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Thermodynamics Notes
Introduction
These notes on Thermodynamics were produced as a result of discussions at the GCE Physics
Inset sessions in autumn 2003 and 2004. The document contains much material, which is
intended to clarify terms and conventions in Thermodynamics, which are often
misunderstood. Examples of this are:



the distinction between heat and internal energy
the use of a sign convention to indicate direction of energy transfer
the use of Work in other than mechanical situations e.g. electrical work
The last section of the notes on the 2nd Law of Thermodynamics and the idea of entropy are
included for interest only and clearly outside the core specification.
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Systems
The laws of thermodynamics apply to systems. A system is just about anything which we can
imagine to be defined by a boundary. At A level the favourite system – because it’s simple
but not trivial – is a sample of ideal gas in a cylinder fitted with a piston. Other examples of
systems are:
•
•
•
•
•
•
a rubber band
a mixture of oxygen and hydrogen in a sealed vessel
a lamp filament
a star
a black hole
the electromagnetic radiation inside a furnace.
The last example is included because it started a revolution in Physics. By applying
thermodynamics (and its sister science, statistical mechanics), the German physicist Max
Planck was led to conclude that the radiation was emitted from, and absorbed by, the walls of
the furnace in discrete packets of energy. From this idea grew Quantum Theory, which has
guided most of the discoveries in Physics from 1900 to the present day.
The Thermodynamic State of a System
For many systems, including gases, any physical property you
can think of can be expressed in terms of just two variables.
For gases, pressure and volume are often chosen as these two
variables. They are said to determine the state of the system.
So a particular point on a plot of p against V represents a
particular state of a sample of gas.
If the system is n moles of an ideal gas, then we have:
pV = nRT
This is called the equation of state of an ideal gas. It gives the property, temperature, in terms
of pressure and volume.
[We could just as well say that eq.1 gives the pressure in terms of volume and temperature. In
other words we can choose which two variables in terms of which we express the system’s
(other) properties.]
Different systems have different equations of state. For example, the equation of state of a
rubber band, relates tension, length and temperature. This equation is beyond A-Level!
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The Internal Energy of a System
The internal energy, U, of a system is the sum of the potential and kinetic energies of its
particles.
[We shall not include gravitational energy or energy due to nuclear forces. This is because
they don’t change in the cases we shall be dealing with, and might just as well be zero.]
The Internal energy of an ideal gas
For an ideal monatomic gas, the forces between molecules are negligible (except during
collisions) and this means that the potential energy can be taken as zero. The internal energy
is, then, simply the sum of the kinetic energies of the molecules. So for N molecules each of
mass m and with mean square speed c 2 :
U  N  12 mc 2
But, using Newton’s Laws, it can be shown that pressure  volume is given by:
pV  13 Nmc 2
So:
U  32 pV
For a given pressure and volume U can, then, have only one value. We say that the system’s
internal energy is a function of the system’s state. This is true of all systems. A special
feature of an ideal gas is that U can be expressed in terms of the number n of moles and the
temperature T alone. Using pV = nRT (eq.1) to define T, we have:
U  32 nRT
eq.2
Internal energy of a real gas
The internal energy of a real gas depends on its volume as well as its temperature. This is
because of forces which act between the molecules. These are fairly negligible for a gas at
lowish density, such as oxygen at room temperature and pressure, which is nearly ideal in
behaviour. A-Level questions on gases will assume ideal behaviour!
Even at low densities the factor of 3/2 in eq.2 is only correct for monatomic gases. For oxygen
and other diatomic molecules (molecules consisting of two atoms bonded together) the factor
is usually close to 5/2. This is because the molecules have kinetic energy of rotation as well as
ordinary translational movement through space.
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Internal energy of a rubber band
A molecule of rubber is a long chain of atoms. The bonds
between segments of the chain allow rotation of the segments,
of the type suggested by the top diagram.
This means that the chain is constantly changing shape. The
most frequently occurring shapes are ‘folded up’ tangles (see
lower diagram). Even an approximately straight chain would be
fantastically improbable – unless tension is applied to the
rubber.
The internal energy of a rubber band includes the kinetic
energy of random rotation of the segments. The band’s internal
energy hardly changes at all when it is stretched, provided its
temperature is kept constant. How is this like ideal gas
behaviour?
Changing the internal energy of a system
A system can gain internal energy by having work done on it or by having heat flowing into
it.
A system can lose internal energy by doing work on something external to it, or by having
heat flowing out of it.
Indeed, in accordance with the Law of Conservation of Energy:
Gain in system’s
internal energy
Net heat flowing into
system
=
+
Net work done on
system
eq. 3a
Each of the three terms in this equation can be either positive or negative. For example, the
net heat flowing into a system is negative if, in fact, heat flows out of the system.
Work
The basic definition is:
work done by a
force
=
magnitude of
force

distance moved by force
in direction of force
Here are some examples of systems doing work or having work done on them:
Work done by a rubber band
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eq.4
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When we increase the length of a rubber band by a small length x by applying a pull F,
then:
Work done on band = Fx
We could equally well write this as
Work done by band = – Fx
To see the point of this, suppose we allow the band to contract, exerting a pull F. In this case
x is negative so the work done by the band is positive, which makes sense.
If we stretch the band a lot, then F will change significantly during the stretching. In this case
we add together all the bits of work, Fx, which means adding together the areas of all the
narrow strips under the graph (see above), from the initial extension x1 up to the final
extension, x2. So:
Work done on (or by) band = area under Force-extension graph
eq.5
A positive amount of work is done on the band if x is increasing; a positive amount of work is
done by the band if x is decreasing.
Work done by a gas
When a gas exerting a pressure p expands by a small volume V,
then:
Work done by gas = pV
[This is, in fact, just a more convenient way of writing Fx, in which F is the force on the
piston and x is the distance it moves.]
We could equally well write
Work done on gas = –pV
If we push the piston in a little way V is negative so a positive amount of work is done on
the gas – as expected, since we’ve had to do the pushing.
If the gas expands a lot, then p will change significantly. In this case we add together all the
bits of work, pV, which means adding together the areas of all the narrow strips under the
graph (see diagram), from the initial volume V1 up to the final volume, V2. So:
Work done by gas = area under pressure - volume graph
eq.6
A positive amount of work is done by the gas if V is increasing; a positive amount of
work is done on the gas if V is decreasing.
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Frictional Work
Suppose we push or pull a file across a piece of metal. We are doing work because we are
exerting a force which is moving in the same direction as the force itself. The system on
which we are doing work is the metal block (and the file).
An important feature of frictional work is that it is irreversible – the block won’t push or pull
the file and move it backwards! Compare with the reversible work done on the rubber band or
the gas when we push in the piston.
Electrical Work
If a potential difference is applied across a resistor (e.g. a wire) and a current I flows for a
time t:
Electrical work done on resistor = VIt
eq.7
What has VIt to do with force  distance? It is not very difficult to show that the sum of the
forces on the free electrons caused by the electric field set up by V, multiplied by the mean
distance which they move in time t as a result, is equal to VIt. [See E-fields in PH4]
The electrical work done on a resistor is irreversible. The resistor won’t be able to do work
pushing the electrons back the other way.
The electrical work done on a pure inductor or a capacitor is, however, reversible.
Heat
Heat is energy flowing from a region of higher temperature to a region of lower
temperature because of the temperature difference.
Compare with the flow of charge due to a potential difference…
If there is no temperature difference, no heat will flow. [In the electrical case, if there is no
potential difference, no charge will flow.]
It takes time for a finite amount of heat to flow, though the rate of flow is greater the greater
the temperature difference (other things being the same).
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Thermodynamics: Applying the ideas
Example 1: Rapid expansion of an ideal gas
Think about the terms on the right hand side of eq.3a:
Gain in system’s
internal energy
=
Net heat flowing into
system
+
Net work done on
system
What about the heat? If the expansion is really rapid the heat flow will be (almost) zero.
What about the work? As the gas expands work is done by the gas. The last term in the
equation is negative. Another way of seeing this is to write the equation as:
Gain in system’s
internal energy
=
Net heat flowing into
system

Net work done by
system
eq.3b
Whichever way we look at it, because the heat term is zero, the right hand side of the
equation is negative, so the gain in internal energy of the gas is negative, i.e. the internal
energy decreases. So the gas temperature falls (see eq.2). It really does happen…

Try this… Squeeze some air down to a quarter of its volume or less in a large plastic
syringe. Wait a few seconds. Let the piston move outwards quickly, doing work on
your hand (i.e. not simply letting go of it). Feel the coolness of the syringe.

Suppose you let the trapped gas return to room temperature, keeping its volume
constant. Draw a line on the graph above to show what happens to the pressure.

When you squashed the gas in the first place what happened to the temperature and
why? Assume you squashed it very quickly. You notice the effect when using a
bicycle pump. In a diesel engine rapid squashing of air and fuel causes the
temperature of the mixture to rise so high that the fuel ignites (without needing a
spark).
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Example 2: Slow expansion of an ideal gas
This time heat will have time to flow. For each small increase in volume the gas temperature
will drop a little (Example 1) and heat will flow in from the surroundings – limiting further
temperature drop. If the expansion is really slow (and the cylinder walls conduct heat well)
the temperature drop is negligible, so the expansion is isothermal.
An isothermal change is a change at constant temperature.
Now look at eq.3b. Re-arranging it slightly we have:
Net heat flowing
into system
=
Gain in system’s
internal energy
+
Net work done by
system
eq.3c
Since, in this case, the temperature does not change, there is no change in the internal energy
of the gas (see eq.2). So
Net heat flowing
into system
=
Net work done by
system
This work can, in principle, be used to lift weights, generate electricity, propel a vehicle and
so on. The process has its uses! [Unfortunately the gas pressure drops until no more useful
work can be done. But if we could restore the gas to its original state, we could do the same
useful thing all over again. The problem is that restoring the state simply by pushing the
piston back in again (slowly) requires us to put in the same amount of work the gas gave out
when it expanded! The solution is to use a cycle of changes – see page 10.]
We see from this example that…
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Constant temperature is not at all the same thing as no heat flow.
Heat flowing into a system does not necessarily make it hotter (raise its temperature).
Example 3: A hot gas cools
Suppose we have a sample of hot gas confined in a cylinder by a piston. If we let the gas cool
down at constant volume (that is with the piston held in one place), no work is done. So
adapting eq.3c,
Heat flowing out
of gas
=
Decrease in internal
energy of gas
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Can’t we say, then, that the hot gas was ‘storing heat’? No, because the hot gas could have
cooled down (i.e. lost internal energy) without giving out heat. It could have done work
instead, as in Example 1. If we had let the piston move out quickly, doing work, then:
Work done by gas
•
=
Decrease in internal
energy of gas
The energy stored by the gas is internal energy. It isn’t heat and it isn’t work. Its
decrease could result in heat being given out or in work being done (or both).
Heat and work are both energy in transit. Thinking of them as ‘stored’ in a system
leads to confusion in thermodynamics!
Example 4: A spark ignites a mixture of hydrogen and oxygen in a vessel
A chemical reaction occurs and the contents of the vessel turn wholly or partly to steam.
There is a sudden huge rise in the temperature and pressure in the vessel. What has happened
to the internal energy of the system (i.e. the contents of the vessel)?
Has heat flowed into or out of the system? Hardly at all in the short time of the chemical
reaction. The spark itself contributes negligible energy. Afterwards, heat will flow out.
Has work been done or by the system? Not during the reaction. Almost immediately
afterwards, the vessel may shatter and the pieces acquire kinetic energy as the expanding gas
does work on them.
So eq.3 (any version: 3a, 3b or 3c) shows that during the reaction there is no change in the
internal energy. How can this be, when the system gets hotter? The increase in molecular
kinetic energy is at the expense of a decrease in chemical potential energy as the atoms bond
with different partners. The overall internal energy is unchanged – at first.
Example 5: Quick extension of a rubber band
We do work on the band. Hardly any heat will flow out of the band if we do the work
quickly. So the internal energy will rise. Since, as for an ideal gas, this is mainly kinetic in
form, the band’s temperature will rise.

Try it, using a fairly stout band. Feel its temperature immediately (on your lips?)

Wait a few seconds, so that heat can flow out of the band into the air, and the band
regain room temperature. Now let the band contract, doing work on you. Feel it again.
Example 6: A filament lamp
Suppose we connect the lamp to a supply giving a steady p.d. of
12·0V. The filament is initially at room temperature and the
current is 30A. The filament gets hotter, and finally achieves a
steady ‘working’ temperature. It now takes a current of 4.0A.
[The current is lower because the resistance of the filament is
higher.]
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
How much does the filament’s internal energy increase during the first millisecond?
The filament will still be very nearly at room temperature, so negligible heat will flow
from the system (the filament) during this time.
Using eq.3a (or 3b or 3c – they’re all equivalent):
Gain in filament’s
internal energy
=
Net work done on
filament
 VIt
 12  0 V  30 A  0  0010s

 0  36 J
How much heat does the filament give out in 1.0 ms at its working temperature?
Since the filament’s temperature is steady it is no longer gaining internal energy. So:
Net heat flowing
out of filament
Net work done on
filament
=
 VIt
 12  0 V  4  0 A  0  0010s
 0  048 J
The First Law of Thermodynamics
We’ve already used it – many times – without giving it a name. Equations 3a, 3b and 3c
(where it is understood that internal energy is a function of the system’s state) are all ways of
writing it. In symbols we can use eq.3c and write:
U = Q  W
Gain in system’s
internal energy
=
Net heat flowing
into system
+
Net work done by
system
U is a change in a property of the system, its internal energy, U. A positive value of U
means an increase in U; a negative value means a decrease in U. As we said earlier, for a
given system U is a function of the system’s state. For example, for n moles of an ideal
monatomic gas, U is given by 3/2nRT (= 3/2pV).
Q is heat entering the system from (hotter) surroundings. A negative value of Q means heat
leaving the system (to cooler surroundings). We don’t have a ‘’ in front of Q, because heat
flow is not a change in heat. It’s energy in transit. It is not a function of the system’s state.
W is work done by the system. A negative value of W means a positive amount of work done
on the system. We don’t have a ‘’ in front of W for the same reason as for Q; namely, W is
not a change in work. It’s energy in transit. Work is not a function of the system’s state.
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As practice in using the symbols, state whether Q, U, W are positive, negative or zero in the
following cases. The first line of the table is filled in as an example. The empty space is for
your own examples.
Process
A gas expands very quickly, doing work
U
Q
W

0
+
An ideal gas expands into a vacuum very
quickly, doing no work. Note: this is an
irreversible process.[Answers a bit repetitive]
An ideal gas expands isothermally, doing work
Water is being heated in an electric kettle.
Take the system as being the whole kettle
(including heating elements) and contents.
Water is being heated in an electric kettle.
Take the system as being just the water.
Cyclic processes – this theme is taken up in the PH5 Energy option.
Suppose an ideal gas is taken through the cycle of changes ABCDA shown above on the left.
Irrespective of exactly what is going on in the individual stages (AB, BC, CD, DA), or the
exact shapes of the curves, we can draw some general conclusions…

When the gas has undergone one cycle and is back at A, its internal energy is the
same as it was originally, that is U = 0. This is because internal energy is a function
of state (see page 2), and the gas is back in the same state as originally.
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
Over the cycle as a whole, the gas has done a positive amount of work. This is
because the area under ABC represents the gas doing a positive amount of work. The
area under CDA represents work being done on the gas (see page 4), but this area is
smaller.

Thus applying the First Law, we see that for the cycle as a whole, there has been a net
flow of heat into the gas.
Some heat has been turned into work, and this will happen each time the cycle repeats.
Question: When this was first done on an industrial scale, what (non-ideal) gas was used?
Go through the argument again for the right hand diagram, where the cycle is anticlockwise.
What are the differences?
A Special Cycle
Suppose AB in the left hand diagram represents an isothermal expansion in which an ideal
gas pushes a piston (see page 7, Example 2). Since the gas does work, and U = 0, heat is
taken in. If the compression, CD, is also isothermal, heat is given out, but less than is taken in
along AB. Since pV is less along CD than along AB, the isothermal process CD is at a lower
temperature than the isothermal process AB.
[Note that the surroundings must be slightly hotter than the gas for heat to flow in along AB,
and slightly cooler than the gas for heat to flow out along CD. So different environments (e.g.
an oven and a cold water bath) must be provided for these stages.]
Let us now suppose that BC and DA occur so rapidly that no heat enters or leaves the gas in
these stages. In this case, applying the First Law to the complete cycle:
Net work out over a
cycle
=
heat taken in along AB (at
higher temperature)

heat given out along CD (at
lower temperature)
Refrigerators
If we run the same special cycle backwards (right hand diagram above), heat flows out along
AB, a smaller quantity of heat flows in along CD, and a net quantity of work has to be put in
over the cycle. So…
Net work in over a
cycle
=
heat given out along AB
(at higher temperature)

heat taken in along CD (at
lower temperature)
In an ordinary domestic fridge, you can feel the heat coming out from the pipework at the
back. A fluid capable of evaporating and condensing (not, then, an ideal gas) circulates in the
pipes. The heat is taken in from anything placed in the fridge, (and also comes in through the
walls and door of the fridge). The work is supplied electrically.
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The Second Law of Thermodynamics (just for interest)
For the ‘special cycle’ on page 11, not all the heat taken in during the isothermal process AB
could be used to produce work. Some heat had to be ‘excreted’ at a lower temperature
(during the process CD). In fact, no exceptions have ever been found to the rule that…
It is not possible, by any cyclic process using any substance, to take in a quantity of heat at
one temperature and turn it all into work.
This is one version of the Second Law of Thermodynamics.
If we could take in heat at one temperature and turn it all into work, life would be a lot
simpler. For example we might be able to take heat out of the polar ice caps (so slowing
down their melting) and get out electrical work. Of course using (that is converting) the
electrical work would give heat, but we could turn this into work. Dream on!
It is possible to show from the Second Law that a very simple rule relates the quantities of
heat taken in and given out to the temperatures in the special cycle of page 11.
heat taken in along AB TAB

heat given out along CD TCD
This applies for any substance, not just an ideal gas, undergoing the special cycle. The one
proviso is that each stage in the process has to be exactly reversible.
Example: Suppose AB is an isothermal at 400 K, and CD an isothermal at 300 K. If the
substance takes in 100 J of heat along AB, it must give out 75 J of ‘waste’ heat along CD.
The net output of work over the cycle will be 25 J. We say that the efficiency of the cycle is
25%. In practice the efficiency will be lower because of irreversible effects like friction
(page 5).
We can increase the cycle’s efficiency by making TAB larger or TCD smaller. There are clearly
technical limits to how large we can make TAB – components will melt! But what about
making TCD small? The problem here is that we must have some body of large heat capacity
outside the system, at a temperature marginally less than TCD, in order that heat will flow out
of the system during the process CD. That’s why, for large-scale practical purposes, TCD can’t
be much less than 300 K. A modern power station has an efficiency of less than 40%. This
means that for every joule of heat supplied from burning fuel (or from nuclear fission) more
than 0.6 J is given out as low temperature heat. It usually goes to waste. Remember this if
you use an electric heater at home!
Degradation of Energy (for interest)
Could we not use the heat excreted at TCD as the input to another ‘heat engine’? Yes, but note
that it would be even more difficult to get a decent yield of work out of that engine, since
there would be even less scope for a large difference between the TAB and TCD for that engine.
In this sense heat at low temperature is low-grade energy: it is less capable of being
converted to work. That’s why we don’t bother to try to re-use the waste heat from (say) the
backs of fridges and television sets. There is a tendency, in fact, for energy always to degrade
– think about effects of friction and electrical resistance.
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There is a deeper, microscopic, interpretation of the degradation of energy. When energy is
delivered to a system as heat it increases the disorderliness of energy distribution among the
particles of the system. This disorderliness (measured by the quantity entropy) makes energy
less available for conversion.
As an example, consider again Example 2, the isothermal expansion of an ideal gas. After the
expansion, the gas has the same amount of internal energy as originally, but the particles are
now moving about all over the place – well, in a larger volume than before! And we know
that it is harder to get the expanded gas to do work.
Work is energy transferred in its least degraded form. The force and the distance moved are
in a single direction; the energy transfer is orderly.
GCE Physics – Teacher Guidance
4 December 2007
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