing water)
fixed vanes
Work done by the falling mass is given by
W = Ep = mg∆h
Figure 301 Schematic diagram of
Joule’s paddlewheel experiment.
= 2.0 kg × 9.8 m s-2 × 100 m
= 1.96 × 10 3 J
Joule arranged the vanes of the paddlewheel so that they
would not interfere with the particles of water set in
motion. He didn’t want to bruise or damage the water
particles so that they might “bleed” heat.
In 1849 he published his results in which he reported
“…the quantity of heat produced by friction of bodies,
whether solid or liquid, is always proportional to the
quantity of [energy] expended.”
Heat energy produced is∆given by
Q = m × c × ∆T
= 10 g × 1 cal × 46.8 °C
∆
= 4.68 × 102 calories
Mechanical equivalent of heat is given by
3
“…the quantity of heat capable of increasing the
temperature of a pound of water by 1 ° Fahrenheit requires
for its evolution the expenditure of a mechanical energy
represented by the fall of 772 pound through the distance
of one foot”.
1.96 × 10 J
W
----- = --------------------------------2
Q
4.68 × 10 cal
= 4.19 J cal-1
he mechanical equivalent of heat for water is 4.2 J cal -1.
Joule found that about 4.2 joules of work would yield one
calorie of heat or that the quantity of heat required to
raise the temperature of one gram of water by 1 °C is one
calorie.
A modern day value for the mechanical equivalent of heat
is 4.18605 joules = 1 calorie.
he experiments proved beyond doubt that mechanical
work can produce heat and as such no caloric luid can
be created or destroyed. Furthermore, Joule reasoned that
the temperature increase must be related to the energy of
the microscopic motions of the particles.
Finally, a paradigm shit in our way of reasoning had again
proved that science is not the ultimate truth.
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CHAPTER 3
3.1 Thermal cOncePTs
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3.1.1 State that temperature determines the
direction of thermal energy transfer
between two objects.
3.1.2 State the relation between the Kelvin and
Celsius scales of temperature.
3.1.3 State that the internal energy of a
substance is the total potential energy and
random kinetic energy of the molecules of
the substance.
3.1.4 Explain and distinguish between the
macroscopic concepts of temperature,
internal energy and thermal energy (heat).
3.1.5 Define the mole and molar mass.
3.1.6 Define the Avogadro constant.
© IBO 2007
3.1.1 TemPeraTure and Thermal
energy Transfer
Everyone seems to have a feel for the concept of heat
because there is so much talk in everyday conversation
of how hot or how cold it is. We endure the seasons of
the year wanting either to cool our bodies or heat up our
surroundings. We are aware of how diicult it was for
our ancestors to keep warm, and our present dwellings
are designed and insulated to suit the climate. Our
consumption of electrical and chemical energy for heating
and other purposes is a continual concern. We have
become aware that increased global warming could spell
the end of the world, as we know it.
But what is the diference between heat and temperature
in physics? hey are deinitely not the same physical
quantity. If you ill a cup and a dish with hot water at the
same temperature, and then place ice cubes into each, the
dish full of hot water can melt more ice cubes than the
cup of hot water even though the water in each was at the
same temperature. he dish containing the larger amount
of hot water has a greater mass of water as well as a greater
amount of heat or thermal energy. A greater mass infers
a greater number of water molecules and more thermal
energy infers that these molecules would have a greater
overall amount of energy. We cannot see the interaction
76
of the water molecules at the microscopic level because
we cannot see atoms. However, we can observe and
monitor the temperature change by using a macroscopic
temperature-measuring device.
hermal energy is the kinetic energy of the component
particles of an object and is measured in joules. Heat is the
thermal energy that is absorbed, given up or transferred
from one object to another.
Temperature is a scalar quantity that gives an indication of
the degree of hotness or coldness of a body. Alternatively,
temperature is a macroscopic property that measures
the average kinetic energy of particles on a deined scale
such as the Celsius or Kelvin scales. he chosen scale
determines the direction of thermal energy transfer
between two bodies in contact from the body at higher
temperature to that of lower temperature. Eventually,
the two bodies will be in thermal equilibrium when they
acquire the same temperature in an isolated system. It will
be deduced later in this text that thermal energy cannot
be transferred from a body at lower temperature to that of
higher temperature.
3.1.2 TemPeraTure scales
here is no instrument that directly measures the amount of
thermal energy a body gives of or absorbs. A property that
varies with temperature is called a thermometric property.
his property can be used to establish a temperature scale
and construct a thermometer. hermometers are made
using the thermometric properties of a substance such as the
expansion of a liquid, the electrical resistance of a wire.
A typical laboratory thermometer as shown in Figure 302
contains a liquid such as mercury or coloured alcohol. he
expansion of alcohol is six times greater than mercury.
Alcohol thermometers are safer and can be used at lower
temperatures than mercury which turns to a solid below
-38.9 °C. Its disadvantage is that it boils above 78.5 °C. To
make the thermometer sensitive, it has a narrow bore tube
and a large bulb. he bulb is made of thin glass so that
heat can be transferred quickly between the bulb liquid
and the material being observed. here is a vacuum above
the thermometer liquid and it can move easily along the
glass bore.
Thermal Physics
points). For example, diferent thermometers will give
diferent values for the boiling point of zinc (907 °C).
100°C
he standard fundamental temperature scale in the SI
system is denoted by the symbol T. and is measured in
Kelvin, K. It is the thermodynamic temperature scale used
in scientiic measurement.
capillary tube
glass
stem
0 °C
mercury or alcohol
thin glass
bulb
he lower ixed point is absolute zero and is assigned a
value of 0 K. his is the point where molecular motion
becomes a minimum – the molecules have minimum
kinetic energy but molecular motion does not cease. he
upper ixed point is the triple point of water. his is the
temperature at which saturated water vapour, pure water
and melting ice are all in equilibrium. For historical
reasons, it is assigned a value of 273.16 K.
T in K = T in °C + 273.16
Figure 302 Typical laboratory thermometer
Exercise
A clinical thermometer as shown in Figure 303 does not
need the temperature range of a laboratory thermometer.
It is designed so that the maximum temperature remains
constant ater the patient’s temperature is taken. It has a
small constriction to stop the mercury lowing back into
the bulb. he mercury is then shaken back into the bulb
ater the temperature has been taken.
1.
At room temperature, an iron rod feels cooler
when held in the hand than wood held in the same
hand. his is because:
A.
B.
35
36
37
38
39
40
41
42
C.
Figure 303 A Clinical Thermometer
D.
In order to calibrate these thermometers, two ixed points
are used to deine the standard temperature interval. he
ice point (the lower ixed point) marked at 0 °C is the
temperature of pure ice at standard atmospheric pressure
and is in thermal equilibrium with the liquid in the bulb.
he steam point (the upper ixed point) marked at 100 °C
is the temperature of steam at standard atmospheric
pressure and is in thermal equilibrium with the liquid in
the bulb. he scale between these values is marked with
even spaces. he Celsius temperature scale named ater
the Swedish astronomer Anders Celsius (1701-1774) is
constructed in such a manner.
Although thermometers constructed using thermometric
properties are useful for everyday use, they are not
accurate enough for scientiic work. Unfortunately, two
thermometers constructed using diferent thermometric
properties do not necessarily agree with each other as they
do not vary linearly over large temperature ranges. (hey
are of course in agreement at the lower and upper ixed
3.1
thermal energy tends to low from the
metal to the wood
wood has a higher speciic heat capacity
than the iron rod
wood has a lower speciic heat capacity than
the iron rod
the iron rod conducts thermal energy better
than the wood
2.
Explain the diference between heat and
temperature.
3.
If you were travelling to Antarctica, deduce
what would be the better thermometer to take
– mercury or alcohol?
4.
State one advantage and one disadvantage of a
i.
ii.
mercury in glass thermometer
constant volume thermometer.
5.
he triple point of water is 273.16 K. Express this
as a Celsius temperature.
6.
Determine the ice point and the steam point of
pure water on the Kelvin scale?
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vacuum
CHAPTER 3
7.
Deine absolute zero.
8.
If normal body temperature is 37.0 °C, what is it
on the thermodynamic temperature scale?
hermal energy of a system is referred to as internal energy
- the sum total of the potential energy and the random
kinetic energy of the molecules of the substance making
up the system.
gas
equilibrium separation
separation, r / m
r0
liquid
ε
–
solid
minimum potential energy = –ε
Figure 305 Potential energy
versus separation of particles
he potential energy is due to
•
•
the energy stored in bonds called bond energy
intermolecular forces of attraction between particles.
he bond energy is a form of chemical potential energy. It
becomes signiicant in chemistry when a chemical reaction
occurs, and bonds are broken and formed.
he intermolecular forces of attraction between particles
is due to the electromagnetic fundamental force since the
gravitational force is too small to be of any signiicance.
Figure 304 indicates how the intermolecular electromagnetic force F between particles varies with the distance
r between their centres.
At distances greater than r0 (less than 2.5 × 10-10 m)
attraction takes place, and at distances closer than r0 the
particles repel. At r0 the particles are in equilibrium. Any
displacement from the equilibrium position results in a
simple harmonic oscillation of a particle or molecule.
Work done = force × distance = F × r = change in
potential energy
∆ Ep
∴ F = ____
r
In other words, the gradient of the potential energy curve
at any point on the curve gives the force that must be
applied to hold the molecules at that separation. We can
classify the phases according to the sizes of the energy ε.
ε , the vibrations occur about ixed
When less than ___
10
positions and the particles are in the solid phase. When
ε , the particles have suicient
approximately equal to ___
10
energy to partly overcome the attractive forces and melting
occurs.
ε , a liquid can form. When greater
When greater than ___
10
than ε, the particles have suicient energy to leave the
liquid and form a gas.
Force, F / N
he kinetic energy is mainly due to the translational,
rotational and vibrational motion of the particles as
depicted in Figure 306.
repulsion
+
nuclear seperation, r / m
10r 0
r0
attraction
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3.1.3 heaT and inTernal energy
Potential energy, U / J
+
–
gaseous phase
solid and liquid phase
Figure 304 Force versus separation of particles
Figure 305 shows the relationship between the potential
energy and the separation r of two molecules. At 0 K, the
average separation of particles centres is r0 and the overall
force is zero. his is the point of minimum potential
energy. Work will need to be done to move the particles
apart and there will be an increase in potential energy.
78
Vibrational
kinetic energy
Rotational
kinetic energy
Translational
kinetic energy
Figure 306 major particle motion and energy
Thermal Physics
3.1.4 macrOscOPic cOncePTs Of
shop you smell the aroma of the cofee due to difusion
caused by the kinetic energy of the air molecules allowing
the aroma molecules to spread.
internal energy and thermal
energy (heat)
A macroscopic property is one that can be observed.
Physical properties such as melting point, boiling point,
density, thermal conductivity, thermal expansion and
electrical conductivity can be observed and measured.
TemPeraTure
At the macroscopic level, temperature is the degree
of hotness or coldness of a body as measured by a
thermometer. hermometers are made using the
thermometric properties of a substance such as:
•
•
•
•
•
•
•
the expansion of a column of liquid in a capillary
tube (laboratory and clinical thermometers).
the electrical resistance of a wire (resistance and
thermister thermometers).
the diference in the rates of expansion of two
metals in contact (bimetallic strips).
the pressure of a gas at constant volume.
the volume of a gas at constant pressure (gases
expand by a greater amount and more evenly than
liquids).
the heating of two metal wires wound together
(thermocouple thermometers rely on the two
metals producing diferent currents).
the colour of a solid heated to high temperatures
(pyrometers).
Temperature is measured in Kelvin, K.
he expansion of solids, liquids and gases is a macroscopic
property that allows us to understand that matter in a
system has potential energy. When you heat a liquid it can
be seen to expand as in a thermometer and this means
that the potential energy of the system is increasing as
the molecules move further apart. he compressibility of
gases allow us to understand that the potential energy of
the molecules is decreasing.
Although the internal energy in the examples above can
never be absolutely determined, the change in internal
energy can be observed.
Thermal energy (heaT)
“Students should understand that the term thermal
energy refers to the non-mechanical transfer of energy
between a system and its surroundings”.
© IBO 2007
herefore, it is incorrect to refer to the thermal energy in a
body. At the macroscopic level, thermal energy (heat) can
be transferred from one body to another by conduction,
convection, radiation or by combinations of these three.
hermal conduction is the process by which a temperature
diference causes the transfer of thermal energy from the
hotter region of the body to the colder region by particle
collision without there being any net movement of the
substance itself.
hermal convection is the process in which a temperature
diference causes the mass movement of luid particles
from areas of high thermal energy to areas of low thermal
energy (the colder region).
inTernal energy
As already mentioned, internal energy is the sum total of
the potential energy and the random kinetic energy of the
molecules of the substance making up a system. In order
to apply the Law of conservation to thermal systems, one
has to assume that a system has internal energy.
At the macroscopic level, it can be observed that molecules
are moving. When pollen (a ine powder produced by
lowers) is sprinkled on the surface of water and the setup is viewed under magniication, it can be seen that the
smaller pollen particles carry out zig-zag motion called
Brownian Motion. heir motion is caused by the kinetic
energy of the larger water molecules. Walking past a cofee
hermal radiation is energy produced by a source because
of its temperature that travels as electromagnetic waves. It
does not need the presence of matter for its transfer.
Conduction can occur in solids, liquids and gases. In
gases it occurs due to the collision between fast and slow
moving particles where kinetic energy is transferred
from the fast to the slow particle. he transfer of energy
is very slow because the particles are far apart relative
to solids and liquids. In liquids, a particle at higher
temperature vibrates about its position with increased
vibrational energy. Because the majority of the particles
are coupled to other particles they also begin to vibrate
more energetically. hese in turn cause further particles
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TemPeraTure
CHAPTER 3
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to vibrate and thermal conduction occurs. his process is
also slow because the particles have a large relative mass
and the increase in vibrations is rather small. In solids, the
transfer can occur in one of two ways. Most solids behave
similarly to liquids.
However, solids are held in their ixed positions more
rigidly than liquids and the transfer of vibrational energy
is more efective. However, again their large masses do not
allow for large energy transfer. If a substance in the solid
or molten form has mobile electrons, as is the case for
metals, these electrons gain energy due to the temperature
rise and their speeds increase much more than those held
in their ixed positions in the lattice. Metals are said to be
good conductors of heat but most other solids are good
insulators. Saucepans for cooking are usually made of
copper or aluminium because these metals conduct heat
quickly when placed on a stove. he handle is made from
a good solid insulator to reduce the conduction of heat.
Generally, liquids and gases are not good thermal
conductors. However, they can transfer heat readily by
convection. Ocean currents, wind and weather patterns
suggest that the mass movement of particles from one area
to another can cause movement of particles on the grand
scale.
Figure 307 shows a potassium permanganate crystal placed
in water inside a convection tube. Heat is applied for a
short period of time and the direction of the purple trail
is noted. Particles in a region of high thermal energy are
further apart and hence their density is lower. In a region
of low thermal energy the particles are closer together and
the region is more dense. As a result, the less dense region
rises as it is pushed out of the way by the more dense
region and a convection current is produced.
Convection tube
water
Heat
Figure 307
Potassium
permanganate
crystal
Convection current.
Another way in which a luid can move is by forced
convection. In this case, a pump or fan system maintains
the movement of a luid. he cooling system in nuclear
reactors operates on this principle.
80
All thermal energy ultimately comes from the Sun in our
solar system. It travels through 150 million km of mostly
empty space. At the Earth’s atmosphere the radiant energy
is mainly relected back into space. However, some is
transmitted and absorbed causing a heating efect. Just as
the Sun emits thermal radiation so does any source that
produces heat such as a light bulb or an electric heater.
hermal radiation is mainly electromagnetic waves in
the infra-red region of the electromagnetic spectrum at
temperatures below 1000 °C. Above this temperature,
wavelengths of the visible and ultra-violet regions are
also detected. Dull black bodies are better absorbers and
radiators than transparent or shiny bodies.
3.1.5,6
The mOle,
mOlar mass and
aVOgadrO’s numBer
he mass of an atom is exceedingly small. For example,
the mass of a luorine atom is 3.16 × 10-23 g and the mass of
the isotope carbon–12 is 1.99 × 10-23 g. Because the masses
of atoms are so small, it is more convenient to describe
the mass of an atom by comparing its mass with those of
other atoms.
In 1961 the International Union of Pure and Applied
Chemistry (IUPAC) deined the masses of atoms relative to
carbon–12 that was assigned a value of 12.0000. herefore,
the relative atomic mass is deined as the mass of an atom
when compared with 1/12 the mass of carbon–12 atom.
Just as the relative atomic mass is used to describe the
masses of atoms, the relative molecular mass is used to
describe the masses of molecules. herefore, the relative
molecular mass is deined as the mass of a molecule when
compared with 1/12 of the mass of a carbon–12 atom.
It is convenient to group things into quantities. For
example, a box of diskettes, a ream of photocopy paper
(500 sheets), a dozen eggs are common groupings. he
SI fundamental unit for the amount of a substance is the
mole (mol).
he mole is the amount of substance that contains as
many elementary particles as there are in 0.012 kg of
carbon–12.
Amadeo Avogadro (1776 – 1856) found that equal volumes
of gases at the same temperature and pressure contained
the same number of particles. One mole of any gas contains
the Avogadro number of particles NA. It is now known
Thermal Physics
that one mole of a gas occupies 22.4 dm3 at 0 °C and
101.3 kPa pressure (STP) and contains 6.02 × 10 23
particles.
1.
64.0 g
m = _________________
= 2 mol
n = __
M ( 16.0 + 16.0 ) g mol–1
2.
he number of oxygen molecules
= 2NA × the number of molecules
If we have one mole of NH4NO3, this contains:
= 6.02 × 1023 × 2
1 mol of NH4+ ions = 6.02 × 10 23 ammonium ions
1 mol of NO3– ions = 6.02 10 23 nitrate ions
2 mol of nitrogen atoms = 1.204 × 1024 nitrogen atoms
4 mol of hydrogen atoms = 2.408 × 1024 hydrogen atoms
3 mol of oxygen atoms = 1.806 × 1024 oxygen atoms
= 1.024 × 1024 molecules.
3.
Volume
= 2 mol × 22.4 dm3
= 44.8 dm3
he amount of substance (the moles) is related to the mass
and the molar mass according to the following equation:
m
n = __
M
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When using the mole, the atoms or molecules should be
clearly stipulated. For example, one mole of copper atoms
contains 6.02 × 10 23 copper atoms. One mole of nitrogen
molecules (N2) contains 6.02 × 10 23 of nitrogen molecules
and 12.04 × 1023 nitrogen atoms.
Solution
4.
m=nM
= 0.75 mol × (12 +16 +16) g mol-1
where
= 33 g
n = amount of a substance in mol, m = the mass in g and
M = the molar mass in g mol-1. he molar mass can be
obtained from the periodic table.
Example
1.
Calculate the number of moles of oxygen
molecules contained in 64.0 g of oxygen gas, O2.
2.
Calculate the number of oxygen molecules inpart
1 of this example.
3.
Determine the volume of oxygen gas that would
be present at STP
4.
Calculate the mass in 0.75 mol of carbon dioxide gas.
81
CHAPTER 3
6.
Exercise
1.
cOre
B.
C.
D.
he number of moles of sodium chloride (NaCl)
in 100g of pure sodium chloride is (Mr of NaCl =
58.5 gmol-1)
A.
B.
C.
D.
3.
B.
C.
D.
8.
from the lower temperature object to the
higher temperature object
half way between the temperatures of the
two objects
from the higher temperature object to the
lower temperature object
in many diferent directions
16
8
4
2
5. he number of molecules present in 0.5 mol SO3 is
A.
B.
C.
D.
3 × 1023
6 × 1023
12 × 1023
24 × 1023
100 g of copper
5.0 g of oxygen molecules
100 g of calcium carbonate, CaCO3
4.4 g of carbon dioxide
13.88 g of lithium
A sample of aluminium sulfate Al2 (SO4)3 has a
mass of 34.2 g. Calculate:
(a)
(b)
11.
2.0 mole of iron, Fe
0.2 mole of zinc, Zn
2.5 mole of carbon dioxide, CO2
0.001 mole of sulfur dioxide, SO2
50 mole of benzene, C6H6
Calculate the amount of subtance (number of
mole) in:
(a)
(b)
(c)
(d)
(e)
10.
Cl2
HCl
CuSO4
Na2CO3
CH4
Calculate the mass of the given amounts of each of
the following substances:
(a)
(b)
(c)
(d)
(e)
9.
3 × 1023
6 × 1023
12 × 1023
24 × 1023
Calculate the approximate molar masses of each of
the following molecules and compounds:
(a)
(b)
(c)
(d)
(e)
A sealed lask contains 16 g of oxygen (mass
number 16) and also 8 g of hydrogen (mass
number 1). he ratio of the number atoms of
hydrogen to the number of atoms of oxygen is
A.
B.
C.
D.
82
7.
5850 mol
0.585 mol
1.71 mol
41.5 mol
Two diferent objects with diferent temperatures
are in thermal contact with one another. When the
objects reach thermal equilibrium, the direction of
transfer of thermal energy will be
A.
4.
the total potential energy stored in the
bonds of a substance
the potential and kinetic energy of
molecules in a substance
the energy stored in bonds and
intermolecular forces of a substance
the translational, rotational and vibrational
motion of particles in the substance
he number of atoms present in 0.5 mol SO3 is
A.
B.
C.
D.
he internal energy of a substance is equal to:
A.
2.
3.2
the number of alumimium ions Al 3+ in the
sample
the number of sulfate ions SO4 2- in the
sample
Classify the following as a macroscopic or
microscopic property of a gas
(a)
(b)
(c)
(d)
(e)
volume
speciic heat capacity
kinetic energy of a particle
pressure
temperature
Thermal Physics
3.2 Thermal PrOPerTies Of maTTer
used on the back of refrigerators because of its low heat
capacity.
Example
3.2.2 Solve problems involving specific heat
capacities and thermal capacities.
3.2.3 Explain the physical differences between
the solid, liquid and gaseous phases in
terms of molecular structure and particle
motion.
he heat capacity of a sphere of lead is 3.2 × 103 JK-1
Determine how much heat can be released if the
temperature changes from 61 °C to 25 °C.
3.2.4 Describe and explain the process of phase
changes in terms of molecular behaviour.
3.2.5 Explain in terms of molecular behaviour
∆ a
why temperature does not change during
⇒∆
∆
phase change.
3.2.6 Distinguish between evaporation and
boiling.
cOre
3.2.1 Define specific heat capacity and thermal
capacity.
Solution
∆Q
= -------- ⇒ ∆Q = Heat capacity × ∆T
× ∆∆T
Heat capacity
3
–1
× ( 61 – 25 ) °C
3
–1
× 36 °C
= 3.2 × 10 JK
= 3.2 × 10 JK
×
×(
×
×
)°
°
= 115200J
3.2.7 Define specific latent heat.
×
5
= 1.2 × 10 J
3.2.8 Solve problems involving specific latent
heats.
© IBO 2007
•
Note that a change in kelvin temperature is the
same as a change in celsius temperature
3.2.1 Thermal caPaciTy
3.2.2 sPecific heaT caPaciTy
When diferent substances undergo the same temperature
change they can store or release diferent amounts of
thermal energy. hey have diferent thermal (heat)
capacities. If a substance has a high heat capacity it will
take in the thermal energy at a slower rate than a substance
with a low heat capacity because it needs more time to
absorb a greater quantity of thermal energy. hey also
cool more slowly because they give out thermal energy at
a slower rate.
Heat capacity does not take into account the fact that
diferent masses of the same substance can absorb or
release diferent amounts of thermal energy.
We deine the thermal (heat) capacity as,
Consider three one kilogram blocks of aluminium, zinc
and lead with the same sized base that have been heated
to the same temperature of 80 °C. hey are quickly placed
on top of a large block of candle wax for a time period as
shown in Figure 308.
A luminium
zinc
lead
∆Q
Heat capacity = ___ J K -1
∆T
∆Q is the change in thermal energy in joules J
CA NDL E WAX
∆T is the change in temperature in kelvin degrees K.
Water is used in car cooling systems and heating systems
because of its high heat capacity. A metal heat sink is
Figure 308 Front on view of the
metal blocks after a period of time.
83
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CHAPTER 3
he aluminium block melts the most wax and the lead
melts the least. herefore, the metals of the same mass
give out diferent amounts of thermal energy in a certain
time period. his can be explained from a microscopic
viewpoint. he kilogram unit masses have diferent
numbers of particles of diferent types and masses. he
metal blocks were given the same amount of thermal
energy when they were heated to 80 °C. When the thermal
energy gained by each metal is distributed amongst its
particles, the average energy change of each particle will
be diferent for each metal.
Example
Determine how much heat can be released if the
temperature changes from 61 °C to 25 °C.
Solution
Using the fact that ∆Q = m.c.∆T, we have,
To obtain a characteristic value for the heat capacity
of materials, equal masses of the materials must be
considered. he physical property that includes the mass
is called the speciic heat capacity of a substance c.
Speciic heat capacity or speciic heat is the heat capacity
per unit mass. It is deined as the quantity of thermal
energy required to raise the temperature of one kilogram
of a substance by one degree Kelvin.
∆Q
c = _____
m∆T
∆Q = 0.650 kg × 9.1 × 102 J kg-1 K-1 × (80 – 20) K
= 3.549 × 104 J
= 4 × 104 J
hat is, 4 × 104 J of heat is released.
Example
∆Q = m × c × ∆T
∆Q = the change in thermal energy required to produce a
temperature change J
m = mass of the material in kilograms kg
∆T = the temperature change K
Note that ∆T is always positive because heat always
transfers from the higher temperature region to the lower
temperature region.
For gases, the molar heat capacity at constant volume Cv
and the molar heat capacity at constant pressure Cp are
more commonly used. Molar heat capacity is the quantity
of heat required to raise the temperature of one mole of the
gas by one degree Kelvin under the constant condition.
Figure 309 shows the speciic heat capacity for some
common substances at room temperature (except ice)
Substance
Lead
Mercury
Zinc
Brass
Copper
Speciic heat
J kg -1 K -1
1.3 × 102
1.4 × 102
3.8 × 102
3.8 × 102
3.85 × 102
Substance
Iron
Aluminium
Sodium
Ice
Water
Speciic heat
J kg -1 K -1
4.7 × 102
9.1 × 102
1.23 × 103
2.1 × 103
4.18 × 103
Figure 309 Specific heat of some common substances
84
An active solar heater is used to heat 50 kg of water initially
at a temperature of 12 °C. If the average rate that thermal
energy is absorbed in a one hour period is 920 J min-1 ,
determine the equilibrium temperature ater one hour
Solution
Quantity of heat absorbed in one hour = 920 J min-1 ×
60 min = 5.52 × 104 J
Using the fact that ΔQ = m c ΔT , we have
5.52 × 104 J = 5.0 × 101 kg × 4.18 × 103 J kg-1 K-1 × (Tf – 12)K
5.52 × 104 J = 2.09 × 105 JK-1 × (Tf – 12) °C
5.52 × 104 J = 2.09 × 105 Tf – 2.51 × 106 J
2.09 × 105 Tf = 5.52 × 104 J + 2.51 × 106 J
2.553 × 106 J
Tf = ______________
2.09 × 105 ( J K–1 )
Tf = 12.26 °C
∴ Tf = 12 °C
Thermal Physics
exTensiOn - meThOds TO deTermine
sPecific heaT caPaciTy
A calorimeter is a useful piece of equipment for
investigations in hermal Physics because it allows masses
at diferent temperatures to be mixed with minimum
energy loss to the surroundings. It is used for direct and
indirect methods in determining the speciic heat capacity
of a substance. (he name of the instrument is derived
from the Imperial unit, the calorie.)
he calorimeter is also insulated with lagging materials
such as wool or polystyrene to reduce heat loss due to
conduction and convection.
Ater the power supply is switched of, the temperature
should continue to rise for a period, and then level out for
an ininite time. However, heat is lost to the surroundings,
and the maximum temperature that could be achieved,
in theory, is never reached. Instead appreciable cooling
occurs. One method used to estimate the theoretical
maximum temperature is to use a cooling correction
curve as shown in Figure 310.
{Note that cooling correction is not required in the syllabus
but is included for possible extended essays.}
stirrer
joulemeter
copper vessel
liquid
heating coil
Temperature, °C
thermometer
theoretical curve
θ3
θ2
θ1
∆θ (= correction)
θ ( = θ2 – θ1 )
actual curve
A1
A2
A
∆θ = -----1- × θ , so that θ3 = θ2 + ∆ θ
A2
lagging
room
temp
Figure 310 Calorimeter being used to measure the
heating effect of a current
Figure 310 illustrates the use of a calorimeter to determine
the speciic heat capacity of a liquid, in this case water.
he heating coil is used to convert electrical energy to
thermal energy. he electrical energy can be measured by
a joulemeter or by using a voltmeter/ammeter circuit. he
duration of time of electrical input is noted.
he thermal energy gained by the calorimeter cup and the
water is equal to the electrical energy lost to the calorimeter
cup and water.
Electrical energy lost =
V × I × t = [m × c × ∆T]calorimeter cup + [m × c × ∆T]water
where V is the potential diference across the heating coil
in volts V and I is the current in the amperes A.
he speciic heat capacity of the calorimeter cup is obtained
from published values. he other quantities are recorded
and the speciic heat capacity of the water is calculated.
In calorimeter investigations, heat losses to the
surroundings need to be minimised. It is normal to polish
the calorimeter cup to reduce loss of heat due to radiation.
t
2t
3t
Time, minutes
Figure 311 Graph of cooling correction.
A cooling correction is based on Newton’s Law of Cooling.
It states that the rate of loss of heat of a body is proportional
to the diference in temperature between the body and its
surroundings (excess temperature). A full explanation of
this Law will not be given. If the power supply is switched of
at time 2t minutes, then the temperature should continue
to be recorded for a further t minutes. he correction to the
temperature θ can be obtained from the graph as shown.
he inal temperature is then given as the inal temperature
of the thermometer plus the correction θ.
Another direct electrical method used to determine the
speciic heat capacity of a metal is shown in Figure 311.
An immersion heater is placed into a metal block. he
hole for the heater is lubricated with oil to allow even
heat transmission. he electrical energy lost to the block
is recorded for a given period of time and the speciic
heat of the metal is calculated. Cooling correction is more
important in this case because the temperatures under
which the investigation is carried out could be much
higher than was the case when using a calorimeter.
85
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herefore the temperature has not changed considering
the degree of uncertainty in the measurements.
CHAPTER 3
We also have that,
thermometer
V
immersion
heater
cOre
= (3.0 kg) (3.85 × 102 J kg -1 K -1) (90.0 –Tf) K
A
metal block
hermal energy lost by the copper
Low voltage
supply
hermal energy gained by the water
lagging
= (2.0 kg) (4.18 × 103 J kg -1 K -1) (Tf – 20.0) K
Figure 312 Electrical Method Using
An Immersion Heater And A Metal Block.
hermal energy gained by the cup
= (0.21 kg) (9.1 × 102 J kg -1 K -1) (Tf – 20.0) K
A common indirect method to determine the speciic
heat capacity of a solid and liquids is called the method of
mixtures. In the case of a solid, a known mass of the solid
is heated to a certain temperature, and then transferred
to a known mass of liquid in a calorimeter whose speciic
heat capacity is known. he change in temperature is
recorded and the speciic heat of the solid is calculated
from the results obtained. In the case of a liquid, a hot
solid of known speciic heat is transferred to a liquid of
unknown speciic heat capacity.
1.04 × 105 – 1.155 × 103 Tf
= (8.36 × 103 Tf – 1.67 × 105) +(1.91 × 102 Tf – 3.82 × 103)
hat is,
– 9.71 × 103 Tf = – 2.75 × 105
Giving
Tf = 28.3 °C
Example
he inal temperature of the water is 28 °C
A block of copper of mass 3.0 kg at a temperature of
90 °C is transferred to a calorimeter containing 2.00 kg of
water at 20 °C. he mass of the copper calorimeter cup is
0.210 kg. Determine the inal temperature of the water.
Exercise
1.
Solution
he amount of thermal energy required to raise
the temperature of 1.53 × 103g of water from 15 K
to 40 K is
A.
B.
C.
D.
he thermal energy gained by the water and the calorimeter
cup will be equal to the thermal energy lost by the copper.
3.3
1.6 × 107 J
1.6 × 105 J
4.4 × 107 J
4.4 × 105 J
hat is, [mc∆T]copper = [mc∆T]calorimeter cup + [mc∆T]cup
2.
he speciic heat capacity of a metal block of mass
m is determined by placing a heating coil in it, as
shown in the diagram above.
he block is electrically heated for time t and the
maximum temperature change recorded is Δθ. he
constant ammeter and voltmeter readings during
the heating are I and V respectively. he electrical
energy supplied is equal to VIt.
86
9.
If 2.93 × 106 J is used to raise the temperature of
water from 288 K to 372 K, calculate the mass of
water present.
10.
5.4 × 106 J of energy is required to heat a 28 kg
mass of steel from 22 °C to 450 °C. Determine the
speciic heat capacity of the steel.
11.
A piece of iron is dropped from an aeroplane at
a height of 1.2 km. If 75% of the kinetic energy
of the iron is converted to thermal energy on
impact with the ground, determine the rise in
temperature.
12.
If 115 g of water at 75.5 °C is mixed with 0.22 kg of
water at 21 °C, determine the temperature of the
resulting mixture.
13.
Describe an experiment that would allow you to
determine the speciic heat capacity of a metal.
thermometer
V
immersion
heater
A
metal block
lagging
3.
Low voltage
supply
∆θ
he speciic heat
∆θ capacity is best calculated using
which one of the following expressions?
∆θ
m∆θ
A.
c=
∆θ
VI
VI
∆θ
B.
c=
m∆θ
VIt
C.
c = ∆∆θθ
m∆θ
m∆θ
D.
c = ∆θ
VIt
∆∆θθ
∆tθ
6
θt
5.4 × 10 J of∆energy
is required to heat a 28 kg
mass of steel tfrom 22 °C to 450 °C. Determine the
speciic heat capacity of the steel.
4.
Liquid sodium is used as a coolant in some
nuclear reactors. Describe the reason why liquid
sodium is used in preference to water.
5.
6.00 × 102 kg of pyrex glass loses 8.70 × 106 J of
thermal energy. If the temperature of the glass was
initially 95.0 °C before cooling, calculate is its inal
temperature.
(i)
(ii)
(iii)
(iv)
(v)
14.
A heating luid releases 4.2 × 107 Jkg-1 of heat as it
undergoes combustion. If the luid is used to heat
250 dm3 of water from 15 °C to 71 °C, and the
conversion is 65% eicient, determine the mass
of the heating luid that will be consumed in this
process.
15.
A large boulder of 125 kg falls of a clif of height
122 m into a pool of water containing 120 kg of
water. Determine the rise in temperature of the
water. Assume that no water is lost in the entry of
the boulder, and that all the heat goes to the water.
16.
A thermally insulated container of water is
dropped from a large height and collides
inelastically with the ground. Determine the
height from which it is dropped if the temperature
of the water increases by 1.5 °C.
(Take the speciic heat capacity of pyrex glass to be
8.40 × 10 2 J kg -1 K-1)
6.
A piece of wood placed in the Sun absorbs more
thermal energy than a piece of shiny metal of the
same mass. Explain why the wood feels cooler
than the metal when you touch them.
7.
A hot water vessel contains 3.0 dm3 at 45 °C.
Calculate the rate that the water is losing thermal
energy (in joules per second) if it cools to 38 °C
over an 8.0 h period.
8.
Determine how many joules of energy are released
when 870 g of aluminium is cooled from 155 °C to
20 °C.
Sketch the apparatus.
Describe what measurements need to be
made and how they are obtained.
State and explain the equation used to
calculate the speciic heat capacity of the
metal.
Describe 2 main sources of error that are
likely to occur in the experiment.
Is the experimental value likely to be higher
or lower than the theoretical value, if the
experiment was carried out in a school
laboratory? Explain you answer.
87
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Thermal Physics
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CHAPTER 3
17.
A piece of copper is dropped from a height of
225 m. If 75% of its kinetic energy is converted to
heat energy on impact with the ground, calculate
the rise in temperature of the copper. (Use the
table of speciic heat capacities to ind the value for
copper).
18.
5kg of lead shot is poured into a cylindrical
cardboard tube 2.0 m long. he ends of the tube
are sealed, and the tube is inverted 50 times. he
temperature of the lead increases by 4.2 °C. If
the speciic heat of lead is 0.031 kcal kg-1 °C-1,
determine the number of work units in joules that
are equivalent to the heat unit of 1 kilocalorie.
3.2.3 Phase sTaTes
An understanding of thermal energy is based upon a theory
called the moving particle theory or kinetic theory (for
gases) that uses models (Figure 314) to explain the structure
and nature of matter. he basic assumptions of this moving
particle theory relevant to thermal energy are:
•
•
•
•
all matter is composed of extremely small particles
all particles are in constant motion
if particles collide with neighbouring particles, they
conserve their kinetic energy
a mutual attractive force exists between particles
solid
liquid
gas
Figure 314 Arrangement of
particles in solids, liquids and gases
An atom is the smallest neutral particle that represents an
element as displayed in a periodic table of elements. Atoms
contain protons, neutrons and electrons and an array of
other sub-atomic particles. Atomic diameters are of the
order of magnitude 10–10 m. Atoms can combine to form
molecules of substances. In chemistry, the choice of the
terms e.g. ‘atoms, molecules, ions’ are speciic to elements
and compounds. In physics, the word ‘particle’ is used to
describe any of these speciic chemistry terms at this stage
of the course.
As previously mentioned, evidence for the constant motion
of particles can be gained from observation of what is
known as Brownian Motion. If pollen grains from lowers
88
are placed on water and observed under a microscope, the
pollen grains undergo constant random zig-zag motion.
he motion becomes more vigorous as the thermal energy
is increased with heating. A Whitley Bay smoke cell uses
smoke in air to achieve the same brownian motion. In
both cases, the motion is due to the larger particles (water
and air) striking the smaller particles (pollen and smoke)
and causing them to move.
he large number of particles in a volume of a solid, liquid
or gas ensures that the number of particles moving in all
directions with a certain velocity is constant over time.
here would be no gaseous state if the particles were losing
kinetic energy.
A mutual attractive force must exist between particles
otherwise the particles of nature would not be combined
as we know them. Further explanation of this assumption
will be given later in this topic.
Matter is deined as anything that has mass and occupies
space. here are four states of matter which are also called
the four phases of matter – solids, liquids, gases and
plasma. Most matter on Earth is in the form of solids,
liquids and gases, but most matter in the Universe is in the
plasma state. Liquids, gases and plasma are luids.
A plasma is made by heating gaseous atoms and molecules
to a suicient temperature to cause them to ionise. he
resulting plasma consists then of some neutral particles but
mostly positive ions and electrons or other negative ions.
he Sun and other stars are mainly composed of plasma.
he remainder of this chapter will concentrate on the other
three states of matter, and their behaviour will be explained in
terms of their macroscopic and microscopic characteristics
of which some are given in Figures 315 and 316.
Characteristic
Solid
Shape
Deinite
Volume
Deinite
Almost
Compressibility
Incompressible
Difusion
Small
Comparative
High
Density
Liquid
Variable
Deinite
Very slightly
Compressible
Slow
Gas
Variable
Variable
Highly
Compressible
Fast
High
Low
Figure 315 Some macroscopic
characteristics of solids, liquids and gases
Macroscopic properties are all the observable behaviours of
that material such as shape, volume and compressibility.
he many macroscopic or physical properties of a substance
can provide evidence for the nature and structure of that
substance.
Characteristic
Kinetic energy
Potential energy
Mean molecular
Separation (r0)
hermal energy
of particles (ε)
Molecules per m3
Solid
Vibrational
High
Liquid
Gas
Mostly
Vibrational translational
Rotational
Higher
rotational
Some
translational Higher
vibrational
Higher
Highest
r0
r0
10r0
< ε /10
< ε > ε /10
>ε
1028
1028
1025
Figure 316 Some microscopic
characteristics of solids, liquids and gases
Microscopic characteristics help to explain what is
happening at the atomic level, and this part of the model
will be interpreted further at a later stage.
he modern technique of X-ray difraction that will be
studied in detail in a later chapter has enabled scientists
to determine the arrangement of particles in solids. he
particles are closely packed and each particle is strongly
bonded to its neighbour and is held fairly rigidly in a ixed
position to give it deinite shape in a crystalline lattice.
Some patterns are disordered as is the case for ceramics,
rubber, plastics and glass. hese substances are said to be
amorphous. he particles have vibrational kinetic energy
in their ixed positions and the force of attraction between
the particles gives them potential energy.
In liquids the particles are still closely packed and the
bonding between particles is still quite strong. However,
they are not held as rigidly in position and the bonds can
break and reform. his infers that the particles can slowly
and randomly move relative to each other to produce
variable shape and slow difusion. Particles in a liquid
have vibrational, rotational and some translational kinetic
energy due to their higher mean speeds. he potential
energy of the particles in a liquid is somewhat higher than
for a solid because the spacing between the particles is
large.
In gases the particles are widely spaced and the particles
only interact signiicantly on collision or very close
approach. Because of the rapid random zig-zag motion of
the particles, a gas will become dispersed throughout any
container into which it is placed. Difusion (the spreading
out from the point of release) can occur readily. Gases
are compressible because the particles are widely spaced
at a distance much greater than the size of the particles.
he much higher mean speeds are due to an increased
translational kinetic energy of the particles. Gases have
a much higher potential energy than liquids because the
particles are much further apart.
3.2.4 The PrOcess Of Phase
changes
A substance can undergo changes of state or phase changes
at diferent temperatures. Pure substances (elements and
compounds) have deinite melting and boiling points
which are characteristic of the particular pure substance
being examined. For example, oxygen has a melting point
of -218.8 °C and a boiling point of -183 °C at standard
atmospheric pressure.
he heating curve for benzene is illustrated in Figure 317.
A sample of benzene at 25°C is heated in a closed container
and the change in temperature is graphed as a function
of time. he macroscopic behaviour of benzene can be
described using the graph and the microscopic behaviour
can be interpreted from the macroscopic behaviour.
Temperature /°C
L iquid-gas phase
change
boiling 80°C
point
G AS
Solid-liquid
phase change
melting
point 5.5°C
L I QUID
SO L I D
Heating time /min
Figure 317 Heating curve for benzene.
When the solid benzene is heated the temperature
begins to rise. When the temperature reaches 5.5 °C the
benzene begins to melt. Although heating continues the
temperature of the solid – liquid benzene mixture remains
constant until all the benzene has melted. Once all the
benzene has melted the temperature starts to rise until the
liquid benzene begins to boil at a temperature of 80 °C.
With continued heating the temperature remains constant
until all the liquid benzene has been converted to the
gaseous state. he temperature then continues to rise as
the gas is in a closed container.
89
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Thermal Physics
CHAPTER 3
3.2.5 mOlecular BehaViOur and
cOre
Phase changes
The moving particle theory can be used to explain the
microscopic behaviour of these phase changes. When
solid benzene is heated the particles of the solid vibrate
at an increasing rate as the temperature is increased.
The vibrational kinetic energy of the particles increases.
At the melting point a temperature is reached at which
the particles vibrate with sufficient thermal energy to
break from their fixed positions and begin to slip over
each other. As the solid continues to melt, more and
more particles gain sufficient energy to overcome the
forces between particles and over time all the solid
particles change to a liquid. The potential energy of
the system increases as the particles begin to move. As
heating continues the temperature of the liquid rises
due to an increase in the vibrational, rotational and
part translational kinetic energy of the particles. At
the boiling point a temperature is reached at which the
particles gain sufficient energy to overcome the interparticle forces present in the liquid benzene and escape
into the gaseous state. Continued heating at the boiling
point provides the potential energy needed for all the
benzene molecules to be converted from a liquid to a
gas. With further heating the temperature increases
due to an increase in the kinetic energy of the gaseous
molecules due to the larger translational motion.
remaining particles in the liquid has been lowered. Since
temperature is proportional to the average kinetic energy
of the particles, a lower kinetic energy implies a lower
temperature, and this is the reason why the temperature of
the liquid falls as evaporative cooling takes place. Another
way of explaining the temperature drop is in terms of latent
heat. As a substance evaporates, it needs thermal energy
input to replace its lost latent heat of vaporisation and this
thermal energy can be obtained from the remaining liquid
and its surroundings.
A substance that evaporates rapidly is said to be a volatile
liquid. A liquid’s volatility is controlled by a factor known
as its equilibrium vapour pressure. here are forces that
must be overcome before a particle can leave the surface of
a liquid. Diferent liquids exert diferent vapour pressures
that depend on the relative strengths of the intermolecular
forces present in the liquids. Freon used in refrigerators,
and ether, chloroform and ethanol have relatively high
vapour pressures.
he values in Figure 318 compare the vapour pressure of
some liquids at 293 K.
Substance
Ether
Chloroform
Ethanol
Water
Mercury
Vapour pressure / kPa
58.9
19.3
5.8
2.3
0.0002
3.2.6 eVaPOraTiOn and BOiling
Figure 318 Some common vapour pressures
When water is let in a container outside, exposed to the
atmosphere, it will eventually evaporate. Mercury from
broken thermometers has to be cleaned up immediately
due to its harmful efects. Water has a boiling point of
100 °C and mercury has a boiling point of 357 °C. Yet they
both evaporate at room temperature.
he process of evaporation is a change from the liquid
state to the gaseous state that occurs at a temperature
below the boiling point.
he moving particle theory can be applied to understand
the evaporation process. A substance at a particular
temperature has a range of kinetic energies. So in a liquid
at any particular instant, a small fraction of the molecules
will have kinetic energies considerably greater then the
average value. If these particles are near the surface of the
liquid, they may have enough kinetic energy to overcome
the attractive forces of neighbouring particles and escape
from the liquid as a vapour. Now that the more energetic
particles have escaped, the average kinetic energy of the
90
Although the vapour pressure of mercury is much lower
than the other substances listed at room temperature,
some evaporation does occur. Because of its extreme
toxicity any mercury spill is treated seriously. Ether has a
high vapour pressure. If a stream of air is blown through a
sample of ether in a beaker that is placed on a thin ilm of
water, the water will eventually turn to ice.
When overheating occurs in a human on hot days, the
body starts to perspire. Evaporation of the perspiration
results in a loss of thermal energy from the body so that
body temperature can be controlled. Local anaesthetics
with high vapour pressures are used to reduce pain on the
skin. hermal energy lows from the surrounding lesh
causing its temperature to drop and thereby anaesthetises
the area.
A liquid boils when the vapour pressure of the liquid
equals the atmospheric pressure of its surroundings.
As the boiling point is reached, tiny bubbles appear
Thermal Physics
1.
C.
D.
5.
A.
B.
D.
6.
freezing and sublimation
melting and evaporation
evaporation and condensation
sublimation and condensation
watt
the product of the newton and the metre
the quotient of the watt and the second
the product of the joule and the second
Base your answers to Questions 5 and 6 on the
graph below. he graph shows the temperature
of an unknown substance of mass 10.0 kg as heat
is added at a constant rate of 6300 Jmin-1. he
substance is a solid at 0 0C.
20
irst minute to the end of the fourth minute
seventh minute to the end of the
seventeenth minute
seventeenth minute to the end of the twenty
irst minute
twenty second to the end of the twenty ith
minute
50.4 Jkg-1K-1
105 Jkg-1K-1
126 Jkg-1K-1
504 Jkg-1K-1
5.
Give ive macroscopic and ive microscopic
characteristics of the liquid/gas in a butane lighter.
6.
Describe the components of internal energy in
each of the following situations
Which of the following is a unit of thermal
energy?
A.
B.
C.
D.
15
he speciic heat capacity of the substance when it
is solid is:
A.
B.
C.
D.
For a given mass of a certain liquid, the magnitude
of the thermal energy transfer is the same for the
following two processes
A.
B.
C.
D.
10
he internal potential energy of the unknown
substance increases without any change in internal
kinetic energy from the beginning of the:
A.
the potential energy holding the atoms in
ixed positions
the vibrational energy of the atoms
the random translational energy of the
atoms
the rotational energy of the atoms
5
T ime / min
C.
D.
4.
all particles moving in straight lines
smoke particles moving randomly by air
molecules
smoke particles colliding with each other
air molecules in random motion
0
he internal energy of a monatomic gas such as
neon is mainly due to
B.
C.
3.
50
When smoke is strongly illuminated and viewed
under a microscope it is possible to observe
A.
B.
2.
3.4
100
cOre
Exercise
150
Temperature / oC
throughout the liquid. If the vapour pressure of the bubble
is less than the atmospheric pressure the bubbles are
crushed. However a point is reached when the pressures
are equal. he bubble will then increase in size as it rises to
the surface of the liquid.
(a)
(b)
(c)
air at room temperature
a jar of honey
a melting ice cream.
7.
Explain the diference between heat, thermal
energy and temperature.
8.
Does a block of ice contain any heat? Explain your
answer fully.
9.
Draw a fully labelled cooling curve for the
situation when steam at 110 °C is converted to ice
at –25 °C.
91
cOre
CHAPTER 3
10.
(a)
(b)
11.
he temperatures of the same volume of air and
water are raised by a small amount. Explain why
a diferent amount of heat is required for each
process.
12.
If you increase the heat under a pot of water in
which you are boiling potatoes, will the potatoes
be cooked faster?
13.
14.
15.
16.
Convert 63 °C to Kelvin
Convert 52 K to degrees Celsius
If you wanted to cool a bottle of sot drink at the
beach, would you be better to wrap a wet towel
around it or to put it into the seawater? Explain
your answer.
Why is it important not to stand in a draught ater
vigorous exercise?
Describe and explain the process of evaporative
cooling in terms of its microscopic properties.
A kettle made of stainless steel containing water
is heated to a temperature of 95 °C. Describe
the processes of thermal energy transfer that are
occurring in the stainless steel kettle and the water.
3.2.7 laTenT heaT
he thermal energy which a particle absorbs in melting,
evaporating or sublimating or gives out in freezing,
condensing or sublimating is called latent heat because it
does not produce a change in temperature. See Figure 320.
sublimation
SOL I D
evaporation
L I QUI D
freezing
∆Q = mL
∆Q is the quantity of heat absorbed or released during the
phase change in J,
m is the mass of the substance in kg and
L is the latent heat of the substance in J kg -1
L could be the latent heat of fusion Lf, the latent heat of
vaporisation Lv or the latent heat of sublimation Ls. he
latent heat of fusion of a substance is less than the latent
heat of vaporisation or the latent heat of sublimation. More
work has to be done to reorganise the particles as they
increase their volume in vaporisation and sublimation
than the work required to allow particles to move from
their ixed position and slide over each other in fusion.
Figure 321 lists the latent heat of some substances.
Substance Melting Latent heat Boiling Latent heat of
point of fusion point Vaporisation
K
105 J kg-1
K
105 J kg-1
55
159
600
1356
273
0.14
1.05
0.25
1.8
3.34
90
351
1893
2573
373
2.1
8.7
7.3
73
22.5
GAS
condensation
T H E R M A L E NE R G Y GI V E N OUT
Figure 320 Macroscopic
transformations between states of matter.
Sublimation is a change of phase directly from a solid to
a gas or directly from a gas to a solid. Iodine and solid
carbon dioxide are examples of substances that sublime.
92
he quantity of heat required to change one kilogram of
a substance from one phase to another is called the latent
heat of transformation.
Oxygen
Ethanol
Lead
Copper
Water
T H E R M A L E NE R G Y A DDE D
melting
When thermal energy is absorbed/released by a body,
the temperature may rise/fall, or it can remain constant.
If the temperature remains constant then a phase change
will occur as the thermal energy must either increase the
potential energy of the particles as they move further apart
or decrease the potential energy of the particles as they
move closer together. If the temperature changes, then the
energy must increase the kinetic energy of the particles.
Figure 321 Some Latent Heat Values
Example 1
Calculate the heat energy required to evaporate 5.0 kg of
ethanol at its boiling point.
he latent heat of vaporisation can be found using a selfjacketing vaporiser as shown in Figure 323. he liquid
to be vaporised is heated electrically so that it boils at a
steady rate. he vapour that is produced passes to the
condenser through holes labelled H in the neck of the
inner lask. Condensation occurs in the outer lask and
the condenser.
Solution
Given that m = 5.0 kg and Lv = 8.7 × 105 J kg -1.
We then have,
= 4.35 × 106 J = 4.4 × 106 J
A
he heat energy required for the vaporisation is 4.4 × 106 J.
V
Example 2
H
H
Va pour acting as
a jacket
L iquid under
investigation
Determine the heat energy released when 1.5 kg of gaseous
water at 100 °C is placed in a freezer and converted to ice
at -7 0C.
H eating coil
Wa ter outflow
Solution
C ondenser
he energy changes in this process can be represented as
shown in Figure 322.
C old water
Using
C ollecting vessel
Q = mLV + mc∆TWATER + mLf + mc∆TICE
C ondensed vapour
Figure 323 Latent heat of vaporisation apparatus.
= m [LV + c∆TWATER+ Lf + c∆TICE]
= 1.5 [22.5 × 105+ (4180 × 100) + 3.34 × 105 + (2100 × 7)]
= 4.52 × 106 J
hat is, the energy released is 4.5 × 106 J or 4.5 MJ.
T H E R M A L E NE R G Y R E L E A SE D
L atent heat of fusion
I C E at
–7°C
L atent heat of evapor ation
L I Q U I D water at
0 °C
0 °C
100 °C
Eventually, the temperature of all the parts of the apparatus
becomes steady. When this steady state is reached, a
container of known mass is placed under the condenser
outlet for a measured time t, and the measured mass of the
condensed vapour m is determined. he heater current I
is measured with the ammeter ‘A’ and potential diference
V is measured with a voltmeter ‘V’. hey are closely
monitored and kept constant with a rheostat.
he electrical energy supplied is used to vaporise the liquid
and some thermal energy H is lost to the surroundings.
G ASE OU S water at
100 ° C
herefore:
V1I1t = m1LV + H
Specific heat I ce
Figure 322
Specific heat water
Energy released in steam-ice change.
In order to eliminate H from the relationship, the process
is repeated using a diferent heater potential diference
and current. he vapour is collected for the same time t
he rate of vaporisation will be diferent but the heat lost
to the surroundings will be the same as each part of the
93
cOre
Thermal Physics
CHAPTER 3
apparatus will be at the same temperature as it was with
the initial rate vaporisation.
4.
Determine the amount of thermal energy that is
required to melt 35 kg of ice at its melting point.
herefore:
5.
A 5.0 × 102 g aluminium block is heated to 350 °C.
Determine the number of kilograms of ice at 0 °C
that the aluminium block will melt as it cools.
6.
Steam coming from a kettle will give you a nastier
burn than boiling water. Explain why.
7.
An immersion heater can supply heat at a rate of
5.2 × 102 J s -1. Calculate the time that it will take
to completely evaporate 1.25 × 10-1 kg of water
initially at a temperature of 21 °C?
8.
A 3.45 kg sample of iron is heated to a temperature
of 295 °C and is then transferred to a 2.0 kg copper
vessel containing 10.0 kg of a liquid at an initial
temperature of 21.0 °C. If the inal temperature of
the mixture is 31.5 °C, determine the speciic heat
capacity of the liquid?
9.
A mass of dry steam at 1.0 × 102 °C is blown over
a 1.5 kg of ice at 0.0 °C in an isolated container.
Calculate the mass of steam needed to convert the
ice to water at 21.5 °C.
10.
A freezer in a refrigerator takes 2.00 hours to
convert 2.15 kg of water initially at 21.5 °C to just
frozen ice. Calculate the rate at which the freezer
absorbs heat.
11.
Describe an experiment to determine the speciic
heat capacity of an unknown metal. Sketch the
apparatus used and describe what measurements
are made. State the main sources of error and
explain how they can be minimised.
12.
Calculate how much thermal energy is released
when 1.2 kg of steam at 100 °C is condensed to
water at the same temperature.
(Lv = 2.25 × 106 Jkg-1)
13.
Determine how much energy is released when
1.5 kg of gaseous water at 100 °C is placed in a
freezer and converted to ice at –7 °C . (the speciic
heat capacity of ice is 2.1 × 103 J kg-1 K-1).
14.
Describe an experiment that can be used to
determine the latent heat of vaporisation of a
liquid.
cOre
V2I2t = m2LV + H
By subtracting the two equations:
(V1I1 – V2I2)t = (m1 – m2)LV
From this equation, the value of the latent heat
of vaporisation of the unknown substance can be
determined.
Exercise
1.
he speciic latent heat of fusion of ice is the heat
required to
A.
B.
C.
D.
2.
Average kinetic
energy
Average potential
energy
constant
increases
increases
constant
increases
constant
decreases
constant
hermal energy is transferred to a mass of water
in four steps. Which one of the four steps requires
the most thermal energy?
A.
B.
C.
D.
94
raise the temperature of ice from 0 °C to 10 °C
change 1 dm3 of ice at 0 °C to water at 0 °C
change 1kg of ice at 0 °C to water at 0 °C
change the temperature of 1 kg by 10 °C
A substance changes from liquid to gas at its
normal boiling temperature. What change, if
any, occurs in the average kinetic energy and the
average potential energy of its molecules?
A.
B.
C.
D.
3.
3.5
5 °C to 20 °C
15 °C to 35 °C
75 °C to 90 °C
95 °C to 101 °C
Thermal Physics
3.2.9 Define pressure.
3.2.10 State the assumptions of the kinetic model
Of an ideal gas.
3.2.11 State that temperature is a measure of
the average random kinetic energy of the
molecules of an ideal gas.
3.2.12 Explain the macroscopic behaviour of an
ideal gas in terms of a molecular model.
© IBO 2007
3.2.9 Pressure
Investigations into the behaviour of gases involve
measurement of pressure, volume, temperature and mass.
Experiments use these macroscopic properties of a gas to
formulate a number of gas laws.
certain limited conditions but they can condense to liquids,
then solidify if the temperature is lowered. Furthermore,
there are relatively small forces of attraction between
particles of a real gas, and even this is not allowable for
an ideal gas.
Most gases, at temperatures well above their boiling points
and pressures that are not too high, behave like an ideal gas.
In other words, real gases vary from ideal gas behaviour at
high pressures and low temperatures.
When the moving particle theory is applied to gases it is
generally called the kinetic theory of gases. he kinetic
theory relates the macroscopic behaviour of an ideal gas to
the behaviour of its molecules.
he assumptions or postulates of the moving particle
theory are extended for an ideal gas to include
•
•
In 1643 Torricelli found that the atmosphere could
support a vertical column of mercury about 76 cm high
and the irst mercury barometer became the standard
instrument for measuring pressure. he pressure unit
760 mm Hg (760 millimetres of mercury) represented
standard atmospheric pressure. In 1646, Pascal found that
the atmosphere could support a vertical column of water
about 10.4 m high.
•
•
•
•
For our purposes in this section, pressure can be deined
as the force exerted over an area.
•
Pressure = Force / Area
•
Gases consist of tiny particles called atoms
(monatomic gases such as neon and argon) or
molecules.
he total number of molecules in any sample of a
gas is extremely large.
he molecules are in constant random motion.
he range of the intermolecular forces is small
compared to the average separation of the
molecules.
he size of the particles is relatively small compared
with the distance between them.
Collisions of short duration occur between
molecules and the walls of the container and the
collisions are perfectly elastic.
No forces act between particles except when they
collide, and hence particles move in straight lines.
Between collisions the molecules, obey Newton’s
Laws of motion.
P = F /A
he SI unit of pressure is the pascal Pa.
1 atm = 1.01 × 105 Nm-2 = 101.3 kPa = 760 mmHg
3.2.10 The KineTic mOdel Of an
Based on these postulates the view of an ideal gas is one
of molecules moving in random straight line paths at
constant speeds until they collide with the sides of the
container or with one another. heir paths over time are
therefore zig-zags. Because the gas molecules can move
freely and are relatively far apart, they occupy the total
volume of a container.
ideal gas
An ideal gas is a theoretical gas that obeys the ideal gas
equation exactly. Real gases conform to the gas laws under
he large number of particles ensures that the number of
particles moving in all directions is constant at any time.
95
cOre
KineTic mOdel Of an ideal gas
CHAPTER 3
3.2.11 TemPeraTure and aVerage
randOm KineTic energy
cOre
Temperature is a measure of the average random kinetic
energy of an ideal gas.
At the microscopic level, temperature is regarded
as the measure of the average kinetic energy per
molecule associated with its movements. For gases,
it can be shown that the average kinetic energy,
1
3
E k = --- mv 2 = --- kT where k = Boltzmann constant
2
2
2
∴v ∝ T
∴ ∝
he term average kinetic energy is used because, at a
particular temperature diferent particles have a wide
range of velocities, especially when they are converted to a
gas. his is to say that at any given temperature the average
speed is deinite but the velocities of particular molecules
can change as a result of collision.
Figure 325 shows a series of graphs for the same gas
at three diferent temperatures. In 1859 James Clerk
Maxwell (1831-1879) and in 1861 Ludwig Boltzmann
(1844-1906) developed the mathematics of the kinetic
theory of gases. he curve is called a Maxwell-Boltzmann
speed distribution and it is calculated using statistical
mechanics. It shows the relationship between the relative
number of particles N in a sample of gas and the speeds v
that the particles have when the temperature is changed.
(T3 > T2 > T1)
he graphs do not show a normal distribution as the
graphs are not bell-shaped. hey are slightly skewed to
the let. he minimum speed is zero at the let end of
the graphs. At the right end they do not touch the x-axis
because a small number of particles have very high speeds.
N
T1
T2
he peak of each curve is at the most probable speed vp
a large number of particles in a sample of gas have their
speeds in this region. When the mathematics of statistical
mechanics is applied it is found that mean squared speed
vav2 is higher than the most probable speed. Another
quantity more oten used is called the root mean square
speed Vrms and it is equal to the square root of the mean
squared speed.
v rms =
v
2
he root mean square is higher than the mean squared
speed.
Other features of the graphs show that the higher the
temperature, the more symmetric the curves becomes.
he average speed of the particles increases and the peak
is lowered and shited to the right. he areas under the
graphs only have signiicance when N is deined in a
diferent way from above.
Figure 326 shows the distribution of the number of
particles with a particular energy N against the kinetic
energy of the particles E k at a particular temperature. he
shape of the kinetic energy distribution curve is similar to
the speed distribution curve and the total energy of the
gas is given by the area under the curve.
N
T1
T2
T2 > T1
Ek
Figure 326 Distribution of kinetic energies
for the same gas at different temperatures.
he average kinetic energy of the particles of all gases is
the same. However, gases have diferent masses. Hydrogen
molecules have about one-sixteenth the mass of oxygen
molecules and therefore have higher speeds if the average
kinetic energy of the hydrogen and the oxygen are the same.
T3 > T2 > T1
T3
Because the collisions are perfectly elastic there is no loss
in kinetic energy as a result of the collisions.
v mp v v rms
v ms
–1
Figure 325 Maxwell-Boltzmann speed distribution for
the same gas at different temperatures.
96
Thermal Physics
an ideal gas
Robert Boyle (1627-1691) discussed that the pressure of a
gas at constant temperature is proportional to its density.
He also investigated how the pressure is related to the
volume for a ixed mass of gas at constant temperature.
Boyle’s Law relates pressure and volume for a gas at ixed
temperature.
Boyle’s Law for gases states that the pressure of a ixed mass
of gas is inversely proportional to its volume at constant
temperature.
changed with temperature. Gay-Lussac (1778–1850)
published more accurate investigations in 1802.
A very simple apparatus to investigate Charles’ Law is
shown in Figure 328. A sample of dry air is trapped in a
capillary tube by a bead of concentrated sulfuric acid. he
capillary tube is heated in a water bath and the water is
constantly stirred to ensure that the whole air column is at
the same temperature.
thermometer
capillary tube
bead of acid
(e.g. sulfuric acid)
water bath
air column
1 ⇔ PV = constant
P α __
V
When the conditions are changed, with the temperature
still constant
H E A T
P1V1 = P2V2
Figure 328 Apparatus for Charles’ law.
When a pressure versus volume graph is drawn for the
collected data a hyperbola shape is obtained, and when
pressure is plotted against the reciprocal of volume a
straight line is obtained. See Figure 327.
pressure, P mm Hg
pressure, P mm Hg
volume, V cm 3
he investigation should be carried out slowly to allow
thermal energy to pass into or out of the thick glass walls
of the capillary tube. When the volume and temperature
measurements are plotted, a graph similar to Figure 328
is obtained.
V cm3
he readings of P and V must be taken slowly to maintain
constant temperature because when air is compressed, it
warms up slightly.
PV
1 cm–3
V
P
–273
0
0 100
273 373
T °C
T K
Figure 327 (a), (b) and)c) pressure-volume graphs.
Figure 329 Variation of volume with temperature.
he pressure that the molecules exert is due to their
collisions with the sides of the container. When the
volume of the container is decreased, the frequency of the
particle collisions with the walls of the container increases.
his means that there is a greater force in a smaller area
leading to an increase in pressure. he pressure increase
has nothing to do with the collisions of the particles with
each other.
Note that from the extrapolation of the straight line that
the volume of gases would be theoretically zero at –273 °C
called absolute zero. he scale chosen is called the Kelvin
scale K.
he Charles (Gay-Lussac) Law of gases states that:
he volume of a ixed mass of gas at constant pressure
is directly proportional to its absolute (Kelvin)
temperature.
In 1787 Jacques Charles (1746–1823) performed
experiments to investigate how the volume of a gas
97
cOre
3.2.12 macrOscOPic BehaViOur Of
CHAPTER 3
cOre
he volume of a ixed mass of gas increases by 1/273
of its volume at 0 °C for every degree Celsius rise in
temperature provided the pressure is constant.
he variation in pressure as the temperature is changed
is measured and graphed. A typical graph is shown in
Figure 331.
Pressure, P
kPa
his can also be stated as:
V
V α T ⇒ V = kt so that ___1 = k
T1
herefore,
V1 ___
V
___
= 2
T1 T2
As the temperature of a gas is increased, the average kinetic
energy per molecule increases. he increase in velocity
of the molecules leads to a greater rate of collisions, and
each collision involves greater impulse. Hence the volume
of the gas increases as the collisions with the sides of the
container increase.
he Pressure (Admonton) Law of Gases states that:
Experiments were similarly carried out to investigate the
relationship between the pressure and temperature of a
ixed mass of various gases.
he pressure of a ixed mass of gas at constant volume
is directly proportional to its absolute (Kelvin)
temperature.
he essential parts of the apparatus shown in Figure 330
are a metal sphere or round bottomed lask, and a
Bourdon pressure gauge. he sphere/lask and bourdon
gauge are connected by a short column of metal tubing/
capillary tube to ensure that as little air as possible is at a
diferent temperature from the main body of enclosed gas.
he apparatus in Figure 330 allows the pressure of a ixed
volume of gas to be determined as the gas is heated.
B ourdon gauge
counter–balance
metal stem
thermometer
retort
stand
air enclosed
in a metal
sphere
Figure 330
98
Pressure law apparatus.
–273
0
0 100
273 373
T °C
T K
Figure 331 Variation of pressure with temperature.
P1
- = k
P∝∝ T⇔⇔ P = kT
∴∴----T1
herefore,
P
P
-----1- = -----2T2
T1
As the temperature of a gas is increased, the average kinetic
energy per molecule increases. he increase in velocity of
the molecules leads to a greater rate of collisions, and each
collision involves greater impulse. Hence the pressure of
the gas increases as the collisions with the sides of the
container increase.