A Bad Homework Problem about Entropy

Some disciplines in physics have been around for so long that it seems like all of the possible homework and exam problems have been tried out. There is a theory that these problems are stored away in the libraries of fraternities and sororities, ripe for the study of any student.

When this theory is brought up, it's always taken for granted that learning how to do the work on the exam is necessarily a bad thing. I'll just say that this isn't the only possible reaction to learning that students know how to do the assigned problems.

Thermodynamics is one of those disciplines. All of thermodynamics has basically been figured out since the 19th century. Thermodynamics is extremely important to understand in many fields, especially chemistry and biochemistry, so students still have to take this old, dusty subject. So the exam problems get repeated, and books probably copy each other. Upon occaision, however, one comes across something different. The "something different" is usually stupid. For instance,

[The instructor's] desk typically gets to the point where there are a lot of papers randomly distributed through out (sic) it. Every month or so I clean and organize the papers. ... treat the papers on my desk as a [two-dimensional] ideal gas. Each paper has an area a and the whole desk has an area A. We'll say that the papers are ideal particles ... this ideal gas [is] at constant temperature and thus assume [that the energy is constant] ... If I can clean my desk in a reversible manner, how much work will it take?

To someone not in a thermodynamics course (or someone who hasn't taken one in the past), this rightly seems like gibberish. To parse it, let's restate it with the trickier parts in bold:

[The instructor's] desk typically gets to the point where there are a lot of papers randomly distributed through out (sic) it. Every month or so I clean and organize the papers. ... treat the papers on my desk as a [two-dimensional] ideal gas. Each paper has an area a and the whole desk has an area A. We'll say that the papers are ideal particles ... this ideal gas [is] at constant temperature and thus assume [that the energy is constant] ... If I can clean my desk in a reversible manner, how much work will it take?

The first concept to understand is an ideal gas. This is a hypothetical material that is a very good model for real gases (like air) at moderate temperatures and pressures (room temperature and atmospheric pressure are not bad). The way the problem is stated, this ideal gas is composed of particles (the paper) that are like the atoms and molecules in a real gas, in that they don't interact much--this is what is meant by ideal particles. (I say "don't interact much" rather than the more common "don't interact" because particles in an ideal gas can crash into one another, like billiards balls, and still have the properties of an ideal gas.)

But this ideal gas is a little funny, since it exists in two dimensions instead of the usual three. One can imagine a model for such a gas, for instance with two panes of glass a tiny bit separated. The gas inside might behave like a two-dimensional ideal gas. However, it's not important that this system even be realizable: it's a model, and using the equations taught in the course the students should be able to compute the properties of such a system, even if it is weird.

What do the areas have to do with this problem? Well, the trick (in the course) to solving this problem was to recognize this system as a lattice model and being able to write down its entropy. This is a little equation, due to Boltzmann (who killed himself, probably because he was tired of stupid homework problems)

S = k ln W

which is an equation which relates microscopic properties--in particular, the combinatoric factor W--with a macroscopic property, the entropy S. This is a little like calculating the macroscopic energy of a system by adding up all the microscopic energy of all the constituent particles, but a little trickier, since it involves weirder stuff, like a logarithm and a number k, which is actually just a constant that guarantees that we measure temperature with a certain scale, like Fahrenheit (although no one does that) and energy with a certain scale, like kilowatt-hours (although no one does that, either, at least not in science classes. Ask an engineer). W is often called "the number of ways of arranging the system."

To understand this, imagine the lattice model as a muffin tin with 12 divets for muffins. (The analogy will work for cupcakes, if one prefers.) There are 12 different ways to "arrange the system" if there is one muffin. If there are 2 muffins, the problem is a little trickier, so we make it a little easier by saying the muffins are "ideal particles" and can occupy the same hole in the muffin tin (since ideal particles don't interact). Working through the math, there are 12-to-the-2 power, or 144, different ways of arranging 2 muffins in a 12-hole muffin tin. This is W for muffins.

Generalizing this, if there are m holes in the muffin tin, and n muffins, there are m-raised-raised-to-the-n-power different arrangements.

Don't worry if the math gets too nutty here. The important part is to know that, given a certain interpretation of the problem, it's solvable. (But since I'm a sicko I'm going to outline the math anyway.)

For the papers on a desk, the number of "holes" is A/a since one can fit exactly this many sheets of paper, each of area a, on a desk with area A. There are N papers, so for the papers on a desk,

W = (A/a)-to-the-N-power

so the entropy can be calculated:

S = N k ln (A/a)

But why calculate the entropy? Well, for this system, the work is the temperature T times the entropy change in entropy as we take all the randomly distributed papers and put them in a stack in a specific spot. Stacking the papers up, and putting them in one specific spot is like saying, in the final, clean state, is that there is 1 arrangement in the final state, so its entropy is zero (since the logarithm of 1 is zero). So the change in entropy as we take all the papers and stack them in one, specific spot is just

change in S = 0 - N k ln (A/a) - 0 = - N k ln (A/a) (it's a negative number)

and as I said before, the work requred is just the temperature times this number (it doesn't look like a number, but if one plugs in values for things like the area, it is!).

But what do we mean by work? This is what the question asks for, and the key (given in the statement of the problem) is that the energy is constant. Energy cannot be created or destroyed (it's "conserved") so if there were an energy change it would have to be because the system had heat removed or work done on it; those are the only two ways that energy can change. In our case, the energy change is zero, so if there were heat taken out of the two-dimensional gas, it would equal to the negative of the work. (Since if that were true, the heat and work would add up to zero, and since the heat and work are zero together, the energy change is zero.)

This is not the first place the problem starts to get silly. There is no heat added to the papers. Their temperature does not change. Someone just made up a stupid homework problem as if there were heat being removed from the system. If this were a two-dimensional ideal gas, and not stacks of papers, we would have to remove heat from the system in order to get all the particles to stick together. In fact, this is just like what happens when a gas gets cooled enough: it starts to condense into a liquid (usually). In our case, we're pretending to suck heat out of the randomly arranged papers. If we can calculate this pretend heat, we can calculate the work that one would have to do to clean the desk, at least if the heat is removed reversibly, which for our purposes it just means that the cleaning is done as carefully as possible, without wasting any energy or motion.

The heat sucked out turns out to be the entropy change times the temperature. Therefore the reversible (no wasting) work would be the negative of this, which is

work = negative of heat = - T x (change in S) = N k T ln (A/a)

The important part to recognize here is that the work done depends on the number of papers, N, which makes sense since one should have to do more work to clean up more papers, but other than that this answer is nonsense for cleaning up a desk (although it's correct for taking all the particles in a two-dimensional gas and squeezing them into one specific spot, as long as the squeezing is done reversibly).

For one thing, it depends on the temperature, which is silly. The papers do have a temperature; one could measure it with, among other things, an infrared thermometer. It seems wrong that it would take more work to clean the papers at 50 ° F than at 40 ° F, and that's because it is wrong. Temperature is the average kinetic energy of particles, in this case the particles that constitute the papers, not the papers themselves! Papers that are just sitting on a desk, in order or not, don't have kinetic energy and so they don't have a temperature: rather, their constituent particles have kinetic energy and so have a temperature.

What this does is just confuse people. Some students will get it. But there is no reason for instructors to confuse people on purpose. Physics is already hard enough without people running around making it more confusing. The above discussion about "temperature" of the particles in a piece of paper and "temperature of the papers" should seem confusing to the reader. That's because it is. The papers don't have a temperature because they're just sitting there on a desk. The particles that make up a paper, on the other hand, do have a temperature (as can be verified with a thermometer) since, on a microscopic scale, they're jostling around a little bit (but not a lot--if they jostled around too much the paper would fly apart or burn. In fact, this is one of the reasons why paper burns when its temperature is raised. The other reason is the presence of oxygen).

And for another thing, the approach to calculating the work is just plain wrong. An everyday reading of the problem would be something like, "How much do I have to exert myself to clean up some papers?" This could be measured in terms of the number of food calories burned (how much food energy was used by a person) who cleaned up the paper. Rather easier would be the question of the minimum amount of work needed to pick up N papers. This means fighting against gravity since, in principle, the papers have mass, and they need to be picked up and moved.

To do this much better calculation, we break down the steps to pick up one piece of paper. First, we have to pick up the piece of paper off of the table. To pick it up, we have to exert a force to overcome gravity and make the paper move up. It turns out that the work required to do this is just the mass of the paper, times how far we lifted up the paper, times the gravitational acceleration (which is about 10 meters per second per second, but the important thing is that it's a number that we can look up). For instance, if the paper had mass 1 gram, and we picked it up and moved it 1 centimeter above the desk, then we would have to do

1 g times 1 cm times 10 meters per second per second = 0.0001 Joules

A Joule is just a unit of energy; it's the same as about one-four-thousandth (1/4000) of a food calorie. A grape has about 4 food calories, so using the energy of one grape, one could (in principle) pick about 1000 one-gram papers 1 cm off of the table.

The next step to cleaning up the papers, after picking them up, is moving them over to the stack. This actually takes very little work, because the only force we have to fight against is gravity, and gravity works downward, not sideways. That is, there is no force that resists motion of a paper sideways other than its own inertia, which just means that we have to give it a little push to get it to move. Then it moves to our stack, then we have to give it another little push to make it stop moving. This will depend on the mass, too, and the distance the paper moves. The total amount of work will depend on how hard we push, and if we push just a tiny bit, then we can pretend that the work to move the paper sideways is negligible (zero).

The last step, after moving the paper, is dropping it onto the stack. This costs the cleaner no energy, since gravity takes care of moving the paper down.

To clean up N papers, then, depends on just a few things, among them the mass of each paper, and the height to which we pick up each paper off of the table. If the papers all have the same mass, m, and we pick them all up the same distance, h, from the table, then the work really done is

work = N m g h

which isn't that great of an answer, either, but at least it makes sense! Notice there's no temperature or k or area of the table A in this answer. Why should those values matter to the amount of work done? The fact is that they don't! But it makes intuitive sense that if the papers are heavier (m is bigger), then it would take more work. The "thermodynamics" answer doesn't depend on how massive the papers are, which doesn't make sense.

Furthermore, the real answer depends on how high we pick up the papers, h. It makes sense that if we pick up each paper 1 mile above the desk that the work will be more. It's much more efficient to pick them up a little bit (like .5 or 1 cm) than a lot. Again, the fake thermodynamics answer doesn't take this action into consideration. (Unless you pretend that we can slide the papers, h = 0, in which case the fake-thermo answer is still horribly wrong.)

What's more, if we went to a planet with stronger gravity, it should take more work to clean a desk, too. Stronger gravity means g is higher, and you can see that if g is higher the real answer predicts that more work will be required to clean.

The point of all this is that answers should make sense. The two-dimensional ideal gas answer for cleaning desks doesn't make sense, since it does not depend on obvious things like the mass of the papers and gravity, which are intuitively what should matter when we're doing something like cleaning. But things like collections of papers sitting on a desk don't have a "temperature" since they're not moving around. So it makes no sense, intuitively, for the answer to have anything to do with temperature.

Maybe this is why Boltzmann killed himself--he was being driven crazy by stupid homework problems in thermodynamics! Or maybe people weren't carefully checking their answers with their intuition.