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This is the next instalment in Phelonius’ Pedantic Solutions which aims to tackle something I wish I had when I was doing my physics degree: some step-by-step solutions to problems that explained how each piece was done in a painfully clear way (sometimes I was pretty thick and missed important points that are maybe obvious to me now, looking back). I have realized I could have benefited from a more integrated approach to examples, where whatever tools that are needed are explained as the solution is presented, so that is at the core of these attempts. I have also been told all my life that the best way to learn something is to try to teach it to someone else, and so this is part of my attempt to become more agile and comfortable with the ideas and tools needed to do modern physics. If you don’t understand something, please ask, because I haven’t explained it well enough and others will surely be confused as well.
Things covered:
- the Ideal Gas Law equation
- basic problem solving techniques
As I make my way through Schroeder’s “Thermal Physics” (for reasons to become clear later), I am reading all the questions to make sure I think I know how to solve them (I’m not solving them all) and ran across one I couldn’t do in my head right away. It’s a fun one though. Problem 1.11: “Rooms A and B are the same size, and are connected by an open door. Room A, however, is warmer (perhaps because its windows face the sun). Which room contains the greater mass of air? Explain carefully.” I knew the answer off the top of my head from experience, but how to “explain carefully”? What was interesting is that some assumptions needed to be made to arrive at an answer, and that’s the problem-solving part and why I thought I would share my reasoning.
At first, trying to reason it in my head brought the “math meme” to mind: “If you have 4 pencils and I have 7 apples, how many pancakes will fit on the roof? Purple, because aliens don’t wear hats”, but when I sat down with a pencil, it soon sorted itself out. To start, we are given one whole equation to work with by this point in the book, the Ideal Gas Law:
\[PV = nRT\]
Luckily, that’s enough to be able to answer the question mathematically (which, one would hope, is a careful enough explanation to satisfy the question’s request). This equation states that pressure, \(P\), times volume, \(V\), is equal to the number of moles of gas within the given volume, \(n\), times a constant, \(R\), times the temperature, \(T\) (in Kelvins). A mole is simply a count, specifically there are about \(6.02\times 10^{23}\) molecules (or atoms, or whatever is being counted) in a mole of a substance (whatever its phase, be it gaseous, liquid, or solid, for instance). \(R\) is about \(8.31\,J\cdot mol^{-1}\cdot K^{-1}\), which is just a compact way of writing:
\[R = 8.31\frac{J}{mol^1\cdot K^1} = 8.31\frac{J}{mol\cdot K}\]
(since anything to a negative exponent is the reciprocal of that thing without the negative exponent, and anything to the power of 1 is just that thing). And Kelvins must be used to specify temperature when doing thermal physics (not entirely true, but a good practice: changes in temperature can use any scale, but specifications of absolute temperature must use an absolute scale like Kelvin). A change of one degree Celsius is the same as a change of one Kelvin, but the zero for the Kelvin scale is absolute zero (nothing can be colder) whereas the zero for Celsius is the freezing point of water (which is \(273.16 K\), fyi). Pro tip: apparently one is never supposed to say “degrees Kelvin” like we say “degrees Celcius”, it’s just “Kelvin”. I don’t know why.
So let’s see what we know…
The temperature of one room is higher than the temperature of the other room, and we are told \(T_A > T_B\) (the temperature of room A is greater than, or higher, than room B). The volumes of the two rooms are equal, so \(V_A = V_B\). \(R\) is a constant, so that’s easy since it will be same everywhere (which is a deep observation, by the way, reflecting that the laws of physics are the same everywhere in the universe). We are left with pressure and the number of moles to figure out. The first thing to realize is that the number there is of something in a given volume determines how heavy it is, so the more of it that there is, the more mass it has, so it is going to be the relationship between \(n_A\) (the amount of gas in room A) and \(n_B\) (the amount of gas in room B) that is going to answer the question.
What about pressure? The more of something there is an a given volume, the higher the pressure (try putting a balloon in a coffee can and blowing into it… the pressure will eventually be too high to blow into it anymore), but that’s a bit of a red herring on this question. Since we need to figure out what’s going on with pressure, we have to figure out the relationship between \(P_A\) and \(P_B\). What’s more useful here is to realize that air will move from an area of high pressure to an area of low pressure in an attempt to reach a state of equilibrium. For weather, air moves from areas of high pressure to areas of low pressure, in a phenomenon we call wind. The same happens moving between rooms that have different pressures: the air from the higher pressure room will blow into the lower pressure room. The thing is, in our simple system, there are two closed volumes with an opening between them, so the amount of air in total is constant: there is no place for it to go, no fans to keep bringing in new air or vents to let out existing air. If one room has a different pressure than the other, there will be a movement of air into the lower pressure room. If the wind kept blowing forever, we would have unlimited clean energy and the world’s energy problems would be solved with just a pair of rooms, but this is not the case. Since our system can’t be some magical perpetual motion machine, the wind must stop and it does so because the pressures in the two rooms must eventually be the same! Even if one room changes its pressure (say, when the sun comes up in the morning into the windows of room A), the air is free to flow into the other room and eventually they balance out. So, and this is the key to solving the problem, \(P_A = P_B\). We now have only one pair of variables, \(n_A\) and \(n_B\) left to figure out, which is where we needed to be.
We then have two equations, one for room A and one for room B:
\[R = \frac{P_A\cdot V_A}{n_A\cdot T_A}\,\,\,\,\,\text{and}\,\,\,\,\, R = \frac{P_B\cdot V_B}{n_B\cdot T_B}\]
or more compactly:
\[R = \frac{P_A V_A}{n_A T_A}\,\,\,\,\,\text{and}\,\,\,\,\, R = \frac{P_B V_B}{n_B T_B}\]
which we can do by re-arranging \(PV = nRT\) for each of the rooms (essentially divide both sides by \(nT\) to get \(R = PV/nT\). And since \(R = R\), we can then write:
\[\frac{P_A V_A}{n_A T_A} = \frac{P_B V_B}{n_B T_B}\]
But we know that \(V_A = V_B\) and \(P_A = P_B\), so substituting on the right hand side, we get:
\[\frac{P_A V_A}{n_A T_A} = \frac{P_A V_A}{n_B T_B}\]
Multiplying both sides by \((P_A V_A)^{-1}\):
\[\frac{1}{P_A V_A}\cdot\frac{P_A V_A}{n_A T_A} = \frac{1}{P_A V_A}\cdot\frac{P_A V_A}{n_B T_B} \rightarrow \frac{1}{n_A T_A} = \frac{1}{n_B T_B}\]
We know the relationship between the temperatures, so we want them on one side of the equation, and we want that relationship to tell us about the relationship between the number of moles of gas in each room, so we want those on the other side of the equation. This can be done by multiplying both sides by \(T_A\cdot n_B\):
\[T_A n_B\cdot\frac{1}{n_A T_A} = T_A n_B\cdot\frac{1}{n_B T_B}\rightarrow \frac{n_B}{n_A} = \frac{T_A}{T_B}\]
Since we know \(T_A > T_B\), then this equation says that the ratio between \(n_B\) and \(n_A\) must be the same, and therefore \(n_B > n_A\). So, there are more molecules (a greater number of moles) in room B than in room A. Since more molecules is heavier (more massive) than less molecules, then the air in room B (the cooler room) has a greater mass than the air in room A (the warmer room). No aliens or pancakes are required.
When flying aircraft, pilots are both trained to know and experience it pretty quickly themselves once they’re behind the controls. On cold days, your plane will take off sooner from the runway and climb faster than it will on warm days. Cold air has more air molecules per unit of volume, so there is just that much more for the wings to push against for a given forward velocity. As an aside, flying on cold days is usually a lot less “bumpy” than on warm days because there’s usually less convection in the air on cold days, but that’s another story.
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