Category: Physics (page 1 of 2)

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Good effort. Almost had it.

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Thermo Thursday: Compression of an Ideal Gas

Thermo Thursday returns. But on a Tuesday! WHO COULD HAVE SUSPECTED?


This section introduces isothermal and adiabatic compression. Isothermal compression is compression that occurs so slowly that no heat is added to the gas. For isothermal compression of an ideal gas, the temperature remains constant, so you can use the ideal gas law, PV=NkT, and the integral equation for work done during compression, W=-\int _{{V}_{i}}^{{V}_{f}}{P(V)dV}, to derive that {W}_{isothermal}=NkT\ln {\frac {{V}_{i}}{{V}_{f}}}. Since work is being done but the temperature is not changing, heat must be flowing out of the gas, in an amount equal to the work being done. On a PV graph, an isothermal compression takes the shape of a concave-up hyperbola.

Adiabatic compression, in contrast, happens so fast that no heat escapes from the gas during the process. Thus, for adiabatic compression, \Delta U = W.  The PV curve for this starts on a lower temperature isotherm and ends on a higher temperature isotherm. Adiabat To find an equation for the shape of that curve, we look at the equipartition of energy theorem, U = \frac {f}{2}NkT, where f is the number of degrees of freedom per molecule. The infinitesimal change in energy along the curve is then given by dU=\frac {f}{2}NkdT. If we assume the compression is quasistatic, then from the equation for work we know that dU=-PdV. (We can say this because we previously established that in adiabatic compression the entire change in energy comes from work.) This gives \frac {f}{2}NkdT=-PdV. Now you can plug in the ideal gas law for P and do some canceling to get \frac {f}{2}\frac {dT}{T}=-\frac {dV}{V} \rightarrow \frac {f}{2}\ln {\frac {{T}_{f}}{{T}_{i}}}=-\ln{\frac{{V}_{f}}{{V}_{i}}}.  This ends up simplifying down to V{T}^{{f}/{2}}=C for some constant C. If you are looking for pressure instead of temperature, you can use the ideal gas law to rewrite this {V}^{\gamma}P=D for some constant D, where \gamma = {(f+2)}/{f} and is called the adiabatic constant.

Problem 1.35: Derive {V}^{\gamma}P=constant from V{T}^{{f}/{2}}=constant.

I’m going to use a subscript on the constant term to show when that side of the equations changes. Let’s start with V{T}^{{f}/{2}}={C}_{0}. I want to get this equation in terms of pressure, so we use the ideal gas law to say that T=\frac {PV}{Nk}. This gives us
Raising both sides of the equation to the power 2/f  and moving the constant terms to the right side gives
I can then combine the volume terms,

Problem 1.36: In the course of pumping up a bicycle tire, a liter of air at atmospheric pressure is compressed adiabatically to a pressure of 7 atm. (Air is mostly diatomic nitrogen and oxygen.)

(a) What is the final volume of this air after compression?

I will treat the air as a diatomic ideal gas for these calculations, which means the air molecules have five degrees of freedom. This gives me \gamma =7/5. For adiabatic compression, {V}_{i}^{\gamma}{P}_{i}={V}_{f}^{\gamma}{P}_{f}, so
which for an initial pressure of 1 atmosphere and volume of 1 liter, and final pressure of 7 atmospheres, gives a final volume of 0.25 liters.

(b) How much work is done in compressing the air?

The total change in energy is the heat added or lost plus the work done, \Delta U = Q + W. Since this is adiabatic compression, there is no heat lost to the environment, so the entire change in energy is due to work. From the equipartition of energy theorem and the ideal gas law I can write \Delta U= \frac {f}{2} \Delta (PV), which for this system gives
\Delta U = \frac {5}{2} ({P}_{f}{V}_{f}-{P}_{i}{V}_{i})
Plugging in {V}_{f}=0.25 L, {P}_{f}=7 atm, {V}_{i}=1.0 L, {P}_{i}=1 atm, converting to SI units, and calculating gives an energy added due to work of 189.5 J.

(c) If the temperature of the air is initially 300 K, what is the temperature after compression?

I can use the equation V{T}^{{f}/{2}}=constant to say that
and so
{T}_{f}={(\frac {{V}_{i}{T}_{i}^{\frac{5}{2}}}{{V}_{f}})}^{\frac {2}{5}}.
Plugging in the values and calculating this out gives a final temperature of 522 K.

Problem 1.37: In a Diesel engine, atmospheric air is quickly compressed to about 1/20 of its original volume. Estimate the temperature of the air after compression, and explain why a Diesel engine does not require spark plugs.

Since the air is compressed “quickly,” I will assume adiabatic compression, and since it’s air that’s being compressed it has the same degrees of freedom as the previous problem. So I can use the last equation from part (c) above, plug in {V}_{f}=\frac {1}{20}{V}_{i}, and simplify to get
{T}_{f}={(20{T}_{i}^{\frac {5}{2}})}^{\frac {2}{5}} \approx 3.3{T}_{i}.
So if the initial temperature of the air is 300 K, then after compression the temperature will rise to around 990 K. Since the ignition temperature of Diesel fuel is 483 K, the air after compression is hot enough to ignite it without a spark.

Problem 1.38: Two identical bubbles of gas form at the bottom of a lake, then rise to the surface. Because the pressure is much lower at the surface than at the bottom, both bubbles expand as they rise. However, bubble A rises very quickly so not heat is exchanged between it and the water. Meanwhile, bubble B rises slowly (impeded by a tangle of seaweed), so that it always remains in thermal equilibrium with the water (which has the same temperature everywhere). Which of the two bubbles is larger by the time they reach the surface? Explain your reasoning fully.


IsoAdiaBubble A undergoes adiabatic compression (in this case, expansion), while bubble B undergoes isothermal compression. Initially, the bubbles are identical, so their pressures and volumes are equal. For isothermal compression, {P}_{B}{V}_{B}=C where C is a constant, and for adiabatic compression {P}_{A}{V}_{A}^{\gamma}=C. The constants C must be the same, because the bubbles are initially identical. Since \gamma > 1, {V}_{A}<{V}_{B}. Thus, the bubble that undergoes isothermal compression, bubble B, is larger when it reaches the surface. This can be seen by visual inspection of a graph depicting an isotherm and an adiabat. Notice that for the same initial conditions, the volume of the isotherm rises faster than that of the adiabat.

Thermo Thursday: Finishing Compression Work

It’s the end of the semester and I’ve got other projects I need to be working on, so this week I’m just going to do the last problem from the compression work section and leave the adiabats and isotherms for next time.

PVdiagramProblem 1.34: An ideal diatomic gas, in a cylinder with a movable piston, undergoes the rectangular cyclic process shown in the figure to the right. Assume that the temperature is always such that rotational degrees of freedom are active, but vibrational modes are “frozen out.” Also assume that the only type of work done on the gas is quasistatic compression-expansion work. (a) For each of the four steps A through D, compute the work done on the gas, the heat added to the gas, and the change in the energy content of the gas. Express all answers in terms of {P}_{1}, {P}_{2}, {V}_{1}, and {V}_{2}. (Hint: Compute \Delta U before Q, using the ideal gas law and the equipartition theorem.)

Step A: This is a diatomic gas with no vibration modes, so for this system the equipartition theorem says that U = \frac {5}{2}PV. For this step, \Delta U = \frac{5}{2}({P}_{2}-{P}_{1}){V}_{1}. Since the volume of the gas doesn’t change in this step, there is no work being done on the gas. Recalling that \Delta U = W + Q, that means that Q = \frac{5}{2}({P}_{2}-{P}_{1}){V}_{1}.

Step B: here \Delta U = \frac{5}{2}{P}_{2}({V}_{2}-{V}_{1}). The volume is increasing, so the work done on the gas is negative, W=-{P}_{2}({V}_{2}-{V}_{1}). Since W is negative, that means that the heat added must be the total energy change plus the amount of energy subtracted by negative work. Q=\Delta U - W = \frac{7}{2}{P}_{2}({V}_{2}-{V}_{1}).

Step C: this time \Delta U = \frac{5}{2}({P}_{1}-{P}_{2}){V}_{2}. Note that this is a negative quantity, the system has lost energy. Again, there is no volume change, so no work is being done on the gas. Thus the entire energy change is due to the system losing heat energy, so Q=\frac{5}{2}({P}_{1}-{P}_{2}){V}_{2}.

Step D: in the last step, \Delta U = \frac{5}{2}{P}_{1}({V}_{1}-{V}_{2}). This is another negative quantity. But this time, the work being done on the gas is positive, W=-{P}_{1}({V}_{1}-{V}_{2}). Since the work done on the gas is positive, and the energy change is negative, it must be losing even more heat energy than it is gaining in work energy. Here Q= \Delta U - W = \frac{7}{2}{P}_{1}({V}_{1}-{V}_{2}).

(b) Describe in words what is physically being done during each of the four steps; for example, during step A, heat is added to the gas (from an external flame or something) while the piston is held fixed.

The author went ahead and did step A for me. In step B, the volume of the gas increases, so the piston is being drawn out (or, more likely, pushed out by internal pressure). But this doesn’t result in any decrease in pressure, which means that the gas is also being heated as the volume increases. In step C, the system just loses heat energy, so the system is being cooled while the piston is held fixed. In step D the piston is compressed, but the cylinder is still being cooled so the pressure doesn’t change.

(c) Compute the net work done on the gas, the net heat added to the gas, and the net change in energy of the gas during the entire cycle. Are the results as you expected? Explain briefly.

The net change in energy is zero, that’s what makes it a cycle. For the work over the whole cycle, that’s {W}_{cycle}=-({P}_{2}-{P}_{1})({V}_{2}-{V}_{1}). That’s a negative quantity, meaning that over the course of the whole cycle work was done on the environment. Since the energy change is zero, the heat added perfectly balances the work, so {Q}_{cycle}=({P}_{2}-{P}_{1})({V}_{2}-{V}_{1}). That’s a positive quantity. As mentioned at the end of the last Thermo Thursday, this cycle takes in heat energy and converts it to work done on the environment. So these results are as I expected.

Thermo Thursday: Chapter 1.5: Compression Work, continued

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?

PVwaterI’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%.

Fig1Problem 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.

Thermo Thursday: chapter 1.5: Compression Work

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.)

Thermo Thursday: Chapter 1.4: Heat and Work

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.

Thermo Thursday: 1.2 continued, and 1.3 Equipartition of Energy

More student meetings today, so I’m not sure how far I’ll get with this (I type these up as I work the problems), but I want to finish the last problem from two weeks ago because it ends up in a nice equation. I don’t know how to solve it, though, without invoking something that, weirdly, the book doesn’t introduce until after the problem: the equipartition of energy theorem. So let’s start there.

The equipartition of energy theorem states that the average energy of any quadratic degree of freedom, at temperature T, is given by \frac {1}{2}kT. A quadratic degree of freedom is any way a particle can move whose equation of motion is quadratic. So examples of this would translational kinetic energy in each individual spatial dimension (e.g. \frac {1}{2}m{v}_{x}^{2}), rotational kinetic energy (e.g. \frac {1}{2}I{\omega}_{x}^{2}), elastic potential energy (\frac {1}{2}{k}_{s}{x}^{2}), etc. So a system of N molecules in which each molecule has f degrees of freedom and no other (non-quadratic) temperature-dependent forms of energy will have a total average thermal energy given by {U}_{thermal}=Nf(\frac {1}{2}kT). For a very large N, as will usually be the case, deviations away from the average will be negligible.

It’s important to note that {U}_{thermal} doesn’t include other sources of energy, such as relativistic rest energy, or energy stored in chemical bonds. So the equipartition theorem is best used to look at changes in energy due to temperature variation, and not things like phase transformations.

So now that I have the equipartition of energy theorem, I can go back and finish problem 1.22 from last time.

(b) It’s not easy to calculate \bar {{v}_{x}}, but a good enough approximation is \sqrt {\bar {{v}_{x}^{2}}}, where the bar now represents an average over all molecules in the gas. Show that \sqrt {\bar {{v}_{x}^{2}}} = \sqrt{\frac {kT}{m}}.

This comes directly from the equipartition of energy theorem, which gives me \frac {1}{2}m\bar {{v}_{x}^{2}} = \frac {1}{2}kT  so \bar {{v}_{x}^{2}} = \frac {kT}{m} and \sqrt {\bar{{v}_{x}^{2}}} = \sqrt {\frac {kT}{m}} .

(c) If we now take away this small part of the wall of the container, the molecules that would have collided with it will instead escape through the hole. Assuming that nothing enters through the hole, show that the number N of molecules inside the container as a function of time is governed by the differential equation \frac {dN}{dt} = -\frac {A}{2V} \sqrt {\frac {kT}{m}}N. Solve this equation (assuming constant temperature) to obtain a formula of the form N(t)=N(0){e}^{{-t}/{\tau}} where \tau is the “characteristic time” for N (and P) to drop by a factor of e.

In my answer to part (a) I got the expression for the number of molecules colliding with the area of the hole over a given time interval as PA\Delta t/(2m\bar { { v }_{ x } }). Now I will let the time interval become infinitesimal, and rewrite this as \frac {PAdt}{2m\bar {{v}_{x}}}. Now once you poke a hole in the container, what was previously a number of collisions becomes a number of molecules leaving the container, which turns N into a decreasing function of time, with the infinitesimal change in N(t) given by dN = -\frac {PAdt}{2m\bar {{v}_{x}}}.

By using the ideal gas law and bringing the dt to the lefthand side, I can rewrite the above as \frac {dN}{dt} = - \frac {N(t)kTA}{2Vm\bar {{v}_{x}}} (notice that from the ideal gas law is now N(t), a value that is varying with time). As shown before, kT=m\bar {{v}_{x}^{2}}, so I can rewrite this \frac {dN}{dt} = - \frac {N(t)m\bar {{v}_{x}^{2}}A}{2Vm\bar {{v}_{x}}}.

Using the approximation \sqrt {\bar {{v}_{x}^{2}}}=\bar {{v}_{x}} and the result from part (b), I can express the above as \frac {dN}{dt} = - \frac {N(t)\bar {{v}_{x}}A}{2V} = -\frac {N(t)A}{2V}\sqrt {\bar{{v}_{x}^{2}}} =-\frac {N(t)A}{2V}\sqrt {\frac {kT}{m}}. Reorganizing the terms, I get the first sought expression, \frac {dN}{dt} = -\frac {A}{2V}\sqrt {\frac {kT}{m}}N(t). (The textbook expresses that final term as instead of N(t), but I’m pretty sure that it has to be a function of time since we have a derivative of N with respect to t on the other side, so for the sake of clarity I’m expressing it as a function in my answer.)

The above is a separable differential equation, so \int {\frac {dN}{N(t)}} = \int {-\frac {A}{2V}\sqrt {\frac{kT}{m}}dt} \rightarrow \ln {N(t)} = -\frac {A}{2V}\sqrt {\frac{kT}{m}}t + C. Exponentiating both sides gives N(t) = {e}^{c}{e}^{-\frac {A}{2V}\sqrt {\frac{kT}{m}}t} = B{e}^{-\frac {A}{2V}\sqrt {\frac{kT}{m}}t} for some constant B. Setting t=0 gives N(0)=B, so if we plug that in and we let \tau = \frac {2V}{A\sqrt{\frac{kt}{m}}}, then we get the sought equation, N(t)=N(0){e}^{-\frac {t}{\tau}}.

The rest of the parts of this problem are just using the equations for N(t) and \tau to solve how long it takes air to escape through a given hole in a given container, or figure out how big a hole was given how long it takes a tire to deflate. Things like that. The most interesting one is part (f):

(f) In Jules Verne’s  Round the Moon, the space travelers dispose of a dog’s corpse by quickly opening a window, tossing it out, and closing the window. Do you think they can do this quickly enough to prevent a significant amount of air from escaping? Justify your answer with some rough estimates and calculations.

In the preliminary chapter of Round the Moon, the spacecraft is described as a shell with an interior diameter of 84 inches (108 inch diameter with 12 inch thick walls), giving a radius of approximately one meter. (Given that the work was originally in French, I would bet that in the original the diameter is two meters and a conversion was made for the English translation.) Since there are many compartments in the shell, one above another, and they have to hunt “a long time” to find the dog in the upper compartment, I will say the shell is 90 feet long with 10 compartments, which would make it 1/10 as long as the gun it as fired from. Since they had to hunt for the dog, I assume all the compartments are open to each other, and since it makes things simpler I declare the shell to be cylindrical (even though it isn’t). If you plug in these values you get a volume of approximately 86 cubic meters. Assume the area of the window is 0.2 cubic meters and the shell is filled with oxygen, giving a mass of 5 x 10^-26 kg. Recalling that \tau = \frac {2V}{A\sqrt{\frac{kt}{m}}}, using these numbers and room temperature gives a characteristic time of about 3 seconds. That’s the amount of time for the air in the capsule to reduce by a factor of e, so after 3 seconds more than half the oxygen is gone. Throwing the body of Satellite the dog into space is a bad idea.

For chapter 1.3, the problems are all plug-and-play scenarios with the equipartition of energy theorem, except for

Problem 1.25: List all the degrees of freedom, or as many as you can, for a molecule of water vapor. (Think carefully about the various ways in which the molecule can vibrate.)


A water molecule is shaped like this, so to my mind it has nine quadratic degrees of freedom. There are the three translational degrees in the x, y, and z axes, and the three rotational degrees as well. (If it were radially symmetric there would be fewer rotational degrees, but it’s not.) Then there is the possibility of vibration in those bonds. One vibration mode would have the bond distance between the oxygen and the hydrogens oscillating (imagine holding the oxygen and pulling the hydrogens away and letting them bounce back). One would have the bond angles oscillating (imagine holding the oxygen and waving it up and down so the hydrogens flap around it). And one would be a torsional oscillation (imagine holding onto the hydrogens and twisting the oxygen back and forth). Those are the ones I can think of, and Googling tells me I have the number right. There’s a chance my physical descriptions are off, though. Chemistry, bonds; these were never my strong suit. I’m just imagining the molecule as a toy made of balls and springs and trying to figure it out from that.


Thermo Thursday: chapter 1.1

1.1: Thermal Equilibrium This chapter introduces the several basic definitions of temperature, starting with the operational (“temperature is what you measure with a thermometer”) and ending with a qualitative theoretical definition (“Temperature is a measure of the tendency of an object to spontaneously give up energy to its surroundings. When two objects are in thermal contact, the one that tends to spontaneously lose energy is at the higher temperature.”) Some time is spent on what exactly thermal contact entails–that the objects are in some way able to exchange energy, and that this energy is called heat. When the two objects have been in contact long enough, we learn that they are considered to be in thermal equilibrium, and that the time required to come to thermal equilibrium is the relaxation time. Curiously, Schroeder doesn’t say in the definition that the two objects are in thermal equilibrium when they’ve reached the same temperature. This is probably because of the way he defined relaxation time, which gets a footnote that some authors consider the relaxation time as the time required for the temperature difference between the two objects to decrease by a factor of e. The most interesting part of the section is an analogy of thermal equilibrium to other kinds of equilibrium. Schroeder gives the example that if you pour cold cream into a hot cup of coffee, the relaxation time of the cream/coffee system is very short, but the relaxation time for the coffee to come to thermal equilibrium with the room is still very long. But he notes that there are more kinds of equilibrium on display in the example; the cream in the coffee reaches diffusive equilibrium when it is blended with the coffee such that its molecules have no greater tendency to move in one direction than another. Also defined is mechanical equilibrium, when large-scale motions can take place, but no longer do. Whereas the exchanged quantity to reach thermal equilibrium is energy, the exchanged quantity for diffusive equilibrium is particles, and for mechanical equilibrium, volume. The rest of the section describes how thermometers work, and introduces the concepts of absolute zero and the kelvin scale. These things are sufficiently basic that I’m not going to synopsize them.

I don’t think I’ll always work every problem in a chapter, but to start out I’m going to.

Problem 1.1: The Fahrenheit temperature scale is defined so that ice melts at 32°F and water boils at 212°F. (a) Derive the formulas for converting from Fahrenheit to Celsius and back. (b) What is absolute zero on the Fahrenheit scale?

These are two linear scales, meaning that every degree F is the same size as every other degree F, and equivalently for every degree C. But F degrees and C degrees are different sizes, and the scales don’t start in the same place. So first I’ll figure out how do describe the size of a degree F in terms of a degree C, or the number of degrees F per degrees C. The magnitude of the temperature change between where water freezes and where it boils is the same regardless of scale, so let be the size of a degree Fahrenheit and C be the size of a degree Celsius. Then,

( 212F - 32F ) = (100C - 0C).

Solving this equation gives us

F = \frac { 100 }{ 180 } C = \frac { 5 }{ 9 } C.

So one degree F is only five ninths the size of a degree C. So if we have a number of degrees Celsius, we have to divide by five ninths (same as multiplying by nine fifths) to get the equivalent number of degrees Fahrenheit. But there’s still the issue that the scales don’t start in the same place. When we’re at 0° C, we’re still at 32° F. So we need to add a constant factor of 32 to the equation above to adjust for the different zero-point, giving us a temperature conversion equation

{ T }_{ F }=\frac { 9 }{ 5 } { T }_{ C }+32 and { T }_{ C }=\frac { 5 }{ 9 } ({ T }_{ F }-32). (Solution to part a.)

For part b, we know from the chapter that absolute zero is -273°C, so

{ T }_{ F }=\frac { 9 }{ 5 } (-273)+32=-459. (Solution to part b.)

Problem 1.2: The Rankine temperature scale (abbreviated °R) uses the same size degrees as Fahrenheit, but measured up from absolute zero like kelvin (so Rankine is to Fahrenheit as kelvin is to Celsius). Find the conversion formula between Rankine and Fahrenheit, and also between Rankine and kelvin. What is room temperature on the Rankine scale?

Degrees Fahrenheit and Degrees Rankine are the same size, the scales just start in a different place. We know from the previous problem what absolute zero in °F is, so

{ T }_{ R }={ T }_{ F }+459.

Rankine and kelvin start in the same place, but have different degree sizes. The sizes are the same as for Fahrenheit and Celsius respectively, so we already know that one degree Rankine is five ninths the size of one kelvin. (I’ll note here that, for no reason I’m aware of, it’s considered improper to say “degree kelvin.”) So again we divide our kelvins by five ninths to get the number of degrees Rankine, giving us a temperature conversion equation

{ T }_{ R }=\frac { 9 }{ 5 } { T }_{ K }.

Room temperature is defined to be 300 K, so by the equation above, room temperature is also 540°R.

Problem 1.3: Determine the kelvin temperature for each of the following: (a) human body temperature; (b) the boiling point of water (at standard pressure of 1 atm); (c) the coldest day you can remember; (d) the boiling point of liquid nitrogen (-196°C); (e) the melting point of lead (327°C).

(a) 98.6°F = 37°C = 310 K. (b) 100°C = 373 K. (c) -10°F ≈ -23.3°C = 249.6 K. (d) -196°C = 77 K. (e) 327°C = 600 K.

Problem 1.4: Does it ever make sense to say that one object is “twice as hot” as another? Does it matter whether one is referring to Celsius or kelvin temperatures? Explain.

The book hasn’t explicitly defined “heat” yet, so I’m going to interpret this problem to be asking just about temperature scales. Without a defined scale, saying “twice as hot” is meaningless. Twice as many degrees Celsius is a much greater difference in temperature than twice as many degrees Fahrenheit. Also, suppose our first object is at 0°C. Within the Celsius scale, you can’t define what temperature an object “twice as hot” would be. On the kelvin scale, excepting the special case of absolute zero, the phrase “twice as hot” does always have a mathematically coherent definition. (I’m not sure that it has a physically coherent definition, but the book hasn’t gotten to specific enough definitions for me to make that argument.)

Problem 1.5: When you’re sick with a fever and you take your temperature with a thermometer, approximately what is the relaxation time?

I’m not sure what the intention with this problem is. Actually calculating this would both require information that hasn’t been introduced yet, and defining a specific type of thermometer. But I don’t see anything in the chapter to provide a basis for approximation. From anecdotal experience, the relaxation time (i.e. how long it takes for an oral thermometer to go from room temperature to the temperature of the inside of my mouth) is on the order of one minute.

Problem 1.6: Give an example to illustrate why you cannot accurately judge the temperature of an object by how hot or cold it feels to the touch.

The key here is judging accurately. Imagine that you touch an object that is at exactly your body temperature. Your body and the object are in thermal equilibrium, so there will be no transfer of energy, and you will feel no temperature difference. So you know that it’s the same temperature as you are, but what temperature is that? You can’t find out by touching yourself, as you are always in thermal equilibrium with yourself. You can estimate your own temperature, but that wouldn’t be accurate. Since body temperature is variable, it’s impossible to know where the “feels hot, feels cold” scale starts without recourse to a thermometer. (There’s also the issue that beyond certain temperature ranges the tissues of your body with do the sensing will be damaged, making you unable to feel temperature at all.)

Problem 1.7: When the temperature of liquid mercury increases by one degree Celsius (or one kelvin), its volume change increases by one part in 5500. The fractional increase in volume per unit change in temperature (when the pressure is held fixed) is called the thermal expansion coefficient, β:

\beta =\frac { { \Delta V }/{ V } }{ \Delta T }

(where V is volume, T is temperature, and the Δ signifies a change, which in this case should really be infinitesimal if β is to be well defined). So for mercury, β = 1/5500 K^-1 = 1.81 x 10 ^-4 K^-1. (The exact value varies with temperature, but between 0°C and 200°C the variation is less than 1%.)

(a) Get a mercury thermometer,


estimate the size of the bulb at the bottom, and then estimate what the inside diameter of the tube has to be in order for the thermometer to work as required. Assume that the thermal expansion of the glass is negligible.

Screw getting an thermometer. Let’s just assert that the bulb at the bottom has volume V of one cubic centimeter. Let’s also assert, because I’m making things up and I can, that the degree markings are a height h ofone millimeter apart. The amount of mercury that rises up the tube radius with a 1 K increase in temperature, then, is

\beta (V)(\Delta T)=\pi { r }^{ 2 }h

Solve for the radius of the cylinder,

r=\sqrt { \frac { \beta (V)(\Delta T)}{\pi h}} =0.02. So the radius is 0.002 cm. (Frustratingly, I haven’t been able to make the unit symbols parse in the LaTeX plugin, but the dimensional analysis checks out.

(b) The thermal expansion coefficient of water varies significantly with temperature: It is 7.5 x 10^-4 K^-1 at 100°C, but decreases as the temperature is lowered until it becomes zero at 4°C. Below 4°C it is slightly negative, reaching a value of -0.68×10^-4 K^-1 at 0°C. (This behavior is related to the fact that ice is less dense than water.) With this behavior in mind, imagine the process of a lake freezing over, and discuss in some detail how this process would be different if the thermal expansion coefficient of water were always positive.

Okay, this one is interesting. Imagine a lake getting colder and colder. The positive thermal expansion coefficient means that, above 4°C, the colder the water is the denser it is, so the coldest water will sink to the bottom. (This matches my intuition that the bottom of a swimming pool is colder than the surface.) Let this behavior continue until, hypothetically, we have an entire lake in which all the water is at 4°C. Then let some of the water in that lake get colder. Because the thermal expansion coefficient turns negative now, the colder water will be less dense, and will start rising to the top. That means that by the time the first water gets down to 0°C and freezes, it will be at the surface, and the water below will be warmer. So lakes freeze from the surface down. If the thermal expansion was always positive, I would expect lakes to freeze from the bottom up.

Problem 1.8: For a solid, we also define the linear thermal expansion coefficient, α, as the fractional increase in length per degree:

\alpha =\frac { { \Delta L }/{ L } }{ \Delta T } .

(a) For steel, α is 1.1 x 10^-5 K^-1. Estimate the total variation in the length of a 1-km steel bridge between a cold winter night and a hot summer day.

Let’s say that a hot summer day is 43°C, and a cold winter night is 0°C. For an L of 1km, we have

\alpha (L)(\Delta T)=\alpha (1)(43)=0.00047

in units of kilometers, which is just under half a meter.

(b) The dial thermometer in Figure 1.2 uses a coiled metal strip made of two different metals laminated together. Explain how this works.

The two different metals have different thermal expansion coefficients. Assume that they are the same length when laminated together at some thermal equilibrium. When the temperature changes, the strips will not be able to expand linearly because they are attached to each other and changing by different lengths. As such, there will be a lateral displacement; the laminated strip will curl. This curling is proportional to the change in temperature.

(c) Prove that the volume thermal expansion coefficient of a solid is equal to the sum of its linear expansion coefficients in the three directions: \beta ={\alpha}_{ x }+{\alpha}_{y}+{\alpha}_{z}. (So for an isotropic solid, which expands the same in all directions, β=3α.)

Recall that \Delta V={ V }_{ final }-{ V }_{ initial }. If {V}_{initial}=xyz, then it follows that

{ V }_{ final }=(x+\Delta x)(y+\Delta y)(z+\Delta z)\\ =xyz+\Delta xyz+x\Delta yz+xy\Delta z+x\Delta y\Delta z+\Delta xy\Delta z+\Delta x\Delta yz+\Delta x\Delta y\Delta z\\ \approx xyz+\Delta xyz+x\Delta yz+xy\Delta z

That last step works because all of the terms with multiple deltas are so tiny compared to the other terms that they are physically negligible, so I can drop them. This gives me

{\Delta V=V }_{ final }-{ V }_{ initial }=xyz+\Delta xyz+x\Delta yz+xy\Delta z-xyz\\ =\Delta xyz+x\Delta yz+xy\Delta z

Which I can plug into the equation for β to get

\beta =\frac { { \Delta V }/{ V } }{ \Delta T } =\frac { { (\Delta xyz+x\Delta yz+xy\Delta z) }/{ (xyz) } }{ \Delta T } =\frac { ({ \Delta x }/{ x })+({ \Delta y }/{ y })+({ \Delta z }/{ z }) }{ \Delta T }\\ ={ \alpha}_{ x }+{ \alpha}_{ y }+{ \alpha}_{ z }

thus proving the theorem.

Introducing Thermo Thursday


Once upon a time, long before I ever moved to Iowa or started teaching creative writing courses, I thought I was going to be a physicist. But I was the kind of physics student who occasionally got tired of doing math all day and would take a week off to write a story or make a wireframe sculpture or do large scale papercraft. Anything creative rather than analytical. I eventually followed those impulses out of the physical sciences and into my current position.

Now, though, I find my situation reversed; I spend my days thinking about literature and aesthetics, and discover that I am the kind of creative writing professor who intermittently longs to dust off his calculus. From time to time I’ve been pulling some of my favorite physics and math textbooks off the shelf and recreationally doing a couple of problems, but I’ve noticed that just jumping in and solving things is getting harder. My physics and math is atrophying from disuse. So I’ve decided to get more systematic about this. I’m going to set aside a day of the week for brushing up on the subjects that most interest me, starting with thermodynamics. Welcome to Thermo Thursday.

I’ll be using the book An Introduction to Thermal Physics by Daniel V. Schroeder, which I remember thinking highly of back when it was assigned in college. Every Thursday I’ll work through some sections of the book, synopsize the key concepts and work some problems here. (This will hopefully also give me a chance to brush up my LaTeX skills.) My math may be rusty enough that I reach a point where I have to stop the physics and refresh my fundamentals on more basic concepts, but I’ll deal with that problem when I get to it. Let’s get started.

Introduction to the Holographic Principle

Inspired by a conversation with a friend, and my growing interest in Erik Verlinde’s entropic construction of gravity, here is a video of Raphael Bousso of UC Berkeley giving a fairly non-technical lecture explaining the holographic principle.  Dr. Bousso does a good job of building up to concepts like Planck length and Schwarzschild radius (a term I don’t recall him actually using in the video) from simple principles.

An aside: I don’t recall having been explicitly introduced to the idea that the density of a black hole decreases as its mass increases before.  In fact I’m not sure that black hole density ever came up at all in my modern physics class, which is the only place I  worked with Swarzschild radii.  It’s obvious once presented, but when I got to that part of the video I was shocked to find I had been completely unaware of such a fundamental characteristic of black holes.