THERMAL PHYSICS
Thermal Physics
Thermodynamics
10.2
Processes
10.3
The second law of thermo-dynamics and entropy
10.1 Thermodynamics
10.1.1 State the equation of state for an ideal gas.
10.1.2 Describe the difference between an ideal
gas and a real gas.
10.1.3 Describe the concept of the absolute zero
of temperature and the Kelvin scale of
temperature.
10.1.4 Solve problems using the equation of state
of an ideal gas.
© IBO 2007
10.1.1 The ideal gas equaTion
A
n ideal gas is a theoretical gas that obeys the equation
of state of an ideal gas exactly. hey obey the equation
pV = RT when there are no forces between molecules at all
pressures, volumes and temperatures.
Remember from Avogadro’s hypothesis that one mole
of any gas contains the Avogadro number of particles NA
equal to 6.02 × 1023 particles. It also occupies 22.4 dm3 at
0 °C and 101.3 kPa pressure (STP).
he internal energy of an ideal gas would be entirely kinetic
energy as there would be no intermolecular forces between
the gaseous atoms. As temperature is related to the average
kinetic energy of the atoms, the kinetic energy of the atoms
would depend only on the temperature of the ideal gas.
10
ahl
10.1
From the combined gas laws, we determined that:
PV/T = k
or
PV = kT
If the value of the universal gas constant is compared for
diferent masses of diferent gases, it can be demonstrated
that the constant depends not on the size of the atoms but
rather on the number of particles present (the number of
moles). hus for n moles of any ideal gas:
PV/nT=R
P V = nRT
his is called the ‘equation of state’ of an ideal gas, where R
is the universal gas constant and is equal to 8.31 J mol‑1 K‑1.
he equation of state of an ideal gas is determined from
the gas laws and Avogadro’s law.
10.1.2 real and ideal gases
An ideal gas is a theoretical gas that obeys the ideal gas
equation exactly. Real gases conform to the gas laws under
certain limited conditions but they can condense to liquids,
then solidify if the temperature is lowered. Furthermore,
there are relatively small forces of attraction between
particles of a real gas, and even this is not allowable for
an ideal gas.
A real gas obeys the gas laws at low pressure and at high
temperature (above the temperature at which they liquefy),
and we refer to such a gas as an ideal gas.
273
CHAPTER 10
Pressure / P kPa
Most gases, at temperatures well above their boiling points
and pressures that are not too high, behave like an ideal gas.
In other words, real gases vary from ideal gas behaviour at
high pressures and low temperatures.
10.1.3 absoluTe zero of
TemPeraTure
–273
0
When the variation in volume as a function temperature
is plotted for an ideal gas, a graph similar to Figure 1001
is obtained.
T °C
K
0
100
273 373
Figure 1002 Variation of pressure with temperature
Volume / V m3
he Pressure (Admonton) Law of Gases states that:
he pressure of a ixed mass of gas at constant volume is
directly proportional to its its temperature.
ahl
P
P ∝ T ⇔ P = kT ∴-----1- = k
T1
0
100
273 373
T °C
K
herefore,
P1
P2
------ = ----T1
T2
Figure 1001 Variation of volume with temperature
he volume of a ixed mass of gas at constant pressure
is directly proportional to its absolute (Kelvin)
temperature.
his can also be stated as:
pressure, P / mm Hg
he Charles (Gay‑Lussac) Law of gases states that:
When a pressure versus volume graph is drawn for the
collected data, a hyperbola shape is obtained, and when
pressure is plotted against the reciprocal of volume a
straight line is obtained. See Figure 1003.
pressure, P / mm Hg
Note that from the extrapolation of the straight line that the
volume of gases would be theoretically zero at –273.15 °C
called absolute zero. he scale chosen is called the Kelvin
scale K and this is the fundamental unit of thermodynamic
temperature.
volume, V / cm3
PV
–273
0
1 /cm–3
V
P
Figure 1003 Pressure-volume graphs
he volume of a ixed mass of gas increases by 1 / 273.15
of its volume at 0 °C for every degree Celsius rise in
temperature provided the pressure is constant.
Boyle’s Law for gases states that the pressure of a ixed
mass of gas is inversely proportional to its volume at
constant temperature.
When the variation in pressure as a function temperature
is plotted for an ideal gas, a graph similar to Figure 1002
is obtained.
P ∝ --1- ⇔ P V = constant
V
When the conditions are changed, with the temperature
still constant
274
THERMAL PHYSICS
P 1 V1 = P 2 V2
(1.01×105Nm-1)×(molarmass/1.25kgm-3)=1×(8.31J
mol-1K-1)×(0+273K)
Sothatthemolarmass=
10.1.4 using The equaTion of sTaTe
1×8.31Jmol-1K-1×273K×1.25kgm-3÷1.01×105Nm-1
n=39.46mol
=28×10-3kgmol-1
heidealgasisheliumwithamolarmasof2.8 × 10-2 kg mol-1.
Exercise
of an ideal gas
Example 1
1.
A weather balloon of volume 1.0 m3 contains helium at
a pressure of 1.01 × 105 N m‑2 and a temperature of 35oC.
What is the mass of the helium in the balloon if one mole
of helium has a mass of 4.003 × 10‑3 kg?
If the average translational kinetic energy EK at a
temperature T of helium (molar mass 4 g mol‑1),
then the average translational kinetic energy
of neon (molar mass 20 g mol‑1) at the same
temperature would be:
A.
B.
C.
D.
Solution
2.
Usetheequation,PV=nRT,wehave
(1.01×105Nm-1)×(1.0m3)=n×(8.31Jmol-1K-1)×(35+273K)
hen,themassofhelium=(39.46mol)×(4.003×10-3kgmol-1)
1/5 EK
5 EK
√ 5 EK
EK
A sample of gas is contained in a vessel at 20 °C
at a pressure P. If the pressure of the gas is to be
doubled and the volume remain constant, the gas
has to be heated to:
A.
B.
C.
D.
Sothat,n=39.46mol
10.1
40 °C
293 °C
586 °C
313 °C
=0.158kg.
3.
Real gases behave most like ideal gases at
hemassofheliumintheballoonis0.16 kg.
A.
B.
C.
D.
Example 2
An ideal gas has a density of 1.25 kg m‑3 at STP. Determine
the molar mass of the ideal gas.
4.
he Kelvin temperature of an ideal gas is a
measure of:
A.
Solution
Usetheequation,PV=nRT,withV=molarmass/density.
ForImole
low temperatures and high pressures
high temperatures and low pressures
low temperatures and low pressures
high temperatures and high pressures
B.
C.
D.
the average potential energy of the gas
molecules
the average speed of the gas molecules
the average pressure of the gas molecules
the average kinetic energy of the gas
molecules
275
ahl
Although the pressure and the reciprocal of volume have
a directly proportional linear plot, it is the irst volume‑
temperature graph that is used to deine absolute zero.
Although diferent samples of an ideal gas have diferent
straight‑line variations, they still extrapolate back to
absolute zero.
CHAPTER 10
5.
Two identical containers A and B contain an ideal
gas under the diferent conditions as shown below.
CONTAINER A
CONTAINER B
N moles
3N moles
Temperature T
Pressure PA
Temperature T / 3
ahl
6.
3
2
3⁄2
2⁄3
he internal volume of a gas cylinder is
3.0 × 10‑2 m3. An ideal gas is pumped into
the cylinder until the pressure is 15 MPa at a
temperature of 25 °C.
(a)
(b)
(c)
(d)
Determine the number of moles of the gas
in the cylinder
Determine the number of gas atoms in the
cylinder
Determine the average volume occupied by
one atom of the gas
Estimate the average separation of the gas
atoms
7.
A cylinder of an ideal gas with a volume of 0.2 m3
and a temperature of 25 °C contains 1.202 × 1024
molecules. Determine the pressure in the cylinder.
8.
(a)
(b)
(c)
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.
Pressure PB
he ratio PA : PB would be:
A.
B.
C.
D.
10.2 Processes
State what is meant by the term idealgas.
In terms of the kinetic theory of gases, state
what is meant by an idealgas.
Explain why the internal energy of an ideal
gas is kinetic energy only.
10.2.3 Identify the first law of thermodynamics
as a statement of the principle 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.
© IBO 2007
10.2.1 WorK done in volume
change
Consider a mass of gas with pressure p enclosed in
a cylinder by a piston of cross‑sectional area A as in
Figure 1005.
surface area,A
pressure, p
cylinder
F
Δl
Figure 1005 Expansion of a gas at constant pressure
276
THERMAL PHYSICS
he pressure, p, on the piston = force per unit area
volume, temperature and change in internal energy in
determining the state of a system.
So that,
herefore, the force on the piston, F, is given by
F = pA
Suppose the piston is moved a distance l when the gas
expands. Normally, if the gas expands, the volume
increases and the pressure decreases, as was determined
from Boyle’s Law for ideal gases in the previous section.
However, if the distance l is a small Δl, then the pressure
can be considered constant. If the pressure is constant then
the force F will be constant. he work done by the gas is:
Heat can be transferred between a system and its
environment because of a temperature diference. Another
way of transferring energy between a system and its
environment is to do work on the system or allow work to
be done by the system on the surroundings.
In order to distinguish between thermal energy (heat) and
work in thermodynamic processes
•
•
ΔW= FΔ l = pA Δ l since pressure p = Force F / Area A
If a system and its surroundings are at diferent
temperatures and the system undergoes a process,
the energy transferred by non‑mchanical means is
referred to as thermal energy (heat).
Work is deined as the process in which
thermal energy is transferred by means that are
independent of a temperature diference.
= pΔV since volume ΔV = Al
hat is, (work done / J) = (pressure / Nm‑2) × (volume change / m3)
So that,
∆W = p ⋅ ∆V = p ( V2 – V1 )
he sign of the work done by the gas depends on whether
volume change is positive or negative. When a gas expands,
as is the case for Figure 1005, then work is done by the
gas, and the volume increases. As V is positive, then W is
positive.
his equation is also valid if the gas is compressed. In the
compression, work is done on the gas and the volume is
decreased. herefore, ΔV is negative which means that W
will be negative. From the irst law of thermodynamics
this means that positive work is done on the gas.
10.2.2 and 10.2.3 firsT laW of
Thermodynamics
hermodynamics is the name given to the study of
processes in which thermal energy is transferred as
heat and as work. It had its foundations with engineers
in the 19th century who wanted to know what were the
limitations of the Laws of Physics with regard to the
operation of steam engines and other machines that
generate mechanical energy. hermodynamics treats
thermal energy from the macroscopic point of view in that
it deals with the thermodynamic variables of pressure,
In thermodynamics the word system is used oten.
A system is any object or set of objects that is being
investigated. he surroundings will then be everything in
the Universe apart from the system. For example, when a
volume of gas in a cylinder is compressed with a piston,
then the system is the cylinder‑gas‑piston apparatus and
the surroundings is everything else in the Universe. A
closed system is one in which no mass enters or leaves the
system. It is an isolated system if no energy of any kind
enters or leaves the system. Most systems are open systems
because of the natural dynamic processes that occur in the
Universe.
In Chapter 3, internal energy U was deined as the sum
total of the potential energy and kinetic energy of the
particles making up the system. From a microscopic
viewpoint, the internal energy of an ideal gas is due to
the kinetic energy of the thermal motion of its molecules.
here are no intermolecular forces and thus there cannot
be any increase in potential energy. herefore a change in
the temperature of the gas will change the internal energy
of the gas.
From the macroscopic point of view of thermodynamics,
one would expect that the internal energy of the system
would be changed if:
•
•
•
•
work is done on the system
work is done by the system
thermal energy is added to the system
thermal energy is removed from the system
Internal energy is a property of the system that depends on
the “state” of the system. In thermodynamics, a change of
277
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p = F
--A
CHAPTER 10
state of an ideal gas occurs if some macroscopic property
of the system has changed eg. phase, temperature,
pressure, volume, mass, internal energy.
Heat and work can change the state of the system but they
are not properties of the system. hey are not characteristic
of the state itself but rather they are involved in the
thermodynamic process that can change the system from
one state to another.
he absolute value for internal energy is not known. his
does not cause a problem as one is mainly concerned
with changes in internal energy, denoted by ΔU, in
thermodynamic processes.
p
p
A rea = Work done
= p( V 2 – V 1 )
pconst
isobaric
(i.e., “same pressure” process)
V2
V1
V
V
Figure 1006 Work done by a gas
expanding at constant pressure
An isobaric transformation requires a volume change at
constant pressure, and for this to occur, the temperature
needs to change to keep the pressure constant.
p = constant, or V/T = constant.
ahl
he irst law of thermodynamics is a statement of the Law
of Conservation of Energy in which the equivalence of
work and thermal energy transfer is taken into account. It
can be stated as:
the heat added to a closed system equals the change in
the internal energy of the system plus the work done by
the system.
hat is,
Q = ∆U + W = ∆U + p ∆V
or,
∆U = Q – W
where ‘+Q’ is the thermal energy added to the system and
‘+W’is the work done by the system.
If thermal energy leaves the system, then Q is negative. If
work is done on the system, then W is negative.
For an isolated system, then W = Q = 0 and ΔU = 0.
For an isobaric expansion, work is done by the system
so ΔW is positive. hermal energy is added to cause the
expansion so ΔQis positive. his means that ΔUmust be
positive. For an isobaric compression, all terms would be
negative.
isochoric (isovolumetric) processes
A graph of pressure as a function of volume change when
the volume is kept constant is shown in Figure 1007. Such
a process is said to be isochoric. When the volume is kept
ixed, the curve of the transformation is said to be an
isochore.
p
isochoric
(i.e., “same volume” process)
V
10.2.4 changes of sTaTe of an
Figure 1007 A Constant volume transformation
ideal gas
isobaric processes
A graph of pressure as a function of volume change when
the pressure is kept constant is shown in Figure 1006. Such
a process is said to be isobaric. Note that the work done by
the gas is equal to the area under the curve.
Note that the work done by the gas is equal to zero as
ΔV = 0. here is zero area under the curve on a p–V
diagram. However, the temperature and pressure can both
change and so such a transformation will be accompanied
by a thermal energy change.
V = constant, or p/T = constant.
For an isochoric expansion, no work is done by the system
so ΔWis zero. hermal energy leaves the system so ΔQis
negative. his means that ΔU must be negative. For an
isobaric expansion, ΔW is zero, and ΔQand ΔU are positive.
278
THERMAL PHYSICS
isothermal processes
A thermodynamic process in which the pressure and the
volume are varied while the temperature is kept constant
is called an isothermal process. In other words, when an
ideal gas expands or is compressed at constant temperature,
then the gas is said to undergo an isothermal expansion or
compression.
Figure 1008 shows three isotherms for an ideal gas at
diferent temperatures where
Consider an ideal gas enclosed in a thin conducting vessel
that is in contact with a heat reservoir, and is itted with a
light, frictionless, movable piston. If an amount of heat Q
is added to the system which is at point A of Figure 1008,
then the system will move to another point on the graph, B.
he heat taken in will cause the gas to expand isothermally
and will be equivalent to the mechanical work done by
the gas. Because the temperature is constant, there is no
change in internal energy of the gas.
hat is,
∆ T = 0 and ∆U = 0 ⇒ Q = W
T1 < T2 < T3.
isothermal process
If the gas expands isothermally from A to B and then
returns from B to A following exactly the same path
during compression, then the isothermal change is said
to be reversible. he conditions described above would
follow this criterion.
ahl
p
T1 T2 T 3
A
adiabatic processes
B
Figure 1008 Isotherms for an ideal gas
An adiabatic expansion or contraction is one in which no
heat Q is allowed to low into or out of the system. For the
entire adiabatic process, Q = 0.
he curve of an isothermal process represents a Boyle’s
Law relation
To ensure that no heat enters or leaves the system during
an adiabatic process it is important to
T=constant, or pV = constant = nRT
•
he moles of gas n, the molar gas constant R, and the
absolute temperature T are constant.
•
For an isothermal expansion, temperature is constant so
ΔUis zero. Work is done by the system so ΔWis positive.
his means that ΔUmust be positive.
he compression stroke of an automobile engine is
essentially an adiabatic compression of the air‑fuel mixture.
he compression occurs too rapidly for appreciable heat
transfer to take place.
V
In order to keep the temperature constant during an
isothermal process
•
•
the gas is assumed to be held in a thin container
with a high thermal conductivity that is in contact
with a heat reservoir – an ideal body of large mass
whose temperature remains constant when heat
is exchanged with it. eg. a constant–temperature
water bath.
the expansion or compression should be done
slowly so that no eddies are produced to create hot
spots that would disrupt the energy equilibrium of
the gas.
make sure that the system is extremely well
insulated.
carry out the process rapidly so that the heat does
not have the time to leave the system.
In an adiabatic compression the work done on the gas will
lead to an increase in the internal energy resulting in an
increase in temperature.
∆U = Q – W but Q = 0 ⇒ ∆U = – W
In an adiabatic expansion the work done by the gas will
lead to a decrease in the internal energy resulting in a
decrease in temperature.
Figure 1009, shows the relationship that exists between an
adiabatic and three isothermals. Note that the adiabatic
curve is steeper than the isotherm AB because the adiabatic
process has to occur rapidly so that the heat does not have
279
CHAPTER 10
time to leave the system. he gas expands isothermally
from point A to point B, and then it is compressed
adiabatically from B to C. he temperature increases as a
result of the adiabatic process from T1 to T3. If the gas is
then compressed at constant pressure from the point C to
A, the net amount of work done on the gas will equal the
area enclosed by ABC.
p
(a)
(c)
p
V
Work done by gas
in expanding
T1 T2 T 3
p
V
Work done on gas
as it is compressed
V
Net work done
by gas
Figure 1011 Pressure volume diagrams
adiabatic
C
A
(b)
p
10.2.5 and 10.2.6 WorK and The
isothermal process
Thermodynamic cycle
B
V
ahl
Figure 1009 An isothermal and adiabatic process
For an adiabatic compression, no heat enters or leaves
the system so ΔQ is zero. Work is done on the system so
ΔWis negative. his means that ΔUmust be negative. For
an isobaric expansion, ΔQ is zero, and ΔW and ΔU are
positive.
In Figure 1010 the area ABDE = work done by the gas
during isothermal expansion.
he area ACDE = work done by the gas during an adiabatic
expansion.
A
isothermal
B
E
•
•
It must work in cycles to be useful.
he cyclic engine must have more than one heat
reservoir.
A thermodynamic cycle is a process in which the system
is returned to the same state from which it started. hat is,
the initial and inal states are the same in the cyclic process.
A cycle for a simple engine was shown in Figure 5.4. he
net work done in the cycle is equal to the area enclosed by
the cycle.
p
adiabatic
A thermodynamic engine is a device that transforms
thermal energy to mechanical energy (work) or mechanical
energy to thermal energy such as in refrigeration and air‑
conditioning systems. Cars, steam trains, jets and rockets
have engines that transform fuel energy (chemical energy)
into the kinetic energy of their motion. In all presently
manufactured engines, the conversion is accompanied
by the emission of exhaust gases that waste some of the
thermal energy. Consequently, these engines are not very
eicient as only part of the thermal energy is converted to
mechanical energy. An engine has two crucial features:
C
D
V
Figure 1010 Work done during expansion of a gas
he work done by a gas and the work done on a gas can
be seen using the following graphical representation of a
pressure–volume diagram (Figure 1011).
Suppose a piston was placed on a heat reservoir, such as
the hot plate of a stove. hermal energy is supplied by
the thermal reservoir, and work is done by the gas inside
the piston as it expands. But this is not an engine as it
only operates in one direction. he gas cannot expand
indeinitely, because as the volume of the piston increases,
the pressure decreases (Boyle’s Law). Some point will be
reached when the expanding gas will not be able to move
the piston. For this simple engine to function, the piston
must eventually be compressed to restore the system to its
original position ready to do work.
For a cycle to do net work, thermal contact with the original
heat reservoir must be broken, and temperatures other
than that of the original heat reservoir must play a part in
280
THERMAL PHYSICS
heat engines and heat pumps
A heat engine is any device that converts thermal energy
into work. Examples include petrol and diesel engines,
fossil‑fuelled (coal, oil and natural gas) and nuclear power
plants that use heat exchangers to drive the blades in
turbines, and jet aircrat engines.
Although we cannot convert all the random motion
associated with the internal energy into useful work, we
can at least extract some useful work from internal energy
using a heat engine. To make a heat engine, we need a
source of heat such as coal, petrol (gasoline), diesel fuel,
aviation fuel or liquid petroleum gas (LPG) and a working
luid. A working luid is a substance that undergoes a
thermodynamic change of state and in the process does
work on the surroundings. Common working luids
are water that is heated and converted to steam as used
in power stations and petrol‑air mixtures as used in car
engines.
Most heat engines contain either pistons with intake and
exhaust valves that work in thermodynamic cycles as in
a car engine, or rotating turbines used in heat exchanger
systems in power stations.
he purpose of a heat engine is to convert as much of
the heat input QH from a high temperature reservoir into
work. Figure 1012 shows an energy low diagram of a
typical heat engine.
Hot reservoir at T H
Fluid enters carrying energy,
Q H , at temperature,T H .
QH
ENGINE
Fluid leaves carrying energy,
Q L = Q H – W,
at temperature, TL .
W
Engine does useful
work, W.
QL
Cold reservoir at T L
Figure 1012 Energy flow diagram of a heat engine
he heat input QH is represented as coming from the
high temperature reservoir TH which is maintained at a
constant temperature. hermal energy QL is taken from
the hot reservoir. his thermal energy is used to do work
in the heat engine. hen thermal energy can be given to
the low temperature reservoir TL without increasing its
temperature. If a “perfect” engine completed a cycle, the
change in internal energy ∆U would be zero because all
the heat would be converted to work. However, there is no
perfect heat engine and the low diagram in Figure 1009
is more the reality. At this stage, we will assume that the
change in internal energy is zero. From the First Law of
hermodynamics
∆ U = 0 = ∆ Q – W, so that W = ∆Q
hat is,
QH – Q L = W
hus for a cycle, the heat added to the system equals the
work done by the system plus the heat that lows out at
lower temperature.
An ideal gas can be used as a heat engine as in the simple
cycle in Figure 1013.
p (kPa)
6
2
C
D
B
A
3
4
Figure 1013
10
V (m )
Behaviour of an ideal gas
281
ahl
the process. In the above example, if the piston is returned
to its original position while in contact with the hot plate,
then all the work that the gas did in the expansion will
have to be used in the compression. On a p–V diagram,
one would draw an isotherm for the expansion and an
isotherm for the compression lying on top of the expansion
isotherm but in the opposite direction. herefore, the area
enclosed by the cycle would be zero. However, if the gas is
compressed at a lower temperature the internal pressure
of the system will be lower than during the expansion.
Less work will be needed for the compression than was
produced in the expansion, and there will be net work
available for transformation to mechanical energy.
CHAPTER 10
From A to B, the gas is compressed (volume decreases)
while the pressure is kept constant – an isobaric
compression. he amount of work done by the gas is given
by the area under the 2 kPa isobar.
•
he mixture burns rapidly and the hot gases then
expand against the piston in the power stroke.
he exhaust valve is opened as the piston moves
upwards during the exhaust stroke, and the cycle
begins again.
•
Using the fact that W= pΔV , we have that
intake stroke
Gas vapor
and
mixture
intake valve
open
W = 2 kPa × (4 – 10) m3
From B to C, the volume is kept constant as the pressure
increases – an isochoric increase in pressure. his can be
achieved by heating the gas. Since ΔV = 0, then no work is
done by the gas, W = 0.
ahl
compression stroke
intake valve
closed
exhaust
closed
valve
piston
= –1.2 × 104 J
From C to D, the gas expands (volume increases) while
the pressure is kept constant – an isobaric expansion. he
amount of work done by the gas is given by the area under
the 6 kPa isobar.
exhaust
valve
closed
crankshaft
crankshaft
ignition
exhaust
closed
valve
intake valve
closed
exhaust
spent fuel
gases
exhaust
open
valve
intake valve
closed
power stroke
intake valve
closed
exhaust
valve
closed
Now, we have that W= pΔV , so that
W = 6 kPa × (10 – 4) m3
= 3.6 × 104 J
Figure 1014 Four-stroke internal combustion engine
From D to A, the gas is cooled to keep the volume constant
as the pressure is decreased – an isochoric decrease in
pressure. Again ΔV = 0 and no work is done by the gas,
W = 0.
hat is, the net work done by the gas is therefore
3.6 × 104 J – 1.2 × 104 J = 2.4 × 104 J.
The internal combustion engine
Figure 1014 shows a series of schematic diagrams for the
cycle of an internal combustion engine as used in most
automobiles.
•
•
•
With the exhaust valve closed, a mixture of petrol
vapour and air from the carburettor is sucked into
the combustion chamber through the inlet valve as
the piston moves down during the intake stroke.
Both valves are closed and the piston moves up
to squeeze the mixture of petrol vapour and air
to about 1/8 th its original volume during the
compression stroke.
With both valves closed, the mixture is ignited by a
spark from the spark plug.
Motor cars usually have four or six pistons but ive and eight
cylinders are also common. he pistons are connected by
a crankshat to a lywheel which keeps the engine turning
over during the power stroke. Automobiles are about 25%
eicient.
Any device that can pump heat from a low‑temperature
reservoir to a high‑temperature reservoir is called a heat
pump. Examples of heat pumps include the refrigerator
and reverse cycle air‑conditioning devices used for
space heating and cooling. In the summer component
of Figure 1012, the evaporator heat exchanger on the
outside extracts heat from the surroundings. In the winter
component, the evaporator heat exchanger is inside
the room, and it exhausts heat to the inside air. In both
cases, thermal energy is pumped from a low‑temperature
reservoir to a high‑ temperature reservoir.
WINTER
SUMMER
room
room
condenser
TH
evaporator
TL
evaporator
TL
condenser
TH
Figure 1015 A reverse cycle heat pump
282
THERMAL PHYSICS
Figure 1016 shows the energy low that occurs in a heat
pump cycle. By doing work on the system, heat QL is added
from the low temperature TL reservoir, being the inside of
the refrigerator. A greater amount of heat QH is exhausted
to the high temperature TH reservoir.
High temperature reservoir at T H
A refrigerator is a device operating in a cycle that is
designed to extract heat from its interior to achieve or
maintain a lower temperature inside. he heat is exhausted
to the surroundings normally at a higher temperature. A
typical refrigerator is represented in Figure 1018.
W
A volatile liquid called Freon is circulated in a closed
system of pipes by the compressor pump, and, by the
process of evaporative cooling, the vaporised Freon is
used to remove heat. Evaporative cooling was discussed
in Chapter 3.
QL
Low temperature reservoir at T L
Energy flow diagram for a heat pump
An ideal gas can be used as a heat pump as in the simple
cycle in Figure 1017.
p(kPa)
he compressor maintains a high‑pressure diference
across a throttling valve. Evaporation of the Freon occurs
in several loops called the evaporator pipes that are
usually inside the coldest part of the fridge. As the liquid
evaporates on the low‑pressure, low‑temperature side,
heat is added to the system. In order to turn from a liquid
to a gas, the Freon requires thermal energy equal to the
latent heat of vaporisation. his energy is obtained from
the contents of the fridge.
insulation
B
evaporative pipes
throttling valve
C
liquid
freezer
compartment
3
V (m )
condenser
pipes
Figure 1017 An ideal gas as a heat pump
Because the cycle is traced in an anticlockwise direction,
the net work done on the surroundings is negative.
HEAT IN
HEAT OUT
heat in
Freon gas
cold food
cooling fins
compressor
pump
vapour at very high
temperature
Figure 1018 The typical small refrigerator
On the high‑pressure, high‑temperature side of the
throttling valve, thermal energy is removed from the
system. he vaporised Freon in the compressor pipes is
compressed by the compressor pump, and gives up its
283
ahl
A motor driving a compressor pump provides the means
by which a net amount of work can be done on the system
for a cycle. Even though refrigerator cabinets are well
insulated, heat from the surroundings leaks back inside.
he compressor motor can be heard to switch on and of
as it pumps this heat out again.
QH
Figure 1016
The refrigerator – a heat pump
CHAPTER 10
latent heat of vaporisation to the air surrounding the
compressor pipes. he heat ins act as a heat sink to radiate
the thermal energy to the surroundings at a faster rate.
he ins are painted black and they have a relatively large
surface area for their size.
The carnot engine
Before the First Law of hermodynamics was even
established, Nicolas Léonard Sadi Carnot (1796‑1832),
a young engineer, was able to establish the theoretical
maximum eiciency that was possible for an engine
working between two heat reservoirs. In 1824, he
formulated that:
ahl
No engine working between two heat reservoirs can be
more eicient than a reversible engine between those
reservoirs.
Carnot argued that if thermal energy does low from a cold
body to a hot body then work must be done. herefore, no
engine can be more eicient than an ideal reversible one
and that all such engines have the same eiciency. his
means that if all engines have the same eiciency then
only a simple engine was needed to calculate the eiciency
of any engine.
he net work is the area enclosed by ABCDA. In the case
given, the Carnot engine is working in a clockwise cycle
ABCDA. hermal energy is absorbed by the system at the
high temperature reservoir TH and is expelled at the low
temperature reservoir TL. Work is done by the system as
it expands along the top isotherm from A to B, and along
the adiabat from B to C. Work is then done on the system
to compress it along the bottom isotherm from C to D and
along the let adiabat from D to A.
he eiciency of the Carnot cycle depends only on the
absolute temperatures of the high and low temperature
reservoirs. he greater the temperature diference, the
greater the eiciency will be.
As a result of the Carnot eiciency, many scientists list a
hird Law of hermodynamics which states:
It is impossible to reach the absolute zero of temperature,
0 K.
he eiciency of the Carnot cycle would be 100% if the
low temperature reservoir was at absolute zero. herefore
absolute zero is unattainable.
10.2.7 solving Problems on sTaTe
Consider an ideal perfectly insulated, frictionless
engine that can work backwards as well as forwards.
he p–V diagram would have the form of that shown in
Figure 1019.
VB
p
maximum temperature
A
VC
QH
TH
isothermal expansion
VD
Solution
Volume
compression
VA
Q=0
adiabatic
compression
B
VC
VB
adiabatic expansion
Q=0
isothermal
compression
V
Volume
expansion
VA
D
1
VA
C
minimum temperature
V2
VC
Volume
compression
VD
V
Usingtheformula,Q=ΔU+W,wehavethat340J=ΔU
+(-22)J
340J=ΔU+(–22)J
sothatΔU=340J+22J
=362J
QL
TC
Figure 1019 The Carnot engine
284
Example 1
If 22 J of work is done on a system and 3.4 × 102 J of heat
is added, determine the change internal energy of the
system.
Volume
expansion
VA
changes of a gas
hat is, the change in internal energy of the system is
3.6 × 102 J.
THERMAL PHYSICS
Example 2
Solution
(a)
hefuel-airmixtureentersthepistonatpointA.
hecompressionABiscarriedoutrapidlywithno
heatexchangemakingitanadiabaticcompression.
heignitionandcombustionofthegasesintroduces
aheatinputQHthatraisesthetemperatureat
constantvolumefromBtoC.hepowerstrokeisan
adiabaticexpansionfromCtoD.hermalenergy
QLleavesthesystemduringtheexhauststroke,and
coolingoccursatconstantvolumefromDtoA.
(b)
heFigureBelowshowsthechangesthatoccurfor
eachprocessinthecycle.
Solution
UsingtheformulaW=pΔV,wehavethat
W=202.6kPa×(12–6.0)dm3
=202.6×103Pa×(12–6.0)×10-3m3
=1.216×103J
p
constant
pressure
QH
C
hat is, the work done by the gas in the expansion is
1.2 × 103 J.
maximum temperature
adiabatic expansion
constant
D
volume (V 1)
QL
adiabatic
compression
A
minimum temperature
Example 3
A thermal system containing a gas is taken around a cycle
of a heat engine as shown in the Figure below.
(a)
Starting at point A, describe the cycle.
(b)
Label the diagram fully showing the maximum
and minimum temperature reservoirs.
(c)
ahl
6.0 dm3 of an ideal gas is at a pressure of 202.6 kPa. It is
heated so that it expands at constant pressure until its
volume is 12 dm3. Determine the work done by the gas.
V1
(c)
V2
V
henetworkisrepresentedbytheenclosedarea
ABCD.Ifweassumethattheareaisapproximately
arectanglewithsidesof4×105Paand200cm3,we
have:
4×105Nm-2×200×10-6m3=80J.
Estimate the amount of work done in each cycle.
Example 4
5
8
p x 10 Pa C
For the compression stroke of an experimental diesel
engine, the air is rapidly decreased in volume by a factor
of 15, the compression ratio. he work done on the air‑fuel
mixture for this compression is measured to be 550 J.
Q
D
4
Q
B
(a)
What type of thermodynamic process is likely to
have occurred?
(b)
What is the change in internal energy of the air‑
fuel mixture?
(c)
Is the temperature likely to increase or decrease?
A
100
300
V (cm 3)
285
CHAPTER 10
4.
Solution
(a)
Becausethecompressionoccursrapidlyappreciable
heattransferdoesnottakeplace,andtheprocesscan
beconsideredtobeadiabatic,Q=0.
(b)
ΔU=Q–W=0–(–550)J
herefore,thechangeininternalenergyis550J.
(c)
hetemperaturerisewillbeverylargeresultingin
thespontaneousignitionoftheair-fuelmixture.
he Figure below shows the variation of pressure
p with volume Vduring one complete cycle of a
simple heat engine.
A
p
X
C
B
Y
0
0
V
he total work done is represented by the area:
ahl
Exercise
1.
10.2
A.
B.
C.
D.
An ideal gas was slowly compressed at constant
temperature to one quarter of its original volume.
In the process, 1.5 × 103 J of heat was given of.
5.
he Figure below shows the variation of the
pressure p with volume V of a gas during one cycle
of the Otto engine.
p–V
p
he change in internal energy of the gas was
A.
B.
C.
D.
2.
1.5 × 103 J
0J
–1.5 × 103 J
6.0 × 103 J
III.
A.
B.
C.
D.
D
Q
A
the mean speed of the molecules increases
the molecules collide with each other more
frequently
the rate of collision with the sides of the
cylinder increases.
V1
V2
A.
B.
C.
D.
II only
III only
I and II only
II and III only
AB
CD
BC and CD
AB and CD
An ideal gas in a thermally insulated cylinder is
compressed rapidly. he change in state would be:
A system absorbs 100 J of thermal energy and
in the process does 40 J of work. he change in
internal energy is:
A.
B.
C.
D.
A.
B.
C.
D.
isochoric
isothermal
adiabatic
isobaric
V
During which process does the gas do external work?
6.
286
C
When an ideal gas in a cylinder is compressed at
constant temperature by a piston, the pressure
of the gas increases. Which of the following
statement(s) best explain the reason for the
pressure increase?
I.
II.
3.
X+Y
X–Y
X
Y
60 J
40 J
100 J
140 J
THERMAL PHYSICS
Work is done when the volume of an ideal gas
increases. During which of the following state
processes would the work done be the greatest?
A.
B.
C.
D.
8.
14.
isochoric
isothermal
isobaric
adiabatic
How much heat energy must be added at
atmospheric pressure to 0.50 kg of ice at 0 °C to
convert it to steam at 100 °C?
Helium gas at 312 K is contained in a cylinder
itted with a movable piston. he gas is initially
at 2 atmospheres pressure and occupies a volume
of 48.8 L. he gas expands isothermally until
the volume is 106 L. hen the gas is compressed
isobarically at that inal pressure back to the
original volume of 48.8L. It then isochorically
returns back to its original pressure. Assuming
that the helium gas behaves like an ideal gas
(a)
(b)
9.
If 1.68 × 105 J of heat is added to a gas that
expands and does 8.1 × 105 J of work, what is the
change in internal energy of the gas.
(c)
(d)
10.
11.
12.
6.0 m3 of an ideal gas is cooled at constant normal
atmospheric pressure until its volume is 1/6 th
its original volume. It is then allowed to expand
isothermally back to its original volume. Draw the
thermodynamic process on a p–V diagram.
A system consists of 3.0 kg of water at 75 °C.
Stirring the system with a paddlewheel does
2.5 × 104 J of work on it while 6.3 × 104 J of heat is
removed. Calculate the change in internal energy
of the system, and the inal temperature of the
system.
A gas is allowed to expand adiabatically to four
times its original volume. In doing so the gas does
1750 J of work.
(a)
(b)
(c)
13.
How much heat lowed into the gas?
Will the temperature rise or fall?
What is the change in internal energy of the
gas?
For each of the processes listed in the following
table, supply the symbol +, – , or 0 for each
missing entry.
Process
Isobaric compression of an ideal gas
Isothermal compression of an ideal gas
Adiabatic expansion
Isochoric pressure drop
Free expansion of a gas
Q
W
∆U
+
(e)
(f)
(g)
(h)
15.
(a)
(b)
Calculate the number of moles of helium
gas in the system.
Determine the pressure ater the isothermal
expansion.
Draw a diagram of the thermodynamic
cycle.
Assuming that the isotherm is a diagonal
line rather than a curve, estimate the work
done during the isothermal expansion.
Determine the work done during the
isobaric compression.
Determine the work done during the
isochoric part of the cycle.
Calculate the net work done by the gas.
Calculate the inal temperature of the
helium.
Distinguish between an isothermal process
and an adiabatic process as applied to an
ideal gas.
A ixed mass of an ideal gas is held in a
cylinder by a moveable piston and thermal
energy is supplied to the gas causing
it to expand at a constant pressure of
1.5 × 102 kPa as shown in the Figure below.
thermal energy
piston
he initial volume of the container is 0.040 m3 and
ater expansion the volume is 0.10 m3. he total
energy supplied to the gas during the process is
7.0 kJ.
(i)
(ii)
(iii)
State whether this process is isothermal,
adiabatic or neither of these processes.
Determine the work done by the gas.
Calculate the change in internal energy of
the gas.
287
ahl
7.
CHAPTER 10
16.
his question is about a diesel engine cycle as
shown in the Figure below. Mark on the diagram
each of the state changes that occur at AB, BC, CD
and DA. Identify the maximum and minimum
temperature reservoirs and label QH and QL.
p
Q
B
C
10.3 The second
laW of
Thermodynamics and
enTroPy
D
Q
A
ahl
V1
V2
V
10.3.1 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.
10.3.2 State that entropy is a system property that
expresses the degree of disorder in the
system.
10.3.3 State the second law of thermodynamics in
terms of entropy changes.
10.3.4 Discuss examples of natural processes in
terms of entropy changes.
© IBO 2007
introduction
We are always told to conserve energy. But according to
the First Law of hermodynamics, in a closed system,
energy is conserved, and the total amount of energy in the
Universe does not change no matter what we do. Although
the First Law of hermodynamics is correct, it does not
tell the whole story.
How oten have you seen a videotape played in reverse
sequence. Views of water lowing uphill, demolished
buildings rising from the rubble, people walking
backwards. In none of the natural Laws of Physics studied
so far have we encountered time reversal. If all of these
Laws are obeyed, why then does the time‑reversed
sequence seem improbable? To explain this reversal
paradox, scientists in the latter half of the nineteenth
century came to formulate a new principle called the
Second Law of hermodynamics. his Law allows us to
determine which processes will occur in nature, and which
will not.
here are many diferent but equivalent ways of stating the
Second Law of hermodynamics. Much of the language
used for the deinitions had its origins with the physicists
288
THERMAL PHYSICS
10.3.1 The second laW of Thermodynamics and TemPeraTure
he second law of thermodynamics implies that thermal
energy cannot spontaneously transfer from a region of low
temperature to a region of high temperature.
The Kelvin – Planck statement
of the second law of
thermodynamics
All attempts to construct a heat engine that is 100%
eicient have failed. he Kelvin – Planck statement of the
Second Law of hermodynamics is a qualitative statement
of the impossibility of certain types of processes.
•
It is impossible for an engine working in a cycle to
transform a given amount of heat from a reservoir
completely into work.
or
•
Not all the thermal energy in a thermal system is
available to do work.
It is possible to convert heat into work in a non‑cyclic
process. An ideal gas undergoing an isothermal expansion
does just that. But ater the expansion, the gas is not in
its original state. In order to bring the gas back to its
original state, an amount of work will have to be done
on the gas and some thermal energy will be exhausted.
he Kelvin– Planck statement formulates that if energy
is to be extracted from a reservoir to do work, a colder
reservoir must be available in which to exhaust a part of
the energy.
The clausius statement of the
second law of thermodynamics
Just as there is no cyclic device that can convert a given
amount of heat completely into work, it follows that
the reverse statement is also not possible. he Clausius
statement of the Second Law of hermodynamics can be
stated as
•
It is impossible to make a cyclic engine whose only
efect is to transfer thermal energy from a colder
body to a hotter body.
here is no perfect refrigerator and no perpetual motion
machine.
10.3.2 enTroPy
Recall that in thermodynamics, a system in an equilibrium
state is characterised by its state variables (p, V, T, U, n …).
he change in a state variable for a complete cycle is zero.
In contrast, the net thermal energy and net work factors
for a cycle are not equal to zero.
In the latter half of the nineteenth century, Rudolf Clausius
proposed a general statement of the Second Law in terms
of a quantity called entropy. Entropy is a thermodynamic
function of the state of the system and can be interpreted
as the amount of order or disorder of a system. As with
internal energy, it is the change in entropy that is important
and not its absolute value.
10.3.3 change in enTroPy
he change in entropy ΔS of a system when an amount
of thermal energy Q is added to a system by a reversible
process at constant absolute temperature T is given by
∆S = Q
--T
he units of the change in entropy are J K–1.
289
ahl
who formulated the Law, and their desire to improve
the eiciency of steam engines. hese statements of the
Second Law of hermodynamics will be developed within
this section.
CHAPTER 10
or
Example
•
A heat engine removes 100 J each cycle from a heat
reservoir at 400 K and exhausts 85 J of thermal energy to
a reservoir at 300 K. Compute the change in entropy for
each reservoir.
he entropy of the Universe increases.
or
•
Natural processes tend to move toward a state of
greater disorder.
10.3.4 naTural Processes and
Solution
change in enTroPy
Since the hot reservoir loses heat, we have that
Although the local entropy may decrease, any process will
increase the total entropy of the system and surroundings,
that is, the universe. Take for example a chicken growing
inside an egg. he entropy of the egg and its contents
decreases because the inside of the egg is becoming more
ordered. However, the entropy of surroundings increases
by a greater factor because the process gives of thermal
energy. So the total energy of the Universe is increasing.
Q
100 J
–1
∆S = –---- = -------------- = 0.25 J K
T
400 K
ahl
For the cold reservoir we have
–1
85 J
Q
∆S = – ---- = -------------- = 0.283 J K
300 K
T
he change in entropy of the hot reservoir is –0.25 J K-1 and
the change in entropy of the cold reservoir is 0.28 J K-1.
he change in entropy of the cold reservoir is greater than
the decrease for the hot reservoir. he total change in
entropy of the whole system equals 0.033 J K ‑1.
hat is,
∆S = ∆S H + ∆S L = – 0.25 + 0.283 = 0.033 J K
–1
So that the net change in entropy is positive.
In this example and all other cases, it has been found
that the total entropy increases. (For an ideal Carnot
reversible cycle it can equal zero. he Carnot cycle was
discussed earlier). his infers that total entropy increases
in all natural systems. he entropy of a given system can
increase or decrease but the change in entropy of the
system ΔSs plus the change in entropy of the environment
ΔSenv must be greater than or equal to zero.
i.e.,
∆S = ∆S S + ∆S env ≥ 0
In terms of entropy, the Second Law of hermodynamics
can be stated as
•
290
he total entropy of any system plus that of its
environment increases as a result of all natural
processes.
In the beginning of Section 10.3, irreversible processes
were discussed. A block of ice can slide down an incline
plane if the frictional force is overcome, but the ice cannot
spontaneously move up the incline of its own accord. he
conversion of mechanical energy to thermal energy by
friction as it slides is irreversible. If the thermal energy
could be converted completely to mechanical energy,
the Kelvin‑Planck statement of the second Law would be
violated. In terms of entropy, the system tends to greater
disorder, and the entropy increases.
In another case, the conduction of thermal energy from
a hot body to a cold body is irreversible. Flow of thermal
energy completely from a cold body to a hot body violates
the Clausius statement of the Second Law. In terms of
entropy, a hot body causes greater disorder of the cold
body and the entropy increases. If thermal energy was
given by a cold body to a hot body there would be greater
order in the hot body, and the entropy would decrease.
his is not allowed by the Second Law.
Irreversibility can also occur if there is turbulence or an
explosion causing a non‑equilibrium state of the gaseous
system. he degree of disorder increases and the entropy
increases.
Entropy indicates the direction in which processes occur.
Hence entropy is oten called the ‘arrow of time’.
A statistical approach to the deinition of entropy was irst
applied by Ludwig Boltzmann. Boltzmann (1844 –1906),
THERMAL PHYSICS
Exercise
1.
Consider 1022 air molecules in a container. At any one
instant, there would be a large number of possibilities for
the position and velocity of each molecule – its microstate
and the molecules would be disordered. Even if there is
some momentary order in a group of molecules due to
chance, the order would become less ater collision with
other molecules. Boltzmann argued that probability is
directly related to disorder and hence to entropy. In terms
of the Second Law of hermodynamics, probability does
not forbid a decrease in entropy but rather its probability
of occurring is extremely low.
B.
C.
D.
If a coin is lipped 100 times, it is not impossible for
the one hundred coins to land heads up, but it is highly
improbable. he probability of rolling 100 sixes from 100
dice is even smaller.
A small sample of a gas contains billions of molecules
and the molecules have many possible microstates. It
is impossible to know the position and velocity of each
molecule at a given point in time. he probability of
these microstates suddenly coming together into some
improbable arrangement is ininitesimal. In reality, the
macrostate is the only measurable part of the system.
he Second Law in terms of probability does not infer that
a decrease in entropy is not allowed but it suggests that the
probability of this occurring is low.
A inal consequence of the Second Law is the heat
degradation of the Universe. It can be reasoned that in any
natural process, some energy becomes unavailable to do
useful work. An outcome of this suggests that the Universe
will eventually reach a state of maximum disorder. An
equilibrium temperature will be reached and no work
will be able to be done. All change of state will cease as all
the energy in the Universe becomes degraded to thermal
energy. his point in time is oten referred to as the ‘heat
death’ of the Universe.
he eiciency of a heat engine is the ratio of
A.
2.
10.3
the thermal energy input to the thermal
energy output
the thermal energy output to the thermal
energy input
the work output to the thermal energy
input
the work output to the thermal energy
output
A heat engine is most eicient when it works
between objects that have a
A.
B.
C.
D.
large volume
large temperature diference
large surface area
small temperature diference
ahl
an Austrian physicist, was also concerned with the “heat
death” of the Universe and irreversibilty. He concluded
that the tendency toward dissipation of heat is not an
absolute Law of Physics but rather a Statistical Law.
3.
he four‑stroke engine is oten said to consist of
the suck, squeeze, bang and blow strokes. Use a
series of diagrams to describe what this means.
4.
Explain the diference between internal and
external combustion engines, and give an example
of each.
5.
A car engine operates with an eiciency of
34% and it does 8.00 × 103 J of work each cycle.
Calculate
(a)
(b)
the amount of thermal energy absorbed per
cycle at the high‑temperature reservoir.
the amount of exhaust thermal energy
supplied to the surroundings during each
cycle.
6.
On a hot day, a person closed all the doors and
windows of the kitchen and decided to leave the
door of the refrigerator open to cool the kitchen
down. What will happen to the temperature of the
room over a period of several hours. Give a full
qualitative answer.
7.
Modern coal‑ired power plants operate at a
temperature of 520 °C while nuclear reactors
operate at a temperature of 320 °C. If the waste
heat of the two plants is delivered to a cooling
reservoir at 21 °C, calculate the Carnot eiciency
of each type of plant.
291
ahl
CHAPTER 10
8.
It takes 7.80 × 105 J of thermal energy to melt a
given sample of a solid. In the process, the entropy
of the system increases by 1740 J K‑‑1. Find the
melting point of the solid in °C.
9.
If 2.00 kg of pure water at 100 °C is poured into
2.00 kg of water at 0 °C in a perfectly insulated
calorimeter, what is the net change in entropy.
(Assume there is 4.00 kg of water at a inal
temperature of 50 °C).
10.
Use the concepts of entropy and the arrow of time
to explain the biological growth of an organism.
11.
You are given six coins which you shake and then
throw onto a table. Construct a table showing the
number of microstates for each macrostate.
12.
Describe the concept of energy degradation in
terms of entropy.
13.
Using an example, explain the meaning of the
term reversal paradox.
14.
What is meant by the ‘heat death’ of the universe?
292