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16.2 Microscopic description of an ideal gas by Benjamin Crowell, Light and Matter licensed under the Creative Commons Attribution-ShareAlike license.

vCalc Companion Formulas | |

vCalc Formulary | 16.3 Entropy |

`epsilon_("ideal")=1 - Q_c/Q_h` | Carnot Efficiency |

`epsilon_("ideal")=1 - T_c/T_h` | Efficiency of a reversible heat engine |

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.

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.

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.

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.

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, `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

`"efficiency"=1?Q_L/Q_H` [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.

m / The beginning of the first expansion stroke, in which the working gas is kept in thermal equilibrium with the hot reservoir.

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

`"efficiency"=1?T_L/T_H`, [efficiency of a Carnot engine]

where `T_L` is the temperature of the cold reservoir and `T_H` 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.)

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.

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 `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. Thus a Carnot engine is the most efficient possible heat engine.

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.

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.

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:

` Delta S = Q/T` [change in entropy when adding heat `Q` to matter at temperature `T`; `DeltaS` is negative if heat is taken out]

A system with a higher grade of energy has a lower entropy.

Since changes in entropy are defined by an additive quantity (heat) divided by a non-additive one (temperature), entropy is additive.

p / The beginning of the second compression stroke, in which mechanical work is absorbed, heating the working gas back up to `T_H`.

The efficiency of a heat engine is defined by

and the efficiency of a Carnot engine is

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.

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`.

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.

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.

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.

? 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.

? 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: `DeltaS?0`.

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 or disorder at the atomic level, and disorder doesn't spontaneously change into order.

To emphasize the fundamental and universal nature of the second law, here are a few examples.

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.

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.

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.

**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 `6.0×10^(23)` 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 `DeltaV`, pushing out a piston. Show that the work done by the gas on the piston is given by `DeltaW=PDeltaV`.

**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: `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°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 426. Note that `g` is not constant with respect to depth --- it equals `Gmr"/"b^3` 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 `C_nF_(2n+2)`. Estimate what is the smallest fluorocarbon molecule (lowest `n`) whose typical velocity would be lower than that of an `N_2` 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 `10^3` 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.

**11**. (a) Atmospheric pressure at sea level is 101 kPa. The deepest spot in the world’s oceans is a valley called the Challenger Deep, in the Marianas Trench, with a depth of 11.0 km. Find the pressure at this depth, in units of atmospheres. Although water under this amount of pressure does compress by a few percent, assume for the purposes of this problem that it is incomprehensible.(b) Suppose that an air bubble is formed at this depth and then rises to the surface. Estimate the change in its volume and radius. Solution, p. 548.

12. Our sun is powered by nuclear fusion reactions, and as a first step in these reactions, one proton must approach another proton to within a short enough range r. This is difficult to achieve, because the protons have electric charge +e and therefore repel one another electrically. (It’s a good thing that it’s so difficult, because otherwise the sun would use up all of its fuel very rapidly and explode.)To make fusion possible, the protons must be moving fast enough to come within the required range. Even at the high temperatures present in the core of our sun, almost none of the protons are moving fast enough.(a) For comparison, the early universe, soon after the Big Bang,had extremely high temperatures. Estimate the temperature T that would have been required so that protons with average energies could fuse. State your result in terms of r, the mass m of the proton, and universal constants.(b) Show that the units of your answer to part a make sense.(c) Evaluate your result from part a numerically, using `r = 10^(?15)` m and `m = 1.7 × 10^(?27)` kg. As a check, you should find that this is much hotter than the sun’s core temperature of ? `10^7` K. Solution, p. 549

13. Object A is a brick. Object B is half of a similar brick. If A is heated, we have `DeltaS = Q"/"T`. Show that if this equation is valid for A, then it is also valid for B. Solution, p. 549

14. Typically the atmosphere gets colder with increasing altitude. However, sometimes there is an inversion layer, in which this trend is reversed, e.g., because a less dense mass of warm air moves into a certain area, and rises above the denser colder air that was already present. Suppose that this causes the pressure `P` as a function of height `y` to be given by a function of the form `P = P_oe^(?ky) (1 + by)`,where constant temperature would give `b = 0` and an inversion layer would give `b > 0`. (a) Infer the units of the constants `P_o, k,` and `b`.

(b) Find the density of the air as a function of `y`, of the constants,and of the acceleration of gravity `g`.

(c) Check that the units of your answer to part `b` make sense. Solution, p. 549

16.2 Microscopic description of an ideal gas by Benjamin Crowell, Light and Matter licensed under the Creative Commons Attribution-ShareAlike license.