Download The first law of thermodynamics

Physics 12
Giancoli Chapter 15
Objectives
The first law of thermodynamics
 Deduce an expression for the work involved in a
volume change of a gas at constant pressure.
 State the first law of thermodynamics.
 Identify the first law of thermodynamics as a
statement of the principle of energy conservation.
 Describe the isochoric (isovolumetric), isobaric,
isothermal and adiabatic changes of state of an ideal
gas.
 Draw and annotate thermodynamic processes and
cycles on P-V diagrams.
 Calculate from a P-V diagram the work done in a
thermodynamic cycle.
 Solve problems involving state changes of a gas.
Objectives
Second law of thermodynamics and entropy
 State that the second law of thermodynamics
implies that thermal energy cannot
spontaneously transfer from a region of low
temperature to a region of high temperature.
 State that entropy is a system property that
expresses the degree of disorder in the system
 State the second law of thermodynamics in
terms of entropy changes.
 Discuss examples of natural processes in terms
of entropy changes.
Thermodynamics
The study of processes in which energy is
transferred as heat and as work.
 Important to define the system that we are
dealing with;
 Everything other than the system shall be
referred to as the “environment” or
“surroundings”.

First law of thermodynamics
Otherwise known as the law of
conservation of energy.
 Relates work/heat to the change of the
internal energy of a system.

What are some forms of energy that we have
considered in the past? How do they
relate to the conservation of energy?
First law of thermodynamics

The internal energy is the sum total of all
the energy of the molecules in a system.
What would happen to the internal energy if
heat was added to the system?
If heat was taken away from the system?
First law of thermodynamics
The change in internal energy ΔU of a
closed system will be equal to the energy
Q added to the system minus the work W
done by the system on the surroundings.
ΔU = Q + W
First law of thermodynamics

Internal energy of an ideal gas
The internal energy U is the sum of the
translational kinetic energies of the
atoms.
First law of thermodynamics
U = N(KEavg)
where N is the number of molecules of gas or
N = nNA
where n is the number of moles of gas and
NA is Avogadro’s number
The average KE of an ideal gas is given by the
equation
KEavg = (3/2)kBT
where kB is Boltzmann’s constant
and T is the temperature of the gas.
First law of thermodynamics
Therefore, the internal energy of an ideal
gas is given by the expression
U = (3/2)NkBT
where
N is the number of molecules of gas
kB is Boltzmann’s constant, and
T is the temperature of the gas.
First law of thermodynamics
ΔU = Q + W
Q is the heat added to or taken away from
the system.
W is the work done by or on the system.
What would negative signs for either Q or W
imply?
First law of thermodynamics
ΔU = Q + W
IN OTHER WORDS…
The change in energy in a system is the
total of the heat and work that goes in and
out of the system.
Signs will make a difference.
First law of thermodynamics
2500 J of heat is added to a system, and
1800 J of work is done on the system.
What is the change in internal energy of
the system?
First law of thermodynamics
4300 J
What would the change in internal energy
be if 2500 J of heat was added to the
system but 1800 J of work is done by the
system?
First law of thermodynamics
700 J
First law of thermodynamics

The first law of thermodynamics is one of
the great laws of physics that can be
proven experimentally and to which no
exceptions have been seen.

For simplicity’s sake, we shall discuss
these laws in the context of gases.
First law of thermodynamics

Consider a fixed mass of an ideal gas
enclosed in a container fixed with a
movable piston: (p 410 Fig 15-1).

If we compress the gas (lowering the
piston), what happens to its pressure and
volume?
First law of thermodynamics
Decreasing the volume will increase the
pressure.
 There is a constant relationship between
pressure and volume when T is constant:
constant
PV = nRT
 A thermodynamic process in which T is
constant is called an isothermal process.
 p 410 Fig 15-2

First law of thermodynamics

What would you expect the P-V graph of
an isothermal process to look like?

How would the graph change if the
process were to occur at a lower
temperature?
First law of thermodynamics

p 410 Fig 15-2

When the gas is compressed, was work
done on the system or by the system?
First law of thermodynamics

The act of compressing the gas is work done
on the system +W.

However, in order for T to remain
CONSTANT the decrease in V requires an
increase in pressure P which is carried out by
the gas.

So what is the change in internal energy of
the system?
First law of thermodynamics
-W = Q
since ΔU = Q + W,
ΔU = 0
This means that there is no change in the
internal energy of the system.
Does this necessarily mean that there was
no work done on or by the system?
First law of thermodynamics
This is an important concept in physics
(wherein the net work being zero does
not necessarily mean that no work was
done ).
 In what other units have we explored this
concept?

First law of thermodynamics
Another thermodynamic process is one in
which no heat is allowed to flow into or
out of the system.
 This is called an adiabatic process.
 In terms of the eq’n for the first law of
thermodynamics, which term is 0?
ΔU = Q + W

First law of thermodynamics

If Q = 0, what is ΔU?
First law of thermodynamics
If Q = 0,
and ΔU = Q + W
then ΔU = W
But what does this mean?
First law of thermodynamics
Adiabatic expansion
Quantity
Q
ΔU
V
P
Value / Effect
0
=Q+W
=0+W
=W
increases
?
T
?
First law of thermodynamics
Conceptual example
When you suddenly expand a rubber band,
what happens when you touch it with
your lips? Explain this in terms of
thermodynamics.
First law of thermodynamics
Conceptual example
The rubber band will feel warmer than
before it was stretched.
When the rubber band is stretched, you do
work on the system (- W). Since it was
done suddenly, heat was essentially not
allowed to leave the system (Q = 0). Thus,
when you touch it with your lips you will
feel an increase in temperature.
Try it at home.
First law of thermodynamics
Comparison of adiabatic and isothermal
compression
Quantity
Q
ΔU
V
P
T
Isothermal
Adiabatic
decrease
decrease
First law of thermodynamics
Comparison of adiabatic and isothermal compression
Quantity
Q
ΔU
V
P
T
Isothermal
W
0
decrease
Adiabatic
0
-W
decrease
increase
constant
increase
increase
In a fuel engine, the fuel and air is compressed so
rapidly (adiabatically) in the fuel piston that the T
increase causes the mixture to ignite.
First law of thermodynamics

Other thermodynamic processes that may
occur are:
 isobaric (constant pressure)
 isochoric or isovolumetric (constant volume)

How would you expect these processes to
appear on a P-V graph?
First law of thermodynamics
So far, we have discussed situations
qualitatively (which is often insufficient in
physics).
 How do we calculate the work done on a
system?
 How have we calculated work before?

First law of thermodynamics
Recall that
W = Fd
In terms of gases,
P=F/A
F = PA
therefore
W = PAd
if Ad = ΔV
then W = P ΔV
First law of thermodynamics
Dimensional analysis:
1 J = 1 Nm
A calculation of PΔV (using SI units) would yield
(Nm-2)(m3)
= Nm
=J
First law of thermodynamics
Comparison of thermodynamic processes
First law of thermodynamics
Comparison of thermodynamic processes
Constant
Isothermal
Isobaric
Isovolumetric
Adiabatic
Important
characteristic
First law of thermodynamics
Comparison of thermodynamic processes
Constant
Isothermal
T
Isobaric
P
Isovolumetric
V
Adiabatic
Q=0
Important
characteristic
Q = -W
ΔU = Q + W
ΔU= Q + PΔV
ΔV = 0
Q = ΔU
Q=0
ΔU = W
First law of thermodynamics
Comparison of adiabatic and isothermal
expansion
Look at the Figure 15-3 on p 411.
In which process was more work done by the
gas?
First law of thermodynamics
Comparison of adiabatic and isothermal
expansion
More work was done by the gas in the
isothermal process.
The average pressure is higher during the
isothermal process.
Work can also be represented graphically by
the area under a P-V curve.
First law of thermodynamics
An ideal gas is slowly compressed at a constant
pressure of 2.0 atm from 10.0 L to 2.0 L (B to
C). In this process, some heat flows out of the
gas and the temperature drops. Heat is then
added to the gas (C to A), holding the volume
constant, and the pressure and temperature
are allowed to rise until the temperature
reaches its original value.
Sketch a P-V graph of the processes involved.
Calculate the total work done by the gas in the
process CBA .
Calculate the total heat flow into the gas.
First law of thermodynamics
Graph:
 BC: isobaric compression (V = 10.0 L to V = 2.0 L,
P = 2.0 atm )
 CA: isovolumetric increase in pressure
Work is done only in the compression CB. In CA,
ΔV = 0 so W = 0. During CB, W = -1.6 x 103 J.
Because the temperature at the beginning and
end of the process is the same, ΔT = 0 so ΔU =
0. Therefore the total heat flow into the gas is
-1.6 x 103 J.
First law of thermodynamics
In an engine, 0.25 moles of an ideal
monatomic gas in the cylinder expands
rapidly and adiabatically against the
piston. In the process, the temperature of
the gas drops from 1150 K to 400 K. How
much work does the gas do?
First law of thermodynamics
2300 J
First law of thermodynamics

Recall the relationship between the heat and
the change in temperature of a substance:
Q = mcΔT
In terms of gases, we use the expression
Q = nCΔT
where n is the number of moles of gas
C is the molar heat capacity of the gas (this may
be expressed as Cv or Cp at constant volume
or pressure)
and ΔT is the change in temperature of the gas
Activity

Worksheet – First law of thermodynamics
Second law of thermodynamics

If a hot object is placed in contact with a
cold object, in which direction will heat
flow?
Second law of thermodynamics
Experience tells us that the heat will flow
from the hot object to the cold object.
 Does heat flow from a colder object to a
hotter object violate the first law of
thermodynamics?

Second law of thermodynamics

Heat flow from cold to hot does NOT
violate the first law of thermodynamics
but it does violate the second law of
thermodynamics:
Heat will flow spontaneously from a hot
object to a cold object; heat will not flow
spontaneously from a cold object to a hot
object.
Second law of thermodynamics

This concept is especially relevant in the
study of heat engines.
Activity

Independent study
Heat engines and the Carnot cycle
Handout
Sections 15-5 and 15-6 of Giancoli