Wk 7 lecture

1. Know the Understand the Zeroth Law of
thermodynamics.
2. Understand the First Law of
thermodynamics.
3. Apply the laws to solving problems.
4. Explain thermodynamics to explain
temperature and state changes in gasses
2
Objectives

The laws of thermodynamics help us to explain the
changes that happen to gasses. They were developed
during the industrial revolution by Ludwig Boltzmann
and his peers.
3
Zeroth Law

sheffield.ac.uk/international-college
• The zeroth law of thermodynamics provides a simple
definition of thermodynamic equilibrium.
• It is known that some property of an object (the
pressure of a gas, the length of a metal rod, the
electrical conductivity of a wire) can change when the
object is heated or cooled.
• If two of these objects (e.g. two metal rods) are
brought into physical contact there is initially a change
in the property of both objects (in this case their length)
but eventually, the change in the property stops.
4
Zeroth Law

• At this point the objects are said to be in thermal
(thermodynamic) equilibrium.
• Thermodynamic equilibrium leads to the large scale
definition of temperature i.e. when two objects are in
thermal equilibrium we say they have the same
temperature.
• During the process of reaching thermal equilibrium,
heat, which is a form of energy, is transferred between
the objects.
5
Zeroth Law

Objects are in thermal contact if heat can flow between
them.
6
Zeroth Law
Definition of heat:
Heat is the energy transferred between objects
because of a temperature difference.

7
Zeroth Law
The zeroth law of thermodynamics:
Two systems in thermal equilibrium with a third
system are in thermal equilibrium with each
other.
In other words: All objects at the same
temperature are in thermal equilibrium with each
other if they are in contact.

8
Zeroth Law

9
Internal Energy
• Kinetic theory assumes that the particles
in a body in any state are in constant
motion and so possess kinetic energy.
• The particles also possess potential
energy due to the separation between
atoms/molecules.
• The internal energy U, of a body (or
system – group of bodies) is the sum of
the kinetic and potential energies of the
particles within it.

10
Internal Energy
The internal energy U of a system remains
constant unless energy is transferred to or from it.
U = energy input – energy output
positive U = energy gained
negative U = energy lost

11
Internal Energy
Whether the body be solid, liquid or a gas the
energy supplied increases the kinetic energy
and potential energy of the molecules and
the temperature rises.
However, the energy supplied during boiling
or melting simply frees or separates the
molecules, increasing their potential energy.
Therefore, during boiling and melting, there is
no change in the kinetic energy of the
molecules so the temperature remains
constant.

12
Internal Energy
Solid → Liquid → Gas
PE ≪ 0 PE < 0 PE = 0
Increasing KE
Increasing PE

13
Internal Energy
𝑇
Tboil
𝑈Solid Melting Liquid Evaporating Gas
Increasing
KE and
PE
Increasing
PE
Increasing
KE and PE
Increasing
PE
Increasing
KE
PE = 0 J i.e
maximum
Tmelt

14
Internal Energy
The internal energy i.e. temperature
(assuming no change of phase takes
place) of a system can be changed in
two ways:
1. Heating
2. Doing ‘work’

15
First Law of Thermodynamics
Heating is energy transfer due to temperature
difference.
Energy will be transferred from A to B until the average
kinetic energy of the particles in B is the same as that
of those in A i.e. until their temperatures are the same.

16
Work is energy transfer by the action of a
force.
• In both of the above, the work done on
the system causes an increase in
internal energy and the temperature
rises.
• For example hitting a metal block with a
hammer or passing a current through a
metal.

17
As already seen, the internal energy U of a
system can only change by working or heating
or both. This is the first law of
thermodynamics.
• Stated mathematically, this law is:
U = W + Q

18
It is important that we use the correct sign
convention when using the first law.
ΔW = Win − Wout so that ΔW > 0 when the
work done on the system by an external
force exceeds the work done by the
system on its surroundings.
SYSTEM
∆𝑈 = ∆𝑄+ ∆𝑊
𝑊in
𝑄in
𝑊out
𝑄out

19
Similarly, Δ𝑄 = 𝑄in − 𝑄out so that Δ𝑄 > 0
when the heat flowing into the system
exceeds the heat loss by the system to its
surroundings.
SYSTEM
∆𝑈 = ∆𝑄+ ∆𝑊
𝑊in
𝑄in
𝑊out
𝑄out

20
Isothermal change is a change that occurs
with no change in temperature.
Thus U = 0
An example of this is slowly stretching an
elastic band. Work is done on the system
by the stretching force but thermal energy
is transferred from the system to the
surroundings. In this case, W = -Q.

21
Adiabatic change is one in which work is
done quickly so there is no time for energy
transfer by heating.
Thus Q = 0
An example of this is stretching a rubber
band very quickly. Work is done on the
system by the stretching force but there is
not enough time for thermal energy to be
transferred. So Q = 0 and W = U. Thus
the temperature of the band increases.

22
Work and Gasses
Work is done on a gas when it is compressed.
The energy received by the gas as a result of
the work done by the external force F is
simply ΔW = −FΔL

23
Work and Gasses
Recall that the gas pressure P is given by
F = PA
So the work done by the external force is
ΔW = −FΔL = −PAΔL
but of course ΔV = AΔL hence
ΔW = −PΔV

24
Work and Gasses
Note that this equation is only true if
pressure P is constant during the change
i.e. process is isobaric.
This work increases the internal energy of
the gas and its temperature rises.
An example of this is a bicycle pump getting
hot as it is used to pump up a tyre.

25
Work and Gasses
The volume of the gas
increases from V1 to V2
The work done ΔW is
negative since it is done by
the gas
ΔW = – P ΔV and is equal
(but opposite in sign) to the
area under the PV graph.
• If the arrow on the isobar is in the opposite direction then
work done is positive since it is done on the gas.

26
Work and Gasses
Isothermal change
The gas is being
compressed.
The shaded area is the
work done ON the gas.
If the arrow on the
isotherm is in the
opposite direction then
the shaded area is the
work done BY the gas.

Work and Gasses
27
Adiabatic change (no
heat transfer) ΔQ = 0
If the gas is decreasing
in volume the shaded
area is the positive work
done on the gas.
The temperature of the
gas increases.
P
T1
T2
V1V2

Second Law of Thermodynamics
28
The Second Law of Thermodynamics is simply the stating
of a fact that should be obvious.
No system is 100% efficient
Every time energy is transferred some energy is
dissipated from the system.
e.g. In a steam train some heat energy from the fire
warms the surroundings rather than the water it is
meant to boil.

29
Summary
The Laws of Thermodynamics
Zeroth
Two systems in thermal equilibrium
with a third system are in thermal
equilibrium with each other.
First
U = W + Q
Second No system is 100% Efficient
Work done by gasses
ΔW = −FΔL ΔW = −PΔV

30
Preparation for the Tutorial
• You need to do the questions for the tutorial. They are
called “wk7 tutorial (Prep Questions)”
• All the required questions can be found on MOLE.

Wk 7 lecture

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