Download Thermodynamics Day I: UU

Thermodynamics Day I:
Effects of heat on matter, Ideal Gases, Heat Engines, Laws of Thermodynamics
I. Basic definitions:
*HEAT (Q): energy transferred from one substance/body to another by virtue of a
difference in temperature. Note that it is not correct to speak of a substance or object as
containing heat. An object may contain what is referred to as internal energy (U). That
energy is only called heat while it is in the process of being transferred from one body or
substance to another. Heat is a name given to a temporary form of energy.
U AQ→U b
There are 3 types of processes by which heat can be transferred from one object to
another, called CONDUCTION, CONVECTION, and RADIATION. You have
perhaps heard of these before. There are different mathematical relationships governing
the rates of heat transfer by each of the three methods. Of the 3, we will be focusing on
the simplest, conduction, in which heat is transferred from one body to another by direct
contact.
*TEMPERATURE (T): Temperature is somewhat harder to define than heat but is best
thought of as a reflection of the average kinetic energy per atom/ molecule of a substance.
It is worth noting that it is not really possible to directly measure the temperature of
anything: all methods of measure the temperature of an object are based upon a
measurement of some other physical property such as
* length of a column of liquid (traditional thermometer)
* electrical resistance (electrical thermometer)
* gas volume (gas thermometer)
* density of water (Galileo thermometer)
* bending of a bimetallic strip (thermostat)
We will see later that in an ideal gas, the average kinetic energy of the atoms/ molecules
3
is directly proportional to the Kelvin Temperature: KEavg = kbT . This implies then
2
3kbT
, where m is the mass in kg of a single atom/ molecule of the gas and kb
that vrms =
m
is the “Boltzmann constant”, which equals 1.38 X 10-23 J/K
THE KELVIN TEMPERATURE SCALE IS OUR MAIN SCALE IN THIS COURSE.
(TK = TC + 273). EQUATIONS SUCH AS THE ONES ABOVE WILL NOT
WORK IN CELSIUS. THE ONLY TIME YOU CAN GET AWAY WITH USING
CELSIUS IS IF THE EQUATION REFERS TO A CHANGE IN TEMPERATURE.
THE CHANGE IN CELSIUS TEMPERATURE OF AN OBJECT OR
SUBSTANCE IS THE SAME AS THE CHANGE IN KELVIN TEMPERATURE
ANYWAY.
II. Effects of Heat on matter:
A. temperature increase or phase change: Q = mc∆T / Q = mLf or mLv
B. thermal expansion of solids: ∆l = αlo∆T
(for liquids and gases, ∆V = βVo∆T)
C. rate of heat conduction: Heat Current: H =
T −T
ΔQ
= kA 2 1
Δt
L
III. Intro to heat engines:
A. basic definition: A heat engine is a device for the purpose of transferring
thermal energy into useful work. There are a many types of heat engines, 2 familiar ones
being steam engines and internal combustion engines. In both cases, a fuel is burned to
generate heat and some of this heat is successfully converted into useful work. Engines
run in “cycles”, which consist of repeating steps. One example is shown below. We will
be considering somewhat more idealized engine cycles than this. We consider the
enclosed gas as the “working substance” in a heat engine and all of the mathematics that
we will do will relate to the changing temperature, volume, pressure, and internal energy
of the gas (get ready to review the ideal gas law!).
B. sign convention: the enclosed gas inside of a heat engine expands and
contracts during different steps of the engine cycle. When it is expanding, it is doing
positive work on the outside world. Alternately, we could say that negative work is being
done on the gas. When the gas is contracting, positive work is being done on the gas and
negative work is being done by the gas. Choosing a sign convention for work is thus a
little tricky. The textbook uses the convention that “W” in thermodynamic
equations means work done on the gas. This is somewhat counterintuitive, as it means
that the work term is negative while the gas is doing useful work and positive while the
gas is being compressed. More on this later.
C. engine efficiency: The efficiency of an engine is defined as the useful work
W
outputted by the engine divided by the heat inputted to the engine: e = out This should
Qin
seem like common sense, as we always define efficiency as output/input. A useful way
of thinking of the energy transfer in an engine is the following overview:
Heat is inputted at the “hot” side of the engine (Qin). Some of this energy is transferred
into useful work (W) while the rest is exhausted (Qout). Some textbooks use QH and QC
instead of Qin and Qout. H and C basically stand for the hot and cold sides of the engine.
By conservation of energy, Qin = Wout + Qout. This may also be written as
QH = [W ] + QC . The Absolute value is used because remember that by convention, in all
the textbook formulas, W is the work done on the gas, not by the gas. Thus the
[W ] or e = QH − QC
efficiency formula may be rewritten as e =
QH
QH
D. Carnot Efficiency:
Sadi Carnot was a French theoretical physicist in the early 1800s who studied
engine efficiency. Strictly through theoretical work, Carnot determined that the ideal
engine would consist of a cycle of 4 reversible processes:
I. an isothermal expansion. “Isothermal” means at constant temperature. In
order to expand or (contract) isothermally, the process must take place slowly and
the gas enclosed gas must be in contact with an outside “heat sink”, which will
keep the gas’ temperature constant. so that the temperature of the enclosed gas
can be kept constant Otherwise, gases cool as they expand and get hotter as they
are compressed. A perfectly isothermal process is probably impossible but a
process can be nearly isothermal.
II. an adiabatic expansion. “Adiabatic” means no heat exchanged with the
outside environment. In order to expand or contract adiabatically, the process
must take place very quickly, so that there is no time for heat to flow into or out
of the enclosed gas.
III. an isothermal compression
IV. an adiabatic compression.
This idealized process represented on the “P-V” diagram (you’ll be seeing a lot more of
those) became known as a “Carnot Cycle”. You are not really responsible for knowing
the steps of the Carnot Cycle. An extremely important result of Carnot’s work, however,
was his finding that in an ideal engine, the heats QH and QC must be proportional to the
operating temperatures TH and TC.
[W ]
QH − QC
T −T
T
, this means that eideal = H C or 1 − C .
QH
QH
TH
TH
T −T C
On the textbook, the equation is written as ec = H
(C for Carnot).
TH
Since e =
or e =
Carnot’s equation is considered incredibly important because it places a theoretical limit
on the efficiency of an engine. For example, a gasoline engine which burns fuel at 800oC
and exhausts at 300o C can achieve, at most, an efficiency of 46.5% (1 – 573/1073). Real
engines can never achieve Carnot ideal efficiency because of friction and such. Well
designed engines may achieve efficiencies as high as 60-80% of the theoretical ideal
(Carnot) value.
Thermodynamics Day II and III:
IV. Ideal Gases
Gases are probably the best understood phase of matter. You probably learned a
good deal about the properties of gases in Chemistry. We will essentially review this
material, approaching it from a slightly different perspective than you did in Chemistry.
A. Gas Quantities: Gases are quantified according to 4 basic quantities:
V (volume, usually in m3)
P (pressure, usually in Pa)
T (temperature, always in K)
and
particles/mole)
N (number of atoms/molecules) or n (number of moles)
(N = n*Na, Avogadro’s number, 6.022X1023
B. Early Gas Laws: The gas laws are mathematical relationships which were
discovered empirically by European scientists during the 17th and 18th centuries. In order
of discovery, they are
P
I. Boyle’s Law (1660): P1V1 = P2V2
or PV = constant
(*assumes that n, T are fixed)
V
II. Charles’ Law (late 1700s):
V1 V2
=
T1 T2
V
V
= const.
T
(*assumes n, P are fixed)
or
III. Gay-Lussac’s Law (early 1800s):
P1 P2
P
or = const.
=
T1 T2
T
(*assumes n, V fixed)
T (K)
P
T(K)
IV. Avogadro’s’ Principle (1811):
V1 V2
V
or = const.
=
n1 n2
n
(*assumes P, T fixed)
V
n
C: *Ideal Gas Law: Boyle’s, Charles’, Gay-Lussac’s, and Avogadro’s laws can
PV PV
PV
be combined to give a “combined gas law”: 1 1 = 2 2 or
= const.
nT
n1T1 n2T2
What is remarkable is that this constant is very nearly the same for all gases! The
constant, as you may remember, is usually called “R”, the universal gas constant, and the
PV
relationship is written as
= R , or more familiarly,
nT
PV = nRT, where R = 8.31 J/mol*K
The equation may also be written in terms of number of molecules instead of number of
moles. In this case, the equation becomes
PV = NkbT , where kb = R/Na = 1.38 X 10-23 J/K,
called “Boltzmann’s constant”
D. Kinetic Molecular Theory of an ideal gas
Real gases are not ideal gases. Ideal gases follow all of the gas laws perfectly,
including the ideal gas law. They also never condense into liquids or solids. Real gases
follow the ideal gas law very closely under conditions of very high Temperature and low
Pressure but poorly under conditions of lower Temperature and higher Pressure. Real
gases, of course, will eventually condense into liquids and/or solids which don’t even
come close to obeying any of the gas laws. Why? Because the ideal gas law is actually
based upon 5 fundamental assumptions that do not apply 100% to any real gas. They are
1. The particles in an ideal gas move in rapid motion with a bell-curve type distribution
of speeds, in random directions and obey the laws of classical mechanics when they
collide with each other or with the walls of the container that they are in.
(random motion, obey Newton’s laws, etc.)
2. The space between the particles in an ideal gas are >> than the size of the particles
themselves. The size of the particles themselves is completely negligible
(Vparticles ≈ 0)
3. The particles in an ideal gas exert no attractive or repulsive forces on each other
except when they collide.
(Fparticle-particle≈0 except during collisions)
4. All collisions between particles and between particles and container walls are
perfectly elastic.
(all collisions perfectly elastic)
5. The average kinetic energy of an ideal gas particle is proportional to the Kelvin
temperature of the gas. For monoatomic gases, more specifically
KEavg. = 3/2 kbT
E. DERIVATION OF THE IDEAL GAS LAW
Starting with the 5 assumptions of the kinetic molecular theory of an ideal gas, the ideal
gas law can actually be proven using some clever physics/mathematics. This is
sometimes called the “particle in a box derivation”.
Picture a single particle in a rectangular box of length L and cross sectional area A. The
particle has a mass m and begins with some speed v, which has components vx, vy, and
vz.
Each time the particle collides with the wall, the collision is elastic. Since the wall is
essentially infinitely massive compared to the particle, this means that the particle will
bounce off the wall with a speed equal and a direction opposite to the way in which it was
traveling before it hit the wall. Actually, only one of the vector components of the
velocity will be reversed, in this case the vx.
The change in momentum ∆p of the particle each time it hits the left wall is given by
∆p = m∆v = m(vx-(-vx)) = 2mvx.
After colliding with the left wall, the particle will bounce back and hit the right wall,
eventually coming back and hitting the left wall again. The time it will take to return to
the left wall is given by
t = ∆x/vx = 2L/vx
The average force the left wall exerts on this particle is given by
Favg = ∆p/∆t =
2mvx mvx 2
=
2L
L
vx
According to Newton’s 3rd law, the magnitude of the average force exerted by the particle
on the wall is the same, Favg = mvx2/L. Assuming that the particle moves in random
direction, then the average force exerted by the particle on any wall should have this
magnitude.
Since Pressure = Force/Area, the pressure by this single particle would be given by
mvx 2
mv 2
F
P= = L = x
A
A
AL
Since AL = the volume of the container, P =
F mvx 2 mvx 2
=
=
A
AL
V
The last expression on the previous page can be rearranged to yield PV = mvx2. So far,
we have assumed only that the particle travels in straight lines, the collisions with the
wall are elastic, and that the mass of the particle is negligible compared to the mass of the
wall. If we further assume that the vx, vy, and vz of the particle will, on average, be the
same, we can take this a little further.
By Pythagorean theorem, the overall speed of the particle “v” is given by
v = v x 2 + v y 2 + vz 2
Therefore,
v 2 = vx 2 + v y 2 + v z 2
Assuming that vx, vy, and vz contribute (on the average), equally to the speed of the
particle, then
v2
2
2
2
2
2
2
Rearranged, vx = .
vx = v y = vz
and
v = 3vx
3
mv 2
Substituting this into the result show in bold at the top of the page yields PV =
3
Since this result is for just a single particle, if our box contains “N” particles, then each
Nmv 2
particle would contribute to the pressure and the result would become PV =
3
By definition, K.E. = ½ mv2. Thus our equation above can be rewritten
2 Nmv 2 2
3
PV =
= N ( K .E.avg ) . Finally, using assumption #5 that KE = kbT , we
3 2
3
2
2 3
achieve the result PV = N ( kbT ) or PV = NkbT !!!
3 2
(Since N = nNA and kb = R/N, this can also be written as the more familiar PV = nRT)
F. More definitions:
INTERNAL ENERGY (U): Internal energy is a catchall for the total energy stored
within the atoms or molecules of a substance. This includes both the kinetic energy of
the atoms/molecules and the potential energy associated with the interatomic/
intermolecular bonding/structure/state. Because of the potential energy part, we can
never strictly define the absolute internal energy of any real substance. We can speak
only of changes in internal energy (∆U). These are of paramount importance in
thermodynamics.
Since the particles of an ideal gas do not attract or repel each other at all, an ideal gas has
no potential energy term and U depends only on the kinetic energy of the atoms, which
3
for a monoatomic gas is related to the temperature by K .E. = kbT . For an ideal
2
3
monoatomic gas consisting of N atoms, U = NkbT . For an ideal gas consisting of n
2
3
3
moles, U = nRT , so ΔU = nRΔT
2
2
WORK (W): You know one definition of work already: W = Fdcosθ. In
thermodynamics, work is also of tremendous importance, particularly for the subject of
heat engines. Heat engines are essentially devices for converting heat energy into work.
Like heat, work is a name given to a temporary form of energy. Work is a transfer of
energy from one body or substance to another not based on a difference in temperature
between the two bodies. This latter part of the definition is to distinguish work from
heat.
When a gas expands, it does positive work on its environment and negative work is done
on it. When a gas is compressed, it does negative work on its environment while positive
work is done on it. It will be shown below that the amount of work done is given by
Won = -Pavg∆V (alternatively, Wby = +Pavg∆V )
By definition, P = F/A, so the force exerted on the gas is given by F = P*A.
When a piston containing an enclosed gas is compressed an amount d, the work done on
the gas is given by W = Fd, which can be rewritten as W = PAd.
The change in volume of the gas ∆V = -Ad. The work done on the gas can therefore be
rewritten as -P∆V.
G. Laws of Thermodynamics
The laws of thermodynamics were pieced together by a number of scientists
during the 19th century. They have been stated in various different ways by different
scientists and are thus not usually accredited to particular authors. The different ways in
which they can be stated often seem like completely different laws to a beginning
student. There is a “zeroth” law of thermodynamics, called the “zeroth” law because it
the most basic but was overlooked until years after the other 2 laws had already been
written.
Zeroth law: If object A is in thermal equilibrium with object B and object B is in thermal
equilibrium with object C, then object A is in thermal equilibrium with object C. Don’t
worry about it.
**1st law: The total energy of an isolated system is always conserved.
In thermodynamics, the 1st law is usually written as
∆U = Q + W
All of these quantities are taken with respect to the enclosed gas: ∆U refers to the change
in internal energy of the gas. Q refers to the amount of heat that flows into the gas from
the environment. W refers to the amount of work done on the gas*. If heat flows into the
gas and/or positive work is done on the gas, the internal energy of the gas will increase,
which implies that the Temperature will as well, since ∆U = 3/2 nR∆T (at least for a
monoatomic gas). Heat flowing out of a gas and/or positive work done by the gas will
result in a decrease in internal energy and in temperature.
*in the past, the convention was used that W referred to the amount of work done by the gas. Using that
convention, the law was written as ∆U = Q – W
2nd law: The entropy of a closed system may never decrease. It may stay the same (for
reversible processes) or increase (for irreversible processes). Since no real process is
perfectly reversible and the only really closed system is the universe as a whole, the 2nd
law may also be stated as “for any real process, the entropy of the universe always
increases”. Entropy, as you may recall, roughly relates to“disorder”. A more precise
definition of entropy relates to the number of “microstates” of a system: S = kblnW,
where W is the “multiplicity” of the system (essentially, the number of possible
microstates corresponding to a given macrostate). To do half-decent calculations with
entropy requires calculus so don’t worry about it.
Implications of the 2nd law include that heat will always flow from an object at higher
Temperature into an object at lower temperature, never the other way around. Also, there
is no such thing as a perfectly efficient engine or a perfectly efficient refrigerator. Also,
gases will spontaneously expand into empty space but never spontaneously contract to
create empty space. Doubtless, there are many others.
H: Types of Thermodynamic Processes
When a gas is heated/cooled and/or expanded/compressed, various changes occur
in the temperature (T), pressure (P), volume (V), and internal energy (U) of the enclosed
gas. In addition, heat (Q) may be exchanged with the environment and work (W) may be
done. You are expected to be familiar with several idealized types of thermodynamic
processes, called isochoric, isobaric, isothermal, and adiabatic.
Isochoric means constant volume (∆V = 0).
That means that as the gas is heated/cooled,
I. Isochoric.
the pressure increases/decreases
proportionally. Since ∆V = 0, no work is
P
done. Thus, during an isochoric process,
∆U = Q + W = Q + 0 Æ ∆U = Q
V
Isobaric means constant pressure. That
means that as the gas is heated/cooled, the
volume increases/decreases proportionally.
Since the pressure is constant, we can easily
calculate the work done on the gas using
II. Isobaric
P
W = -P∆V
V
III. Isothermal
P
V
Isothermal means that the temperature stays
constant. This means that as the gas
expands/contracts, the pressure
decreases/increases inversely to the volume
change. Since the internal energy of an
ideal gas depends only on temperature,
∆U = 0 for an isothermal process. Thus the
1st law becomes
∆U = 0 = Q + W Æ Q = -W
IV. Adiabatic
P
V
(adiabatic curve is steeper than isothermal)
Adiabatic means that no heat is exchanged
with the environment (Q = 0). An adiabatic
expansion is sometimes also referred to as a
free expansion. During an adiabatic
expansion, the temperature of the gas drops
as internal energy is used to do work on the
environment. During an adiabatic
compression, the temperature increases.
The pressure, volume, and temperature all
change during an adiabatic process and
calculus is needed to calculate them. The
1st law, though, becomes
∆U = Q + W = 0 + W Æ ∆U = W
I. Engine cycles
Probably the most common type of long problem involving thermodynamics
has been a problem involving a P-V diagram for an engine cycle.
Gases are assumed to be ideal, systems are assumed to be closed, and processes are
assumed to be reversible. The gas is assumed to return to its initial conditions at the end
of the cycle. Students are asked to calculate QH, QC, W, ∆U, etc. for various steps in the
cycle and/or for the cycle as a whole. The ideal gas law may be needed to calculate
Temperatures at various points in the cycle. It is important to realize that for any closed
cycle, the change in temperature and thus the change in internal energy of the gas must be
zero, since the gas returns to its original conditions.
∆Ucycle = 0 Æ Qcycle = -Wcycle
Here is an example problem:
a) calculate the work done on the gas during processes AB, BC, and CA and the net
work for one complete cycle.
AB is an isobaric process, so the work done is given by WAB = -P∆V Æ
WAB = -P(VB-VA) = - 20 Pa (3.0 m3 – 1.0 m3) = -40 J (the work done by the gas is +40 J)
BC is an isochoric process. ∆V = 0 so W = 0.
CA is a process which occurs at varying pressure but the pressure varies linearly
with the volume. The work done can thus be easily calculated either by using the average
pressure during the process or by calculating the area under line CA
40 Pa + 20 Pa
)(1.0m3 − 3.0m3 ) = +60 J (the work done by the gas is -60 J)
WCA = −(
2
The work done on the gas for a complete cycle is -40 J + 0 + 60 J = 20 J. This
means that the work done by the gas for a complete cycle is -20 J. This is not a heat
engine, this is a refrigerator. In a heat engine, the gas does net positive work. In a
refrigerator, positive work is done on the gas
b) If the temperature at point A is 250 K, calculate the temperatures at points B and C:
PV/nT is constant. Assuming the system is closed, PV/T is constant. Plugging in the
numbers, TB = 750 K and TC = 1500 K
c) fill in the signs of Q, W, and ∆U for each part of the cycle and for the cycle as a
whole.