Author: Eugene Fischer

I never got around to finishing the problem set last week, so instead of editing the last entry, I’ll just finish it for this week.

Problem 1.32: By applying a pressure of 200 atm, you can compress water to 99% of its usual volume. Sketch this process (not necessarily to scale) on a PV diagram, and estimate the work required to compress a liter of water by this amount. Does the result surprise you?

I’m just going to assume a linear function here and say that the work done is the area under the PV curve. Here that ends up being about 100 J to compress water by 1%.

Problem 1.33: An ideal gas is made to undergo a cyclic process shown in the figure to the right. For each of the steps A, B, C, determine whether each of the following is positive, negative, or zero: (a) the work done on the gas; (b) the change in the energy content added to the gas; (c) the heat added to the gas. Then determine the sign of each of these three quantities for the whole cycle. What does this process accomplish?

For A, the work being done on the gas is (a) negative, by the equation W = -P\Delta V. The energy added to the gas is (b) positive, which I could demonstrate with the equipartition of energy principle, but would rather show by observing that if the gas increases in volume but maintains the same pressure, that means that the frequency of collisions with the container must remain the same as when the gas was at lower volume. The only way for that to happen is if the molecules are moving faster, so energy must have been added to the system. Since the energy of the system is the heat plus the work, and the work is negative and the energy is positive, then the heat added must also be (c) positive (and greater than the absolute value of the work).

For B, the work done on the gas is (a) zero, since there is no change in volume. By equipartition of energy for an ideal gas, U = \frac{3}{2}PV, the energy increase is (b) positive. You can also get this from observing that increased pressure at the same volume means more collisions with the container per unit time, which means that the molecules of the gas must be moving faster, and so must have more energy. Since there is more energy and no work has been done, the heat added is (c) positive.

For C, the work done on the gas must be (a) positive, because the volume decreases. The energy added to the gas is (b) negative, because even though the volume is decreasing, the pressure is going down. Since the work done on the gas is positive and the overall energy change is negative, the heat added to the gas is also (c) negative.

For the whole cycle, the net work must be (a) positive, because the average pressure is higher during step C than during step A. Since the pressure and volume at the end of the cycle are the same as at the start, the net energy change must be (b) zero. And since the work is positive and the energy change is zero, the heat added must be (c) negative. So this cycle takes in energy as work and emits the energy as heat.

There’s another problem here, but it’s basically the same, except it’s a four-step rectangular cycle that goes clockwise, and turns heat added into work done by the gas. It’s late, so I’m not going to go through the details.

This section introduces the equations for the work done on a system by compression, such as when a piston is used to compress a volume of gas. Assuming nearly quasistatic compression (that is, compression slow enough that all of the gas can be said to be at a single pressure; in practice, usually any compression slower than the speed of sound in the gas), the work done per infinitesimal change in volume is W = -P\Delta V. When the pressure changes significantly during compression, you have to approximate the process as a series of small compressions, unless you have an equation for the pressure as a function of volume. If you do, then you can integrate, and the equation for work becomes W=-\int _{{V}_{i}}^{{V}_{f}}{P(V)dV} .

Problem 1.31: Imagine some helium in a cylinder with an initial volume of 1 liter and an initial pressure of 1 atm. Somehow the helium is made to expand to a final volume of 3 liters, in such a way that its pressure rises in direct proportion to its volume. (a) Sketch a graph of the pressure vs. volume for this process.

(b) Calculate the work done on the gas during this process, assuming that there are no “other” types of work being done.

Here I actually do have an equation for pressure with respect to volume. Since the relationship is linear, I can say that P = V (\frac {atm}{L}). Plugging into the equation for work gives W = -\frac {atm}{L} \int _{1 (L)}^{3 (L)}{V dV}. This ends up giving –4 atmosphere-liters, which converts to approximately –400 J. The negative sign indicates that, rather than work being done on the gas, it is the gas that is doing work on the environment (presumably by displacing whatever was in the volume it expands to occupy).

(c) Calculate the change in the helium’s energy content during this process.

Helium is a monatomic gas, so it has three quadratic degrees of freedom. By the equipartition of energy theorem, U = \frac {3}{2}NkT, which by the ideal gas law is equivalent to U = \frac {3}{2}PV. The change in energy, then, is \Delta U = \frac {3}{2} (P_{final}V_{final} - P_{initial}V_{initial}). This gives 12 atmosphere-liters, or approximately 1200 J.

(d) Calculate the amount of heat added to or removed from the helium during this process.

The first law of thermodynamics is \Delta U = Q + W, which I can rewrite Q = \Delta U - W. In this case, the work done is -400 J, and the change in energy is 1200 J, so the heat added to the system is 1600 J.

(e) Describe what you might do to cause the pressure to rise as the helium expands.

Normally as volume increased pressure would decrease. (Solve the ideal gas law for pressure and volume ends up in the denominator). To counter that, you would have to increase the temperature. So you could cause the pressure to rise as the helium expands by heating it.

I’ll have to finish these problems another time, as I need to go to the reception for Marina Warner, who is receiving the Truman Capote award. I’ll edit this post later. (Tumblr, you won’t see the edit, you’ll have to click through.)

This section defines heat, “the spontaneous flow of energy from one object to another caused by the difference in temperature between the objects,” and work, any non-heat transfer of energy into or out of a system. It also introduces the first law of thermodynamics, \Delta U = Q + W, where U is the total energy of the system, Q is the energy that enters the system as heat, and W is the energy that enters the system as work. Schroeder points out that these definitions of heat and work are counter to the way we use the terms in common speech. For example, if you rub your cold hands together to warm them up, by these definitions there is no heating being done to the system, because the change in temperature isn’t due to spontaneous energy flow. Rather, your hands are being warmed by work.

Schroeder goes on to introduce the units Joule, calorie (equal to 4.186 J), and kilocalorie (also known as the food calorie, 4186 J); and the three types of heat transfer, conduction (transfer by molecular contact), convection (transfer by the bulk motion of a gas or liquid), and radiation (emission of electromagnetic waves.)

Problem 1.26: A battery is connected in series to a resistor, which is immersed in water (to prepare a hot cup of tea). Would you classify the flow of energy from the batter to the resistor as “heat” or “work”? What about the flow of energy from the resistor to the water?

The flow of energy from the battery to the resistor is work, because the energy transferred by the electrons moving through the resistor is driven by the voltage difference, it doesn’t occur spontaneously due to a temperature difference. The flow of energy from the resistors to the water, though, is heat.

Problem 1.27: Give an example of a process in which no heat is added to a system, but its temperature increases. Then give an example of the opposite: a process in which heat is added to a system but its temperature does not change.

Imagine you have a cylinder of compressed air that is at room temperature. If you open the valve and start letting air out, the pressure within the cylinder will drop. By the ideal gas law, the temperature of the air inside the cylinder will drop proportionally with the pressure. (This is why the propane tanks attached to gas grills occasionally ice up while the grill is in use.) The opposite condition would hold if you pumped air from the room into the cylinder. Then the temperature of the gas in the cylinder would rise proportionally with the pressure, but no heat would be added.

A system where heat is added but the temperature doesn’t change would be one where either the heat energy is being compensated for by work, or where the heat energy is causing a non-temperature change, like a phase change. So for example if you boil water, energy is constantly flowing into the water due to heat, but the temperature stays at 100°C. The energy goes into changing the water from liquid phase to gas phase.

Problem 1.28: Estimate how long it should take to bring a cup of water to boiling temperature in a typical 600-watt microwave oven, assuming that all the energy ends up in the water. (Assume any reasonable initial temperature for the water.) Explain why no heat is involved in this process.

A calorie is the amount of energy needed to raise one gram of water by 1° C. A watt is one joule per second. One cup of water is 236.6 ml, which is also 236.6 grams. If we assume the water is at 21° C, then the the amount of energy needed to bring the water to boiling is 236.6 x 79 = 18691.4 calories, which is 78,242 joule. Dividing by 600, we get 130 seconds to heat the water to boiling. No heat is involved because the microwaves that are physically vibrating the polar water molecules are being generated, not emitted spontaneously. This is work, just like when you rub your hands together so that the friction warms them. This isn’t a spontaneous transfer of energy due to temperature, it’s a driven energy transfer.

Problem 1.29: A cup containing 200 g of water is sitting on your dining room table. After carefully measuring its temperature to be 20° C, you leave the room. Returning ten minutes later, you measure the temperature again and find that it is now 25° C. What can you conclude about the amount of heat added to the water? (Hint: This is a trick question)

You can infer that the net heat added to the water was less than or equal to 1000 calories. It’s possible that no heat was added, and the temperature increase was due entirely to work. It’s possible that some work was done, and also some heat added. It’s possible that more than 1000 calories of heat were added, and then removed again. You can’t know for sure anything other than that there are 1000 new calories of energy, some or all or none of which could have come from heat.