Topic 10: Thermal physics
10.2 Processes
The first law of thermodynamics
10.2.1 Deduce an expression for the work involved
in a volume change of a gas at constant
pressure.
10.2.2 State the first law of thermodynamics.
10.2.3 Identify the first law of thermodynamics
as a statement of energy conservation.
10.2.4 Describe the isochoric (isovolumetric),
isobaric, isothermal, and adiabatic changes
of state of an ideal gas.
10.2.5 Draw and annotate thermodynamic processes
and cycles on P-V diagrams.
10.2.6 Calculate from a P-V diagram the work done
in a thermodynamic cycle.
10.2.7 Solve problems involving state changes of
a gas.
Topic 10: Thermal physics
10.2 Processes
The first law of thermodynamics
Deduce an expression for the work involved in a
volume change of a gas at constant pressure.
Suppose we take a beaker that is filled
with an ideal gas, and stopper it with
a gas-tight cork and a weight, as shown.
The weight F causes a pressure in
the gas having a value given by
P = F/A, where A is the area of the
∆V
cork in contact with the gas.
A
If we now heat up the gas it will
expand against the cork, pushing it
upward:
The dashed red box shows the change
in volume ∆V.
Note that ∆V = Ax. P is constant. Why?
x
Topic 10: Thermal physics
10.2 Processes
The first law of thermodynamics
Deduce an expression for the work involved in a
volume change of a gas at constant pressure.
From the previous slide:
F
P = F/A  F = PA and ∆V = Ax.
The work W done by the gas is just
the force F it exerts on the
weighted cork times the distance x
it moves the cork. Thus
∆V
A
W = Fx = PAx = P∆V.
W = P∆V
Work done by
expanding gas
(constant P)
FYI
If ∆V > 0 (gas expands) then W > 0.
If ∆V < 0 (gas contracts) then W < 0.
x
Topic 10: Thermal physics
10.2 Processes
The first law of thermodynamics
State the first law of thermodynamics.
Consider the previous “system” containing gas,
and heat, and providing mechanical work.
Just as we have done before, we will use Q to
represent a quantity of heat. If heat is added to
our system (as it was) then Q > 0. If heat is
removed from our system then Q < 0.
If the gas expands then W > 0. If the gas
contracts then W < 0.
Finally, we define the change in internal energy
of the system as ∆U.
The first law of thermodynamics relates Q, W and
∆U as follows:
Q = ∆U + W
first law of
thermodynamics
Topic 10: Thermal physics
10.2 Processes
The first law of thermodynamics
Identify the first law of thermodynamics as a
statement of energy conservation.
Q = ∆U + W
first law of
thermodynamics
What the first law shows is that when heat energy
Q is added to a system, some (or all of it) may
be used to change the internal energy ∆U of the
system, and some (or all of it) may be used to
provide mechanical work W.
FYI
Sometimes the first law is expressed ∆U = Q – W.
In this form we see that there are two ways to
change the internal energy of a system:
(1)By passing heat Q through the system boundary.
(2)By doing work W through the system boundary.
Topic 10: Thermal physics
10.2 Processes
The first law of thermodynamics
Identify the first law of thermodynamics as a
statement of energy conservation.
Q = ∆U + W
first law of
thermodynamics
In words, the first law says “Heat added to a
closed system can change its internal energy and
cause it to do work on its environment.”
This is a statement of the conservation of
energy. All of the energy added to the system is
accounted for.
FYI
Recall that ∆U consists of both ∆EP (manifested
in phase change) and ∆EK (temperature change) of
the substance. Lack of internal forces in an
ideal gas means that ∆EP = 0.
Topic 10: Thermal physics
10.2 Processes
The first law of thermodynamics
Describe the isochoric (isovolumetric), isobaric,
isothermal, and adiabatic changes of state of an
ideal gas.
For an ideal gas, the equation of state (PV =
nRT) tells us all of the important things about a
gas through the state variables P,V, n, and T.
When we add (or remove) heat Q from a closed
system containing an ideal gas, or have it do
work W on the external world, its state variables
may change.
We call changing the state of a system a process.
A process in which the state of a gas is changed
without changing its volume is called isochoric.
FYI
Don’t confuse “state” with “phase” in Topic 10.
Topic 10: Thermal physics
10.2 Processes
The first law of thermodynamics
Describe the isochoric (isovolumetric), isobaric,
isothermal, and adiabatic changes of state of an
ideal gas.
A process in which the state of a gas is
changed without changing its volume is called
isochoric.
We have already seen an isochoric process
when we studied the concept of absolute
zero:
p
During an isochoric
10 20
process the
30
temperature and the
0
pressure change.
NOT the volume.
-300
-200 -100
0
100
200
300
T (°C)
Topic 10: Thermal physics
10.2 Processes
The first law of thermodynamics
Describe the isochoric (isovolumetric), isobaric,
isothermal, and adiabatic changes of state of an
ideal gas.
A process in which the state of a gas is
changed without changing its volume is called
isochoric.
PRACTICE:
Show that the first law of thermodynamics reduces
to Q = ∆U for an isochoric process.
SOLUTION:
Recall that the work done by a gas is given by
W = P∆V.
Isochoric means ∆V = 0. Thus W = 0.
The first law, Q = ∆U + W, thus reduces to
Q = ∆U.
Topic 10: Thermal physics
10.2 Processes
The first law of thermodynamics
Describe the isochoric (isovolumetric), isobaric,
isothermal, and adiabatic changes of state of an
ideal gas.
A process in which the state of a gas is
changed without changing its volume is called
isochoric.
EXAMPLE:
Show that for an isolated ideal gas P  T during
an isochoric process.
SOLUTION: Use PV = nRT. Then
P = (nR/V)T
Isolated means n is constant (no gas is added to
or lost from the system).
Then n and V are constant (as is R). Thus
P = (nR/V)T = (CONST)T
P  T. (Isochoric)
Topic 10: Thermal physics
10.2 Processes
The first law of thermodynamics
Describe the isochoric (isovolumetric), isobaric,
isothermal, and adiabatic changes of state of an
ideal gas.
A process in which the state of a gas
is changed without changing its pressure
is called isobaric.
We have already seen such a process,
when we derived the formula for the
∆V
work done by a gas:
A
W = P∆V
Work done by
expanding gas
(constant P)
For an isobaric process the first law
can therefore be written
Q = ∆U + P∆V.
x
Topic 10: Thermal physics
10.2 Processes
The first law of thermodynamics
Describe the isochoric (isovolumetric), isobaric,
isothermal, and adiabatic changes of state of an
ideal gas.
A process in which the state of a gas is changed
without changing its pressure is called isobaric.
EXAMPLE:
Show that for an isolated ideal gas V  T during
an isobaric process.
SOLUTION: Use PV = nRT. Then
V = (nR/P)T
Isolated means n is constant (no gas is added to
or lost from the system).
Then n and P are constant (as is R). Thus
V = (nR/P)T = (CONST)T
V  T. (Isobaric)
Topic 10: Thermal physics
10.2 Processes
The first law of thermodynamics
Describe the isochoric (isovolumetric), isobaric,
isothermal, and adiabatic changes of state of an
ideal gas.
A process in which the state of a gas is changed
without changing its pressure is called isobaric.
PRACTICE:
Show that for an isolated ideal gas W = nR∆T
during an isobaric process.
SOLUTION:
From PV = nRT we can write (if n and P are
constant)
P∆V = nR∆T.
Recall W = P∆V.
Thus
W = nR∆T. (Isobaric)
Topic 10: Thermal physics
10.2 Processes
Why do we wait
before recording
our values?
The first law of thermodynamics
Describe the isochoric (isovolumetric), isobaric,
isothermal, and adiabatic changes of state of an
ideal gas.
If the state of a gas is changed without changing
its temperature the process is called isothermal.
10 20
EXAMPLE: A graduated syringe which is
30
0
filled with air is placed in an ice
bath and allowed to reach the
temperature of the water. Demonstrate
that P1V1 = P2V2. How do we know that the
SOLUTION:
process is isothermal?
Record initial states after a wait:
P1 = 15, V1 = 10, and T1 = 0ºC.
Record final states after a wait:
P2 = 30, V2 = 5, and T2 = 0ºC.
P1V1 = 15(10) = 150 = 30(5) = P2V2.
Topic 10: Thermal physics
10.2 Processes
The first law of thermodynamics
Describe the isochoric (isovolumetric), isobaric,
isothermal, and adiabatic changes of state of an
ideal gas.
If the state of a gas is changed without changing
its temperature the process is called isothermal.
PRACTICE:
Show that for an isolated ideal gas P1V1 = P2V2
during an isothermal process.
SOLUTION:
From PV = nRT we can write (if n and T are
constant)
P1V1 = nRT
P2V2 = nRT.
Thus
P1V1 = nRT = P2V2. (Isothermal)
Topic 10: Thermal physics
10.2 Processes
The first law of thermodynamics
Describe the isochoric (isovolumetric), isobaric,
isothermal, and adiabatic changes of state of an
ideal gas.
If the state of a gas is changed without adding
or losing heat the process is called adiabatic.
PRACTICE:
Show that for an isolated ideal gas W = -∆U
during an adiabatic process.
SOLUTION:
From Q = ∆U + W we can write (if n is constant
and Q is zero)
Q = ∆U + W
0 = ∆U + W
W = -∆U.
FYI
We accomplish this by insulating the container.
Topic 10: Thermal physics
10.2 Processes
The first law of thermodynamics
Draw and annotate thermodynamic processes and
cycles on P-V diagrams.
z
Perhaps you have enjoyed the pleasures
of analytic geometry and the graphing
of surfaces in 3D.
The three variables of a surface are
x, y, and z, and we can describe any
surface using the "state" variables x
x, y, and z:
The equation "of state" of a sphere is x2 + y2 +
z2 = r2, where r is the radius of the sphere:
FYI
We “built” the 3D sphere with layers of 2D
circles.
We have transformed a 3D surface into a stack of
2D surfaces.
y
P
Topic 10: Thermal physics
10.2 Processes
T
T3 4
T
The first law of thermodynamics
T1 2
Draw and annotate thermodynamic processes and
P
cycles on P-V diagrams.
The three state variables (if n is kept
constant) of a gas are analogous.
We can plot the three variables
P, V, and T on mutually perpendicular axes like this:
We have made layers in T.
Thus each layer has a single
temperature.
T
FYI
Each layer is an isotherm.
The 3D graph (below) can then be redrawn in its
simpler 2D form (above) without loss of
information.
V
V
Topic 10: Thermal physics
10.2 Processes
The first law of thermodynamics
Draw and annotate thermodynamic processes and
cycles on P-V diagrams.
A thermodynamic process involves moving from one
state to another state. This could involve
changing any or even all of the state variables
(P, V, or T).
EXAMPLE: In the P-V graph shown,
identify each process type as
A
P
ISOBARIC, ISOTHERMAL, OR
ISOVOLUMETRIC (isochoric).
SOLUTION:
B
AB is isothermal (constant T).
C
BC is isobaric (constant P).
CA is isochoric (constant V).
FYI
The purple line could be an adiabatic process.
V
Topic 10: Thermal physics
10.2 Processes
The first law of thermodynamics
Draw and annotate thermodynamic processes and
cycles on P-V diagrams.
A thermodynamic cycle is a set of processes which
ultimately return a gas to its original state.
EXAMPLE: A fixed quantity of a gas
P
undergoes a cycle by changing
between the following three states:
B
C
8
3
State A: (P = 2 Pa, V = 10 m )
State B: (P = 8 Pa, V = 10 m3)
State C: (P = 8 Pa, V = 25 m3)
2
A
Each process is a straight line,
and the cycle goes like this:
10
25
ABCA. Sketch the complete
cycle on a P-V diagram.
SOLUTION:
Scale your axes and plot your points…
V
Topic 10: Thermal physics
10.2 Processes
The first law of thermodynamics
Calculate from a P-V diagram the work done in a
thermodynamic cycle.
A thermodynamic cycle is a set of processes which
ultimately return a gas to its original state.
EXAMPLE: A fixed quantity of a gas
P
undergoes the cycle shown here
(from the last example):
B
C
8
(a) Find the work done during
the process AB.
(b) Find the work done during
2
A
the process BC.
SOLUTION: Use W = P∆V.
10
25
(a) From A to B: ∆V = 0. Thus the W = 0.
(b) From B to C: ∆V = 25 – 10 = 15; P = 8.
Thus W = P∆V = 8(15) = 120 J.
V
Topic 10: Thermal physics
10.2 Processes
The first law of thermodynamics
Calculate from a P-V diagram the work done in a
thermodynamic cycle.
A thermodynamic cycle is a set of processes which
ultimately return a gas to its original state.
EXAMPLE: A fixed quantity of a gas
P
undergoes the cycle shown here
(from the last example):
B
C
8
(c) Find the work done during
the process CA.
SOLUTION: (c) Use W = Area under
2
A
the P-V diagram.
Observe that ∆V is negative when
10
25
going from C (V = 25) to A (V = 10).
Observe that P is NOT constant so W  P∆V.
W = Area = -[ (2)(15) + (1/2)(6)(15) ] = -75 J.
V
Topic 10: Thermal physics
10.2 Processes
The first law of thermodynamics
Calculate from a P-V diagram the work done in a
thermodynamic cycle.
A thermodynamic cycle is a set of processes which
ultimately return a gas to its original state.
EXAMPLE: A fixed quantity of a gas
P
undergoes the cycle shown here
(from the last example):
B
C
8
(d) Find the work done during
the cycle ABCA.
SOLUTION: (d) Just total up the
2
A
work done in each process.
WAB = 0 J.
10
25
FYI
WBC = +120 J.
Work is done on the
WCA = -75 J.
external environment
Wcycle = 0 + 120 – 75 = +45 J.
during each cycle.
V
Topic 10: Thermal physics
10.2 Processes
The first law of thermodynamics
Calculate from a P-V diagram the work done in a
thermodynamic cycle.
A thermodynamic cycle is a set of processes which
ultimately return a gas to its original state.
PRACTICE:
P
Find the total work done if the
previous cycle is reversed.
B
C
8
SOLUTION:
We want ACBA.
WAC = Area
A
= +[ (2)(15) + (1/2)(6)(15) ] 2
= +75 J.
10
25
FYI
WCB = P∆V = 8(10–25) = 120 J.
Reversing the
WBA = 0 J (since ∆V = 0).
cycle reverses the
Wcycle = 75 - 120 + 0 = 45 J.
sign of the work.
V
Topic 10: Thermal physics
10.2 Processes
The first law of thermodynamics
Calculate from a P-V diagram the work done in a
thermodynamic cycle.
A heat engine is any device which converts heat
energy into mechanical work.
To make a heat engine we need a source of heat
energy (coal, gas, etc.) and a
hot reservoir
at TH
working fluid which undergoes
thermodynamic change of state
QH
causing work do be done on the
external environment.
engine W
Common working fluids are water
(made into steam) and petrol-air
mixtures (ignited).
QL
The energy flow diagram of a
cold reservoir
heat engine is shown here:
at TL
Topic 10: Thermal physics
10.2 Processes
The first law of thermodynamics
Calculate from a P-V diagram the work done in a
thermodynamic cycle.
EXAMPLE: An internal combustion engine is an
example of a heat engine that does work on the
environment. A four-stroke engine
is animated here.
http://www.animatedengines.com/otto.html
http://chemcollective.org/activities/simulations/
engine
Topic 10: Thermal physics
10.2 Processes
The first law of thermodynamics
Calculate from a P-V diagram the work done in a
thermodynamic cycle.
EXAMPLE: From the animation of the second link on
the previous slide, show
the direction of each
process with an arrow.
SOLUTION:
View the animation. Or
reason that POSITIVE work
must be done by the cycle.
FYI
Remember “work done BY
system” is positive, and
“work done ON system” is
negative.
Topic 10: Thermal physics
10.2 Processes
The first law of thermodynamics
Calculate from a P-V diagram the work done in a
thermodynamic cycle.
EXAMPLE: From the same animation, label the
COMPRESSION, POWER,
EXHAUST, AND INTAKE
COMPRESSION STROKE
STROKES. Which stroke has
POWER STROKE
a positive area under it?
Which has a negative area?
EXHAUST STROKE
SOLUTION:
INTAKE STROKE
A STROKE involves a
change in volume.
Since ∆V > 0 for the
power stroke it has a
positive area.
The compression stroke
has a negative area.
Topic 10: Thermal physics
10.2 Processes
The first law of thermodynamics
Calculate from a P-V diagram the work done in a
thermodynamic cycle.
EXAMPLE: From the same animation, show that the
work done in a cycle is
positive.
COMPRESSION STROKE
SOLUTION:
POWER STROKE
The work done during the
power stroke is positive
EXHAUST STROKE
(sketched in red).
INTAKE STROKE
The work done during the
compression stroke is
negative (sketched in
purple).
There is more positive
than negative. Thus W > 0.
W = area between graphs.
Topic 10: Thermal physics
10.2 Processes
The first law of thermodynamics
Calculate from a P-V diagram the work done in a
thermodynamic cycle.
A heat pump is any device which can move heat
from a low temperature reservoir to a high
temperature reservoir.
A refrigerator is an example
hot reservoir
at TH
of a heat pump.
A common working fluid is Freon.
QH
The Freon vaporizes inside the
fridge removing QL = LV.
pump
W
The Freon condenses outside the
fridge releasing QH = Lf.
QL
FYI
The compressor does the work ON
the system.
cold reservoir
at TL
Topic 10: Thermal physics
10.2 Processes
The first law of thermodynamics
Solve problems involving state changes of a gas.
”Thermally insulated” means that heat can not
enter or leave the system.
Thus Q = 0.
This is the definition of an adiabatic process.
Topic 10: Thermal physics
10.2 Processes
The first law of thermodynamics
Solve problems involving state changes of a gas.
W = P∆V
= 1105(-3)
= -3106 J.
Or you can
just find
the area
under the
process.
Topic 10: Thermal physics
10.2 Processes
The first law of thermodynamics
Solve problems involving state changes of a gas.
Adiabatic (and
insulated) both
mean that Q = 0.
∆V < 0 means
W < 0 (-).
The first law says Q = ∆U + W. Then
0 = ∆U + (-)  ∆U > 0.
Topic 10: Thermal physics
10.2 Processes
The first law of thermodynamics
Solve problems involving state changes of a gas.
At A:
PV = 25
= 10.
At B:
PV = 52
= 10.
At C:
PV = 6.82
= 13.6.
Isothermal means T is constant.
PV = nRT then becomes PV = CONST.
Thus process AB is the isothermal one.
Topic 10: Thermal physics
10.2 Processes
The first law of thermodynamics
Solve problems involving state changes of a gas.
The difference in work is the area between
the graphs.
Topic 10: Thermal physics
10.2 Processes
The first law of thermodynamics
Solve problems involving state changes of a gas.
This can only
be an estimate.
A tally of
small
rectangles
should
suffice.
There are
about 170 of
them.
18
16
14
12
12
10
9
9
8
8
77
6
5
54
4
Each rectangle has a work value given by
W = PV = (0.1105)(0.110-3) = 1 J.
Thus the difference in work is 170 J.
19
Topic 10: Thermal physics
10.2 Processes
The first law of thermodynamics
Solve problems involving state changes of a gas.
AC is the
adiabatic
compression.
The first
law says
Q = ∆U + W.
Adiabatic
means Q = 0.
∆U = -W.
Since W < 0 during the compression of a gas,
∆U = -W > 0. Thus the temperature increases.
Topic 10: Thermal physics
10.2 Processes
The first law of thermodynamics
Solve problems involving state changes of a gas.
The gas only does external work when it expands.
QR and SP are isochoric (isovolumetric) so W = 0.
PQ is a compression so W < 0.
RS is an expansion so W > 0.
Topic 10: Thermal physics
10.2 Processes
The first law of thermodynamics
Solve problems involving state changes of a gas.
NOT adiabatic since Q  0 (Q = 8103 J).
From PV = nRT we see that
PiVi = 1.2105(0.05) = 6000 J = nRTi.
PfVf = 1.2105(0.10) = 12000 J = nRTf.
Thus T changes and the process is NOT isothermal.
So the process is NEITHER.
Topic 10: Thermal physics
10.2 Processes
The first law of thermodynamics
Solve problems involving state changes of a gas.
From W = P∆V we see that
W = 1.2105(0.10 - 0.05) = 6.0103 J.
Topic 10: Thermal physics
10.2 Processes
The first law of thermodynamics
Solve problems involving state changes of a gas.
From the first law: Q = ∆U + W, or
8.0103 = ∆U + 6.0103
∆U = 2.0103 J.
Topic 10: Thermal physics
10.2 Processes
The first law of thermodynamics
Solve problems involving state changes of a gas.
Work done ON a gas is a compression.
Only process PQ has a decreasing volume.
Topic 10: Thermal physics
10.2 Processes
The first law of thermodynamics
Solve problems involving state changes of a gas.
For an ideal gas, ISOTHERMAL means ∆U = 0.
From the first law of thermodynamics we have
Q = ∆U + W
Q = 0 + W
Q = W = 2500 J.
Topic 10: Thermal physics
10.2 Processes
The first law of thermodynamics
Solve problems involving state changes of a gas.
From the first law we have
q = ∆U + w.
Keep T constant with ICE BATH.
Keep V constant with FIXED CONTAINER.
Keep q zero with INSULATION.
Topic 10: Thermal physics
10.2 Processes
The first law of thermodynamics
Solve problems involving state changes of a gas.
A heat pump transfers heat from
a cold reservoir to a hot one.
Since this is not the natural
direction of flow, work must be
done ON the system to make this
happen.
QH
W
QL
It is clear from the
Topic 10: Thermal physics diagram that both AB
10.2 Processes
and CD are isobaric
(P = CONST).
The first law of thermodynamics
Solve problems involving state changes of a gas.
From
phase
From
phase
CD evaporation occurs. During
changes ∆T = CONST (isothermal).
AB compression occurs. During
changes ∆T = CONST (isothermal).
Topic 10: Thermal physics
10.2 Processes
The first law of thermodynamics
Solve problems involving state changes of a gas.
Wherever W > 0 heat is removed from the
cold reservoir. Thus BC and CD.
Wherever W < 0 heat is forced into the hot
reservoir. Thus DA and AB.
Topic 10: Thermal physics
10.2 Processes
The first law of thermodynamics
Solve problems involving state changes of a gas.
From the diagram Ein = W + Qc.
Eout = QH.
From Ein = Eout we have W + QC = QH.
Topic 10: Thermal physics
10.2 Processes
Wherever W > 0 heat
is removed from the
cold reservoir.
Thus where ∆V > 0.
The first law of thermodynamics
Solve problems involving state changes of a gas.
A
A refrigerator is a heat
pump. Work must be done
ON it. Thus W < 0.
Area inside curves must
be NEGATIVE.
B
A B
P = CONST
Topic 10: Thermal physics
10.2 Processes
The first law of thermodynamics
Solve problems involving state changes of a gas.
For each square:
W = P∆V
W = (1105)(0.110-3)
W = 10 J.
There are about 47 squares.
Thus W = 47(10 J) = 470 J (done ON substance).