take .png files generated from .svg out of version control

[light-and-matter.git] / sn / ch05 / ch05.rbtex

<%
  require "../eruby_util.rb"
%>
<%
  chapter(
    '05',
    %q{Thermodynamics},
    'ch:thermo',
    '',
    {'opener'=>''}
  )
%>

\epigraphlong{$S=k \log W$}{Inscription on the tomb of Ludwig Boltzmann, 1844-1906.
Boltzmann originated the microscopic theory of thermodynamics.}

In a developing country like China, a refrigerator is the
mark of a family that has arrived in the middle class, and a
car is the ultimate symbol of wealth. Both of these
are \emph{heat engines}: devices for 
converting between heat and other
forms of energy.\index{heat engine}\index{engine!heat}
Unfortunately for the Chinese, neither is a very efficient
device. Burning fossil fuels has made China's big cities the
most polluted on the planet, and the country's total energy
supply isn't sufficient to support American levels of energy
consumption by more than a small fraction of China's
population. Could we somehow manipulate
energy in a more efficient way?

Conservation of energy is a statement that the total amount of
energy is constant at all times, which encourages us to
believe that any energy transformation can be undone ---
indeed, the laws of physics you've learned so far don't even
distinguish the past from the future.
If you get in a car and drive around the block,
the net effect is to consume some of the energy you paid
for at the gas station,  using it to heat the neighborhood.
There would not seem to be any fundamental physical principle
to prevent you from recapturing all that heat and using it again
the next time you want to go for a drive. More modestly,
why don't engineers design a car engine so that
it recaptures the heat
energy that would otherwise be wasted via
the radiator and the exhaust?

Hard experience, however, has shown that designers of more
and more efficient engines run into a brick wall at a
certain point. The generators that the electric company uses
to produce energy at an oil-fueled plant are indeed much
more efficient than a car engine, but even if one is willing
to accept a device that is very large, expensive, and
complex, it turns out to be impossible to make a perfectly
efficient heat engine --- not just impossible with present-day
technology, but impossible due to a set of fundamental physical
principles known as the science of \emph{thermodynamics}.\index{thermodynamics}
And thermodynamics isn't just a pesky set of constraints on 
heat engines. Without thermodynamics, there is no way to explain
the direction of time's arrow --- why we can remember the past
but not the future, and why it's easier to break Humpty Dumpty than
to put him back together again.

<% begin_sec("Pressure and Temperature",0) %>\index{pressure}\index{temperature}
When we heat an object, we speed up the mind-bogglingly
complex random motion of its molecules. One method for
taming complexity is the conservation laws, since they tell
us that certain things must remain constant regardless of
what process is going on. Indeed, the law of conservation of
energy is also known as the first law of thermodynamics.\index{thermodynamics!first law of}\label{first-law-of-thermodynamics}

But as alluded to in the introduction to this chapter,
conservation of energy by itself is not powerful enough to
explain certain empirical facts about heat. A second way to
sidestep the complexity of heat is to ignore heat's
atomic nature and concentrate on quantities like
temperature and pressure that tell us about a system's
properties as a whole. This approach is called
macroscopic in contrast to the microscopic method of attack.
Pressure and temperature were fairly well understood in the
age of Newton and Galileo, hundreds of years before there
was any firm evidence that atoms and molecules even existed.

Unlike the conserved quantities such as mass, energy, momentum,
and angular momentum, neither pressure nor temperature is
additive. Two cups of coffee have twice the heat
energy of a single cup, but they do not have twice the
temperature. Likewise, the painful pressure on your eardrums
at the bottom of a pool is not affected if you insert or
remove a partition between the two halves of the pool.

We restrict ourselves to a discussion of pressure in fluids
at rest and in equilibrium. In physics, the term ``fluid''
is used to mean either a gas or a liquid. The important
feature of a fluid can be demonstrated by comparing with a
cube of jello on a plate. The jello is a solid. If you shake
the plate from side to side, the jello will respond by
shearing, i.e., by slanting its sides, but it will tend to
spring back into its original shape. A solid can sustain
shear forces, but a fluid cannot. A fluid does not resist a
change in shape unless it involves a change in volume.

<% begin_sec("Pressure") %>
If you're at the bottom of a pool, you can't relieve the
pain in your ears by turning your head. The water's
force on your eardrum is always the same, and is always
perpendicular to the surface where the eardrum contacts the water. If your
ear is on the east side of your head, the water's force is
to the west. If you keep your ear in the same spot while turning
around so your ear is on the north, the force will still be the same in
magnitude, and it will change its direction so that it is
still perpendicular to the eardrum: south. This shows that
pressure has no direction in space, i.e., it is a scalar. The direction
of the force is determined by the orientation of the surface on which
the pressure acts, not by the pressure itself. A fluid flowing over a surface can also
exert frictional forces, which are parallel to the surface, but the present
discussion is restricted to fluids at rest.

Experiments also show that a fluid's force on a surface is
proportional to the surface area. The vast force of the water
behind a dam, for example, in proportion to the
dam's great surface area. (The bottom of the dam experiences
a higher proportion of its force.)

Based on these experimental results, it appears that the
useful way to define pressure is as follows.
The pressure of a fluid at a given point is defined as $F_\perp/A$,
where $A$ is the area of a small surface inserted in the fluid
at that point, and $F_\perp$ is the component of the fluid's force on
the surface which is perpendicular to the surface.
(In the case of a moving fluid,
fluid friction forces can act parallel to the
surface, but we're only dealing with stationary fluids, so there is
only an $F_\perp$.)

<% marg(30) %>
<%
  fig(
    'pressuregauge',
    %q{%
      A simple pressure gauge consists of a cylinder open at one
      end, with a piston and a spring inside. The depth to which
      the spring is depressed is a measure of the pressure. To
      determine the absolute pressure, the air needs to be pumped
      out of the interior of the gauge, so that there is no air
      pressure acting outward on the piston. In many practical
      gauges, the back of the piston is open to the atmosphere, so
      the pressure the gauge registers equals the pressure of the
      fluid minus the pressure of the atmosphere.
    }
  )
%>
<% end_marg %>
This is essentially how a pressure gauge works. The reason
that the surface must be small is so that there will not be
any significant difference in pressure between one part of it
and another part.  The SI units
of pressure are evidently $\nunit/\munit^2$, and this combination
can be abbreviated as the pascal, 1 Pa=1 $\nunit/\munit^2$.\index{pascal!unit} The
pascal turns out to be an inconveniently small unit, so car
tires, for example, normally have pressures imprinted on
them in units of kilopascals.

\begin{eg}{Pressure in U.S. units}
In U.S. units, the unit of force is the pound, and the unit
of distance is the inch. The unit of pressure is therefore
pounds per square inch, or p.s.i. (Note that the pound is
not a unit of mass.)
\end{eg}

\begin{eg}{Atmospheric pressure in U.S. and metric units}
\egquestion
A figure that many people in the U.S. remember is
that atmospheric pressure is about 15 pounds per square
inch. What is this in metric units?

\eganswer
\begin{align*}
        (\text{15 lb})/(\text{1 in}^2)
                &= \frac{68\ \nunit}{(0.0254\ \munit)^2}\\
                 &=   1.0\times10^5\ \nunit/\munit^2 \\
                 &=  100\ \zu{kPa}
\end{align*}
\end{eg}

<% begin_sec("Only pressure differences are normally significant.") %>
If you spend enough time on an airplane,
the pain in your ears subsides. This is because your body
has gradually been able to admit more air into the cavity
behind the eardrum. Once the pressure inside is equalized
with the pressure outside, the inward and outward forces on
your eardrums cancel out, and there is no physical sensation
to tell you that anything unusual is going on. For this
reason, it is normally only pressure differences that have
any physical significance. Thus deep-sea fish are perfectly
healthy in their habitat because their bodies have enough
internal pressure to cancel the pressure from the water in
which they live; if they are caught in a net and brought to
the surface rapidly, they explode because their internal
pressure is so much greater than the low pressure outside.

\begin{eg}{Getting killed by a pool pump}
\egquestion
My house has a pool, which I maintain myself. A
pool always needs to have its water circulated through a
filter for several hours a day in order to keep it clean.
The filter is a large barrel with a strong clamp that holds
the top and bottom halves together. My filter has a
prominent warning label that warns me not to try to open the
clamps while the pump is on, and it shows a cartoon of a
person being struck by the top half of the pump. The
cross-sectional area of the filter barrel is 0.25 $\zu{m}^2$. Like
most pressure gauges, the one on my pool pump actually reads
the difference in pressure between the pressure inside the
pump and atmospheric pressure. The gauge reads 90 kPa. What
is the force that is trying to pop open the filter?

\eganswer
If the gauge told us the absolute pressure of the
water inside, we'd have to find the force of the water
pushing outward and the force of the air pushing inward, and
subtract in order to find the total force. Since air
surrounds us all the time, we would have to do such a
subtraction every time we wanted to calculate anything
useful based on the gauge's reading. The manufacturers of
the gauge decided to save us from all this work by making it
read the difference in pressure between inside and outside,
so all we have to do is multiply the gauge reading by the
cross-sectional area of the filter:
\begin{align*}
         F         &=  PA\\
                        &=  (90\times10^3\ \nunit/\munit^2)( 0.25\ \munit^2)\\                        &= 22000\ \nunit
\end{align*}
That's a lot of force!
\end{eg}

The word ``suction'' and other related words contain a
hidden misunderstanding related to this point about pressure
differences. When you suck
water up through a straw, there is nothing in your mouth
that is attracting the water upward. The force that lifts
the water is from the pressure of the water in the cup. By
creating a partial vacuum in your mouth, you decreased the
air's downward force on the water so that it no longer
exactly canceled the upward force.

<% end_sec() %>
<% begin_sec("Variation of pressure with depth") %>
The pressure within a fluid in equilibrium can only depend
on depth, due to gravity. If the pressure could vary from
side to side, then a piece of the fluid in between, \figref{sidetoside},
would be subject to unequal forces from the parts of the
fluid on its two sides. Since fluids do not exhibit shear
forces, there would be no other force that could keep this
piece of fluid from accelerating. This contradicts the
assumption that the fluid was in equilibrium.

<% marg(0) %>
<%
  fig(
    'sidetoside',
    %q{%
      This doesn't happen. If pressure could vary horizontally
      in equilibrium, the cube of water would accelerate
      horizontally. This is a contradiction, since we assumed the
      fluid was in equilibrium.
    }
  )
%>
\spacebetweenfigs
<%
  fig(
    'funkycontainer',
    %q{%
      The pressure is the same at all the points marked with
      dots.
    }
  )
%>
\spacebetweenfigs
<%
  fig(
    'toptobottom',
    %q{%
      This does happen. The sum of the forces from the
      surrounding parts of the fluid is upward, canceling the
      downward force of gravity.
    }
  )
%>
<% end_marg %>
<% self_check('solidspressureside',<<-'SELF_CHECK'
How does this proof fail for solids?
  SELF_CHECK
  ) %>

To find the variation with depth, we consider the vertical
forces acting on a tiny, imaginary cube of the fluid having infinitesimal
height $\der y$ and areas $\der A$ on the top and bottom. Using positive
numbers for upward forces, we have
\begin{align*}
        P_{bottom}\der A - P_{top}\der A - F_g  =  0 \qquad .
\end{align*}
The weight of the fluid is $F_g = mg = \rho Vg = \rho\der A\der y\:g$, where $\rho$ is
the density of the fluid, so the difference in pressure is
\begin{multline*}
        \der P  =  -\rho g \der y \qquad . 
\hfill\shoveright{\text{[variation in pressure with depth for}}\\
\hfill\shoveright{\text{a fluid of density $\rho$ in equilibrium;}}\\
\hfill\shoveright{\text{positive $y$ is up.]}}
\end{multline*}
A more elegant way of writing this is in terms
of a dot product, $\der P = \rho\vc{g}\cdot\der\vc{y}$, which automatically
takes care of the plus or minus sign, depending on the relative
directions of the \vc{g} and $\der\vc{y}$ vectors, and avoids
any requirements about the coordinate system.

The factor of $\rho$ explains why we notice the difference in
pressure when diving 3 m down in a pool, but not when going
down 3 m of stairs. The equation only tells us the
difference in pressure, not the absolute pressure. The
pressure at the surface of a swimming pool equals the
atmospheric pressure, not zero, even though the depth is
zero at the surface. The blood in your body does not even
have an upper surface.

In cases where $g$ and $\rho$ are independent of depth, we can
integrate both sides of the equation to get everything in terms of
finite differences rather than differentials: $\Delta P  =  -\rho g \Delta y$.

<% self_check('pressureconst',<<-'SELF_CHECK'
In which of the following situations is the equation 
$\Delta  P  =  -\rho  g \Delta  y$ valid? Why?\\
(1) difference in pressure between a tabletop and the feet
(i.e., predicting the pressure of the feet on the floor)\\
(2) difference in air pressure between the top and bottom of
a tall building\\
(3) difference in air pressure between the top and bottom of
Mt. Everest\\
(4) difference in pressure between the top of the earth's
mantle and the center of the earth\\
(5) difference in pressure between the top and bottom of an
airplane's wing
  SELF_CHECK
  ) %>

\begin{eg}{Pressure of lava underneath a volcano}
\egquestion
A volcano has just finished erupting, and a pool
of molten lava is lying at rest in the crater. The lava has
come up through an opening inside the volcano that connects
to the earth's molten mantle. The density of the lava is 4.1
$\zu{g}/\zu{cm}^3$. What is the pressure in the lava underneath the base
of the volcano, 3000 m below the surface of the pool?

\eganswer
\begin{align*}
        \Delta P         &=  \rho g\Delta y\\
                        &= ( 4.1\ \zu{g}/\zu{cm}^3)( 9.8\ \munit/\sunit^2)(3000\ \zu{m})\\
                        &= ( 4.1\times10^6\ \zu{g}/\zu{m}^3)
                                        ( 9.8\ \munit/\sunit^2)(3000\ \zu{m})\\
                        &= ( 4.1\times10^3\ \zu{kg}/\zu{m}^3)
                                        ( 9.8\ \munit/\sunit^2)(3000\ \zu{m})\\
                        &= 1.2\times10^8\ \nunit/\munit^2 \\
                        &= 1.2\times10^8\ \zu{Pa}
\end{align*}
This is the difference between the pressure we want to find and
atmospheric pressure at the surface. The latter, however, is tiny compared
to the $\Delta P$ we just calculated, so what we've found is essentially the
pressure, $P$.
\end{eg}

\begin{eg}{Atmospheric pressure}
Gases, unlike liquids, are quite compressible, and at a given temperature, the
density of a gas is approximately proportional to the pressure. The proportionality
constant is discussed on page \pageref{subsec:pvnrt}, but for now let's just
call it $k$, $\rho= kP$. Using this fact, we can find the variation of
atmospheric pressure with altitude, assuming constant temperature:
\begin{align*}
        \der P &= -\rho g\der y\\
        \der P &= - kPg\der y\\
        \frac{\der P}{ P} &= - kg\der y\\
        \zu{ln}\: P &= - kgy+\text{constant} \qquad \text{[integrating both sides]}\\
         P &= (\text{constant}) e^{- kgy} \qquad \text{[exponentiating both sides]}
\end{align*}
Pressure falls off exponentially with height. There is no sharp cutoff to the
atmosphere, but the exponential factor gets extremely small by the time you're
ten or a hundred miles up.
\end{eg}

 % 
<% end_sec() %>
<% end_sec() %>
<% marg(10) %>
<%
  fig(
    'fever',
    %q{%
      We have to wait for the thermometer to equilibrate its
      temperature with the temperature of Irene's armpit.
    }
  )
%>
<% end_marg %>
<% begin_sec("Temperature") %>
<% begin_sec("Thermal equilibrium") %>
We use the term temperature casually, but what is it
exactly? Roughly speaking, temperature is a measure of how
concentrated the heat energy is in an object. A large,
massive object with very little heat energy in it has a low
temperature.

But physics deals with operational definitions, i.e.,
definitions of how to measure the thing in question. How do
we measure temperature? One common feature of all
temperature-measuring devices is that they must be left for
a while in contact with the thing whose temperature is being
measured. When you take your temperature with a fever
thermometer, you are waiting for the mercury inside to come
up to the same temperature as your body. The thermometer
actually tells you the temperature of its own working fluid
(in this case the mercury). In general, the idea of
temperature depends on the concept of thermal equilibrium.
When you mix cold eggs from the refrigerator with flour that
has been at room temperature, they rapidly reach a
compromise temperature. What determines this compromise
temperature is conservation of energy, and the amount of
energy required to heat or cool each substance by one
degree. But without even having constructed a temperature
scale, we can see that the important point is the phenomenon
of thermal equilibrium itself: two objects left in contact
will approach the same temperature. We also assume that if
object A is at the same temperature as object B, and B is at
the same temperature as C, then A is at the same temperature
as C. This statement is sometimes known as the zeroth law of
thermodynamics,\index{thermodynamics!zeroth law of}\label{zeroth-law-of-thermodynamics}
so called because after the first, second, and third
laws had been developed, it was realized that there was
another law that was even more fundamental.

<% marg(10) %>
<%
  fig(
    'otters',
    %q{%
      Thermal equilibrium can be prevented. Otters
      have a coat of fur that traps air bubbles for insulation.
      If a swimming otter was in thermal equilibrium with cold water, it would be dead.
      Heat is still conducted from the otter's body to the water, but much more slowly than
      it would be in a warm-blooded animal that didn't have this special adaptation.
    }
  )
%>
\spacebetweenfigs
<%
  fig(
    'hot-air-balloon',
    %q{%
      A hot air balloon is inflated. Because of thermal
      expansion, the hot air is less dense than the surrounding cold air, and
      therefore floats as the cold air drops underneath it and pushes it up
      out of the way.
    }
  )
%>
<% end_marg %>
<% end_sec() %>
<% begin_sec("Thermal expansion") %>
The familiar mercury thermometer operates on the principle
that the mercury, its working fluid, expands when heated and
contracts when cooled. In general, all substances expand and
contract with changes in temperature. The zeroth law of
thermodynamics guarantees that we can construct a
comparative scale of temperatures that is independent of
what type of thermometer we use. If a thermometer gives a
certain reading when it's in thermal equilibrium with
object A, and also gives the same reading for object B, then
A and B must be the same temperature, regardless of the
details of how the thermometers works.

What about constructing a temperature scale in which every
degree represents an equal step in temperature? The Celsius
scale has 0 as the freezing point of water and 100 as its
boiling point. The hidden assumption behind all this is that
since two points define a line, any two thermometers that
agree at two points must agree at all other points. In
reality if we calibrate a mercury thermometer and an alcohol
thermometer in this way, we will find that a graph of one
thermometer's reading versus the other is not a perfectly
straight $y=x$ line. The subtle inconsistency becomes a
drastic one when we try to extend the temperature scale
through the points where mercury and alcohol boil or freeze.
Gases, however, are much more consistent among themselves in
their thermal expansion than solids or liquids, and the
noble gases like helium and neon are more consistent with
each other than gases in general. Continuing to search for
consistency, we find that noble gases are more consistent
with each other when their pressure is very low.

As an idealization, we imagine a gas in which the atoms interact only
with the sides of the container, not with each other. Such a gas is perfectly
nonreactive (as the noble gases very nearly are), and never
condenses to a liquid (as the noble gases do only at
extremely low temperatures). Its atoms take up a negligible
fraction of the available volume. Any gas can be made to behave very much
like this if the pressure is extremely low, so that the atoms hardly ever
encounter each other. Such a gas is called an ideal gas, and we define
the Celsius scale in terms of the volume of the gas in a thermometer
whose working substance is an ideal gas maintained at a
fixed (very low) pressure, and which is calibrated at 0 and
100 degrees according to the melting and boiling points of
water. The Celsius scale is not just a comparative scale but
an additive one as well: every step in temperature is equal,
and it makes sense to say that the difference in temperature
between 18 and $28\degcunit$ is the same as the difference between 48
and 58.

<% marg(90) %>
<%
  fig(
    'gasthermometer',
    %q{%
      A simplified version of an ideal gas thermometer. The
      whole instrument is allowed to come into thermal equilibrium
      with the substance whose temperature is to be measured, and
      the mouth of the cylinder is left open to standard pressure.
      The volume of the noble gas gives an indication of
      temperature.
    }
  )
%>
\spacebetweenfigs
<%
  fig(
    'abszeroextrap',
    %q{%
      The volume of 1 kg of neon gas as a function of
      temperature (at standard pressure). Although neon would
      actually condense into a liquid at some point, extrapolating
      the graph gives to zero volume gives the same temperature as
      for any other gas: absolute zero.
    }
  )
%>
<% end_marg %>

<% end_sec() %>
<% begin_sec("Absolute zero and the Kelvin scale") %>
We find that if we extrapolate a graph of volume versus
temperature, the volume becomes zero at nearly the same
temperature for all gases: -273\degcunit. Real gases will all
condense into liquids at some temperature above this, but an
ideal gas would achieve zero volume at this temperature,
known as absolute zero. The most useful temperature scale in
scientific work is one whose zero is defined by absolute
zero, rather than by some arbitrary standard like the
melting point of water. The temperature scale used
universally in scientific work, called the Kelvin scale, is
the same as the Celsius scale, but shifted by 273 degrees to
make its zero coincide with absolute zero. Scientists use
the Celsius scale only for comparisons or when a change in
temperature is all that is required for a calculation. Only
on the Kelvin scale does it make sense to discuss ratios of
temperatures, e.g., to say that one temperature is twice as
hot as another.\index{temperature!macroscopic definition}
\index{temperature!Celsius}\index{temperature!Kelvin}\index{temperature!absolute zero}
\index{thermometer}\index{Celsius (unit)}\index{kelvin (unit)}

\begin{eg}{Which temperature scale to use}
\egquestion
You open an astronomy book and encounter the
equation
\begin{equation*}
        (\text{light emitted}) = (\text{constant}) \times  T^ 4
\end{equation*}
for the light emitted by a star as a function of its surface
temperature. What temperature scale is implied?

\eganswer
The equation tells us that doubling the
temperature results in the emission of 16 times as much
light. Such a ratio only makes sense if the Kelvin scale is
used.
\end{eg}

Although we can achieve as good an approximation to an ideal
gas as we wish by making the pressure very low, it seems
nevertheless that there should be some more fundamental way
to define temperature. We will construct a more fundamental
scale of temperature in section \ref{sec:mic-entropy}.

<% end_sec() %>
<% end_sec() %>
\startdqs

\begin{dq}\label{dq:heat-cond-propto}
Figure \subfigref{dq-heat-cond-propto}{1} shows objects 1 and 2, each with a certain temperature $T$
and a certain amount of thermal energy $E$. They are connected by a
thin rod, so that eventually they will reach thermal equilibrium.
We expect that the rate at which heat is transferred into object 1
will be given by some equation $\der E_1/\der t=k(\ldots)$, where
$k$ is a positive constant of proportionality and ``$\ldots$'' is some expression that depends
on the temperatures. Suppose that the following six forms are
proposed for ``\ldots:''
\begin{enumerate}
\item $T_1$
\item $T_2$
\item $T_1-T_2$
\item $T_2-T_1$
\item $T_1/T_2$
\item $T_2/T_1$
\end{enumerate}
Give physical reasons why five of these are not possible.
\end{dq}
<% marg(100) %>
<%
  fig(
    'dq-heat-cond-propto',
    %q{%
      Discussion questions \ref{dq:heat-cond-propto}-\ref{dq:heat-cond-series}.
    }
  )
%>
<% end_marg %>

\begin{dq}
How should the rate of heat conduction in \subfigref{dq-heat-cond-propto}{2} compare with
the rate in \subfigref{dq-heat-cond-propto}{1}?
\end{dq}

\begin{dq}\label{dq:heat-cond-series}
The example in \subfigref{dq-heat-cond-propto}{3} is different from the preceding ones
because when we add the third object in the middle, we don't necessarily know the intermediate temperature. We could
in fact set up this third object with any desired initial temperature.
Suppose, however, that the flow of heat is \emph{steady}. For example, the $36\degunit$ object could
be a human body, the $0\degunit$ object could be the air on a cold day, and the object in between could
be a simplified physical model of the insulation provided by clothing or
body fat. Under this assumption, what is the intermediate temperature?
How does the rate of heat conduction compare in the two cases?
\end{dq}

\begin{dq}
Based on the conclusions of questions \ref{dq:heat-cond-propto}-\ref{dq:heat-cond-series}, how should the rate
of heat conduction through an object depend on its length and cross-sectional area? If all the linear dimensions
of the object are doubled, what happens to the rate of heat conduction through it? How would this apply if we
compare an elephant to a shrew?
\end{dq}

<% end_sec %>
<% begin_sec("Microscopic Description of an Ideal Gas",nil,'microscopicidealgas') %>
<% begin_sec("Evidence for the kinetic theory") %>
Why does matter have the thermal properties it does? The
basic answer must come from the fact that matter is made of
atoms.
How, then, do the atoms give rise to the bulk properties we
observe? Gases, whose thermal properties are so simple,
offer the best chance for us to construct a simple
connection between the microscopic and macroscopic worlds.

A crucial observation is that although solids and liquids
are nearly incompressible, gases can be compressed, as when
we increase the amount of air in a car's tire while hardly
increasing its volume at all. This makes us suspect that the
atoms in a solid are packed shoulder to shoulder, while a
gas is mostly vacuum, with large spaces between molecules.
Most liquids and solids have densities about 1000 times
greater than most gases, so evidently each molecule in a gas
is separated from its nearest neighbors by a space something
like 10 times the size of the molecules themselves.

If gas molecules have nothing but empty space between them,
why don't the molecules in the room around you just fall to
the floor? The only possible answer is that they are in
rapid motion, continually rebounding from the walls, floor
and ceiling. In section \ref{atomicphenomenasection}
 I have already given some of the evidence for
the kinetic theory of heat, which states that heat is the kinetic energy of
randomly moving molecules. This theory was proposed by
Daniel Bernoulli in 1738, and met with considerable
opposition because it seemed as though the molecules in a
gas would eventually calm down and settle into a thin film
on the floor. There was no precedent for this kind of
perpetual motion. No rubber ball, however elastic, rebounds
from a wall with exactly as much energy as it originally
had, nor do we ever observe a collision between balls in
which none of the kinetic energy at all is converted to heat
and sound. The analogy is a false one, however. A rubber
ball consists of atoms, and when it is heated in a
collision, the heat is a form of motion of those atoms. An
individual molecule, however, cannot possess heat. Likewise
sound is a form of bulk motion of molecules, so colliding
molecules in a gas cannot convert their kinetic energy to
sound. Molecules can indeed induce vibrations such as sound
waves when they strike the walls of a container, but the
vibrations of the walls are just as likely to impart energy
to a gas molecule as to take energy from it. Indeed, this
kind of exchange of energy is the mechanism by which the
temperatures of the gas and its container become
equilibrated.

 % 
<% end_sec() %>
<% begin_sec("Pressure, volume, and temperature",nil,'pvnrt') %>
A gas exerts pressure on the walls of its container, and in
the kinetic theory we interpret this apparently constant
pressure as the averaged-out result of vast numbers of
collisions occurring every second between the gas molecules
and the walls. The empirical facts about gases can be
summarized by the relation
\begin{equation*}
        PV \propto nT   , \qquad \text{[ideal gas]}
\end{equation*}
which really only holds exactly for an ideal gas. Here $n$ is
the number of molecules in the sample of gas.

\begin{eg}{Volume related to temperature}
The proportionality of volume to temperature at fixed
pressure was the basis for our definition of temperature.
\end{eg}

\begin{eg}{Pressure related to temperature}
Pressure is proportional to temperature when volume is held
constant. An example is the increase in pressure in a car's
tires when the car has been driven on the freeway for a
while and the tires and air have become hot.
\end{eg}

We now connect these empirical facts to the kinetic theory
of a classical ideal gas. For simplicity, we assume that the gas is
monoatomic (i.e., each molecule has only one atom), and that
it is confined to a cubical box of volume $V$, with $L$ being
the length of each edge and $A$ the area of any wall. An atom
whose velocity has an $x$ component $v_x$ will collide regularly
with the left-hand wall, traveling a distance $2L$ parallel to
the $x$ axis between collisions with that wall. The time between collisions
is $\Delta t=2L/v_x$, and in each collision the $x$ component of
the atom's momentum is reversed from $-mv_x$  to $mv_x$. The total
force on the wall is
\begin{equation*}
        F =  \sum \frac{\Delta p_{x,i}}{\Delta t_i}  \qquad \text{[monoatomic ideal gas]} \qquad ,
\end{equation*}
where the index $i$ refers to the individual atoms.
Substituting $\Delta p_{x,i}=2mv_{x,i}$ and $\Delta t_i=2L/v_{x,i}$, we
have
\begin{equation*}
                F        =  \sum \frac{mv_{x,i}^2}{L} \qquad \text{[monoatomic ideal gas]} \qquad .
\end{equation*}
The quantity $mv_{x,i}^2$ is twice the contribution to the
kinetic energy from the part of the atoms' center of mass
motion that is parallel to the $x$ axis. Since we're assuming
a monoatomic gas, center of mass motion is the only type of
motion that gives rise to kinetic energy. (A more complex
molecule could rotate and vibrate as well.) If the quantity
inside the sum included the $y$ and $z$ components, it would be
twice the total kinetic energy of all the molecules. Since we expect
the energy to be equally shared among $x$, $y$, and $z$ motion,\footnote{This equal sharing will be justified more
rigorously on page \pageref{equipartition}.} the quantity
inside the sum must therefore equal 2/3 of the total kinetic
energy, so
\begin{equation*}
        F        =   \frac{2K_{total}}{3L} \qquad \text{[monoatomic ideal gas]} \qquad .
\end{equation*}
Dividing by $A$ and using $AL=V$, we have
\begin{equation*}
        P        =   \frac{2K_{total}}{3V} \qquad \text{[monoatomic ideal gas]} \qquad .
\end{equation*}
This can be connected to the empirical relation $PV \propto nT$ if we
multiply by $V$ on both sides and rewrite $K_{total}$ as $n\bar{K}$,
where $\bar{K}$ is the average kinetic energy per
molecule:
\begin{equation*}
        PV        =          \frac{2}{3}n\bar{K} \qquad \text{[monoatomic ideal gas]} \qquad .
\end{equation*}
For the first time we have an interpretation of
temperature based on a microscopic description of matter: in
a monoatomic ideal gas, the temperature is a measure of the
average kinetic energy per molecule. The proportionality
between the two is $\bar{K}=(3/2)kT$, where the constant of
proportionality $k$, known as Boltzmann's constant, has a
numerical value of $1.38\times10^{-23}\ \junit/\kunit$. In terms of
Boltzmann's constant, the relationship among the bulk
quantities for an ideal gas becomes
\begin{equation*}
        PV        =  nkT \qquad , \qquad \text{[ideal gas]}
\end{equation*}
which is known as the ideal gas law. \index{ideal gas law}\index{Boltzmann's constant}
Although I won't prove it here, this equation applies to all ideal gases, even
though the derivation assumed a monoatomic ideal gas in a cubical box.
 (You may have seen it
written elsewhere as $PV=NRT$, where $N=n/N_A$ is the number of
moles of atoms, $R=kN_A$, and $N_A=6.0\times10^{23}$, called
Avogadro's number, is essentially the number of hydrogen
atoms in 1 g of hydrogen.)\index{Avogadro's number}

\begin{eg}{Pressure in a car tire}
\egquestion
After driving on the freeway for a while, the air
in your car's tires heats up from $10\degcunit$ to $35\degcunit$. How much
does the pressure increase?

\eganswer
The tires may expand a little, but we assume this
effect is small, so the volume is nearly constant. From the
ideal gas law, the ratio of the pressures is the same as the
ratio of the absolute temperatures,
\begin{align*}
          P_2/ P_1
                &= T_2/ T_1\\
                &=(308\ \kunit)/(283\ \kunit)\\
                &= 1.09 \qquad ,\\
\end{align*}
or a 9\% increase.
\end{eg}

\startdqs

\begin{dq}\label{dq:he-xe}
Compare the amount of energy needed to heat 1 liter of helium by 1 degree with the energy needed
to heat 1 liter of xenon. In both cases, the heating is carried out in a sealed vessel that doesn't
allow the gas to expand. (The vessel is also well insulated.)
\end{dq}
 % 
\begin{dq}
Repeat  discussion question \ref{dq:he-xe} if the comparison is 1 kg of helium versus 1 kg of xenon (equal
masses, rather than equal volumes).
\end{dq}
 % 
\begin{dq}
Repeat  discussion question \ref{dq:he-xe}, but now compare 1 liter of helium in a vessel of
constant volume with the same amount of helium in a vessel that allows expansion beyond the initial
volume of 1 liter. (This could be a piston, or a balloon.)
\end{dq}
<% end_sec() %>
<% end_sec() %>

\pagebreak

<% begin_sec("Entropy as a Macroscopic Quantity") %>
<% begin_sec("Efficiency and grades of energy",0) %>
Some forms of energy are more convenient than others in
certain situations. You can't run a spring-powered
mechanical clock on a battery, and you can't run a
battery-powered clock with mechanical energy. However, there
is no fundamental physical principle that prevents you from
converting 100\% of the electrical energy in a battery into
mechanical energy or vice-versa. More efficient motors and
generators are being designed every year. In general, the
laws of physics permit perfectly efficient conversion within
a broad class of forms of energy.

Heat is different. Friction tends to convert other forms of
energy into heat even in the best lubricated machines. When
we slide a book on a table, friction brings it to a stop and
converts all its kinetic energy into heat, but we never
observe the opposite process, in which a book spontaneously
converts heat energy into mechanical energy and starts
moving! Roughly speaking, heat is different because it is
disorganized. Scrambling an egg is easy. Unscrambling it is
harder.

We summarize these observations by saying that heat is a
lower grade of energy than other forms such as mechanical
energy.

Of course it is possible to convert heat into other forms of
energy such as mechanical energy, and that is what a car
engine does with the heat created by exploding the air-gasoline
mixture. But a car engine is a tremendously inefficient
device, and a great deal of the heat is simply wasted
through the radiator and the exhaust. Engineers have never
succeeded in creating a perfectly efficient device for
converting heat energy into mechanical energy, and we now
know that this is because of a deeper physical principle
that is far more basic than the design of an engine.

\pagebreak[4]

<%
  fig(
    'turbine',
    %q{%
      1. The temperature difference between the hot and cold
      parts of the air can be used to extract mechanical energy,
      for example with a fan blade that spins because of the
      rising hot air currents.
      2. If the temperature of the air is first allowed to become
      uniform, then no mechanical energy can be extracted. The
      same amount of heat energy is present, but it is no longer
      accessible for doing mechanical work.
    },
    {
      'width'=>'wide',
      'sidecaption'=>true
    }
  )
%>

<% end_sec() %>
<% begin_sec("Heat engines") %>

Heat may be more useful in some forms than in others, i.e.,
there are different grades of heat energy. In figure \figref{turbine}/1,
the difference in temperature can be used to extract
mechanical work with a fan blade. This principle is used in
power plants, where steam is heated by burning oil or by
nuclear reactions, and then allowed to expand through a
turbine which has cooler steam on the other side. On a
smaller scale, there is a Christmas toy, \figref{angel-chimes}, that consists of a
small propeller spun by the hot air rising from a set of
candles, very much like the setup shown in figure \figref{turbine}.

<% marg(43) %>
<%
  fig(
    'angel-chimes',
    %q{%
      A heat engine. Hot air from the candles rises through the fan blades, and makes
      the angels spin.
    }
  )
%>
\spacebetweenfigs
<%
  fig(
    'carnot',
    %q{Sadi Carnot (1796-1832)}
  )
%>
<% end_marg %>

In figure \figref{turbine}/2, however, no mechanical work can be extracted
because there is no difference in temperature. Although the
air in \figref{turbine}/2 has the same total amount of energy as the air in
\figref{turbine}/1, the heat in \figref{turbine}/2 is a lower grade of energy, since
none of it is accessible for doing mechanical work.

In general, we define a heat engine as any device that takes
heat from a reservoir of hot matter, extracts some of the
heat energy to do mechanical work, and expels a lesser
amount of heat into a reservoir of cold matter. The
efficiency of a heat engine equals the amount of useful work
extracted, $W$, divided by the amount of energy we had to pay
for in order to heat the hot reservoir. This latter amount
of heat is the same as the amount of heat the engine
extracts from the high-temperature reservoir, $Q_H$. (The
letter $Q$ is the standard notation for a transfer of heat.)
By conservation of energy, we have $Q_H=W+Q_L$, where $Q_L$ is the
amount of heat expelled into the low-temperature reservoir,
so the efficiency of a heat engine, $W/Q_H$, can be rewritten
as
\begin{equation*}
        \text{efficiency}        =  1-\frac{Q_L}{Q_H} \qquad . \qquad \text{[efficiency of any heat
engine]}
\end{equation*}

It turns out that there is a particular type of heat engine,
the Carnot engine,\index{Carnot engine}\index{engine!Carnot} which, although not 100\% efficient, is
more efficient than any other. The grade of heat energy in a
system can thus be unambiguously defined in terms of the
amount of heat energy in it that cannot be extracted even by
a Carnot engine.

How can we build the most efficient possible engine? Let's
start with an unnecessarily inefficient engine like a car
engine and see how it could be improved. The radiator and
exhaust expel hot gases, which is a waste of heat energy.
These gases are cooler than the exploded air-gas mixture
inside the cylinder, but hotter than the air that surrounds
the car. We could thus improve the engine's efficiency
by adding an auxiliary heat engine to it, which would
operate with the first engine's exhaust as its hot reservoir and the air as
its cold reservoir. In general, any heat engine that expels
heat at an intermediate temperature can be made more
efficient by changing it so that it expels heat only at the
temperature of the cold reservoir.

<% marg(100) %>
<%
  fig(
    'carnota',
    %q{%
       The beginning of the first expansion stroke, in which
      the working gas is kept in thermal equilibrium with the hot
      reservoir.
    }
  )
%>
\spacebetweenfigs
<%
  fig(
    'carnotb',
    %q{%
      The beginning of the second expansion stroke, in which
      the working gas is thermally insulated. The working gas
      cools because it is doing work on the piston and thus losing
      energy.
    }
  )
%>
\spacebetweenfigs
<%
  fig(
    'carnotc',
    %q{%
      The beginning of the first compression stroke. The
      working gas begins the stroke at the same temperature as the
      cold reservoir, and remains in thermal contact with it the
      whole time. The engine does negative work.
    }
  )
%>
\spacebetweenfigs
<%
  fig(
    'carnotd',
    %q{%
      The beginning of the second compression stroke, in which
      mechanical work is absorbed, heating the working gas back up
      to $T_H$.
    }
  )
%>
<% end_marg %>

Similarly, any heat engine that absorbs some energy at an
intermediate temperature can be made more efficient by
adding an auxiliary heat engine to it which will operate
between the hot reservoir and this intermediate temperature.

Based on these arguments, we define a Carnot engine as a
heat engine that absorbs heat only from the hot reservoir
and expels it only into the cold reservoir. Figures \figref{carnota}-\figref{carnotd}
show a realization of a Carnot engine using a piston in a
cylinder filled with a monoatomic ideal gas. This gas, known
as the working fluid, is separate from, but exchanges energy
with, the hot and cold reservoirs. 
As proved on page \pageref{eg:carnot-efficiency},
this particular Carnot engine has an
efficiency given by
\begin{equation*}
        \text{efficiency}        =  1 - \frac{T_L}{T_H}  \qquad , \qquad \text{[efficiency of a Carnot engine]} 
\end{equation*}
where $T_L$ is the temperature of the cold reservoir and $T_H$ is
the temperature of the hot reservoir.

Even if you do not wish to dig into the details of the
proof, the basic reason for the temperature dependence is
not so hard to understand. Useful mechanical work is done on
strokes \figref{carnota} and \figref{carnotb}, in which the gas expands. The motion of
the piston is in the same direction as the gas's force on
the piston, so positive work is done on the piston. In
strokes \figref{carnotc} and \figref{carnotd}, however, the gas does negative work on
the piston. We would like to avoid this negative work, but
we must design the engine to perform a complete cycle.
Luckily the pressures during the compression strokes are
lower than the ones during the expansion strokes, so the
engine doesn't undo all its work with every cycle. The
ratios of the pressures are in proportion to the ratios of
the temperatures, so if $T_L$ is 20\% of $T_H$, the engine is 80\%
efficient.

We have already proved that any engine that is not a Carnot
engine is less than optimally efficient, and it is also true
that all Carnot engines operating between a given pair of
temperatures $T_H$ and $T_L$ have the same efficiency. (This can be
proved by the methods of section \ref{sec:mic-entropy}.) Thus a Carnot engine
is the most efficient possible heat engine.

<% end_sec() %>
<% begin_sec("Entropy") %>
We would like to have some numerical way of measuring the
grade of energy in a system. We want this quantity, called
entropy, to have the following two properties:

<% marg(0) %>
<%
  fig(
    'waterwheel',
    %q{%
      Entropy can be understood using the metaphor of a water
      wheel. Letting the water levels equalize is like letting the
      entropy maximize. Taking water from the high side and
      putting it into the low side increases the entropy. Water levels in this
      metaphor correspond to temperatures in the actual definition of entropy.
    }
  )
%>
<% end_marg %>
(1) Entropy is additive. When we combine two systems and
consider them as one, the entropy of the combined system
equals the sum of the entropies of the two original systems.
(Quantities like mass and energy also have this property.)

(2) The entropy of a system is not changed by operating a
Carnot engine within it.

It turns out to be simpler and more useful to define changes
in entropy than absolute entropies. Suppose as an example
that a system contains some hot matter and some cold matter.
It has a relatively high grade of energy because a heat
engine could be used to extract mechanical work from it. But
if we allow the hot and cold parts to equilibrate at some
lukewarm temperature, the grade of energy has gotten worse.
Thus putting heat into a hotter area is more useful than
putting it into a cold area. Motivated by these
considerations, we define a change in entropy as
follows:\index{entropy!macroscopic definition}
\begin{multline*}
        \Delta S  =  \frac{Q}{T}        \qquad \shoveright{\text{[change in entropy when adding}}\\                                                \shoveright{\text{heat $Q$ to matter at temperature $T$;}}\\
                                                {\text{$\Delta S$ is negative if heat is taken out]}}
\end{multline*}
A system with a higher grade of energy has a lower entropy.

\begin{eg}{Entropy is additive.}
Since changes in entropy are defined by an additive quantity
(heat) divided by a non-additive one (temperature), entropy
is additive.
\end{eg}

\begin{eg}{Entropy isn't changed by a Carnot engine.}\label{eg:carnotnoentropychange}
The efficiency of a heat engine is defined by
\begin{equation*}
        \text{efficiency}  =  1 -  Q_L/ Q_H  \qquad  ,
\end{equation*}
and the efficiency of a Carnot engine is
\begin{equation*}
        \text{efficiency}  =  1 -  T_L/ T_H \qquad  ,
\end{equation*}
so for a Carnot engine we have $Q_L/ Q_H =  T_L/ T_H$,
which can be rewritten as
$Q_L/ T_{L} =  Q_{H}/ T_H$.
The entropy lost by the hot reservoir is therefore
the same as the entropy gained by the cold one.
\end{eg}

\begin{eg}{Entropy increases in heat conduction.}
When a hot object gives up energy to a cold one,
conservation of energy tells us that the amount of heat lost
by the hot object is the same as the amount of heat gained
by the cold one. The change in entropy is $- Q/ T_{H}+ Q/ T_L$, which
is positive because $ T_L< T_H$.
\end{eg}

\begin{eg}{Entropy is increased by a non-Carnot engine.}
The efficiency of a non-Carnot engine is less than 1 -
$ T_L/ T_H$, so $Q_L/ Q_{H} >  T_{L}/ T_H$
 and $Q_L/ T_{L} >  Q_{H}/ T_H$. This means that
the entropy increase in the cold reservoir is greater than
the entropy decrease in the hot reservoir.
\end{eg}

\begin{eg}{A book sliding to a stop}
A book slides across a table and comes to a stop. Once it stops, all its
kinetic energy has been transformed into heat. As the book and
table heat up, their entropies both increase, so the total entropy increases
as well.
\end{eg}

All of these examples involved closed systems, and in all of
them the total entropy either increased or stayed the same. It
never decreased. Here are two examples of schemes for
decreasing the entropy of a closed system, with explanations
of why they don't work.

\begin{eg}{Using a refrigerator to decrease entropy?}
\egquestion
A refrigerator takes heat from a cold area and
dumps it into a hot area. (1) Does this lead to a net
decrease in the entropy of a closed system? (2) Could you
make a Carnot engine more efficient by running a
refrigerator to cool its low-temperature reservoir and
eject heat into its high-temperature reservoir?

\eganswer
(1) No. The heat that comes off of the radiator
coils is a great deal more than the heat the fridge removes from
inside; the difference is what it costs to run your fridge.
The heat radiated from the coils is so much more than the
heat removed from the inside that the increase in the
entropy of the air in the room is greater than the decrease
of the entropy inside the fridge. The most efficient
refrigerator is actually a Carnot engine running in reverse,
which leads to neither an increase
nor a decrease in entropy.

(2) No. The most efficient refrigerator is a reversed Carnot
engine. You will not achieve anything by running one Carnot
engine in reverse and another forward. They will just cancel
each other out.
\end{eg}

\begin{eg}{Maxwell's demon}
\egquestion
Maxwell imagined a pair of rooms, their air being initially in
thermal equilibrium, having a partition across the middle
with a tiny door. A miniscule demon is posted at the door
with a little ping-pong paddle, and his duty is to try to
build up faster-moving air molecules in room B and slower
moving ones in room A. For instance, when a fast molecule is
headed through the door, going from A to B, he lets it by,
but when a slower than average molecule tries the same
thing, he hits it back into room A. Would this decrease the
total entropy of the pair of rooms?

\eganswer
No. The demon needs to eat, and we can think of his body as
a little heat engine, and his metabolism is less efficient
than a Carnot engine, so he ends up increasing the entropy
rather than decreasing it.
\end{eg}

Observations such as these lead to the following hypothesis,
known as the second law of thermodynamics:\index{thermodynamics!second law of}\label{second-law-of-thermodynamics}

\begin{important}
The entropy of a closed system always increases, or at best
stays the same: $\Delta S\ge0$.
\end{important}

At present our arguments to support this statement may seem
less than convincing, since they have so much to do with
obscure facts about heat engines. In
the following section we will find a more satisfying
and fundamental explanation for the continual increase in entropy. 
To emphasize the fundamental and universal nature of the second law,
here are a few exotic examples.

\begin{eg}{Entropy and evolution}
A favorite argument of many creationists who don't believe
in evolution is that evolution would violate the second law
of thermodynamics: the death and decay of a living thing
releases heat (as when a compost heap gets hot) and lessens
the amount of energy available for doing useful work, while
the reverse process, the emergence of life from nonliving
matter, would require a decrease in entropy. Their argument
is faulty, since the second law only applies to closed
systems, and the earth is not a closed system. The earth is
continuously receiving energy from the sun.
\end{eg}

\begin{eg}{The heat death of the universe}
Living things have low entropy: to demonstrate this fact,
observe how a compost pile releases heat, which then
equilibrates with the cooler environment. We never observe
dead things to leap back to life after sucking some heat
energy out of their environments! The only reason life was
able to evolve on earth was that the earth was not a closed
system: it got energy from the sun, which presumably gained
more entropy than the earth lost.

\enlargethispage{-\baselineskip}

Victorian philosophers spent a lot of time worrying about
the heat death of the universe: eventually the universe
would have to become a high-entropy, lukewarm soup, with no
life or organized motion of any kind. Fortunately (?), we
now know a great many other things that will make the
universe inhospitable to life long before its entropy is
maximized. Life on earth, for instance, will end when the
sun evolves into a giant star and vaporizes our planet.
\end{eg}

\begin{eg}{Hawking radiation}\index{black hole}\index{Hawking radiation}
Any process that could destroy heat (or convert it into
nothing but mechanical work) would lead to a reduction in
entropy. Black holes are supermassive stars whose gravity is
so strong that nothing, not even light, can escape from them
once it gets within a boundary known as the event horizon.
Black holes are commonly observed to suck hot gas into them.
Does this lead to a reduction in the entropy of the
universe? Of course one could argue that the entropy is
still there inside the black hole, but being able to
``hide'' entropy there amounts to the same thing as being
able to destroy entropy.

The physicist Steven Hawking was bothered by this question,
and finally realized that although the actual stuff that
enters a black hole is lost forever, the black hole will
gradually lose energy  in the form of light emitted from
just outside the event horizon. This light ends up
reintroducing the original entropy back into the universe at
large.
\end{eg}

\enlargethispage{-\baselineskip}

\startdqs

\begin{dq}
In this discussion question, you'll think about a car engine in terms
of thermodynamics. Note that an internal combustion engine doesn't fit
very well into the theoretical straightjacket of a heat engine. For
instance, a heat engine has a high-temperature heat reservoir
at a single well-defined temperature, $T_H$. In a typical car engine,
however, there are several very different temperatures you could
imagine using for $T_H$: the temperature of the engine block
($\sim100\degcunit$), the walls of the cylinder ($\sim250\degcunit$), or
the temperature of the exploding air-gas mixture ($\sim1000\degcunit$,
with significant changes over a four-stroke cycle). Let's use
$T_H\sim1000\degcunit$.

Burning gas supplies heat energy $Q_H$ to your car's engine. The
engine does mechanical work $W$, but also expels heat $Q_L$
into the environment through the radiator and the exhaust. Conservation
of energy gives
\begin{equation*}
        Q_H = Q_L+W \qquad ,
\end{equation*}
and the relative proportions of $Q_L$ and $W$ are usually about
90\% to 10\%. (Actually it depends quite a bit on the type of car, the
driving conditions, etc.)

(1) $Q_L$ is obviously undesirable: you pay for it, but all it does is
heat the neighborhood. Suppose that engineers do a really good job
of getting rid of the effects that create $Q_L$, such as friction. Could
$Q_L$ ever be reduced to zero, at least theoretically?

(2) A gallon of gas releases about 140 MJ of heat $Q_H$ when burned.
Estimate the change in entropy of the universe due to running a typical
car engine and burning one gallon of gas. (You'll have to estimate how
hot the environment is. For the sake of argument, assume that the work
done by the engine, $W$, remains in the form of mechanical energy,
although in reality it probably ends up being changed into heat when you step
on the brakes.) Is your result consistent with the second law of
thermodynamics?

(3) What would happen if you redid the calculation in \#2, but assumed
$Q_L=0$? Is this consistent with your answer to \#1?
\end{dq}

\begin{dq}
When we run the Carnot engine in figures \figref{carnota}-\figref{carnotd}, there are four parts
of the universe that undergo changes in their physical states: the hot reservoir, the cold reservoir,
the working gas, and the outside world to which the shaft is connected in order to do physical work.
Over one full cycle, discuss which of these parts gain entropy, which ones lose entropy, and
which ones keep the same entropy. During which of the four strokes do these changes occur?
\end{dq}

<% end_sec() %>
<% end_sec() %>
<% begin_sec("Entropy as a Microscopic Quantity",nil,'mic-entropy') %>
<% begin_sec("A microscopic view of entropy") %>
To understand why the second law of thermodynamics is always true, we need
to see what entropy really means at the microscopic level. An example that is
easy to visualize is the free expansion of a monoatomic
gas. Figure \figref{freeexpansion}/1 shows a box
in which all the atoms of the gas are confined on one side. We very quickly remove
the barrier between the two sides, \figref{freeexpansion}/2, and some time later, the
system has reached an equilibrium, \figref{freeexpansion}/3. Each snapshot
shows both the positions and the momenta of the atoms, which is enough
information to allow us in theory to extrapolate the behavior of the system into the future,
or the past. However, with a realistic number of atoms, rather than just six, this
would be beyond the computational power of any computer.\footnote{Even with smaller
numbers of atoms, there is a problem with this kind of brute-force computation, because
the tiniest measurement errors in the initial state would end up having large effects
later on.}

<% marg(79) %>
<%
  fig(
    'freeexpansion',
    %q{A gas expands freely, doubling its volume.}
  )
%>
\spacebetweenfigs
<%
  fig(
    'fluctuation',
    %q{%
      An unusual fluctuation in the distribution of the atoms
      between the two sides of the box. There has been no external manipulation
      as in figure \figref{freeexpansion}/1.
    }
  )
%>
<% end_marg %>
But suppose
we show figure \figref{freeexpansion}/2 to a friend without any further information,
and ask her what she can say about the system's behavior in the future. She doesn't
know how the system was prepared. Perhaps, she thinks, it was just a strange
coincidence that all the atoms happened to be in the right half of the box
at this particular moment. In any case, she knows that this unusual situation
won't last for long. She can predict that after the passage of any significant amount of time,
a surprise inspection is likely to show roughly
half the atoms on each side. The same is true if you ask her to say what
happened in the past. She doesn't know about the barrier, so as far as she's concerned,
extrapolation into the past is exactly the same kind of problem as extrapolation into
the future. We just have to imagine reversing all the momentum vectors, and then
all our reasoning works equally well for backwards extrapolation. She would conclude,
then, that the gas in the box underwent an unusual fluctuation, \figref{fluctuation},
and she knows that
the fluctuation is very unlikely to exist very far into the future, or to have existed very
far into the past.

What does this have to do with entropy? Well, state \figref{freeexpansion}/3
has a greater entropy than state \figref{freeexpansion}/2. It would be easy to
extract mechanical work from \figref{freeexpansion}/2, for instance by letting
the gas expand while pressing on a piston rather than simply releasing it suddenly
into the void. There is no way to extract mechanical work from state \figref{freeexpansion}/3.
Roughly speaking, our microscopic description of entropy relates to the \emph{number
of possible states}. There are a lot more states like \figref{freeexpansion}/3 than there
are states like \figref{freeexpansion}/2. Over long enough periods of time --- long
enough for equilibration to occur --- the system gets mixed up, and is about equally
likely to be in any of its possible states, regardless of what state it was initially in.
We define some number that describes an interesting property of the whole system,
say the number of atoms in the right half of the box, $R$.
A high-entropy value of $R$ is one like $R=3$, which allows many possible states.
We are far more likely to encounter $R=3$ than 
a low-entropy value like $R=0$ or $R=6$. 
<%
  fig(
    'space-junk',
    %q{%
      Earth orbit is becoming cluttered with space junk, and the pieces can be thought of as the ``molecules'' comprising an exotic kind
      of gas. These image shows the evolution of a cloud of debris arising from a 2007 Chinese test of an anti-satellite
      rocket. Panels 1-4 show the cloud five minutes, one hour, one day, and one month after the impact. The entropy
      seems to have maximized by panel 4.
    },
    {
      'width'=>'wide',
      'sidecaption'=>true,
      'floatpos'=>'t'
    }
  )
%>

 % 
<% end_sec() %>
<% begin_sec("Phase space") %>
There is a problem with making this description of entropy into a mathematical
definition. The problem is that it refers to the number of possible states, but that number
is theoretically infinite. To get around the problem, we coarsen our description of the
system. For the atoms in figure \figref{freeexpansion}, we don't really care exactly where
each atom is. We only care whether it is in the right side or the left side. If a particular atom's
left-right position is described by a coordinate $x$, then the set of all possible values
of $x$ is a line segment along the $x$ axis, containing an infinite number of points.
We break this line segment down into two halves, each of width $\Delta x$, and we
consider two different values of $x$ to be variations on the same state if they both
lie in the same half. For our present purposes, we can also ignore completely the
$y$ and $z$ coordinates, and all three momentum components, $p_x$, $p_y$, and
$p_z$.

<% marg(60) %>
<%
  fig(
    'twobytwo',
    %q{The phase space for two atoms in a box.}
  )
%>
\spacebetweenfigs
<%
  fig(
    'threedimphasespace',
    %q{The phase space for three atoms in a box.}
  )
%>
<% end_marg %>
Now let's do a real calculation. Suppose there are only two atoms in the box,
with coordinates $x_1$ and $x_2$. We can give all the relevant information
about the state of the system by specifying one of the cells in the grid
shown in figure \figref{twobytwo}. This grid is known as the 
 \emph{phase space} of the system. The lower right cell, for instance,
describes a state in which atom number 1 is in the right side of the box
and atom number 2 in the left. Since there are two possible states with
$R=1$ and only one state with $R=2$, we are twice as likely to observe
$R=1$, and $R=1$ has higher entropy than $R=2$.

Figure \figref{threedimphasespace} shows a corresponding calculation
for three atoms, which makes the phase space three-dimensional. Here,
the $R=1$ and 2 states 
are three times more likely than $R=0$ and 3.
Four atoms would require a four-dimensional phase space, which
exceeds our ability to visualize.
Although our present example doesn't require it, a phase space
can describe momentum as well as position, as shown in figure
\figref{phasespace}.
 In general, a phase space for a monoatomic gas
has six dimensions per atom (one for each coordinate
and one for each momentum component).

 % 
<% end_sec() %>
<% begin_sec("Microscopic definitions of entropy and temperature") %>
Two more issues need to be resolved in order to make a microscopic
definition of entropy.

First, if we defined entropy as the number of
possible states, it would be a multiplicative quantity, not an additive one:
if an ice cube in a glass of water has $M_1$ states available to it,
and the number of states available to the water is $M_2$, then the
number of possible states of the whole system is the product 
$M_1 M_2$. To get around this problem, we take the natural logarithm
of the number of states, which makes the entropy additive
because of the property of the logarithm $\ln (M_1 M_2) = \ln M_1 + \ln M_2$.

<% marg(90) %>
<%
  fig(
    'phasespace',
    %q{%
      A phase space for a single atom in one dimension,
      taking momentum into account.
    }
  )
%>
\spacebetweenfigs
<%
  fig(
    'boltzmann-tomb',
    %q{Ludwig Boltzmann's tomb, inscribed with his equation for entropy.}
  )
%>
<% end_marg %>

The second issue is a more trivial one. The concept of entropy was
originally invented as a purely macroscopic quantity, and the
macroscopic definition $\Delta S = Q/T$, which has units of
J/K, has a different calibration than would result from defining
$S=\ln M$. The calibration constant we need turns out to be
simply the Boltzmann constant, $k$.

\mythmhdr{Microscopic definition of entropy}
The entropy of a system is $S = k \ln M$, where $M$ is the number
of available states.\footnote{This is the same relation as the one on
Boltzmann's tomb, just in a slightly different notation.}\index{entropy!microscopic definition}

This also leads to a more fundamental definition of
temperature. Two systems are in thermal equilibrium when
they have maximized their combined entropy through the
exchange of energy. Here the energy possessed by one part of the
system, $E_1$ or $E_2$, plays the same role as the variable
$R$ in the examples of free expansion above.
A maximum of a function occurs when the
derivative is zero, so the maximum entropy occurs when
\begin{equation*}
         \frac{\der\left(S_1+S_2\right)}{\der E_1} = 0 \qquad .
\end{equation*}
We assume the systems are only able to exchange heat
energy with each other, $\der E_1=-\der E_2$, so
\begin{equation*}
        \frac{\der S_1}{\der E_1} = \frac{\der S_2}{\der E_2} \qquad ,
\end{equation*}
and since the energy is being exchanged in the form of heat
we can make the equations look more familiar if we write $\der Q$
for an amount of heat to be transferred into either
system:
\begin{equation*}
        \frac{\der S_1}{\der Q_1} = \frac{\der S_2}{\der Q_2} \qquad .
\end{equation*}
In terms of our previous definition of entropy, this is
equivalent to $1/T_1=1/T_2$, which makes perfect sense since the
systems are in thermal equilibrium. According to our new
approach, entropy has already been defined in a fundamental
manner, so we can take this as a definition of temperature:
\begin{equation*}
   \frac{1}{T} = \frac{\der S}{\der Q} \qquad ,
\end{equation*}
where $\der S$ represents the increase in the system's entropy
from adding heat $\der Q$ to it.\index{temperature!microscopic definition}

<% begin_sec("Examples with small numbers of atoms") %>
Let's see how this applies to an ideal, monoatomic gas with a small
number of atoms. To start with, consider the phase space available
to one atom.
Since we assume the atoms in an ideal gas are noninteracting, 
their positions relative to each other are really
irrelevant. We can therefore
enumerate the number of states available to each atom just
by considering the number of momentum vectors it can have,
without considering its possible locations. The relationship
between momentum and kinetic energy is $E=(p_x^2+p_y^2+p_z^2)/2m$,
so if for a fixed value of its energy, we arrange all of an
atom's possible momentum vectors with their tails at the
origin, their tips all lie on the surface of a sphere in phase space with
radius $|\vc{p}|=\sqrt{2mE}$. The number of possible
states for that atom is proportional to the sphere's surface area,
which in turn is proportional to the square of the sphere's radius,
$|\vc{p}|^2=2mE$.  

Now consider two atoms. For any given way of sharing the energy between the
atoms, $E=E_1+E_2$, the number of possible combinations of
states is proportional to $E_1E_2$. The result is shown in figure
\figref{entropygrapha}. The greatest number of combinations occurs when
we divide the energy equally, so an equal division gives maximum entropy.

<% marg(123) %>
<%
  fig(
    'entropygrapha',
    %q{%
      A two-atom system has the highest number of available
      states when the energy is equally divided. Equal energy
      division is therefore the most likely possibility at any
      given moment in time.
    }
  )
%>
\spacebetweenfigs
<%
  fig(
    'entropygraphb',
    %q{%
      When two systems of 10 atoms each interact, the graph of
      the number of possible states is narrower than with only one
      atom in each system.
    }
  )
%>
<% end_marg %>
By increasing the number of atoms, we get a graph whose peak
is narrower, \figref{entropygraphb}. With more than one atom in each system,
the total energy is
$E=(p_{x,1}^2+p_{y,1}^2+p_{z,1}^2+p_{x,2}^2+p_{y,2}^2+p_{z,2}^2+...)/2m$. With $n$ atoms, a
total of $3n$ momentum coordinates are needed in order to
specify their state, and such a set of numbers is like a
single point in a $3n$-dimensional space (which is impossible
to visualize). For a given total energy $E$, the possible
states are like the surface of a $3n$-dimensional sphere, with
a surface area proportional to $p^{3n-1}$, or $E^{(3n-1)/2}$. The graph in
figure \figref{entropygraphb}, for example, was calculated according to the
formula $E_1^{29/2}E_2^{29/2}=E_1^{29/2}(E-E_1)^{29/2}$.  

Since graph \figref{entropygraphb} is narrower than graph
\figref{entropygrapha}, the fluctuations in energy sharing are
smaller. If we inspect the system at a random moment in time,
the energy sharing is very unlikely to be more lopsided than
a 40-60 split. Now suppose that, instead of 10 atoms interacting
with 10 atoms, we had a $10^{23}$ atoms interacting with
$10^{23}$ atoms. The graph would be extremely narrow, and
it would be a statistical certainty that the energy sharing would
be nearly perfectly equal. This is why we never observe a cold
glass of water to change itself into an ice cube sitting in some
warm water!

By the way, note that although we've redefined
temperature, these examples show that things are coming out consistent
with the old definition, since we saw that the old definition of
temperature could be described in terms of the average energy per
atom, and here we're finding that equilibration results in each subset of the atoms
having an equal share of the energy.

<% end_sec() %>
<% begin_sec("Entropy of a monoatomic ideal gas") %>
Let's calculate the entropy of a monoatomic ideal gas of $n$ atoms. 
This is an important example because it allows us to show that
our present microscopic treatment of thermodynamics is
consistent with our previous macroscopic approach, in which
temperature was defined in terms of an ideal gas thermometer.

The number of possible locations for each atom is $V/\Delta x^3$,
where $\Delta x$  is the size of the space cells in phase space. The
number of possible combinations of  locations for the atoms
is therefore $(V/\Delta x^3)^n$.

The possible momenta cover the
surface of a $3n$-dimensional sphere, whose radius is $\sqrt{2mE}$, and whose surface
area is therefore proportional to $E^{(3n-1)/2}$. In terms of
phase-space cells, this area corresponds to
$E^{(3n-1)/2} / \Delta p^{3n}$ possible combinations of momenta,
multiplied by some constant of proportionality which depends
on $m$, the atomic mass, and $n$, the number of atoms.
To avoid having to calculate this constant of
proportionality, we limit ourselves to calculating the part
of the entropy that does not depend on $n$, so the resulting
formula will not be useful for comparing entropies of ideal
gas samples with different numbers of atoms.

The final result for the number of available states
is
\begin{equation*}
        M  =  \left(\frac{V}{\Delta x^3}\right)^n\:\frac{E^{(3n-1)/2}}{\Delta p^{3n-1}}
                        \qquad , \qquad \text{[function of $n$]}
\end{equation*}
so the entropy is
\begin{equation*}
        S  =  nk \ln V + \frac{3}{2}nk\ln E 
                + \text{(function of $\Delta x$, $\Delta p$, and $n$)} \qquad ,
\end{equation*}
where the distinction between $n$ and $n-1$ has been ignored.
Using $PV=nkT$ and $E=(3/2)nkT$, we can also rewrite this as
\begin{equation*}
        S  =  \frac{5}{2} nk \ln T - nk \ln P + \ldots   ,        \qquad \text{[entropy of a monoatomic ideal gas]}
\end{equation*}
where ``$\ldots$'' indicates terms that may depend on $\Delta x$, $\Delta p$, $m$, and
$n$, but that have no effect on comparisons of gas samples
with the same number of atoms.

<% self_check('idealgasentropy',<<-'SELF_CHECK'
Why does it make sense that the temperature term has a
positive sign in the above example, while the pressure term
is negative? Why does it make sense that the whole thing is
proportional to $n$?
  SELF_CHECK
  ) %>

To show consistency with the macroscopic approach to thermodynamics,
we need to show that these results are consistent with the behavior of
an ideal-gas thermometer. Using the new definition $1/T=\der S/\der Q$,
we have $1/T=\der S/\der E$, since transferring an amount of heat $\der Q$
into the gas increases its energy by a corresponding amount. Evaluating
the derivative, we find $1/T=(3/2)nk/E$, or $E=(3/2)nkT$, which is the
correct relation for a monoatomic ideal gas.

\begin{eg}{A mixture of molecules}\label{eg:mix-two-gases}
\egquestion Suppose we have a mixture of two different monoatomic gases, say helium and argon.
How would we find the entropy of such a mixture (say, in terms of $V$ and $E$)?
How would the energy be shared between the two types of molecules, i.e., would a more massive argon atom have
more energy on the average than a less massive helium atom, the same, or less?

\eganswer
Since entropy
is additive, we simply need to add the entropies of the two types of atom. However,
the expression derived above for the entropy omitted the dependence on the mass $m$ of the atom,
which is different for the two constituents of the gas, so we need to go back and figure out how
to put that $m$-dependence back in. The only place where we threw away $m$'s was when we
identified the radius of the sphere in momentum space with $\sqrt{2mE}$, but then threw away
the constant factor of $m$. In other words, the final result can be generalized merely by
replacing $E$ everywhere with the product $mE$. Since the log of a product is the sum of the logs,
the dependence of the final result on $m$ and $E$ can be broken apart into two different terms,
and we find 
\begin{equation*}
  S=nk \ln V +\frac{3}{2}nk\ln m+\frac{3}{2}nk\ln E+\ldots
\end{equation*}
The total entropy of the mixture
can then be written as 
\begin{multline*}
  S =n_1k\ln V +n_2k \ln V +\frac{3}{2}n_1k\ln m_1+\frac{3}{2}n_2k\ln m_2 \\
    +\frac{3}{2}n_1k\ln E_1+\frac{3}{2}n_2k\ln E_2+\ldots
\end{multline*}

Now what about the energy sharing? If the total energy is $E=E_1+E_2$, then the most ovewhelmingly probable
sharing of energy will the the one that maximizes the entropy. Notice that the dependence of the entropy on
the masses $m_1$ and $m_2$ occurs in terms that are entirely separate from the energy terms. If we want to maximize
$S$ with respect to $E_1$ (with $E_2=E-E_1$ by conservation of energy), then we differentiate $S$ with respect to
$E_1$ and set it equal to zero. The terms that contain the masses don't have any dependence on $E_1$, so their
derivatives are zero, and we find that the molecular masses can have no effect on the energy sharing. Setting the
derivative equal to zero, we have
\begin{align*}
  0 &= \frac{\partial}{\partial E_1} \left(n_1k\ln V +n_2k \ln V +\frac{3}{2}n_1k\ln m_1+\frac{3}{2}n_2k\ln m_2\right. \\
    &  \qquad \qquad  +\left.\frac{3}{2}n_1k\ln E_1+\frac{3}{2}n_2k\ln (E-E_1)+\ldots\right) \\
    &= \frac{3}{2}k \left( \frac{n_1}{E_1} -  \frac{n_2}{E-E_1}  \right) \\
  0 &= \frac{n_1}{E_1} -  \frac{n_2}{E-E_1} \\
   \frac{n_1}{E_1} &=  \frac{n_2}{E_2} \qquad .
\end{align*}
In other words, each gas gets a share of the energy in proportion to the number of its atoms,
and therefore every atom gets, on average, the same amount of energy, regardless of its mass.
The result for the average energy per atom is exactly the same as for an unmixed gas, $\bar{K}=(3/2)kT$.
\end{eg}

<% end_sec() %>
<% begin_sec("Equipartition") %>\label{equipartition}
Example \ref{eg:mix-two-gases} is a special case of a more general statement called the equipartition theorem.\index{equipartition theorem}
Suppose we have only one argon atom, named Alice, and one helium atom, named Harry.
Their total kinetic energy is $E=p_x^2/2m+p_y^2/2m+p_z^2/2m+{p'}_x^2/2m'+{p'}_y^2/2m'+{p'}_z^2/2m'$, where the primes
indicate Harry. We have six terms that all look alike. The only difference among them is that the constant factors
attached to the squares of the momenta have different values, but we've just proved that those differences don't matter.
In other words, if we have any system at all whose energy is of the form $E=(\ldots)p_1^2+(\ldots)p_2^2+\ldots$, with any
number of terms, then each term holds, on average, the same amount of energy, $\frac{1}{2}kT$.
We say that the system consisting of Alice and Harry has six degrees of freedom.\index{degree of freedom}
It doesn't even matter whether the things being
squared are momenta: if you look back over the logical steps that went into the argument, you'll see that none of them
depended on that. In a solid, for example, the atoms aren't free to wander around, but they can vibrate from side to
side. If an atom moves away from its equilibrium position at $x=0$ to some other value of $x$, then its electrical
energy is $(1/2)\kappa x^2$, where $\kappa$ is the spring constant (written as the Greek letter kappa to distinguish it
from the Boltzmann constant $k$). We can conclude that each atom in the solid, on average, has $\frac{1}{2}kT$ of energy
in the electrical energy due to its $x$ displacement along the $x$ axis, and equal amounts for $y$ and $z$.
This is known as equipartition, meaning equal partitioning,
or equal sharing. The equipartition theorem says that if the expression for the energy looks like a sum of squared
variables, then each degree of freedom has an average energy of $\frac{1}{2}kT$. Thus, very generally, we can
interpret temperature as the average energy per degree of freedom (times $k/2$).
<% end_sec() %>

<% begin_sec("An unexpected glimpse of the microcosm") %>
You may have the feeling at this point that of course Boltzmann was right about the literal
existence of atoms, but only very sophisticated experiments could vindicate him definitively.
After all, the microscopic and macroscopic definitions of entropy are equivalent, so it might
seem as though there was no real advantage to the microscopic approach. Surprisingly, very
simple experiments are capable of revealing a picture of the microscopic world, and there is
no possible macroscopic explanation for their results.

<% marg(30) %>
<%
  fig(
    'clement-desormes',
    %q{%
      An experiment for determining the shapes of molecules.
    }
  )
%>
<% end_marg %>
In 1819, before Boltzmann was born, Cl\'{e}ment and Desormes did an experiment like the one shown in
figure \figref{clement-desormes}. The gas in the flask is pressurized using the syringe. This heats
it slightly, so it is then allowed to cool back down to room temperature. Its pressure is measured
using the manometer. The stopper on the flask is popped and then immediately reinserted. Its pressure
is now equalized with that in the room, and the gas's expansion has cooled it a little, because it
did mechanical work on its way out of the flask, causing it to lose some of its internal energy $E$.
The expansion is carried out quickly enough so that there is not enough time for any significant amount of
heat to flow in through the walls of the flask before the stopper is reinserted.
The gas is now allowed to come back up to room temperature (which takes a much longer time),
and as a result regains a fraction $b$ of its original overpressure.
 During this constant-volume reheating,
we have $PV=nkT$, so the amount of pressure regained is a direct indication of how much the gas cooled down
when it lost an amount of energy $\Delta E$. 
<%
  fig(
    'molecular-shapes',
    %q{% 
         The differing shapes of a helium atom (1), a nitrogen molecule (2), and a difluoroethane molecule (3) have surprising
         macroscopic effects.
    },
    {
      'width'=>'wide',
      'sidecaption'=>true
    }
  )
%>

If the gas is monoatomic, then we know what to expect for this relationship between energy and temperature: $\Delta E=(3/2)nk\Delta T$,
where the factor of 3 came ultimately from the fact that the gas was in a
three-dimensional space, \subfigref{molecular-shapes}{1}. Moving in this space, each molecule can have momentum in the x, y, and
z directions. It has three degrees of freedom.
What if the gas is not monoatomic? Air, for example, is made of diatomic molecules,
\subfigref{molecular-shapes}{2}. There is a subtle difference between the two cases. An individual atom of
a monoatomic gas is a perfect sphere, so it is exactly the same no matter how it is oriented. Because of this perfect symmetry, there is
thus no way to tell whether it is spinning or not, and in fact we find that it can't rotate. The diatomic gas, on the other hand,
can rotate end over end about the x or y axis, but cannot rotate about the z axis, which is its axis of symmetry.
It has a total of five degrees of freedom. A polyatomic molecule with a more complicated, asymmetric shape,
\subfigref{molecular-shapes}{3}, can rotate about all three axis, so it has a total of six degrees of freedom.

Because a polyatomic molecule has more degrees of freedom than a monoatomic one, it has more possible states for
a given amount of energy. That is, its entropy is higher for the same energy. From the definition of temperature,
$1/T=\der S/\der E$, we conclude that it has a lower temperature for the same energy. In other words, it is
more difficult to heat $n$ molecules of difluoroethane than it is to heat $n$ atoms of helium. When the
Cl\'{e}ment-Desormes experiment is carried out, the result $b$ therefore depends on the shape of the molecule!
Who would have dreamed that such simple observations, correctly interpreted, could give us this kind of
glimpse of the microcosm?

Lets go ahead and calculate how this works.\label{pt-insulated-expansion}
Suppose a gas is allowed
to expand without being able to exchange heat with the rest
of the universe. 
The loss of thermal energy from the gas equals the work it
does as it expands, and
using the result of homework problem \ref{hw:work-pdv} on
page \pageref{hw:work-pdv}, the work done
in an infinitesimal expansion equals $P\der V$, so
\begin{equation*}
        \der E + P \der V = 0 \qquad .
\end{equation*}
(If the gas had not been insulated, then there would have
been a third term for the heat gained or lost by heat
conduction.) 

From section \ref{sec:microscopicidealgas} we have $E=(3/2)PV$ for a
monoatomic ideal gas. More generally, the equipartition theorem tells us that the 3 simply needs to be replaced with the
number of degrees of freedom $\alpha$, so $\der E=(\alpha/2)P\der V+(\alpha/2)V\der P$, and the
equation above becomes
\begin{equation*}
        0       =  \frac{\alpha+2}{2}P\der V+\frac{\alpha}{2}V\der P \qquad .
\end{equation*}
Rearranging, we have
\begin{equation*}
        (\alpha+2)\frac{\der V}{V}      =  -\alpha\frac{\der P}{P} \qquad .
\end{equation*}
Integrating both sides gives
\begin{equation*}
        (\alpha+2) \ln V        =  -\alpha \ln P + \text{constant}   ,
\end{equation*}
and taking exponentials on both sides yields
\begin{equation*}
        V^{\alpha+2} \propto P^{-\alpha} \qquad .
\end{equation*}

We now wish to reexpress this in terms of pressure and
temperature. Eliminating $V\propto(T/P)$ gives
\begin{equation*}
        T \propto   P^b   ,
\end{equation*}
where $b=2/(\alpha+2)$ is equal to 2/5, 2/7, or 1/4, respectively, for
a monoatomic, diatomic, or polyatomic gas.

\begin{eg}{Efficiency of the Carnot engine}\label{eg:carnot-efficiency}
As an application, we now prove the result claimed earlier for the efficiency
of a Carnot engine. First consider the work done during the constant-temperature
strokes. Integrating the equation $\der W=P\der V$,
 we have $W = \int P \der V$. Since the thermal energy of
an ideal gas depends only on its temperature, there is no
change in the thermal energy of the gas during this
constant-temperature process. Conservation of energy
therefore tells us that work done by the gas must be exactly
balanced by the amount of heat transferred in from the
reservoir.
\begin{align*}
        Q       &=  W \\
                &=   \int P \der V
\end{align*}
For our proof of the efficiency of the Carnot engine, we
need only the ratio of $Q_H$ to $Q_L$, so we neglect constants of
proportionality, and simply subsitutde $P\propto T/V$, giving
\begin{equation*}
        Q \propto \int \frac{T}{V} \der V 
         \propto T  \ln \frac{V_2}{V_1} 
         \propto T \ln \frac{P_1}{P_2} \qquad .
\end{equation*}

The efficiency of a heat engine is
\begin{equation*}
        \text{efficiency}       =  1 - \frac{Q_L}{Q_H} \qquad .
\end{equation*}
Making use of the result from the previous proof for a
Carnot engine with a monoatomic ideal
gas as its working gas, we have
\begin{equation*}
        \text{efficiency}       =   1-\frac{T_L\:\ln(P_4/P_3)}{T_H\:\ln(P_1/P_2)}  ,
\end{equation*}
where the subscripts 1, 2, 3, and 4 refer to figures \ref{fig:carnota}--\ref{fig:carnotd} on
page \pageref{fig:carnota}. We have shown above
that the temperature is proportional to $P^b$ on the insulated
strokes 2-3 and 4-1, the pressures must be related by
$P_2/P_3=P_1/P_4$, which can be rearranged as $P_4/P_3=P_1/P_2$, and we
therefore have
\begin{equation*}
        \text{efficiency}       =  1 - \frac{T_L}{T_H} \qquad .
\end{equation*}
\end{eg}
<% end_sec() %>

<% end_sec() %>
<% begin_sec("The arrow of time, or ``This way to the Big Bang''") %>\index{Big Bang!and the arrow of time}
Now that we have a microscopic understanding of entropy, what does that tell us about
the second law of thermodynamics? The second law
defines a forward direction to time, ``time's arrow.'' The microscopic
treatment of entropy, however, seems to have mysteriously sidestepped
that whole issue. A graph like figure \figref{fluctuation} on page
\pageref{fig:fluctuation}, showing a fluctuation
away from equilibrium, would look just as natural if we flipped it over to
reverse the direction of time. After all, the basic laws of physics are
conservation laws, which don't distinguish between past and future.
Our present picture of entropy suggests that we restate the second
law of thermodynamics as follows: low-entropy states are short-lived.
An ice cube can't exist forever in warm water. We no longer have to
distinguish past from future.\index{thermodynamics!second law of}\index{time!arrow of}

But how do we reconcile this with our strong psychological sense of
the direction of time, including our ability to remember the past but
not the future? Why do we observe ice cubes melting in water, but
not the time-reversed version of the same process?

The answer is that there is no past-future asymmetry in the laws
of physics, but there is a past-future asymmetry in the universe.
The universe started out with the Big Bang.  (Some of the evidence for
the Big Bang theory is given on page \pageref{bigbang}.) The early
universe had a very low entropy, and low-entropy states are short-lived.
What does ``short-lived'' mean here, however? Hot coffee left in a paper
cup will equilibrate with the air within ten minutes or so. Hot coffee in a
thermos bottle maintains its low-entropy state for much longer,
because the coffee is insulated by a vacuum between the inner and outer
walls of the thermos. The universe has been mostly vacuum for a long time,
so it's well insulated. Also, it takes billions of years for a low-entropy
normal star like our sun to evolve into the high-entropy cinder known
as a white dwarf.

The universe, then, is still in the process of equilibrating,
and all the ways we have of telling the past from the future are really just
ways of determining which direction in time points toward the Big Bang,
i.e., which direction points to lower entropy. The psychological arrow of
time, for instance, is ultimately based on the thermodynamic arrow. In some
general sense, your brain is like a computer, and computation has
thermodynamic effects. In even the most efficient possible computer, for example,
erasing one bit of memory
decreases its entropy from $k \ln 2$ (two possible states) to $k \ln 1$
(one state), for a drop of about $10^{-23}$ J/K. One way of determining the
direction of the psychological arrow of time is that forward in psychological
time is the direction in which, billions of years from now, all consciousness
will have ceased; if consciousness was to exist forever in the universe, then
there would have to be a never-ending
decrease in the universe's entropy. This can't happen, because low-entropy states
are short-lived.

Relating the direction of the thermodynamic arrow of time to the existence of
the Big Bang is a satisfying way to avoid the paradox of how the second law
can come from basic laws of physics that don't distinguish past from future.
There is a remaining mystery, however: why did our universe have a Big Bang
that was low in entropy? It could just as easily have been a maximum-entropy
state, and in fact the number of possible high-entropy Big Bangs is vastly greater
than the number of possible low-entropy ones. The question, however, is probably
not one that can be answered using the methods of science. All we can say is that
if the universe had started with a maximum-entropy Big Bang, then we wouldn't be
here to wonder about it. A longer, less mathematical discussion of these concepts,
along with some speculative ideas, is given in ``The Cosmic Origins of Time's
Arrow,'' Sean M. Carroll, Scientific American, June 2008, p. 48.

 % 
<% end_sec() %>
<% begin_sec("Quantum mechanics and zero entropy") %>
The previous discussion would seem to imply that absolute
entropies are never well defined, since any calculation of
entropy will always end up having terms that depend on $\Delta p$
and $\Delta x$. For instance, we might think that cooling an ideal
gas to absolute zero would give zero entropy, since there is
then only one available momentum state, but there would
still be many possible position states. We'll see later in this book,
however, that the quantum mechanical
uncertainty principle makes it impossible to know the
location and position of a particle simultaneously with perfect accuracy.
The best we can do is to determine them with an accuracy such that
the product $\Delta p\Delta x$ is equal to a constant called
Planck's constant. According to quantum physics, then, there
is a natural minimum size for rectangles in phase space, and
entropy can be defined in absolute terms. Another way of
looking at it is that according to quantum physics, the gas
as a whole has some well-defined ground state, which is its
state of minimum energy. When the gas is cooled to absolute
zero, the scene is not at all like what we would picture in
classical physics, with a lot of atoms lying around
motionless. It might, for instance, be a strange
quantum-mechanical state called the Bose-Einstein
condensate, which was achieved for the first time recently
with macroscopic amounts of atoms. Classically, the gas has many
possible states available to it at zero temperature, since the
positions of the atoms can be chosen in a variety of ways.
The classical picture is a bad approximation under these
circumstances, however.
Quantum mechanically there is only one ground state,
in which each atom is spread out over the available
volume in a cloud of probability. The entropy is therefore
zero at zero temperature. This fact, which cannot be understood
in terms of classical physics, is known as the third law of
thermodynamics.\index{thermodynamics!third law of}\label{third-law-of-thermodynamics}

<% end_sec() %>
<% begin_sec("Summary of the laws of thermodynamics") %>

Here is a summary of the laws of thermodynamics:
\index{thermodynamics!laws of!summarized}
\index{thermodynamics!zeroth law of}
\index{thermodynamics!first law of}
\index{thermodynamics!second law of}
\index{thermodynamics!third law of}

\begin{description}
\item[The zeroth law of thermodynamics (page \pageref{zeroth-law-of-thermodynamics})] If
object A is at the same temperature as object B, and B is at
the same temperature as C, then A is at the same temperature
as C.
\item[The first law of thermodynamics (page \pageref{first-law-of-thermodynamics})] Energy is \linebreak[4] con\-served.
\item[The second law of thermodynamics (page \pageref{second-law-of-thermodynamics})] The entropy of a closed system always increases, or at best
stays the same: $\Delta S\ge0$.
\item[The third law of thermodynamics (page \pageref{third-law-of-thermodynamics})] The entropy of a system approaches zero as its temperature approaches
absolute zero.
\end{description}

\noindent From a modern point of view, only the first law deserves to be called a fundamental law of physics.
Once Boltmann discovered the microscopic nature of entropy,
the zeroth and second laws could be understood as statements about probability: a system containing a large
number of particles is overwhelmingly likely to do a certain thing, simply because the number of possible ways
to do it is extremely large compared to the other possibilities. The third law is also now understood to be
a consequence of more basic physical principles, but to explain the third law, it's not sufficient simply to know
that matter is made of atoms: we also need to understand the quantum-mechanical nature of those atoms,
discussed in chapter \ref{ch:quantum}.
Historically, however, the laws of thermodynamics were discovered in the eighteenth century, when the
atomic theory of matter was generally considered to be a hypothesis that couldn't be tested experimentally.
Ideally, with the publication of Boltzmann's work on entropy in 1877, the zeroth and second laws would
have been immediately demoted from the status of physical laws, and likewise the development of quantum
mechanics in the 1920's would have done the same for the third law.

<%
  fig(
    'otto-cycle',
    %q{%
      The Otto cycle. 1. In the exhaust stroke, the piston expels the burned air-gas
      mixture left over from the preceding cycle. 
      2. In the intake stroke, the piston sucks in fresh air-gas mixture. 3. In the compression stroke, the
      piston compresses the mixture, and heats it. 4. At the beginning of the power stroke, the spark plug fires,
      causing the air-gas mixture to burn explosively and heat up much more. The heated mixture expands, and does
      a large amount of positive mechanical work on the piston. An animated version can be viewed in the
      Wikipedia article ``Four-stroke engine.''
    },
    {
      'width'=>'wide',
      'sidecaption'=>true
    }
  )
%>

<% end_sec() %>
<% end_sec() %>
<% begin_sec("More about Heat Engines",0) %>

So far, the only heat engine we've discussed in any detail has been a fictitious Carnot engine,
with a monoatomic ideal gas as its working gas. As a more realistic example,
figure \figref{otto-cycle} shows one full cycle of a cylinder in a standard gas-burning automobile engine.
This four-stroke cycle is called the Otto cycle,\index{Otto cycle}\index{engine!Otto cycle}\index{engine!automobile}
after its inventor, German engineer Nikolaus Otto.\index{Otto, Nikolaus}
The Otto cycle is more complicated than a Carnot cycle, in a number of ways:

\begin{itemize}
\item The working gas is physically pumped in and out of the cylinder through valves, rather than being
sealed and reused indefinitely as in the Carnot engine.
\item The cylinders are not perfectly insulated from the engine block, so heat energy is lost from
each cylinder by conduction. This makes the engine less efficient that a Carnot engine, because heat
is being discharged at a temperature that is not as cool as the environment.
\item Rather than being heated by contact with an external heat reservoir, the air-gas mixture inside
each cylinder is heated by internal combusion: a spark from a spark plug burns the gasoline, releasing heat.
\item The working gas is not monoatomic. Air consists of diatomic molecules ($\zu{N}_2$ and $\zu{O}_2$),
and gasoline of polyatomic molecules such as octane ($\zu{C}_8\zu{H}_{18}$).
\item The working gas is not ideal. An ideal gas is one in which the molecules never interact with one
another, but only with the walls of the vessel, when they collide with it. In a car engine, the molecules
are interacting very dramatically with one another when the air-gas mixture explodes (and less dramatically
at other times as well, since, for example, the gasoline may be in the form of microscopic droplets rather
than individual molecules).
\end{itemize}

This is all extremely complicated, and it would be nice to have some way of understanding and visualizing
the important properties of such a heat engine without trying to handle every detail at once. A good method
of doing this is a type of graph known as a P-V diagram. As proved in homework problem \ref{hw:work-pdv}, the equation $\der W=F\der x$ for mechanical work
can be rewritten as $\der W=P\der V$ in the case of work done by a piston. Here $P$ represents the pressure of the working gas,
and $V$ its volume. Thus, on a graph of $P$ versus $V$, the area under the curve represents the work done. When the gas
expands, $\der x$ is positive, and the gas does positive work. When the gas is being compressed, $\der x$ is negative, and
the gas does negative work, i.e., it absorbs energy. Notice how, in the diagram of the Carnot engine in the top panel of
figure \figref{pv-for-carnot-and-otto}, the cycle goes clockwise around the curve, and therefore
the part of the curve in which negative work is being done (arrowheads pointing to the left) are below the ones in which
positive work is being done. This means that over all, the engine does a positive amount of work. This net work equals the area under
the top part of the curve, minus the area under the bottom part of the curve, which is simply the area enclosed by the curve.
Although the diagram for the Otto engine is more complicated, we can at least compare it on the same footing with the Carnot
engine. The curve forms a figure-eight, because it cuts across itself. The top loop goes clockwise, so as in the case of the Carnot
engine, it represents positive work. The bottom loop goes counterclockwise, so it represents a net negative contribution to the work.
This is because more work is expended in forcing out the exhaust than is generated in the intake stroke.
<% marg(0) %>
<%
  fig(
    'pv-for-carnot-and-otto',
    %q{P-V diagrams for a Carnot engine and an Otto engine.}
  )
%>
<% end_marg %>

To make an engine as efficient as possible, we would like to make the loop have as much area as possible. What is it that
determines the actual shape of the curve? First let's consider the constant-temperature expansion stroke that forms the
top of the Carnot engine's P-V plot. This is analogous to the power stroke of an Otto engine. Heat is being sucked in
from the hot reservoir, and since the working gas is always in thermal equilibrium with the hot reservoir, its temperature
is constant. Regardless of the type of gas, we therefore have $PV=nkT$ with $T$ held constant, and thus $P\propto V^{-1}$
is the mathematical shape of this curve --- a $y=1/x$ graph, which is a hyperbola. This is all true regardless of whether
the working gas is monoatomic, diatomic, or polyatomic.
(The bottom of the loop is likewise of the form $P\propto V^{-1}$, but with a smaller constant of proportionality due
to the lower temperature.)

Now consider the insulated expansion stroke that forms the right side of the curve for the Carnot engine.
As shown on page \pageref{pt-insulated-expansion}, the relationship between pressure and temperature in an
insulated compression or expansion is  $T \propto   P^b$,
with $b=2/5$, 2/7, or 1/4, respectively, for
a monoatomic, diatomic, or polyatomic gas. For $P$ as a function of $V$ at constant $T$, the ideal gas law gives
$P\propto T/V$, so $P\propto V^{-\gamma}$, where $\gamma=1/(1-b)$ takes on the values
5/3, 7/5, and 4/3. The number $\gamma$ can be interpreted as the ratio $C_P/C_V$,
where $C_P$, the heat capacity at constant pressure, is the amount of heat required to raise the temperature
of the gas by one degree while keeping its pressure constant, and $C_V$ is the corresponding quantity under
conditions of constant volume.\index{heat capacity!at constant pressure}\index{heat capacity!at constant volume}

\begin{eg}{The compression ratio}\label{eg:compression-ratio}
Operating along a constant-temperature stroke, the amount of mechanical work done by a heat engine can be calculated
as follows:
\begin{align*}
  PV &= nkT \\
\intertext{Setting $c=nkT$ to simplify the writing,}
  P &= cV^{-1} \\
  W &= \int_{V_i}^{V_f} P \der V \\
    &= c \int_{V_i}^{V_f} V^{-1} \der V \\
    &= c \ln V_f - c \ln V_i \\
    &= c \ln (V_f/V_i)
\end{align*}
The ratio $V_f/V_i$ is called the \emph{compression ratio} of the engine, and higher values result in more
power along this stroke. Along an insulated stroke, we have $P\propto V^{-\gamma}$, with $\gamma\ne1$,
so the result for the
work no longer has this perfect mathematical property of depending only on the ratio $V_f/V_i$. Nevertheless,
the compression ratio is still a good figure of merit for predicting the performance of any heat engine,
including an internal combustion engine. High compression ratios tend to make the working gas of an
internal combustion engine heat up so much that it spontaneously explodes. When this happens in an Otto-cycle
engine, it can cause ignition before the sparkplug fires, an undesirable effect known as pinging. For this
reason, the compression ratio of an Otto-cycle automobile engine cannot normally exceed about 10. In a diesel
engine, however, this effect is used intentionally, as an alternative to sparkplugs, and compression ratios
can be 20 or more.
\end{eg}

\hspace{0mm plus 20mm}

<% marg(0) %>
<%
  fig(
    'pv-for-sound',
    %q{Example \ref{eg:pv-for-sound},}
  )
%>
<% end_marg %>
\begin{eg}{Sound}\label{eg:pv-for-sound}
Figure \figref{pv-for-sound} shows a P-V plot for a sound wave. As the pressure oscillates up and down,
the air is heated and cooled by its compression and expansion. Heat conduction is a relatively slow
process, so typically there is not enough time over each cycle for any significant amount of heat to
flow from the hot areas to the cold areas. (This is analogous to insulated compression or expansion of
a heat engine; in general, a compression or expansion of this type, with no transfer of heat, is called
\emph{adiabatic}.\index{adiabatic}) The pressure and volume of a particular little piece of
the air are therefore related according to $P\propto V^{-\gamma}$. The cycle of oscillation consists
of motion back and forth along a single curve in the P-V plane, and since this curve encloses zero
volume, no mechanical work is being done: the wave (under the assumed ideal conditions) propagates
without any loss of energy due to friction.

The speed of sound is also related to $\gamma$. See example \ref{eg:speed-of-sound}
on p.~\pageref{eg:speed-of-sound}.
\end{eg}

<% marg(0) %>
<%
  fig(
    'gamma-spring-of-air',
    %q{Example \ref{eg:gamma-spring-of-air}.}
  )
%>
<% end_marg %>
\begin{eg}{Measuring $\gamma$ using the ``spring of air''}\label{eg:gamma-spring-of-air}
Figure \figref{gamma-spring-of-air} shows an experiment that can be used to measure the $\gamma$ of a gas.
When the mass $m$ is inserted into bottle's neck, which has cross-sectional area $A$, the mass
drops until it compresses the air enough so that the pressure is enough to support its weight.
The observed frequency $\omega$ of oscillations about this equilibrium position $y_\zu{o}$ can be used to extract the
$\gamma$ of the gas.
\begin{align*}
  \omega^2 &= \frac{k}{m} \\
         &= \left.-\frac{1}{m}\:\frac{\der F}{\der y}\right|_{y_\zu{o}} \\
         &= \left.-\frac{A}{m}\:\frac{\der P}{\der y}\right|_{y_\zu{o}} \\
         &= \left.-\frac{A^2}{m}\:\frac{\der P}{\der V}\right|_{V_\zu{o}} 
\end{align*}
We make the bottle big enough so that its large surface-to-volume ratio prevents the
conduction of any significant amount of heat through its walls during one cycle, so $P\propto V^{-\gamma}$,
and $\der P/\der V=-\gamma P/V$. Thus,
\begin{align*}
  \omega^2 &= \gamma\frac{A^2}{m}\:\frac{P_\zu{o}}{V_\zu{o}}
\end{align*}
\end{eg}

\begin{eg}{The Helmholtz resonator}\label{eg:helmholtz-resonator}\index{Helmholtz resonator}
When you blow over the top of a beer bottle, you produce a pure tone.
As you drink more of the beer, the pitch goes down. This is similar to
example \ref{eg:gamma-spring-of-air}, except that instead of a solid mass
$m$ sitting inside the neck of the bottle, the moving mass is the air itself.
As air rushes in and out of the bottle, its velocity is highest at the bottleneck,
and since kinetic energy is proportional to the square of the velocity, essentially
all of the kinetic energy is that of the air that's in the neck. In other words,
we can replace $m$ with $AL\rho$, where $L$ is the length of the neck, and $\rho$ is
the density of the air. Substituting into the earlier result, we find that the
resonant frequency is
\begin{align*}
  \omega^2 &= \gamma\frac{P_\zu{o}}{\rho}\:\frac{A}{LV_\zu{o}} \qquad .
\end{align*}
This is known as a Helmholtz resonator.
As shown in figure \figref{stradivarius}, a violin or an acoustic guitar has a Helmholtz resonance,
since air can move in and out through the f-holes. Problem \ref{hw:violin-helmholtz} is a more
quantitative exploration of this.
\end{eg}
<%
  fig(
    'stradivarius',
    'The resonance curve of a 1713 Stradivarius violin, measured by
     Carleen Hutchins. There are a number of different resonance peaks, some strong and some weak;
     the ones near 200 and 400 Hz are vibrations of the wood, but the one near 300 Hz is a resonance
     of the air moving in and out through those holes shaped like the letter F.
     The white lines show the frequencies of the four strings.',
    {
      'width'=>'wide',
      'sidecaption'=>true
    }
  )
%>

We have already seen, based on the microscopic nature of entropy, that any Carnot engine has the same
efficiency, and the argument only employed the assumption that the engine met the definition of a Carnot
cycle: two insulated strokes, and two constant-temperature strokes. Since we didn't have to make any
assumptions about the nature of the working gas being used, the result is evidently true for diatomic
or polyatomic molecules, or for a gas that is not ideal. This result is surprisingly simple and general,
and a little mysterious --- it even applies to possibilities that we have not even considered, such as
a Carnot engine designed so that the working ``gas'' actually consists of a mixture of liquid droplets and vapor,
as in a steam engine. How can it always turn out so simple, given the kind of mathematical complications that
were swept under the rug in example \ref{eg:compression-ratio}? A better way to understand this result
is by switching from P-V diagrams to a diagram of temperature versus entropy, as shown in figure 
\figref{ts-for-carnot}. An infinitesimal transfer of heat $\der Q$ gives rise to a change in entropy
$\der S=\der Q/T$, so the area under the curve on a T-S plot gives the amount of heat transferred.
The area under the top edge of the box in figure \figref{ts-for-carnot}, extending all the way down to the
axis, represents the amount of heat absorbed from the hot reservoir, while the smaller area under the
bottom edge represents the heat wasted into the cold reservoir. By conservation of energy, the area enclosed
by the box therefore represents the amount of mechanical work being done, as for a P-V diagram.
We can now see why the efficiency of a Carnot engine is independent of any of the physical details:
the definition of a Carnot engine guarantees that the T-S diagram will be a rectangular box, and the efficiency
depends only on the relative heights of the top and bottom of the box.
<% marg(0) %>
<%
  fig(
    'ts-for-carnot',
    %q{A T-S diagram for a Carnot engine.}
  )
%>
<% end_marg %>

\backofchapterboilerplate{thermo}
 % 
 % ============================= homework ==============================  

<% end_sec() %><% begin_hw_sec %>

<% begin_hw('stpvol',0) %>__incl(hw/stpvol)<% end_hw() %>

<% begin_hw('work-pdv') %>__incl(hw/work-pdv)<% end_hw() %>

<% begin_hw('he',0) %>
(a) A helium atom contains 2 protons, 2 electrons, and 2
neutrons. Find the mass of a helium atom.  \answercheck\hwendpart
(b) Find the
number of atoms in 1.0 kg of helium.  \answercheck\hwendpart
(c) Helium gas is
monoatomic. Find the amount of heat needed to raise the
temperature of 1.0 kg of helium by 1.0 degree C. (This is known
as helium's heat capacity at constant volume.) \answercheck
<% end_hw() %>

<% begin_hw('cracking-gas-molecules') %>__incl(hw/cracking-gas-molecules)<% end_hw() %>

<% begin_hw('igm') %>__incl(hw/igm)<% end_hw() %>

<% begin_hw('centerofearth') %>
Estimate the pressure at the center of the Earth,
assuming it is of constant density throughout. Note that $g$
is not constant with respect to depth --- as shown in
example \ref{eg:gravityinearth} on page \pageref{eg:gravityinearth},
$g$ equals $Gmr/b^3$ for $r$, the distance from the center,
less than $b$, the earth's radius.\hwendpart
(a) State your result in terms of $G$, $m$, and $b$.\answercheck\hwendpart
(b) Show that your answer from part a has the right units for pressure.\hwendpart
(c) Evaluate the result numerically. \answercheck \hwendpart
(d) Given that the earth's atmosphere is on the order of
 one thousandth the earth's radius, and that the density
of the earth is several thousand times greater than the density of the
lower atmosphere, check that your result is of a reasonable order of
magnitude.
<% end_hw() %>

<% begin_hw('fluorocarbon') %>__incl(hw/fluorocarbon)<% end_hw() %>

<% begin_hw('airconditioner') %>
Refrigerators, air conditioners, and heat pumps are heat engines that
work in reverse.
You put in mechanical work, and the effect is to take heat out of
a cooler reservoir and deposit heat in a warmer one: $Q_L+W=Q_H$.
As with the heat engines discussed previously, the efficiency is
defined as the energy transfer you want ($Q_L$ for a refrigerator or
air conditioner, $Q_H$ for a heat pump) divided by the energy transfer
you pay for ($W$).

Efficiencies are supposed to be unitless, but the
efficiency of an air conditioner is normally given in terms of an EER rating
(or a more complex version called an SEER). The EER is defined
as $Q_L/W$, but expressed in the barbaric units of of Btu/watt-hour.
A typical EER rating
for a residential air conditioner is about 10 Btu/watt-hour, corresponding to an
efficiency of about 3.
The standard temperatures used for testing an air conditioner's efficiency
are $80\degunit F$ ($27\degcunit$) inside and
 $95\degunit F$ ($35\degcunit$) outside.

\noindent (a) What would be the EER rating of a reversed Carnot engine used as
an air conditioner? \answercheck\hwendpart
(b) If you ran a 3-kW residential air conditioner, with an efficiency of 3,
 for one hour, what would
be the effect on the total entropy of the universe? Is your answer
consistent with the second law of thermodynamics? \answercheck
<% end_hw() %>

<% begin_hw('heart-efficiency') %>__incl(hw/heart-efficiency)<% end_hw() %>

<% begin_hw('violin-helmholtz') %>__incl(hw/violin-helmholtz)<% end_hw() %>

<% end_hw_sec %>
<% end_chapter() %>