Name……………………
Class……………..
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Module G484.3
Thermal Physics
NOTES AND QUESTIONS
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Lesson 31 notes - Solids, Liquids and Gases
Objectives
Be able to describe solids, liquids and gases in terms of the spacing, ordering
and motion of atoms or molecules.
Be able to describe a simple kinetic model for solids, liquids and gases.
Outcomes
Be able to describe solids, liquids and gases in terms of the spacing, ordering
and motion of atoms or molecules.
Be able to describe a simple kinetic model for solids, liquids and gases.
3 States of Matter
Matter can exist is three states,
as a solid, a liquid or a gas.
Water shows us these 3 states
brilliantly. The solid form, ice is
hard and rigid, a piece of ice has
a shape of its own. The liquid
form, water is "runny", has no
shape of its own and takes up the
shape of its container. The
density of liquid water is just
slightly higher than that of ice (both at 0oC). Steam is what we think of as
water in its gaseous form, this has a very much larger volume (1600 times)
than the water that produced it, a low density and no shape of its own. (NB
Steam as we see it is not a true gas, the gas produced when water boils is
invisible, steam is a cloud of fine droplets of water vapour).
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Phase Changes
The diagram shows phase
changes between solid,
liquid and gas.
Between each phase
change there is no
temperature change as the
bonds are breaking or
forming. It also shows
Plasma which is an ionized
gas normally formed by
heating it up.
The enthalpy of the system
is the heat energy of an
object.
As molecules inside an object have more enthalpy their vibrations and
therefore their kinetic energy goes up.
The Kinetic Molecular Theory
Particles are in constant motion. In solids the particles are close together and
have limited motion. In a liquid some of the attraction between particles is
overcome which allows the particles more freedom of movement. In a gas
particles attraction between particles is minimized and the particles move
freely throughout the container.
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Standard temperature and pressure (STP on NTP)
This set of conditions which is usually applied to gases is defined as a
temperature of 273 K and a pressure of 760 mm of mercury (1.013x10 5 Pa).
At STP 1 mole of any gas has a volume of 22.4x10-3 m3.
Extension
Burning Candles
Another example of solids, liquids and gases is a burning candle. Carefully
light a candle and watch the surface of the candle and the flame. The flame
melts the solid wax, creating a pool of liquid wax. This creeps up the wick
inside the flame and becomes a gas, which then burns in the flame. The flame
melts more wax and the cycle continues.
Sublimation
A few compounds go from solid to gas without becoming liquid in between.
Solid carbon dioxide (dry ice) becomes carbon dioxide gas without becoming
a liquid, as does iodine.
In some circumstances, snow and ice can sublime – sometimes, wet washing
hung outside freezes solid and then dries without melting. This is called
freeze-drying and is used in food preserving and for making instant foods and
drinks.
Intermolecular Forces
In a solid the molecules are held together by intermolecular forces, the exact
type of force depending on the type of solid.
These forces are divide into four types:
(a) ionic bonds - sodium and chloride ions in sodium chloride;
(b) covalent bonds - shared electrons between atoms;
(c) metallic bonds - free electrons wandering through a metal;
(d) Van der Waal's bonds - electric dipole forces
Structure of Solids
Crystalline solids have specific molecular structures – here are a few
examples:
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Lesson 31 questions – Solids Liquids Gases
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1
a)
Draw a diagram of each state of matter and describe each state in terms
of the spacing, ordering and motion of atoms or molecules.
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b)
Describe what is meant by a Kinetic Molecular Theory for Solids
Liquids and Gases.
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SOME
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2
A Student asks on a Physics blog “If all matter is either solid, liquid, gas or
plasma, what state of matter is a single atom or molecule?” What would
you write in answers to this question?
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Lesson 32 notes – Brownian Motion
Objectives
Be able to describe an experiment that demonstrates Brownian motion and
discuss the evidence for the movement of molecules provided by such an
experiment.
Outcomes
Be able to describe an experiment that demonstrates Brownian motion
Be able to discuss and explain the evidence for the movement of molecules
provided by such an experiment.
The ancients such as the philosophers Democritus
and Lucretius held that matter was composed of
minute particles. They also Maintained that these
particles were in a state of continuous random motion
within solids, liquids and gases. The theory was
therefore called the kinetic theory of matter, after the
Greek work kinema - motion.
It was not until 1827, however, that actual experimental evidence for these
particles existed. This was provided by the Scottish physicist Robert Brown.
He observed a weak solution of milk and later pollen grains in suspension with
a high-powered microscope, and saw that the particles of milk and the pollen
grains showed a violent and random motion. Brown wrongly attributed what
he saw to living organisms, and the true explanation was not given until some
thirty years later when the Frenchman Carbonelle proposed that the motion
was due to the impacts of the liquid molecules on the milk particles or pollen
grains. The motion is now known as Brownian movement.
However the first good explanation of Brownian movement was advanced by
the French scientist Desaulx in 1877: "In my way of thinking the phenomenon
is a result of thermal molecular motion in the liquid environment (of the
particles)."
The Smoke Cell
A simple modern
version of Brown's
experiment is the
smoke cell. A small
cell of air is placed
under a microscope
and illuminated
strongly from the
side. Some smoke
is then blown into it.
Through the
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microscope the particles of smoke can be seen to be in violent random motion
just like Brown's pollen grains. This motion is due to the collisions of the
(invisible) air molecules with the much larger particles of smoke. Heating the
cell makes the smoke particles' motion even more violent due to the increased
velocity of the air molecules.
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Lesson 33 notes - Kinetic Theory and Pressure
Objectives
Be able to define the term pressure and use the kinetic model to explain the
pressure exerted by gases.
Outcomes
Be able to define pressure.
Be able to use the kinetic model to describe pressure.
Know the assumptions made of ideal gases.
Be able to use the kinetic model to explain the pressure exerted by gases.
Be able to select and use the equation p=1/3ρv2.
Be able to derive the equation p=1/3ρv2.
Be able to derive the equation(s) P = 1/3 ρ<c>2 and/or PV = 1/3 mn<c>2.
You must be familiar with Change in Momentum to understand this lesson
fully so read through lessons 2, 3 and 5 first to review.
Gas Pressure
From lesson 21 of G481, Pressure is defined as the force per unit area.
P (Pa) = F (N) / A (m2)
Assumptions of the kinetic theory of an IDEAL GAS.
1
2
3
4
5
6
7
8
A Gas consists of particles called molecules.
The molecules are in constant random motion. As many travelling in
one direction as any other. The centre of mass of the gas is at rest.
Intermolecular forces are negligible.
The duration of collisions between molecules is negligible.
Molecules move with constant velocity in between collisions.
The volume of gas molecules is negligible compared with the volume of
the gas.
All collisions are totally elastic.
Newtonian mechanics can be applied to the collisions.
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The pressure exerted by an ideal
gas
Suppose there are N molecules in a
rectangular box of dimensions a, b and l
and suppose that the molecule has a
velocity v with the components as shown.
vy
b
When the molecule hits the shaded face,
the change in momentum is 2mvx since
the collision is elastic.
The time interval before the same
molecule makes a 2nd collision at the same
face is 2Lvx. Therefore the frequency of
collisions is vx/2L
vx
vz
The force exerted by this molecule on the
face is:
a
L
Rate of change of momentum at the face = 2mvx x vx/2L= mvx2/L
(Since F=Δp/Δt)
Therefore the pressure exerted on the shaded face is by one molecule is:
P=F/A
P = (mvx2/L / (ab)
(ab = A, Area of shaded face)
If we sum N contributions, one from each particle in the box, each contribution
proportional to vx2 for that particle, the sum just gives us N times the average
value of vx2. That is to say,
P  F / A  Nmvx2 / LA  Nmvx2 / V
where there are N particles in a box of volume V. Next we note that the
particles are equally likely to be moving in any of 3 directions, so the average
value of vx2 must be the same as that of vy2 or vz2, and since v2 = vx2 + vy2 +
vz2, it follows that
P  Nmv2 / 3V .
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Extension
Finally to simplify even further we can use the definition for density: ρ=m/V
So the mass of the gas in the box will equal Nm
So, P = ⅓ρ v2
<c> can be used for the mean speed.
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Lesson 33 questions - Kinetic theory and Pressure
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ALL
1. (a)
What do you see when looking at Brownian motion in air?
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(b)
What changes do you see if the air is heated?
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2. Define pressure and give its units.
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3. List six assumptions that you have to make about molecules when deriving the
equation for the kinetic theory of gases.
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4. (a)
A molecule of mass 3x10-26 kg moving at 4x102 ms-1 collides
elastically with a wall. What is the change of momentum of the molecule?
Change in momentum = …………………. Unit………… (3)
(b)
How long would it take this molecule to travel a distance of
0.4 m?
Time = …………….. s (2)
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MOST
5.
(a)
What is meant by the root mean square speed of gas molecules?
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(b)
What is the difference between the root mean square of a group
of gas molecules and their average speed?
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6.
Calculate the root mean square speeds for the following gases at a
5
pressure of 10 Pa:
(a)
air
density 1.29 kgm-3
(b)
root mean square speed = …………….. ms-1 (2)
carbon dioxide
density 1.98 kgm-3
(c)
nitrogen
root mean square speed = …………….. ms-1 (2)
density 1.25 kgm-3
chlorine
root mean square speed = …………….. ms-1 (2)
density 3.21 kgm-3
hydrogen
root mean square speed = …………….. ms-1 (2)
density 0.09 kgm-3
(d)
(e)
root mean square speed = …………….. ms-1 (2)
7.
Calculate the pressure of three samples of air of density 1.29 kgm-3
with different root mean square speeds:
(a) 4.0x102 ms-1
pressure = ……………..Pa (2)
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(b) 5.0x102 ms-1
pressure = ……………..Pa (2)
2
(c) 6.0x10 ms
-1
pressure = ……………..Pa (2)
9.
A sample of gas of volume 0.1 m3 and at a pressure of 2.0x105 Pa is
enclosed in a cylinder. If the root mean square speed of the gas molecules is
5.5x102 ms-1 and the mass of each molecule is 3.5x10-26 kg calculate the
number of gas molecules in the cylinder.
Number of gas molecules = ……………… (3)
10.
A sample of gas at a pressure of 1.5x105 Pa is contained in a cylinder
with a volume of 0.05 m3. If there are 1.2x1025 molecules of gas in the
cylinder with a root mean square speed of 350 ms-1 calculate the mass of one
gas molecule.
Mass of one gas molecule = ………………kg
(3)
SOME
11.
Suppose there are N molecules in a
rectangular box of dimensions a, b and l and
suppose that the molecule has a velocity v with
the components as shown.
vy
b
a)
What is the change in velocity
of the particle when it hits the shaded face if
the collision is totally elastic?
vx
vz
a
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L
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(1)
b)
i)
What is the time interval before the same molecule makes a 2nd
collision at the same face?
(1)
ii)
An therefore what is the frequency of collisions?
(1)
c)
From the definition of Pressure and by finding the rate of change of
momentum for one molecule show that for N molecules:
P  Nmv2 / 3V .
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(6)
Lesson 34 notes – Internal Energy
Objectives
Be able to define internal energy as the sum of the random distribution of
kinetic and potential energies associated with the molecules of a system.
Be able to explain that the rise in temperature of a body leads to an increase
in its internal energy.
Outcomes
Be able to define internal energy as the sum of the random distribution of
kinetic and potential energies associated with the molecules of a system.
Be able to explain that the rise in temperature of a body leads to an increase
in its internal energy.
Be able to describe energy changes in particles in materials as they are
expanded/contracted, heated or cooled.
Internal Energy
The Internal Energy (U) of an object (solid, liquid or gas) is the sum of the
heat energy (Q) plus the work done on the object (W). So:
U=Q+W
It is defined as the sum of the random distribution of kinetic and
potential energies associated with the molecules of a system.
Factors not affecting Internal Energy
It is the energy of the particles inside the object and is not affected by the
objects kinetic or potential energy but by the energy of these particles inside
it.
Factors affecting Internal Energy
The internal energy can be changed by changing the temperature or volume
of the object without changing the amount of particles in the object.
Temperature: If the temperature of a system rises, the molecules will travel
quicker, therefore have more kinetic energy and so the Internal Energy will
increase.
Pressure: If the pressure falls but the temp doesn’t change then there is no
change in the internal energy of a system. In a non-ideal expanding gas the
molecules have to overcome the attraction to each other and therefore some
work must be done on them and therefore the internal energy must increase.
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State: For most materials that change from solid to liquid the volume increases and so
the potential energy of the molecules increases.
Lesson 34 questions – Internal Energy
(
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ALL
1
Describe the arrangement of atoms, the forces between the atoms and the
motion of the atoms in:
a)
a solid
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b)
a liquid
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c)
a gas.
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2
A small amount of gas is trapped inside a container. Describe the motion
of the gas atoms as the temperature of the gas within the container in
increased.
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3
a)
Define the internal energy of a substance.
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b)
The temperature of an aluminium block increases when
it is placed in the flame of a Bunsen burner. Explain why this
causes an increase in its internal energy.
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SOME
c)
An ice cube is melting at a
temperature of 0 °C. Explain whether its internal
energy is increasing or decreasing as it melts at
0°C.
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4
Complete the table below for each of the processes shown. Use the symbol ‘+’
for an increase, the symbol ‘–’ for a decrease and ‘0’ for no change, as appropriate.
(3)
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Lesson 35 notes – Heat and Temperature
Objectives
Be able to explain that thermal energy is transferred from a region of higher
temperature to a region of lower temperature.
Be able to explain that regions of equal temperature are in thermal
equilibrium.
Outcomes
Be able to describe what heat and temperature are and the difference
between them.
Be able to describe that thermal energy is transferred from a region of higher
temperature to a region of lower temperature.
Be able to describe that regions of equal temperature are in thermal
equilibrium.
Be able to explain that thermal energy is transferred from a region of higher
temperature to a region of lower temperature.
Be able to explain that regions of equal temperature are in thermal
equilibrium.
Heat
All objects are made up of particles. These particles are vibrating about all the
time unless they are at absolute zero (-273 degrees Celsius). The vibration of
these particles in an object is known as thermal energy. If the particles vibrate
more, or there are more vibrating particles there will be more thermal energy.
If we compare the thermal energy in a bath to that of a cup of tea at the same
temperature, the bath would have more thermal energy because there are
more particles.
Heat is a measurement of the total energy in a substance. That total energy is
made up of not only of the kinetic energies of the molecules of the substance,
but total energy is also made up of the potential energies of the molecules.
Temperature
Temperature is the measure of how hot something is.
Temperature is related to the average kinetic energy of the molecules of a
substance. If temperature is measured in Kelvin, then this number is directly
proportional to the average kinetic energy of the molecules.
Heat transfer
Thermal energy will always move from an area that is hotter to an area that is
cooler. (This comes from the Second Law of Thermodynamics).
A
HOT
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B
COLD
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After a time A and B will reach the same temperature. Thermal Equilibrium will
be reached at a temperature somewhere in between HOT and COLD. This
temperature will depend on the heat capacities (see lessons 38 and 39) of the
objects A and B. This assumes thee is no other heat loss in the system.
A
WARM
B
WARM
No further heat transfer between A and B occurs when they are at the same
temperature.
The Zeroth Law of Thermodynamics
When two systems are put in contact with each other, there will be a net
exchange of energy between them unless or until they are in thermal
equilibrium, that is, they are at the same temperature. While this is a
fundamental concept of thermodynamics, the need to state it explicitly was not
perceived until the first third of the 20th century, long after the first three
principles were already widely in use, hence the zero numbering.
Extension
The First Law of Thermodynamics states that energy cannot be created or
destroyed; rather, the amount of energy lost in a steady state process cannot
be greater than the amount of energy gained. This is the statement of
conservation of energy for a thermodynamic system. It refers to the two ways
that a closed system transfers energy to and from its surroundings – by the
process of heating (or cooling) and the process of mechanical work. The rate
of gain or loss in the stored energy of a system is determined by the rates of
these two processes. In open systems, the flow of matter is another energy
transfer mechanism, and extra terms must be included in the expression of
the first law.
The First Law in equation form as it applies to a gas is:
Increase in internal energy (dU) = Heat energy supplied (dQ) + Work done on
the gas (dW)
The Second Law of Thermodynamics states that the entropy of an isolated
system not in equilibrium will tend to increase over time, approaching a
maximum value at equilibrium. In other words the second law states "energy
systems have a tendency to increase their entropy rather than decrease it."
This can also be stated as "heat can spontaneously flow from a highertemperature region to a lower-temperature region, but not the other way
around."
Entropy: a thermodynamic quantity representing the amount of energy in a
system that is no longer available for doing mechanical work. The
fundamental idea of an increase of the entropy of a system can give us a way
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of checking the passage of time. Physicists call this the 'arrow of time' that
tells us that time is passing and in 'which direction'. For example if you watch
a film of a pile of books falling over it is easy to tell if the film is being run
backwards since in only one case is the entropy or disorder increasing. Mixing
hot and cold water to give a beaker of lukewarm water shows an entropy
increase - you would not expect the lukewarm water to 'unmix' itself!
Notice the word 'isolated' in the Second Law. Intervention is 'not allowed' to
rebuild the pile of books. However even with human help it is not easy to see
how we could 'unmix' the water.
The third law of thermodynamics is a statistical law of nature regarding
entropy and the impossibility of reaching absolute zero of temperature. The la
states that: As a system approaches absolute zero, all processes cease and
the entropy of the system approaches a minimum value.
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Lesson 36 37 notes – Absolute Zero
Objectives
Be able to describe how there is an absolute scale of temperature that does
not depend on the property of any particular substance (ie the thermodynamic
scale and the concept of absolute zero);
Be able to convert temperatures measured in kelvin to degrees Celsius (or
vice versa): T (K)= θ (°C) + 273.15;
Be able to state that absolute zero is the temperature at which a substance
has minimum internal energy.
Outcomes
Be able to describe how there is an absolute scale of temperature that does
not depend on the property of any particular substance.
Be able to convert temperatures measured in kelvin to degrees Celsius (or
vice versa): T (K)= θ (°C) + 273.15;
Be able to state that absolute zero is the temperature at which a substance
has minimum internal energy.
Thermometers
Since Galileo's invention of the thermometer in
1592, some 32 different scales have been
used to represent temperatures. This entry is a
brief introduction to some of today's most
commonly used scales and how they came into
being.
Fahrenheit (°F)
Daniel Gabriel Fahrenheit 1686 - 1736
Fahrenheit, although born in Germany, spent most of his adult life in The
Netherlands. By profession he was a physicist - working in the field of
meteorological instrumentation. He is credited with the invention of both the
alcohol and mercury thermometers (in 1709 and 1714 respectively). Although
it wasn't until over 150 years later that the mercury thermometer was first
used for what is now its most common usage - an aid to medical diagnosis.
This could have had something to do with the fact that the first medical
thermometers were still over 10cm in length and took over five minutes to
register a temperature.
His original scale was based upon three fixed temperatures:



0° - freezing point of a solution of 1 part (by mass) salt in 1 part water
30° - freezing point of water (later revised to 32°)
90° - temperature of the human body (later adjusted to 96° and finally
to today's figure of 98.2°)
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This resulted in a scale where the difference between the freezing and boiling
points of water was divided into 180 equal parts.
The scale was widely used throughout the English speaking world until the
1970s but since then has gradually been replaced in official usage by the
Celsius scale, with one exception: Fahrenheit continues to be used in both the
USA (officially) and the UK (unofficially) when discussing atmospheric
temperatures.
Centigrade/Celsius (°C)
Anders Celsius 1701 - 1744
Celsius, a Swede, was Professor of Astronomy in Uppsala University,
Sweden from 1730 until his death in 1744.
During this time Anders was actively involved in a great deal of practical
astronomical research: Taking part in an expedition to Lapland in 1736 to
verify Isaac Newton's theory that the world was not actually a perfect sphere.
He was also the first person to connect changes in the Earth's magnetic field
to the phenomenon known as the Aurora Borealis.
Sadly, he died of tuberculosis at the age of just 42.
Two years before his death he published details for a new temperature scale;
to be known as the centigrade scale.
It was based upon two pre-set temperatures:


100° being the freezing point of water.
0° being the boiling point of water at standard atmospheric pressure.
These two figures were reversed after his death - leading to the current
system.
The fact that the difference between the two points was broken up into 100
degrees gave the scale its original name: the Centigrade Scale - this was
finally changed to honour the memory of the astronomer in 1948.
The Celsius scale has now replaced Fahrenheit throughout most of the world.
Other Current Temperature Scales
Kelvin (K)
In 1848, the Scottish physicist, William Thomson (later known as Lord Kelvin)
introduced a new temperature scale. Based upon the Celsius scale, but with
one important difference: The zero point for his scale was set at Absolute
Zero (-273.15°C), meaning that 0°C/32°F is equivalent to 273.15K. This scale
is widely used for astronomical and scientific work.
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Note: Kelvin is the only temperature scale to be expressed without using
'degrees':



70°F - Seventy degrees Fahrenheit
5°C - Five degrees Celsius
300K - Three hundred kelvin
Kelvin is also the only temperature unit that can be used with standard
numerical prefixes (micro, kilo, mega, giga, etc) and is the agreed SI unit of
measurement for temperature.
Absolute Zero
This is defined as 0 K, or –273.15 oC. This is the lowest temperature possible.
In fact the third law of thermodynamics states that it is impossible to actually
reach this temperature. When an object is cooled its internal energy is
reduced and the temperature approaches absolute zero it becomes more and
more difficult to lower the temperature further. You can think of the energy
being reduced in smaller and smaller steps – the steps being rather like those
of a stationary escalator as you near the bottom – they get smaller and
smaller. We are always left with what is known as 'zero point energy' and so
we can define absolute zero as the temperature at which substances have a
minimum internal energy.
This can be found from the pressure law
which satates that temperature is directly
proportortional to the pressure of a gas
in a closed system. (lessons 41 and 42)
If you extrapolate the line back so that
pressure equals zero then you get a
value for absolute zero.
The temperature in deep space is about 3 K above absolute zero and
temperatures as low as 10–6 K have been achieved in the laboratory by
sophisticated means.
The Rankine Scale (°R)
The 'American' equivalent of the Kelvin Scale. The zero point again being
Absolute Zero, however, this time the scale is based upon the Fahrenheit
system, meaning that O°C/32°F is now equivalent to 491.67°R.
Conversion Methods
Fahrenheit to Celsius
Take the Fahrenheit temperature
Subtract 32
Divide by 9
Multiply by 5
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212°F
212 - 32 = 180
180 / 9 = 20
5 x 20 = 100
RAB Plymstock School
Giving
212°F = 100°C
Celsius to Fahrenheit
Take the Celsius temperature
Divide by 5
Multiply by 9
Add 32
Giving
20°C
20 / 5 = 4
4 x 9 = 36
36 + 32 = 68
20°C = 68°F
An Aide-memoire to this Formula
One Researcher's method of remembering this formula is simply drawing a
pair of thermometers side-by-side on a piece of paper, Make 0 on the Celsius
scale align with 32 on the Fahrenheit scale. And 100 with the 212. You have
0-100°C corresponding to 32-212 (or 180) °F. That's a ratio of 100/180 or 5/9.
Then you just have to figure out whether to subtract the 32 first or add it on
after doing the division.
A Few Useful Temperature Equivalents
K
273.15
0
°C
°F
100
212
Boiling point of water
37
98.6
Human body temperature
21
70
0
32
-18
0
-40
-40
-273.15 -459.67
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°R
Notes
Room temperature (approximately)
491.67 Freezing point of water
Cold
Very cold!
0
Absolute Zero
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Lesson 38 39 notes – Specific Heat Capacity
Objectives
Be able to define and apply the concept of specific heat capacity.
Be able to select and apply the equation E = mcΔθ.
Be able to describe an electrical experiment to determine the specific heat
capacity of a solid or a liquid.
Outcomes
Be able to define specific heat capacity.
Be able to select and apply the equation E = mcΔθ to a number of different
situations correctly.
Be able to describe an electrical experiment to determine the specific heat
capacity of a solid or a liquid.
Specific Heat Capacity
Different materials will lose their heat at different rates depending on a factor
called their Specific Heat Capacity.
The specific heat capacity of a material is the amount of energy (J) that
is needed to raise the temperature of 1kg of the substance by 1 degree
C.
Its unit is written J/kg ˚C.
Aluminium has a specific heat capacity of 880J/kg˚C.
Steel has a specific heat capacity of 420J/kg˚C.
Water has a specific heat capacity of 4200 J/kg˚C.
The energy needed to raise the temperature of a material is given by the
equation:
Energy needed (J) = specific heat capacity (J/kg˚C) x mass (kg) x change in
temperature (˚C)
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Lesson 38 39 questions – Specific Heat Capacity
(
/33)……..%
ALL
1
What unit is used for all types of energy?
………………………………………………………………………………………(1)
2
State the equation for the energy needed to raise the temperature of a
particular material of known mass.
…………………………………………………………………………………………
……………………………………………………………………………………… (1)
3
Use the following data to calculate the amounts of energy needed to change
the temperature of the questions below.
Material
Specific heat capacity (J/kg ˚C)
Copper
380
Water
4200
Aluminium
880
Air
1000
a)
2kg of water by 5˚C
…………………………………………………………………………………………
……………………………………………………………………………………… (2)
b)
500g of water by 4˚C
…………………………………………………………………………………………
……………………………………………………………………………………… (2)
c)
100g of aluminium from 20˚C to 30˚C
…………………………………………………………………………………………
…………………………………………………………………………………… (2)
d)
200 g of copper from 60˚C to 10˚C
…………………………………………………………………………………………
……………………………………………………………………………………… (3)
4
A 2kg block of iron is given 10kJ of energy and its temperature rises by 10˚C.
What is the specific heat capacity of iron?
…………………………………………………………………………………………
…………………………………………………………………………………………
…………………………………………………………………………………………
……………………………………………………………………………………… (4)
5
Some cooks make toffee. Essentially, this is a process of boiling down a sugar
solution to concentrate it and then allowing the liquid to cool until it sets. Small
children are usually warned not to touch the cooling toffee for a very long time –
much longer than the cooling for the same volume of pure water in the same vessel.
Why do you think that the cooling period so long?
…………………………………………………………………………………………
…………………………………………………………………………………………
……………………………………………………………………………………… (2)
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6
Show that the energy required to heat the air in your physics laboratory from a
chilly 10 °C to a more comfortable 20 °C is about 3 000 000 J if it has the
following dimensions: 3 m  10 m  10 m.
…………………………………………………………………………………………
…………………………………………………………………………………………
…………………………………………………………………………………………
……………………………………………………………………………………… (4)
7
This question is about the operation of an electrically operated shower.
a)
The water moves at constant speed through a pipe of cross section 7.5
2
x 10 m to a showerhead. See the diagram. The maximum mass of water which flows
per second is 0.090kgs-1.
i)
Show that the maximum speed of the water in the pipe is
-1
1.2ms .
Density of water = 1000kg m-3
-5
ii)
iii)
(2)
The total cross-sectional area of the holes in the head is half
that of the pipe. Calculate the maximum speed of the water as it
leaves the shower head.
Speed = ……………… ms-1 (1)
Calculate the magnitude of the force on the showerhead.
Force = …………….. N (3)
b)
The water enters the heater at the temperature of 15˚C. At the
maximum flow rate of 0.090kgs-3, the water leaves the shower head at a
temperature of 27˚C.
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i) Calculate the rate at which energy is transferred to the water.
Give a suitable unit for your answer.
Specific heat capacity of water = 4200Jkg-1K-1
Rate of energy transfer = ……………… unit …………….. (4)
ii) Suggest a reason that the power of the heater must be greater
than your answer to (b)(i)
…………………………………………………………………………………………
…………………………………………………………………………………………
……………………………………………………………………………………… (1)
iii)
Calculate the maximum possible temperature of the water at the
showerhead when the flow rate is half of the maximum.
Temperature = ………………….˚C (1)
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Lesson 40 notes – Specific Latent Heat
Objectives
Be able to describe what is meant by the terms latent heat of fusion and latent
heat of vaporisation.
Outcomes
Be able to recognise on a temp-time heating curve where latent heat of fusion
and latent heat of vaporisation occurs.
Be able to describe what is meant by the terms latent heat of fusion and latent
heat of vaporisation.
Be able to explain that a rise in temperature of a body leads to an increase in
internal energy.
Be able to use a simple kinetic model to describe melting, boiling and
evaporation.
Be able to describe an experiment that can be used to find the latent heat of
fusion.
If you have a glass of a cool drink, well supplied
with ice, you can expect its temperature to drop
until it is close to 0 ºC. You also can expect (and
can easily check with a thermometer) that it will
remain cold, regardless of the outside
temperature, as long as there remains some
unmelted ice in the drink. Only after all the ice
has melted will the temperature of the drink
begin to rise. Why is this?
When a solid substance changes from the solid phase to the liquid phase,
energy must be supplied in order to overcome the molecular attractions
between the constituent particles of the solid. This energy must be supplied
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externally, normally as heat, and does not bring about a change in
temperature. We call this energy latent heat (the word "latent" means
"invisible"). The latent heat is the energy released or absorbed during a
change of state. With this in mind, we define the specific latent heat of
fusion:
"The specific latent heat of fusion of a substance is the amount of heat
required to convert unit mass of the solid into the liquid without a change in
temperature."
The specific latent heat of fusion of ice at 0 ºC, for example, is 334 kJ.kg-1.
This means that to convert 1 kg of ice at 0 ºC to 1 kg of water at 0 ºC, 334 kJ
of heat must be absorbed by the ice. Conversely, when 1 kg of water at 0 ºC
freezes to give 1 kg of ice at 0 ºC, 334 kJ of heat will be released to the
surroundings.
Latent heats of fusion vary widely, and values should always be accompanied
by the temperatures at which they were measured (these are not necessarily
the normal melting points).
Latent heat of vaporisation:
A change of state from liquid to vapour at constant temperature also requires
the input of energy, called the latent heat of vaporisation. This implies that
while a liquid undergoes a change to the vapour state at the normal boiling
point, the temperature of the liquid will not rise beyond the temperature of the
boiling point.
The latent heat of evaporation is the energy required to overcome the
molecular forces of attraction between the particles of a liquid, and bring them
to the vapour state, where such attractions are minimal.
The definition of the specific latent heat of vaporization is
'The specific latent heat of vaporisation is the amount of heat required to
convert unit mass of a liquid into the vapour without a change in temperature."
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For water at its normal boiling point of 100 ºC, the latent specific latent heat of
vaporization is 2260 kJ.kg-1. This means that to convert 1 kg of water at 100
ºC to 1 kg of steam at 100 ºC, 2260 kJ of heat must be absorbed by the water.
Conversely, when 1 kg of steam at 100 ºC condenses to give 1 kg of water at
100 ºC, 2260 kJ of heat will be released to the surroundings.
Latent heats of vaporization vary widely, and values should always be
accompanied by the temperatures at which they were measured (these are
not necessarily the normal boiling points).
Heating/cooling curves:
The diagram shows the uptake of heat by 1 kg of water, as it passes from ice
at -50 ºC to steam at temperatures above 100 ºC, affects the temperature of
the sample.
A: Rise in temperature as ice absorbs heat.
B: Absorption of latent heat of fusion. Intermolecular bonds are breaking.
C: Rise in temperature as liquid water absorbs heat.
D: Water boils and absorbs latent heat of vaporisation. Intermolecular bonds
are breaking.
E: Steam absorbs heat and thus increases its temperature.
The above is an example of a heating curve. One could reverse the process,
and obtain a cooling curve. The flat portions of such curves indicate the
phase changes.
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This diagram shows the rise in temperature of 200ml of water (red) and 1kg of
steel (blue). Water has a specific heat capacity of 4200 J/Kg˚C and a boiling
point of 100˚C. Its specific latent heat of vaporisation is 2258 kJ/kg. Steel has
a boiling point of 825ºC and specific heat capacity of 450 J/kg°C.
Some values for specific latent heats of fusion and vaporization:
Specific latent heat of
fusion
kJ.kg-1
ºC
Specific latent heat of
vaporization
kJ.kg-1
ºC
Water
334
0
2258
100
Ethanol
109
-114
838
78
Ethanoic acid
192
17
395
118
Chloroform
74
-64
254
62
Mercury
11
-39
294
357
Sulphur
54
115
1406
445
Hydrogen
60
-259
449
-253
Oxygen
14
-219
213
-183
Nitrogen
25
-210
199
-196
Substance
Worked example:
Calculate the amount of heat required to completely convert 50 g of ice at 0
ºC to steam at 100 ºC. The specific heat capacity of water is 4.18 kJ.kg-1.K-1.
The specific latent heat of fusion of ice is 334 kJ.kg-1, and the specific heat of
vaporization of water is 2260 kJ.kg-1.
Answer: Heat is taken up in three stages: 1. The melting of the ice, 2. the
heating of the water, and 3. the vapourization of the water. The heat taken up
in the complete process is the sum of the heat taken up in each stage.
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1. Heat taken up for
converting ice
at 0ºC to water at 0ºC
mass of water x latent heat of fusion
= 0.050 (kg) x 334 (kJ.kg-1)
= 16.7 kJ
2. Heat taken up heating the
water
from 0 ºC to the boiling
point, 100 ºC
mass of water x specific heat capacity x
temperature change
= 0.05 (kg) x 4.18 (kJ.kg-1. K-1)x 100 (º K)
= 20.9 kJ
3. Heat taken up vaporizing
the
water
mass of water x latent heat of vaporization
0.05 (kg) x 2260 kJ.kg-1
= 113 kJ
The sum of these is
16.7 + 20.9 + 113
= 150.6 kJ (151 kJ)
The heater in an electric kettle delivers 1.5 kW of power to 2 kg of water at its boiling
point. The specific latent heat of vaporisation of water is 2.26 x 106 J/kg.
(a) How much energy would be needed to boil off 1 kg of the water?
(b) The kettle is switched on for 100 s.
(i) How much heat energy is delivered to the water in this time.
(ii) How much steam is produced in 100 s?
The Answer
(a) 2.26 x 106 J
(b)(i)
Heat supplied, Eh = power x time
= 1.5 x 103 x 100
= 1.5 x 105 J
(ii)
Eh = m l
m = Eh/l
= 1.5 x 105/2.26 x 106
= 0.066
mass of steam produced = 0.066 kg
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Lesson 40 questions – specific latent heat
(
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ALL
Question 1
Calculate the energy released when (a) 10 g water at 100 °C and (b)
10 g of steam at 100 °C are each spilt on the hand.
Take the specific heat capacity of water to be 4200 J /kg °C and the
specific latent heat of vaporisation of water to be 2.2 MJ / kg.
Assume that the temperature of the skin is 33 °C.
a)………………………………………………………………………………………
…………………………………………………………………………………………
…………………………………………………………………………………………
………………………………………………………………………………………(3)
b)………………………………………………………………………………………
…………………………………………………………………………………………
…………………………………………………………………………………………
……………………………………………………………………………………… (3)
Question 2
When a falling hailstone is at a height of 2.00 km its mass is
2.50 g. It has 49.05 J of gravitational potential energy.
Assuming that all of this potential energy is converted to latent
heat during the fall, calculate the mass of the hailstone on
reaching the ground. Take the specific latent heat of fusion of
ice to be 336000 J / kg.
…………………………………………………………………………………………
…………………………………………………………………………………………
…………………………………………………………………………………………
…………………………………………………………………………………………
………………………………………………………………………………………(3)
Question 3
0.30 kg of ice at 0 °C is added to 1.0 kg of water at 45 °C. What is
the final temperature, assuming no heat exchange with the
surroundings? Take the specific heat capacity of water to be 4200 J /
kg °C and the specific latent heat of fusion of ice to be 340 000 J /
kg.
…………………………………………………………………………………………
…………………………………………………………………………………………
…………………………………………………………………………………………
…………………………………………………………………………………………
……………………………………………………………………………………… (3)
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4.
Why is a scald by steam at 100 oC much more painful than one by water at
100 oC?
Specific latent heat of fusion of ice = 335 000 J kg-1
Specific latent heat of evaporation of water 2.26 MJ kg-1
(3)
5.
How long will it take a 50 W heater to melt 2 kg of ice at 0 oC?
(3)
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Lesson 41 42 notes – Boyle’s Law
Objectives
Be able to state Boyle’s law.
Outcomes
Be able to state Boyle’s law;
Be able to describe an experiment that demonstrates Boyle’s Law.
Boyle’s Law
Boyle’s Law states that the pressure of a gas is inversely
proportional to its volume.
Boyle’s law and number of molecules
Two ways to double gas
pressure
N molecules in volume V
cram in more molecules
increase N
squash the gas
decrease V
piston squashes up same molecules into
half the volume, so doubles the number per
unit volume
N molecules
molecules in box:
pressure due to impacts
of molecules with walls
of box
add extra molecules to double the number,
so double the number per unit volume
same:
number of molecules
per unit volume
number of impacts on
wall per second
pressure
pressure proportional to 1/volume
p  1/V
2N molecules in volume V
pressure proportional to number of molecules
pN
If.... pressure is proportional to number of
impacts on wall per second
Then.... pressure is proportional to number of
molecules per unit volume
and if.... number of impacts on wall per second
is proportional to number of molecules per unit
volume
p = constant  N/V
Boyle’s law in two forms
pV = constant  N
p = constant  N/V
Boyle’s law says that pressure is proportional to crowding of molecules
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Boyle’s law and gas density
1
2
4
3
pressure p
volume V
mass m
density d
Boyle’s law:
compress gas to half volume:
double pressure and density
1
2
4
3
pressure 2p
volume V/2
mass m
density 2d
half as much gas in half volume:
same pressure and density
1
2
4
3
double mass of gas in same volume:
double pressure and density
pressure p
2
3
pressure 2p
volume V
mass 2m
density 2d
volume V/2
mass m/2
density d
push in
piston
1
4
pump in
more air
temperature constant in each
Boyle’s law says that gas pressure is proportional to density
Boyle’s Law Experiment
You need to measure how the volume of a fixed number of particles affects the pressure.
Squeezing these particles into a smaller and smaller volume results in more and more
collisions with the walls, giving a higher pressure. However, you will only get a true
relationship between pressure and volume if the number of particles stays the same: you
need to make sure that no molecules escape. Rough and ready results can be obtained by
using syringes, but these leak, so more precise ways have evolved – sealing off a volume of
gas behind a liquid makes for a good seal. It is the measurement of volume that turns out to
be the difficult one to get right. You may be able to set up a slicker arrangement using
automated data capture, but you will need to take great care to measure the volume, ensuring
the equivalent of a leak-proof syringe, where you can measure the position of the plunger or
piston.
Compressing a gas will warm it up and vice versa, so after changing the volume leave
sufficient time for the temperature to return to its original value.
A traditional solution
sample of air
0
101300 Pa
5
10
15
20
oil to transmit pressure
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Extension
The Pressure Law
Pressure and volume of gases increasing with temperature
Constant volume
Constant pressure
1
2
4
3
pressure p
45.1
45.1
volume V
T/C
T/C
heat gas:
pressure
increases
–273
heat gas:
volume
increases
0
temperature/C
–273
0
temperature/C
Pressure and volume extrapolate to zero at same temperature –273.16 C
So define this temperature as zero of Kelvin scale of temperature, symbol T
0
273
temperature/K
pressure proportional to Kelvin temperature
273
temperature/K
volume proportional to Kelvin temperature
pT
VT
0
Pressure and volume proportional to absolute temperature
Charle’s Law
Charles's Law states that the volume of a given amount of dry ideal gas is
directly proportional to the Kelvin Temperature provided the amount of gas
and the pressure remain fixed.
When we plot the Volume of a gas against the Kelvin temperature it forms a
straight line. The mathematical statement is that the V / T = a constant. For
two sets of conditions the following is a math statement of Charles's Law:
V1 / T1 = V2 / T2
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Lesson 43 notes – Ideal Gases
Objectives
Be able to select and apply pV = constant
T
Be able to state the basic assumptions of the kinetic theory of gases;
Outcomes
Be able to select and apply pV = constant
T
correctly to a number of
situations.
Be able to state the basic assumptions of the kinetic theory of gases.
The Ideal Gas Equation
From lessons 41 and 42 we know that:
Pressure Law: P
T (constant n, V)
So for a fixed number of moles, (see lesson 44 for definition) we could say
that:
pV = constant
T
Assumptions of the kinetic theory of an IDEAL GAS.
1
2
3
4
5
6
7
8
A Gas consists of particles called molecules.
The molecules are in constant random motion. As many travelling in
one direction as any other. The centre of mass of the gas is at rest.
Intermolecular forces are negligible.
The duration of collisions between molecules is negligible.
Molecules move with constant velocity in between collisions.
The volume of gas molecules is negligible compared with the volume of
the gas.
All collisions are totally elastic.
Newtonian mechanics can be applied to the collisions.
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Lesson 44 notes – Moles and the Ideal Gas equation
Objectives
Be able to state that one mole of any substance contains 6.02 × 1023 particles
and that 6.02 × 1023 mol-1 is the Avogadro constant NA.
Be able to select and solve problems using the ideal gas equation expressed
as pV = nRT, where N is the number of atoms and n is the number of moles.
Outcomes
Be able to state that one mole of any substance contains 6.02 × 1023 particles
and that 6.02 × 1023 mol-1 is the Avogadro constant NA.
Be able to state that R is the molar gas constant and be able to select and
use it correctly from a data book.
Be able to show that R has the Units J mol-1 K-1.
Be able to select and solve problems using the ideal gas equation expressed
as pV = nRT, where N is the number of atoms and n is the number of moles.
The mole
The mole (symbol: mol) is
the SI base unit that
measures an amount of
substance. One mole
contains Avogadro's
number (approximately
6.022×1023) things. A mole
is like "a dozen" in that
both are absolute numbers
(having no units) and can
describe any type of
elementary object,
although the mole's use is
usually limited to
measurement of
subatomic, atomic, and molecular structures.
Why?
A mole is the amount of substance of a system which contains as many
elementary entities as there are atoms in 0.012 kilogram (or 12 grams) of
carbon-12. It is useful because we can take any element and 1 mole of it will
equal its atomic mass in grams.
n= mass of sample / molar mass of substance
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The Ideal Gas Equation
Pressure Law: P
T (constant n, V)
And so:
So we can use a constant and make V equal nT/P multiplies by this constant
This is normally written as:
PV = nRT
P is pressure of gas in Pa
V is volume of gas m3
n is the number of moles of gas in mol
R is called the gas constant and has the value 8.314 Jmol-1K-1
(since R=PV/nT
an analysis of the units for PV/nT ≡ Nm-2 m3 /mol K
≡ Nm/mol K
≡ J mol-1 K-1)
T is temperature in K.
Examples
Avogadro's Number (NA)


1 mole of atoms or molecules contains 6.022 x 1023 atoms or
molecules
eg, 1 mole of helium atoms (He) contains 6.022 x 1023 helium atoms
(He)
To find the number of atoms in a known number of moles, multiply the
moles by 6.022 x 1023
eg, 2 moles of helium atoms (He) contains 2 x 6.022 x 1023
= 1.2044 x 1024 helium atoms. (He)
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




To find the moles of atoms, divide the number of atoms by 6.022 x 1023
eg, If we have 4.2154 x 1023 neon atoms, how many moles of neon
atoms are there?
Moles of neon atoms = (4.2154 x 1023) ÷ (6.022 x 1023)
= 0.7 mol
1 mole of molecules contains 6.022 x 1023 molecules.
eg, 1 mole of oxygen molecules (O2) contains 6.022 x 1023 oxygen
molecules (O2).
To find the moles of molecules, multiply the number of molecules by
6.022 x 1023
eg, ½ mole of oxygen molecules (O2) contains ½ x 6.022 x 1023
= 3.011 x 1023 oxygen molecules (O2).
To find the number of molecules, divide the moles of molecules by
6.022 x 1023
eg, If we have 6.022 x 1021 chlorine molecules (Cl2), how many
moles of chlorine molecules are there?
Moles of chlorine molecules = (6.022 x 1021) ÷ (6.022 x 1023)
= 0.01 mol
1 mole of molecules does not necessarily contain 1 mole of atoms of
each element in the formula
eg, 1 mole of HCl WILL contain 1 mole of hydrogen atoms (H) and 1
mole of chlorine atoms (Cl)
eg, 1 mole of HCl contains 6.022 x 1023 hydrogen atoms and 6.022 x
1023 chlorine atoms
eg, 5 moles of oxygen molecules (O2) contains 5 x 2 = 10 moles of
oxygen atoms (O)
eg, 5 moles of oxygen molecules contains 10 x 6.022 x 10 23 = 6.022
x 1024 oxygen atoms.
eg, 1 mole of ammonia molecules (NH3) will contain 1 mole of
nitrogen atoms (N) and 3 moles of hydrogen atoms (H)
1 mole of ammonia molecules contains 6.022 x 1023 nitrogen atoms
and 3 x 6.022 x 1023 = 1.8066 x 1024 hydrogen atoms
Molar Mass

1 mole of a pure substance has a mass in grams equal to its molecular
mass (MM).
eg, 1 mole of Helium (a monatomic gas with the formula He) has a
mass equal to its relative atomic mass, 4.003g
eg, 1 mole of hydrogen gas (a diatomic gas with the formula H2) has
a mass equal to 2 x 1.008 = 2.016g
eg, 1 mole of ammonia gas (NH3) has a mass equal to 14.01 + (3 x
1.008) = 17.034g
eg, 1 mole of water (H2O) has a mass equal to (2 x 1.008) + 16.00 =
18.016g
Ideal Gas Volumes

at S.T.P [0oC (273K), 101.3kPa (1 atm)], an ideal gas has a volume of
22.4L
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



To find the volume of a certain number of moles of gas, multiply the
moles by 22.4L
eg, What is the volume of 2.5 moles of chlorine gas at S.T.P?
Volume of chlorine gas = 2.5 x 22.4 = 56.0L
To find the moles of a certain volume of gas, divide the volume by
22.4L
eg, How many moles of argon are in 3.36L of argon gas at S.T.P?
moles of argon gas = 3.36 ÷ 22.4 = 0.15 mol
at S.L.C [25oC (298K), 101.3kPa (1 atm)], an ideal gas has a volume of
24.47L
eg, To find the volume of a certain number of moles of gas, multiply
the moles by 24.47L
What is the volume of 0.2 moles of hydrogen sulfide gas at S.L.C?
Volume of hydrogen sulfide gas = 0.2 x 24.47 = 4.894L
To find the moles of a certain volume of gas, divide the volume by
24.47L
eg, How many moles of carbon monoxide are in 70.5L of carbon
monoxide gas at S.L.C?
moles of argon gas = 70.5 ÷ 24.47 = 2.881 mol
Example:
If we had 1.0 mol of gas at 1.0 atm of pressure at 0?C (273.15 K), what would
be the volume?
PV = nRT
V = nRT/P
V = (1.0 mol)(0.0821 L atm/mol K)(273 K)/(1.0 atm)
V = 22.41 L

0 ?C and 1 atm pressure are referred to as the standard temperature
and pressure (STP)
The molar volume of an ideal gas (any ideal gas) is 22.4 litres at STP
Example
Nitrate salts (NO3-) when heated can produce nitrites (NO2-) plus oxygen (O2).
A sample of potassium nitrate is heated and the O2 gas produced is collected
in a 750 ml flask. The pressure of the gas in the flask is 2.8 atmospheres and
the temperature is recorded to be 53.6 ?C.
How many moles of O2 gas were produced?
PV = nRT
n = PV/RT
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n = (2.8 atm * 0.75 L) / (0.0821 L atm/mol K * (53.6 + 273)K
n = (2.1 atm L) / (26.81 L atm/mol)
n = 0.078 mol O2 were produced
Relationship Between the Ideal-Gas Equation and the Gas Laws
Boyle's law, Charles's law and Avogadro's law represent special cases of the
ideal gas law

If the quantity of gas and the temperature are held constant then:
PV = nRT
PV = constant
P = constant * (1/V)
P

1/V (Boyle's law)
If the quantity of gas and the pressure are held constant then:
PV = nRT
V = (nR/P) * T
V = constant * T
V

T (Charles's law)
If the temperature and pressure are held constant then:
PV = nRT
V = n * (RT/P)
V = constant * n
V

n (Avogadro's law)
A very common situation is that P, V and T are changing for a fixed
quantity of gas
PV = nRT
(PV)/T = nR = constant
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
Under this situation, (PV/T) is a constant, thus we can compare the
system before and after the changes in P, V and/or T:
Example
A 1 liter sample of air at room temperature (25 ?C) and pressure (1 atm) is
compressed to a volume of 3.3 mls at a pressure of 1000 atm. What is the
temperature of the air sample?
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Lesson 44 questions – Ideal Gas Equation
(
/16)…….%……..
ALL
1
What are the Avagadro Constant and the Mole and how are they linked?
…………………………………………………………………………………………..
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…………………………………………………………………………………………..
……………………………………………………………………………………… (2)
MOST
2
a)
The equation of state of an ideal gas is pV = nRT. Explain why the
temperature must be measured in Kelvin.
…………………………………………………………………………………………..
…………………………………………………………………………………………..
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…………………………………………………………………………………………..
……………………………………………………………………………………… (2)
b)
A meteorological balloon rises through the atmosphere until it expands
to a volume of 1.0 x 106m3, where pressure is 1.0 x 103 Pa. The temperature also falls
from 17˚C to –43˚C.
The pressure of the atmosphere at the Earth’s surface = 1.0 x 105Pa.
Show that the volume of the balloon at take off is about 1.3 x 104m3.
(3)
-3
c)
The balloon is filled with helium gas of molar mass 4.0 x 10 kgmol
at 17˚C at a pressure of 1.0 x 105Pa. Calculate
i)
the number of moles of gas in the balloon
-1
number of moles = ……………. (2)
ii)
the mass of gas in the balloon.
Mass = ………….kg (1)
d)
The internal energy of the helium gas is equal to the random kinetic
energy of all of its molecules. When the balloon is filled at ground level at a
temperature of 17˚C the internal energy is 1900 MJ. Estimate the internal energy of
the helium when the balloon has risen to a height where the temperature is –43˚C
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Internal energy = ……………MJ (2)
3
a)
Very high temperature, for example, the temperature of the solar
corona at half a million degrees, are often stated without a complete unit, i.e. degrees
Celsius of Kelvin.
Suggest why it is unnecessary to give degrees Celsius or Kelvin in this case.
…………………………………………………………………………………………
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…………………………………………………………………………………… (2)
b)
Two students attempt the same experiment to find how air pressure
varies with temperature. They heat identical sealed glass flasks of air, to be
considered as an ideal gas, in an oil bath. The flasks are heated from 300k to 400k
The pressure in flask A rises from atmospheric pressure, p0, as expected, but the
pressure in flask B remains at p0 because the rubber bung is defective and air leaks out
of the flask.
i)
Calculate the pressure in flask A at 400K in terms of p0.
ii)
to flask A at 400K.
f
Pressure = …………………… (2)
Calculate the fraction, f, of gas molecules in flask B compared
=
number of gas molecules in B at 400K
number of gas molecules in A at 400K
f = ………………………. (2)
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Lesson 45 notes – Boltzmann’s Constant
Objectives
Be able to select and solve problems using the ideal gas equation expressed
as pV = NkT and pV = nRT, where N is the number of atoms and n is the
number of moles;
Be able to explain that the mean translational kinetic energy of an atom of an
ideal gas is directly proportional to the temperature of the gas in kelvin;
Be able to select and apply the equation E = 3/2kT for the mean translational
kinetic energy of atoms.
Outcomes
Understand the difference between R and k and when to apply them.
Be able to state that the Boltzmann constant is the gas constant for individual
molecules whilst the molar gas constant is the constant used when dealing
with quantities in moles.
Be able to explain that the mean translational kinetic energy of an atom of an
ideal gas is directly proportional to the temperature of the gas in Kelvin.
Be able to select and apply the equation E = 3/2kT for the mean translational
kinetic energy of atoms correctly in different situations.
Be able to derive equation for the translational KE of an atom in an ideal gas
E = 3/2kT.
The Boltzmann Constant
The Boltzmann constant is used when we are looking at amounts of
molecules rather than amounts of moles.
From lesson 44 we know that
pV = nRT
We can now write that
pV = NkT
where N is the number of atoms and n is the number of moles
So we can see that
nR=Nk
Also
N=nNA
Therefore
k=R/NA
Check that you understand this from the definitions of these symbols.
Translational kinetic energy of an atom of an ideal gas
From lesson 33 we saw that from the kinetic theory, gas pressure is given by:
P = ⅓ρ<c>2
Letting <c> = the mean square speed
which is the same as saying:
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P  Nmv2 / 3V .
From lesson 44 we saw that the Ideal Gas equation is:
PV = nRT
Therefore:
nRT = 1/3 Nm<c>2
Now,
KE = 1/2m v2
Rearranging
3nRT / N = m<c>2
So,
1/2 m<c>2= (3/2) nRT/N
We know that
NA = N/n so n/N = 1/NA
(N is the number of molecules in a volume of gas, n is the amount of gas in
moles, NA is Avagadro’s number).
Therefore:
E = 1/2 m<c>2= (3/2) RT / NA = (3/2)kT
This tells us that the mean kinetic energy of a molecule of an ideal gas is
proportional to the thermodynamic temperature. (lesson 35).
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Lesson 45 questions – Boltzmann Constant
(
Ideal gas constant (R) = 8.31 Jmol-1K-1
mol-1
/44)………%………
Avogadro’s number (NA) = 6.02x1023
1.
Use the data given on this page to calculate the value of the Boltzmann
constant (k)
2.
k = …………… units…………. (3)
Calculate:
(a)
the kinetic energy of an individual gas molecule of mass 3.5x10-26 kg
moving at a speed of 600 ms-1.
(b)
Kinetic energy = …………….. J (3)
the kinetic energy of the gas molecules in a cylinder at a temperature
of 20oC
Kinetic energy = …………….. J (3)
3.
A cylinder contains 2 moles of a gas composed of molecules with a mass of
-26
2.5x10 kg moving with an r.m.s. speed of 500 ms-1. Calculate the temperature of the
gas.
Temperature = ………….. Kelvin (4)
4.
A container of volume 1.5 m3 is full of gas with a density of 1.8 kgm-3 at a
pressure of 1.4x105 Pa.
(a)
Calculate the root mean square velocity of the molecules within the
gas.
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root mean square velocity = ……………… ms-1 (3)
(b)
If the mass of an individual molecule of the gas is 3.0x10-27 kg
calculate the number of molecules in the container
Number of molecules = ……………….. (3)
5.
A deuterium plasma contains deuterium ions of mass 4.9x10-27 kg at a
temperature of 150x106 K.
Calculate:
(a) the r.m.s. speed and
r.m.s. speed = ……………….. ms-1 (3)
(b) the kinetic energy of a deuterium ion in the plasma
Kinetic energy = …………………….. J (3)
6.
The mean kinetic energy of a gas molecule at an absolute temperature T is
given by:
kinetic energy = 3RT/2NA
where R is the molar gas constant and NA is the Avogadro constant.
a)
Calculate the mean kinetic energy of gas atoms at 0 °C.
Kinetic energy = …………………….. J (2)
b)
Determine the speed of carbon dioxide molecules at 0 °C. The molar
mass
of carbon dioxide is 44 g.
Speed = ……………. ms-1 (5)
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c)
Calculate the change in the internal energy of one mole of carbon
dioxide
gas when its temperature increases from 0 °C to 100 °C. [3]
change in the internal energy = ……………… J (3)
SOME
7
from the kinetic theory, gas pressure is given by:
P = ⅓ρ<c>2
a)
State the ideal gas equation for n number of moles
(1)
b)
By using the equation for pressure given by the kinetic theory, the
ideal gas equation and the definition of kinetic energy show that E = 3/2kT for the
mean translational kinetic energy of an atom.
(7)
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b)
What can we deduce about the relationship between the mean
translational kinetic energy of a molecule and the thermodynamic
temperature of a gas?
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…………………………………………………………………………………………
…………………………………………………………………………………………..
……………………………………………………………………………………… (1)
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