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Three cylinders
Three identical cylinders are sealed with identical pistons that
are free to slide up and down the cylinder without friction.
Each cylinder contains ideal gas, and the gas occupies the
same volume in each case, but the temperatures differ. In
each cylinder the piston is above the gas, and the top of each
piston is exposed the atmosphere. In cylinders 1, 2, and 3 the
temperatures are 0°C, 50°C, and 100°C, respectively. How
do the pressures compare?
1. 1>2>3
2. 3>2>1
3. all equal
4. not enough information to say
Free-body diagram of a piston
Sketch the free-body diagram of one of the pistons.
Free-body diagram of a piston
Sketch the free-body diagram of one of the pistons.
The internal pressure, in this case, is
determined by the free-body diagram.
All three pistons have the same
pressure, so the number of moles of
gas must decrease from 1 to 2 to 3.
Introducing the P-V diagram
P-V (pressure versus
volume) diagrams can be
very useful.
What are the units resulting
from multiplying pressure in
kPa by volume in liters?
Rank the four states shown on the diagram based on
their absolute temperature, from greatest to least.
Introducing the P-V diagram
P-V (pressure versus
volume) diagrams can be
very useful.
What are the units resulting
from multiplying pressure in
kPa by volume in liters?
1000
N
3

0.001
m
 1 Nm  1 J
2
m
Rank the four states shown on the diagram based on
their absolute temperature, from greatest to least.
Temperature is proportional to PV, so rank by PV:
2 > 1=3 > 4.
Isotherms
Isotherms are lines of
constant temperature.
On a P-V diagram, isotherms
satisfy the equation:
PV = constant
Thermodynamics
Thermodynamics is the study of systems involving energy in
the form of heat and work.
Consider a cylinder of ideal gas, at room temperature.
When the cylinder is placed in a
container of hot water, heat is
transferred into the cylinder.
Where does that energy go?
The piston is free to move up
or down without friction.
Thermodynamics
The First Law of Thermodynamics
Some of the added energy goes into raising the temperature
of the gas (we call this raising the internal energy). The rest of
it does work, raising the piston. Conserving energy:
Q  Eint  W (the first law of thermodynamics)
Q is heat added to a system (or removed if it is negative)
Eint is the internal energy of the system (the energy
associated with the motion of the atoms and/or molecules),
so Eint is the change in the internal energy, which is
proportional to the change in temperature.
W is the work done by the system.
The First Law is often written as Eint  Q  W
Work
We defined work previously as:
W  F x (true if the force is constant)
F = PA, so:
W  PAx  P V
At constant pressure the work done by the system is the
pressure multiplied by the change in volume.
If there is no change in volume, no work is done.
In general, the work done by the system is the area under the
P-V graph. This is why P-V diagrams are so useful.
Work: the area under the curve
The net work done by the gas is positive in this case,
because the change in volume is positive, and equal to the
area under the curve.
Solving thermodynamics problems
A typical thermodynamics problem involves some process
that moves an ideal gas system from one state to another.
1. Draw a P-V diagram to get some idea what the work is.
2. Apply the First Law of Thermodynamics (this is a
statement of conservation of energy).
3. Apply the Ideal Gas Law.
• the internal energy is determined by the temperature
• the change in internal energy is determined by the change
in temperature
• the work done depends on how the system moves from
one state to another (the change in internal energy does not)
Constant volume (isochoric) process
No work is done by the gas: W = 0. The P-V diagram is a
vertical line, going up if heat is added, and going down if heat
is removed.
Applying the first law:
Q  Eint
For a monatomic ideal gas:
Eint
3
 nRT
2
Q  Eint
3
 nR T
2
Constant pressure (isobaric) process
In this case the region on the P-V diagram is rectangular, so
its area is easy to find. W  P V
For a monatomic ideal gas:
Q  Eint  W
3
nR T  P V
2
3
 nR T  nR T
2
5
 nR T
2

Constant temperature (isothermal) process
No change in internal energy: Eint  0
The P-V diagram follows the isotherm.
Applying the first law, and
using a little calculus:
 Vf 
Q  W  nRT ln  
 Vi 
Worksheet
You have some monatomic ideal gas in a cylinder. The
cylinder is sealed at the top by a piston that can move up or
down, or can be fixed in place to keep the volume constant.
Blocks can be added to, or removed from, the top of the
cylinder to adjust the pressure, as necessary.
Starting with the same initial conditions each time, you do
three experiments. Each experiment involves adding the
same amount of heat, Q.
A – Add the heat at constant pressure.
B – Add the heat at constant temperature.
C – Add the heat at constant volume.
Worksheet
A – Add heat Q to the system at constant pressure.
B – Add heat Q to the system at constant temperature.
C – Add heat Q to the system at constant volume.
Sketch these processes on
the P-V diagram. The circle
with the i beside it
represents the initial state of
the system. One of the
processes is drawn already.
Identify which one, and draw
the other two.
Rank by final temperature
Rank the processes based on the final temperature.
1.
2.
3.
4.
5.
6.
7.
A>B>C
A>C>B
B>A>C
B>C>A
C>A>B
C>B>A
Equal for all three
Rank by final temperature
In process B, the temperature is constant.
In process A, some of the heat added goes to increasing the
temperature.
In process C, all the heat added goes to increasing the
temperature.
C>A>B
Rank by work
Rank the processes based on the work done by the gas.
1.
2.
3.
4.
5.
6.
7.
A>B>C
A>C>B
B>A>C
B>C>A
C>A>B
C>B>A
Equal for all three
Rank by work
In process C, no work is done.
In process A, some of the heat added goes to doing work.
In process B, all the heat added goes to doing work.
B>A>C
Rank by final pressure
Rank the processes based on the final pressure.
1.
2.
3.
4.
5.
6.
7.
A>B>C
A>C>B
B>A>C
B>C>A
C>A>B
C>B>A
Equal for all three
Rank by pressure
In process A, the pressure stays constant.
In process B, the pressure decreases.
In process C, the pressure increases.
C>A>B
Worksheet
A thermodynamic system undergoes a three-step process. An
adiabatic expansion takes it from state 1 to state 2; heat is
added at constant pressure to move the system to state 3;
and an isothermal compression returns the system to state 1.
The system consists of a diatomic ideal gas with CV = 5R/2.
The number of moles is chosen so nR = 100 J/K.
The following information is known about states 2 and 3.
Pressure: P2 = P3 = 100 kPa
Volume: V3 = 0.5 m3
What is the temperature
of the system in state 3?
Apply the ideal gas law
P3V3 (100  103 Pa)  0.5 m3
T3 

 500 K
nR
100 J/K
The system does 20000 J of work in the constant pressure
process that takes it from state 2 to state 3. What is the
volume and temperature of the system in state 2?
The temperature in state 2
What is the temperature of the system in state 2?
1.
2.
3.
4.
5.
200 K
300 K
500 K
700 K
None of the above
Finding work
The system does 20000 J of work in the constant pressure
process that takes it from state 2 to state 3. What is the
volume and temperature of the system in state 2?
For constant pressure, we can use: W  P V  nR T
Finding volume: V  W  20000 J  0.2 m3
P 100000 Pa
V  V3  V2 so V2  V3  V  0.5 m3  0.2 m3  0.3 m3
Finding temperature: (use the ideal gas law, or …)
W
20000 J
T 

 200K
nR
100 J/K
T  T3  T2 so T2  T3  T  500K  200K  300K
Complete the table
For the same system, complete the table. The total work done
by the system in the cycle is –19400 J.
Process
Q
ΔEint
W
1 to 2
2 to 3
+20000 J
3 to 1
Entire
cycle
-19400 J
First fill in all the terms that are zero.
Each row satisfies the First Law of Thermodynamics.
Also remember that Eint  nCV T
Complete the table
For the same system, complete the table. The total work done
by the system in the cycle is –19400 J.
Process
Q
1 to 2
0
ΔEint
2 to 3
W
+20000 J
3 to 1
0
Entire
cycle
0
-19400 J
Q is zero for an adiabatic process.
The change in internal energy is zero for an isothermal
process, and is always zero for a complete cycle.
Complete the table
For the same system, complete the table. The total work done
by the system in the cycle is –19400 J.
Process
Q
1 to 2
0
ΔEint
2 to 3
+20000 J
3 to 1
Entire
cycle
W
0
-19400 J
0
-19400 J
Even the last row has to satisfy the first law:
Q  Eint  W
Complete the table
For the same system, complete the table. The total work done
by the system in the cycle is –19400 J.
Process
Q
1 to 2
0
2 to 3
W
+50000 J +20000 J
3 to 1
Entire
cycle
ΔEint
0
-19400 J
0
-19400 J
Find the change in internal energy for the 2  3 process.
Eint
5
5
 nCV T  nRT  (100 J/K)( 200K)  50000 J
2
2
Complete the table
For the same system, complete the table. The total work done
by the system in the cycle is –19400 J.
Process
Q
ΔEint
1 to 2
0
-50000 J
W
2 to 3
+70000 J +50000 J +20000 J
3 to 1
0
Entire
cycle
-19400 J
0
-19400 J
Rows have to obey the first law.
Columns have to sum to the value for the entire cycle.
Complete the table
For the same system, complete the table. The total work done
by the system in the cycle is –19400 J.
Process
Q
1 to 2
0
ΔEint
W
-50000 J +50000 J
2 to 3
+70000 J +50000 J +20000 J
3 to 1
-89400 J
0
-89400 J
Entire
cycle
-19400 J
0
-19400 J
Rows have to obey the first law.
Columns have to sum to the value for the entire cycle.
A heat engine
A heat engine is a device that uses heat to do work. A
gasoline-powered car engine is a good example.
To be useful, the engine must go through cycles, with work
being done every cycle. Two temperatures are required. The
higher temperature causes the system to expand, doing work,
and the lower temperature re-sets the engine so another
cycle can begin. In a full cycle, three things happen:
Heat QH is added at a relatively high temperature TH.
Some of this energy is used to do work W.
The rest is removed as heat QL at a lower temperature TL.
For the cycle: QH = W + QL (all positive quantities)
Efficiency
In general, efficiency is the ratio of the work done divided by
the heat needed to do the work.
e
Q
W QH  QL

 1 L
QH
QH
QH
The net work done in one cycle
is the area enclosed by the cycle
on the P-V diagram.
Carnot’s principle
Sadi Carnot (1796 – 1832), a French engineer, discovered an
interesting result that is a consequence of the Second Law of
Thermodynamics.
Even in an ideal situation, the efficiency of a heat engine is
limited by the temperatures between which the engine
operates. 100% efficiency is not possible, and most engines,
even in ideal cases, achieve much less than 100% efficiency.
Carnot’s principle:
QL TL

QH TH
Ideal (Carnot) efficiency:
TL
eC  1 
TH
Heat engines running backwards
Refrigerators and air conditioners
are heat engines that run backward.
Work is done on the system to pump
some heat QL from a low
temperature region TL. An amount of
heat QH = QL + W is then removed
from the system at a higher
temperature TH.
(a) Represents the cylinder in a car
engine; while (b) represents a
refrigerator.
A heat pump
If you heat your home using electric heat, 1000 J of
electrical energy can be transformed into 1000 J of heat.
An alternate way of heating is to use a heat pump, which
extracts heat from a lower-temperature region (outside the
house) and transfers it to the higher-temperature region
(inside the house). Let's say the work done in the process
is 1000 J, and the temperatures are TH = 27°C = 300 K and
TL = -13 °C = 260 K. What is the maximum amount of heat
that can be transferred into the house?
1. Something less than 1000 J
2. 1000 J
3. Something more than 1000 J
A heat pump
The best we can do is determined by the Carnot relationship.
QL TL
TL


QL 
 QH
QH TH
TH
Using this in the energy equation gives:
TL
QH  QL  W 
 QH  W
TH

THQH  TLQH  THW
For our numerical example this gives:
QH 
THW
(300 K)(1000 J)

 7500 J
TH  TL
300 K  260 K
This is why heat pumps are much better than electric heaters.
Instead of 1000 J of work going to 1000 J of heat we have
1000 J of work producing 7500 J of heat.