Download Unit 12 - HKU Physics

Unit 12
Laws of Thermodynamics and Thermal Processes
12.1
Temperature scale
12.2
Thermodynamic temperature and absolute zero
12.3
Ideal gas laws
12.4
Kinetic theory
12.5
The internal energy of an ideal gas
12.6
Important terms used in thermodynamics
12.7
The zeroth law of thermodynamics
12.8
The first law of thermodynamics
12.9
Thermal Processes
12.10 Specific heats for an ideal gas
12.11 The second law of thermodynamics
12.12 The third law of thermodynamics
12.1 Temperature scale
The Fahrenheit scale goes from 32o to 212o (a total of 180 divisions).
The Celsius scale goes from 0o to 100o (a total of 100 divisions).
Fahrenheit oF
Celsius oC
Boiling of water
212
100
Freezing of water
32
0
Now, we have 100 Co correspond to 180 Fo, that is, 1Co corresponds to 180/100 Fo, or 1Co
corresponds to (9/5) Fo. Here, we have a formula to relate Co and Fo
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TF = 32 + TC .
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Example
Two thermometers, one marked in Fahrenheit the other in Celsius, are placed in a bath. At
what temperature will both thermometers read the same?
Answer
The same reading will occur at T, where=
T
5
(T − 32) gives T = −40o.
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1
12.2 Thermodynamic temperature and absolute zero
•
Pressure versus temperature for a gas at constant volume
As the gas is cooled, its atoms lose some of their thermal energy; they travel more slowly and
collide with the chamber walls less frequently and less forcefully, As a result, the pressure in
the chamber drops. A plot of pressure versus temperature is shown below. All gases show
straight lines of the same intercept at −273.15oC, but different gases have different slopes.
P
−273.15
•
0
T (oC)
Volume versus temperature for a gas at constant pressure
As the temperature of a gas is lowered, its atoms are less energetic. To keep the pressure
constant, the volume must decrease, which will make it easier for atoms to strike the chamber
walls frequently, thereby sustaining the pressure. All gases shrink toward zero volume at T =
−273.15oC.
V
−273.15
0
T (oC)
•
The −273.15oC is referred as the absolute zero temperature.
•
An absolute temperature is given a SI unit, Kelvin K, in which 0 K corresponds to
−273.15oC; and has the same size as Celsius; therefore 0oC corresponds to 273.15 K.
In general, the temperature in Kelvin relates the Celsius degree by the relation T = TC
+ 273.15.
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12.3 Ideal gas laws
•
The ideal gas
It is a gas of small, hard, randomly moving atoms that bounce off each other and the walls of
their container but otherwise do not interact. The attractive forces (intermolecular forces)
between gas particles are neglected. Two conditions are met:
(1)
The density is low (i.e. the atoms occupy a volume much smaller than that of the
container and the atoms are far apart from each other), and
(2)
The temperature is well above the condensation point.
Remark: The ideal-gas model does not distinguish between a monatomic gas and a diatomic
gas; both are considered as simply small and hard spheres.
With the ideal- gas model, we obtain the following relations.
•
Boyle’s law
Keeping the temperature constant, the volume of a gas varies inversely with the pressure,
which is equivalent to saying that pressure times volume is constant.
V∝
•
1
P
or
PV = constant
Charles’s law
When the pressure is kept constant, the volume of a given amount of any gas varies directly
with the temperature (in Kelvin).
V ∝ T or
•
V
= constant
T
Gay-Lussac’s law
When the volume is kept constant, the absolute pressure of a given amount of any gas varies
directly with the temperature (in Kelvin).
P∝T
•
or
P
= constant
T
The ideal gas law
As the pressure increases with the number of moles of gas molecules linearly, we have P ∝ n ,
where n is the number of moles of gas molecules. From the Boyle’s law and the GayLussac’s law, we have P ∝
conclude that P ∝
1
and P ∝ T respectively. Summing up the relations, we
V
PV
nT
, i.e.
= n × constant . The proportional constant is nearly the same
T
V
3
for all gases at low pressures and is referred as the universal gas constant R, where R =
8.31451 J/(mol K). The ideal gas law is given by
PV = n RT
(The equation of state for an ideal gas).
The Avogadro’s number NA = 6.022 × 1023 is the number of molecules per mole, and so nNA
equals N molecules present in a sample of gas. Now, the ideal gas law can be rewritten as
PV = n RT =
N
R
)T = N k B T ,
RT = N (
NA
NA
where kB is the Boltzmann’s constant and kB = 1.38066 × 10−23 J/K.
Example
A cylindrical flask of cross-sectional area A is fitted with an airtight piston that is free to slide
up and down. Contained within the flask is an ideal gas. Initially the pressure applied by the
piston is 130 kPa and the height of the piston above the base of the flask is 25 cm. When
additional mass is added to the piston, the pressure increases to 170 kPa. Assuming the
system is always at the temperature 290 K, find the new height of the piston.
Answer
Piston
Piston
h1
P1
h2
P2
The system is basically isotherms, that is, the temperature is kept constant. The Boyle’s law is
applied here as PV = constant, or P1V1 = P2V2.
(130×103)(0.25 A) = (170×103)(h2 A)
gives h2 = 19 cm.
Example
A person’s lungs might hold 6.0 L of air at body temperature (310K) and atmospheric
pressure (101 kPa). Given that the air is 21 % oxygen, find the number of oxygen molecules
in the lungs.
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Answer
Assume that the air in the lung is ideal gas. The formula PV = N k B T gives
PV (1.01 × 105 )(6.0 × 10−3 )
=
= 1.4 × 1023 air molecules in the lung. Hence, there are
−23
kBT
(1.381 × 10 )(310)
N
=
(0.21)(1.4 × 1023) oxygen molecules in the lung.
12.4 Kinetic theory
Pressure and temperature of a gas are macroscopic quantities. However, the macroscopic
quantities connect the microscopic quantities, such as the position or velocity of an individual
molecule. The kinetic theory of gas states the connection. We imagine a gas to be made up of
a collection of molecules moving about inside a container of volume V. In particular, we
assume the following:
•
The container holds a very large number N of identical molecules. Each molecule has
a mass m, and behaves as point particle.
•
The molecules move about the container in a random manner. They obey Newton’s
laws of motion at all times.
•
When molecules hit the walls of the container or collide with one another, they
bounce elastically. Other than these collisions, the molecules have no interactions.
(a)
The origin of pressure
Imagine a container that is a cube of length L on a side. Its volume, then, is V = L3. In
addition, consider a given molecule of mass m that moves in the negative x direction toward a
wall. If its velocity is −vx and then vx after rebound from the container’s wall. The change of
momentum of the molecule is given by
∆p x = p f , x − pi , x = mv x − (−mv x ) = 2mv x .
As the time required for this round trip of length 2L is
∆t = 2 L / v x ,
L
−vx
vx
the average force exerted by the wall on the molecule is
F=
mv x2
∆p 2mv x
.
=
=
∆t 2 L / v x
L
The average pressure exerted by the wall is P =
Round trip = 2L
F F mv x2 / L mv x2 mv x2
.
=
=
= 3 =
A L2
V
L2
L
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Since a gas consists of molecules of different velocities, the relation above should be replaced
by P =
m(v x2 ) av
, where ‘av’ represents the average value. Now, the gas molecules moves in
V
a three dimensional space, we have
v av2 = (v x2 ) av + (v y2 ) av + (v z2 ) av = 3(v x2 ) av which yields P =
1 mv av2
1
) or P = ρ v av2 , where N
N(
3
V
3
is the number of molecules in the container and ρ is the density of the gas. Rearrange the
equation, we finally obtain an equation in kinetic theory of gas
PV =
1
N mv av2 .
3
The L.H.S. of the above equation can be expressed in terms of the kinetic energy of
individual molecule, e.g. PV =
1
2 mv 2
2
N mv av2 = N ( av ) = N K av . But, as we know that
3
3
2
3
PV = N k B T , we then have a most important result
K av =
(b)
3
k BT .
2
RMS speed of a gas molecule
The square root of (v 2 ) av is given a special name – the root mean square (rms) speed. As
K av =
3k T
1
3
m(v 2 ) av = k B T , we obtain (v 2 ) av = B , hence
2
2
m
v rms =
3k B T
.
m
The rms speed of the gas molecules can be expressed in terms of the molecular mass M, e.g.
v rms =
(c)
3RT
, since kB = R/NA.
M
The speed distribution of molecules
The distribution of speeds of molecules is governed by the Maxwell speed distribution. The
following figure shows the distribution of speeds of oxygen gas at the temperatures T = 300
K, and T = 1100 K. Note that the most probable speed increases with increasing temperature.
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12.5 The internal energy of an ideal gas
The internal energy U of a substance is the sum of its potential energy and kinetic energy. In
an ideal gas, as there are no interactions between gas molecules, the potential energy is then
zero. As a result, the internal energy U of an ideal gas is represented by its kinetic energy.
U=
Since U =
3
N k BT .
2
3
3
N k B T = (n N A )k B T and R = N A k B , we obtain a similar relation, but this time
2
2
in terms of moles U =
3
nRT .
2
Example
A basketball at 290 K holds 0.95 mol of air molecules. What is the internal energy of the air
in the ball?
Answer
Applying the relation U =
=
U
3
nRT , we have
2
3
=
(0.95)(8.31)(290) 3434 J .
2
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Example
Two identical containers A and B are connected by a tap S that is initially closed. A contains
an ideal gas at a pressure P1 and temperature T1. B contains the same gas at a pressure P2 and
a temperature T2. The taps is then opened. If the temperatures of containers A and B are
maintained and remain unchanged, find the molar ratio of the gas in the two containers. Find
also the pressure of the gas mixture.
Answer
Before opening S, the number of moles of gas in containers A and B:
PV
PV
n1 = 1 and n2 = 2 ,
RT1
RT2
P1
A
where V is the volume of containers A and B.
T1
S
P2
T2
After opening S, the number of moles of gas in containers A (or B):
n1 ' =
n1 '
Therefore, =
n2 '
P 'V
P 'V
and n2 ' =
.
RT2
RT1
P 'V
PV P V P 'V P 'V
RT1 T2
. As n1 + n2 = n1 '+ n2 ' , we have 1 + 2 =
,
+
=
P 'V T1
RT1 RT2
RT1 RT2
RT2
where P’ is the common pressure in the containers.
After simplification, we obtain
P1 P2 P ' P '
+
=
+ . Rearrange the expression
T1 T2 T1 T2
P1 P2
+
T1 T2 PT
+ P2T1
.
=
P' = 1 2
1 1
T1 + T2
+
T1 T2
12.6 Important terms used in thermodynamics
•
Heat: Heat is the energy transferred between objects because of a temperature
difference
•
Thermal equilibrium: Objects that are in thermal contact, but have no heat exchange
between them, are said to be in thermal equilibrium.
•
Temperature: Temperature is the quantity that determines whether or not two objects
will be in thermal equilibrium.
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B
12.7 The zeroth law of thermodynamics
If object A is in thermal equilibrium with object C, and object B is separately in thermal
equilibrium with object C, then objects A and B will be in thermal equilibrium if they are
placed in thermal contact.
12.8 The first law of thermodynamics
The change in a system’s internal energy, ∆U, is related to the heat Q and the work W as
follows:
∆U = Q + W , where W = − ∫ PdV .
Q is positive
System gains heat
Q is negative
System loses heat
W is positive
Work done on system
W is negative
Work done by system
It is noted that if work is done on an insulated system (i.e. Q = 0) its internal energy increases.
The work done W has a positive value when the external world does work on the system (e.g.
compression), and it has a negative value when work is done by the system (e.g. expansion).
∆U = U f − U i = W
On the other hand, if the system does work, and / or heat is removed, its internal energy
decreases.
12.9 Thermal Processes
All processes discussed in this section are assumed to be quasi-static, which means the
processes occur so slowly that at any given time the system and its surroundings are
essentially in equilibrium. Thus, in a quasi-static process, the pressure and temperature are
always uniform throughout the system. If friction or dissipative forces are neglected, the
process is considered to be reversible. For a reversible process, it must be possible to return
both the system and its surroundings to exactly the same states they were in before the
process began.
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(a)
Isothermal processes
Constant-temperature
heat bath
Piston
Piston
Q
Consider the following reversible processes at constant temperature, the piston moves down
without friction. Applying the first law of thermodynamics, ∆U = Q + W , and making use
the property of ideal gas which has its internal energy depends only on the temperature, we
have ∆U = 0 and Q = −W , since the process is carried out at constant temperature. The
work done, W, is positive as work is done on the system. Hence we have a negative Q, that is,
heat is given out by the system to the external world. The reversed process in the figure
below performs the same idea, but this time the work W is negative, as work is done by the
system. Since ∆U = 0 again, the heat Q is then positive or we conclude that heat is gained by
the system from the external world.
Constant-temperature
heat bath
Piston
Piston
Q
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Remarks: PV plots
The PV plot of one mole of an ideal gas at the temperatures 300 K, 500 K, 700 K and 900 K.
Notice that each isotherm has the shape of a hyperbola. As the temperature is increased,
however, the isotherms move farther from the origin. Thus, the pressure corresponding to a
given volume increases with temperature, as one would expect.
In an isothermal expansion from the volume Vi to the volume Vf, the absolute value of the
work done is equal to the shaded area. For n moles of an ideal gas at the temperature T, the
work done on the gas is
Vf
V f nRT
V f dV
W =
− ∫ P dV =
−∫
dV =
− nRT ∫
=
− nRT ln V
Vi
Vi
V
i
V
V
Vf
Vi
 Vf 
=
− nRT ln  
 Vi 
Example
A cylinder contains 0.50 mol of an ideal gas at a temperature of 310 K. As the gas expands
isothermally from an initial volume of 0.31 m3 to a final volume of 0.45 m3, find the amount
of heat that must be added to the gas in order to maintain a constant temperature.
Answer
Vf
Vi
T = constant
Q
Since the process is isothermal, the internal energy of an ideal gas has no change, i.e. ∆U =
0.
The relation ∆U = Q + W gives Q = −W, where W is negative (work is done by the system).
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Now, we know that the gas gains heat from the surrounding (positive Q). As the work done
Vf
W is given by − ∫ PdV , we obtain W as
Vi
Vf
Vf
V f nRT
V f dV
V
− ∫ PdV =
−∫
− nRT ∫
=
− nRT ln V V f =
− nRT ln( ) .
W =
dV =
Vi
Vi
Vi V
i
V
Vi
Substituting n = 0.50 mol, R = 8.31 J/(mol K), Vf = 0.45 m3 and Vi = 0.31 m3, then we obtain
Q =− W = 480 J.
(b)
Adiabatic processes
A process is said to be adiabatic if there are no heat flows into or out of the system. We can
have such process if the system is well insulated. The figure below is an adiabatic
compression, the gas heats up and its pressure increases.
Insulation
Piston
P increases
T increases
Similarly, for an adiabatic expansion, the gas cools down and the pressure decrease.
Example
Describe and explain what happen when a rod is pushed
downward rapidly through the cylinder as shown in the right
figure.
Answer
When a rod that fits snugly into a cylinder is pushed downward
rapidly, the temperature of the gas increases before there is time
for heat to flow out of the system. Thus, the process is essentially
adiabatic. As a result, the temperature of the gas can increase
enough to ignite bits of paper in the cylinder. In a diesel engine
the same principle is used to ignite an air-gasoline mixture.
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Example
An amount of gas is compressed from volume A isothermally
and the PV plot is obtained as shown in curve (ii). If the same
gas is compressed adiabatically, which curve would show the
correct curve? Curve (i) or curve (iii)?
Answer
In an adiabatic compression, no heat change occurs in the process. According to the first law
of thermodynamics, ∆U = Q + W = W , ∆U is positive as W is positive during compression.
The increasing temperature gives an increasing pressure for a given volume of gas. Hence,
curve (iii) is the correct curve.
12.10 Specific heats for an ideal gas
(a)
Constant volume
A heat Q flows into a constant-volume container, which is filled with n mole ideal gas, the
temperature of the gas rises by the amount ∆T, and its pressure increases as well. The heat Q
is given by QV = n CV ∆T, where CV is the molar specific heat of a gas at constant volume. In
other words, we can write
CV =
(b)
QV
.
n∆T
Constant pressure
Similarly, the whole process can be carried out at constant pressure, but this time with the
temperature and volume varies. If a heat Q is added to the gas then its temperature and
volume increase. The molar specific heat of a gas at constant pressure is related by QP = n CP
∆T, where CP is the molar specific heat of a gas at constant pressure. And, we have
CP =
QP
.
n∆T
According to the first law of thermodynamics ∆U = Q + W , we have in process (a) the
internal energy ∆U = Q , since W =− P∆V = −P(0) = 0. But, ∆U =
3
nR∆T , hence we have
2
3
nR∆T
QV
3
CV =
= 2
= R.
n∆T
n∆T
2
13
For process (b), it is a constant-pressure case, as heat is added to the container, the increasing
volume involves a work done W by the gas, where W =− P∆V =−nR∆T . Making using the
relation ∆U = Q + W again and we obtain
Q= ∆U − W=
3
5
nR∆T + nR∆T=
nR∆T .
2
2
5
nR∆T
Q
5
Plug it into the expression of CP, we obtain C P = P = 2
= R.
n∆T
n∆T
2
Now, we conclude that CP is greater than CV, as extra work is required for expansion while
increasing the temperature.
C P − CV = R .
12.11 The second law of thermodynamics
When objects of different temperatures are brought into thermal contact, the spontaneous
flow of heat that results is always from the high temperature object to the low temperature
object. Spontaneous heat flow never proceeds in the reverse direction. The second law of
thermodynamics is more restrictive than the first law; it says that of all processes that
conserve energy, only those that proceed in a certain direction actually occur. In a sense, the
second law enforces a certain “directionality” on the behavior of nature.
12.12 The third law of thermodynamics
The third law of thermodynamics states that there is no temperature lower than absolute zero,
and that absolute zero is unattainable. It is possible to cool an object to temperatures
arbitrarily close to absolute zero – experiments have reached temperatures as low as 2.0×10-8
K – but no object can ever be cooled to precisely 0 K.
To cool an object, you can place it in thermal contact with an object that is colder. Heat
transfer will occur, with your object ending up cooler and the other object ending up warmer.
In particular, suppose you had a collection of objects at 0 K to use for cooling. You put your
object in contact with one of the 0-K objects and your object cools, while the 0-K object
warms slightly. You continue this process, each time throwing away the “warmed up” 0-K
object and using a new one. Each time you cool your object it gets closer to 0 K, without ever
actually getting there.
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