Thermodynamics
From Lm
Conservation Laws by Benjamin Crowell
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Chapter 6 - Thermodynamics
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Thermodynamics
This chapter is optional, and should probably be omitted from a two-semester survey course. It can be covered at any time after chapter 3.
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 heat engines: devices for converting between heat and other forms of energy. 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 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.
a / 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.
b / 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.
c / This does happen. The sum of the forces from the surrounding parts of the fluid is upward, canceling the downward force of gravity.
d / The pressure is the same at all the points marked with dots.
e / We have to wait for the thermometer to equilibrate its temperature with the temperature of Irene's armpit.
f / 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.
g / 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.
h / 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.
i / 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 to zero volume gives the same temperature as for any other gas: absolute zero.
Pressure and 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.
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.
Pressure
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.
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 head 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⊥/A, where A is the area of a small surface inserted in the fluid at that point, and F⊥ is the component of the fluid's force on the surface which is perpendicular to the surface.
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 different in pressure between one part of it and another part. The SI units of pressure are evidently N/m2, and this combination can be abbreviated as the pascal, 1 Pa=1 N/m2. The pascal turns out to be an inconveniently small unit, so car tires, for example, have recommended pressures imprinted on them in units of kilopascals.
Example 1: 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.)
Example 2: Atmospheric pressure in U.S. and metric units
◊ 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?
◊
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.
Example 3: Getting killed by a pool pump
◊ 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 m2. 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?
◊ 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:
That's a lot of force!
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.
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, b, would be subject to unequal forces from the parts of the fluid on its two sides. But fluids do not exhibit shear forces, so 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.
self-check: How does this proof fail for solids? (answer in the back of the PDF version of the book)
To find the variation with depth, we consider the vertical forces acting on a tiny, imaginary cube of the fluid having height Δ y and areas d A on the top and bottom. Using positive numbers for upward forces, we have PbottomΔ A - PtopΔ A - Fg = 0 . The weight of the fluid is Fg = mg = ρ Vg = ρ Δ AΔ y g, where ρ is the density of the fluid, so the difference in pressure is
Failed to parse (unknown function\begin): \begin{multline} \Delta P = -\rho g \Delta 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}
The factor of ρ 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. Note also that 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.
Example 4: Pressure of lava underneath a volcano
◊ 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 g/cm3. What is the pressure in the lava underneath the base of the volcano, 3000 m below the surface of the pool?
◊
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 Δ P we just calculated, so what we've found is essentially the
pressure, P.
Example 5: Atmospheric pressure
This example uses calculus.
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 in section A.2, but for now let's just call it k, ρ= kP. Using this fact, we can find the variation of atmospheric pressure with altitude, assuming constant temperature:
Failed to parse (lexing error): \begin{align} d P &= -\rho g\:d y\\ d P &= - kPg\:d y\\ \frac{d P}{ P} &= - kg\:d y\\ text{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 gets extremely small by the time you're ten or a hundred miles up.
Temperature
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 wait 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, 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.
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°C is the same as the difference between 48 and 58.
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°C. 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 ideal temperature scale for 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.
Example 6: Which temperature scale to use
◊ You open an astronomy book and encounter the equation
for the light emitted by a star as a function of its surface
temperature. What temperature scale is implied?
◊ 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.
j / A space suit (example 11).
Microscopic Description of An Ideal Gas
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 chapter 2, we have already seen 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 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.
Pressure, volume, and temperature
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
Failed to parse (lexing error): PV \propto nT , \qquad \text{[ideal gas]}
which really only holds exactly for an ideal gas. Here n is the number of molecules in the sample of gas.
Example 7: Volume related to temperature
The proportionality of volume to temperature at fixed pressure was the basis for our definition of temperature.
Example 8: 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.
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 vx 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 Δ t=2L/vx, and in each collision the x component of the atom's momentum is reversed from -mvx to mvx. The total force on the wall is
Failed to parse (lexing error): F = \frac{\Delta p_{x,1}}{\Delta t_1}+\frac{\Delta p_{x,2}}{\Delta t_2}+\ldots \qquad \text{[monoatomic ideal gas]} \qquad ,
where the indices 1, 2, … refer to the individual atoms. Substituting Δ px,i=2mvx,i and Δ ti=2L/vx,i, we have
Failed to parse (lexing error): F = \frac{mv_{x,1}^2}{L}+\frac{mv_{x,2}^2}{L}+\ldots \qquad \text{[monoatomic ideal gas]} \qquad .
The quantity mvx,i2 is twice the contribution to the kinetic energy from the part of the atom's 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. By symmetry, it must therefore equal 2/3 of the total kinetic energy, so
Failed to parse (lexing error): F = \frac{2KE_{total}}{3L} \qquad \text{[monoatomic ideal gas]} \qquad .
Dividing by A and using AL=V, we have
Failed to parse (lexing error): P = \frac{2KE_{total}}{3V} \qquad \text{[monoatomic ideal gas]} \qquad .
This can be connected to the empirical relation PV ∝ nT if we multiply by V on both sides and rewrite KEtotal as nKEav, where KEav is the average kinetic energy per molecule:
Failed to parse (lexing error): PV = \frac{2}{3}nKE_{av} \qquad \text{[monoatomic ideal gas]} \qquad .
For the first time we have an interpretation for the 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 KEav=(3/2)kT, where the constant of proportionality k, known as Boltzmann's constant, has a numerical value of Failed to parse (unknown function\kunit): 1.38\times10^{-23}\ \text{J}/\kunit . In terms of Boltzmann's constant, the relationship among the bulk quantities for an ideal gas becomes
Failed to parse (lexing error): PV = nkT \qquad , \qquad \text{[ideal gas]}
which is known as the ideal gas law. 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/NA is the number of moles of atoms, R=kNA, and NA=6.0×1023, called Avogadro's number, is essentially the number of hydrogen atoms in 1 g of hydrogen.)
Example 9: Pressure in a car tire
◊ After driving on the freeway for a while, the air in your car's tires heats up from 10°C to 35°C. How much does the pressure increase?
◊ 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,
Failed to parse (unknown function\kunit): \begin{align} P_2/ P_1 &= T_2/ T_1\\ &=(308\ \kunit)/(283\ \kunit)\\ &= 1.09 \qquad ,\\ \end{align}
or a 9% increase.
Example 10: Earth's senescence
Microbes were the only life on Earth up until the relatively recent advent of multicellular life, and are arguably still the dominant form of life on our planet. Furthermore, the sun has been gradually heating up ever since it first formed, and this continuing process will soon (“soon” in the sense of geological time) eliminate multicellular life again. Heat-induced decreases in the atmosphere's CO2 content will kill off all complex plants within about 500 million years, and although some animals may be able to live by eating algae, it will only be another few hundred million years at most until the planet is completely heat-sterilized.
Why is the sun getting brighter? The only thing that keeps a star like our sun from collapsing due to its own gravity is the pressure of its gases. The sun's energy comes from nuclear reactions at its core, and the net result of these reactions is to fuse hydrogen atoms into helium atoms. It takes four hydrogens to make one helium, so the number of atoms in the sun is continuously decreasing. Since PV=nkT, this causes a decrease in pressure, which makes the core contract. As the core contracts, collisions between hydrogen atoms become more frequent, and the rate of fusion reactions increases.
Example 11: A piston, a refrigerator, and a space suit
Both sides of the equation PV=nkT have units of energy. Suppose the pressure in a cylinder of gas pushes a piston out, as in the power stroke of an automobile engine. Let the cross-sectional area of the piston and cylinder be A, and let the piston travel a small distance Δ x. Then the gas's force on the piston F=PA does an amount of mechanical work W=FΔ x=PAΔ x=PΔ V, where Δ V is the change in volume. This energy has to come from somewhere; it comes from cooling the gas. In a car, what this means is that we're harvesting the energy released by burning the gasoline.
In a refrigerator, we use the same process to cool the gas, which then cools the food.
In a space suit, the quantity PΔ V represents the work the astronaut has to do because bending her limbs changes the volume of the suit. The suit inflates under pressure like a balloon, and doesn't want to bend. This makes it very tiring to work for any significant period of time.
k / 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.
l / 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.
m / The beginning of the first expansion stroke, in which the working gas is kept in thermal equilibrium with the hot reservoir.
n / 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.
o / 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.
p / The beginning of the second compression stroke, in which mechanical work is absorbed, heating the working gas back up to TH.
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.
Entropy
Efficiency and grades of energy
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.
Heat engines
Heat may be more useful in some forms than in other, i.e., there are different grades of heat energy. In figure k, 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 that consists of a small propeller spun by the hot air rising from a set of candles, very much like the setup shown in the figure.
In figure l, however, no mechanical work can be extracted because there is no difference in temperature. Although the air in l has the same total amount of energy as the air in k, the heat in l 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, QH. (The letter Q is the standard notation for a transfer of heat.) By conservation of energy, we have QH=W+QL, where QL is the amount of heat expelled into the low-temperature reservoir, so the efficiency of a heat engine, W/QH, can be rewritten as
Failed to parse (lexing error): \text{efficiency} = 1-\frac{Q_L}{Q_H} \qquad . \qquad \text{[efficiency of any heat engine]}
It turns out that there is a particular type of heat engine,
the Carnot engine, 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.
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 m-p 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. It turns out that this particular Carnot engine has an efficiency given by
Failed to parse (lexing error): \text{efficiency} = 1 - \frac{T_L}{T_H} \qquad , \qquad \text{[efficiency of a Carnot engine]}
where TL is the temperature of the cold reservoir and TH is the temperature of the hot reservoir. (A proof of this fact is given in my book Simple Nature, which you can download for free.)
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 m and n, 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 o and p, 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 TL is 20% of TH, 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 TH and TL have the same efficiency. Thus a Carnot engine is the most efficient possible heat engine.
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:
(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:
Failed to parse (unknown function\begin): \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.
Example 12: Entropy is additive.
Since changes in entropy are defined by an additive quantity (heat) divided by a non-additive one (temperature), entropy is additive.
Example 13: Entropy isn't changed by a Carnot engine.
The efficiency of a heat engine is defined by efficiency = 1 - QL/ QH , and the efficiency of a Carnot engine is efficiency = 1 - TL/ TH , so for a Carnot engine we have QL/ QH = TL/ TH, which can be rewritten as QL/ TL = QH/ TH. The entropy lost by the hot reservoir is therefore the same as the entropy gained by the cold one.
Example 14: 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/ TH+ Q/ TL, which is positive because TL< TH.
Example 15: Entropy is increased by a non-Carnot engine.
The efficiency of a non-Carnot engine is less than 1 - TL/ TH, so QL/ QH > TL/ TH
and QL/ TL > QH/ TH. This means that
the entropy increase in the cold reservoir is greater than the entropy decrease in the hot reservoir.
Example 16: 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.
Examples 14-16 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.
Example 17: Using a refrigerator to decrease entropy?
◊ 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?
◊ (1) No. The heat that comes off of the radiator coils on the back of your kitchen fridge 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.
Example 18: Maxwell's daemon
◊ Physicist James Clerk Maxwell imagined pair of neighboring rooms, their air being initially in thermal equilibrium, having a partition across the middle with a tiny door. A miniscule daemon 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 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?
◊ No. The daemon needs to eat, and we can think of his body as a little heat engine. His metabolism is less efficient than a Carnot engine, so he ends up increasing the entropy rather than decreasing it.
Observation such as these lead to the following hypothesis, known as the second law of thermodynamics:
The entropy of a closed system always increases, or at best
stays the same:
.
At present my arguments to support this statement may seem less than convincing, since they have so much to do with obscure facts about heat engines. A more satisfying and fundamental explanation for the continual increase in entropy was achieved by Ludwig Boltzmann, and you may wish to learn more about Boltzmann's ideas from my book Simple Nature, which you can download for free. Briefly, Boltzmann realized that entropy was a measure of randomness at the atomic level, and randomness doesn't spontaneously change into organization.
To emphasize the fundamental and universal nature of the second law, here are a few examples.
Example 19: 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.
Example 20: The heat death of the universe
Victorian philosophers realized that living things had low entropy, as discussed in example 19, and 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.
Example 21: 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.
Homework Problems
1. (a) Show that under conditions of standard pressure and
temperature, the volume of a sample of an ideal gas depends
only on the number of molecules in it.
(b) One mole is
defined as 6.0×1023 atoms. Find the volume of one mole
of an ideal gas, in units of liters, at standard temperature
and pressure (0°C and 101 kPa). (answer check available at lightandmatter.com)
2. A gas in a cylinder expands its volume by an amount Δ V,
pushing out a piston.
Show that the work done by
the gas on the piston is given by Δ W = PΔ V.
3.
(a) A helium atom contains 2 protons, 2 electrons, and 2
neutrons. Find the mass of a helium atom. (answer check available at lightandmatter.com)
(b) Find the
number of atoms in 1 kg of helium. (answer check available at lightandmatter.com)
(c) Helium gas is
monoatomic. Find the amount of heat needed to raise the
temperature of 1 kg of helium by 1 degree C. (This is known
as helium's heat capacity at constant volume.) (answer check available at lightandmatter.com)
4.
Refrigerators, air conditioners, and heat pumps are heat engines that
work in reverse.
You put in mechanical work, and it the effect is to take heat out of
a cooler reservoir and deposit heat in a warmer one: QL+W=QH.
As with the heat engines discussed previously, the efficiency is
defined as the energy transfer you want (QL for a refrigerator or
air conditioner, QH 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 QL/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° F (27°C) inside and
95° F (35°C) outside.
(a) What would be the EER rating of a reversed Carnot engine used as
an air conditioner? (answer check available at lightandmatter.com)
(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? (answer check available at lightandmatter.com)
5.
(a) Estimate the pressure at the center of the Earth,
assuming it is of constant density throughout.
Use the technique of example 5 on
page 148.
Note that g
is not constant with respect to depth ---
it equals Gmr/b3 for r, the distance from the center,
less than b, the earth's radius.1
State your result in terms of G, m, and b.
(b) Show that your answer from part a has the right units for pressure.
(c) Evaluate the result numerically. (answer check available at lightandmatter.com)
(d) Given that the earth's atmosphere is on the order of
one thousandth the thickness of 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.
∫
6. (a) Determine the ratio between the escape velocities
from the surfaces of the earth and the moon. (answer check available at lightandmatter.com)
(b) The
temperature during the lunar daytime gets up to about 130°C.
In the extremely thin (almost nonexistent) lunar atmosphere,
estimate how the typical velocity of a molecule would
compare with that of the same type of molecule in the
earth's atmosphere. Assume that the earth's atmosphere has
a temperature of 0°C. (answer check available at lightandmatter.com)
(c) Suppose you were to go to the moon
and release some fluorocarbon gas, with molecular formula
CnF2n+2. Estimate what is the smallest fluorocarbon molecule
(lowest n) whose typical velocity would be lower than that
of an N2 molecule on earth in proportion to the moon's lower
escape velocity. The moon would be able to retain an atmosphere made
of these molecules. (answer check available at lightandmatter.com)
7. Most of the atoms in the universe are in the form of gas that is not
part of any star or galaxy: the intergalactic medium (IGM). The IGM
consists of about 10-5 atoms per cubic
centimeter, with a typical temperature of about 103 K.
These are, in some sense, the density and temperature of
the universe (not counting light, or the exotic particles known
as “dark matter”). Calculate the pressure of the universe
(or, speaking more carefully, the typical pressure due to
the IGM).(answer check available at lightandmatter.com)
8. A sample of gas is enclosed in a sealed chamber. The gas consists of
molecules, which are then split in half through some process such as
exposure to ultraviolet light, or passing an electric spark through the
gas. The gas returns to thermal equilibrium with the surrounding
room. How does its pressure now compare with its pressure before the
molecules were split?
r / Problem 9.
9. The figure shows a demonstration performed by Otto von Guericke for Emperor Ferdinand III, in which two teams of horses failed to pull apart a pair of hemispheres from which the air had been evacuated. (a) What object makes the force that holds the hemispheres together? (b) The hemispheres are in a museum in Berlin, and have a diameter of 65 cm. What is the amount of force holding them together? (Hint: The answer would be the same if they were cylinders or pie plates rather then hemispheres.)
10. Even when resting, the human body needs to do a certain amount of mechanical
work to keep the heart beating. This quantity is difficult to define and
measure with high precision, and also depends on the individual and
her level of activity,
but it's estimated to be about 1 to 5 watts. Suppose we consider the human
body as nothing more than a pump. A person who is just lying in bed all
day needs about 1000 kcal/day worth of food to stay alive.
(a) Estimate the person's thermodynamic efficiency as a pump, and (b) compare with
the maximum possible efficiency imposed by the laws of thermodynamics for
a heat engine operating across the difference between a body
temperature of 37°C and an ambient temperature of
22°C. (c) Interpret your answer.\hwans{hwans:heart-efficiency}
Footnotes
theorem tells us that the gravitational field at r is the same as if all the mass existing at greater depths was concentrated at the earth's center. Since volume scales like the third power of distance, this constitutes a fraction (r/b)3 of the earth's mass, so the field
is (Gm/r2)(r/b)3=Gmr/b3.
