Section opens with the ideal gas law in the form that I was taught it in high school, PV=nRT where P is pressure, V is volume, T is temperature in kelvins, n is the number of moles of gas, and R is the ideal gas constant, R = 8.31 J/(mol K). Also from high school, a mole of molecules is Avogadro’s number of them, 6.02 x 10^23.
For physics it is often more useful to talk about the number of molecules, rather than the number of moles, so we multiply n by Avogadro’s number to get N, the number of molecules of gas. This transforms the ideal gas law into PV=NkT where k is Boltzmann’s constant, k = 1.381 x 10^-23 J/K. The section then shows how if you consider an ideal gas one molecule at a time, you can eventually conclude that this constant k is essentially a conversion factor between temperature and molecular energy. From there it introduces the electron-volt (eV) as a more convenient unit than the Joule to talk about the tiny kinetic energies of molecules, and derives the equation for the root-mean-square velocity of a molecule in an ideal gas, {v}_{rms} = \sqrt {\frac {3kT}{m}}.
Interesting problems:
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 has a greater mass of air? Explain carefully.
The rooms are the same size, so the volume of the two rooms is the same. The rooms are connected by an open door, so the pressure of the two rooms is the same. The only variable term in the ideal gas law between the two rooms is temperature. Expressing the ideal gas law in terms of the number of molecules in each room, I get N=\frac {PV} {kT}. Since the temperature term is in the denominator, the room with the higher temperature is going to contain a lower number of gas molecules, and therefore a lower mass of air. So the colder room, room B, has the greater mass of air.
Problem 1.12: Calculate the average volume per molecule for an ideal gas at room temperature and atmospheric pressure. Then take the cube root to get an estimate of the average distance between molecules. How does this distance compare to the size of a molecule like { N }_{ 2 } or { H }_{ 2 }O?
From \frac { V }{ N } =\frac { kT }{ P } we get an average volume per molecule of 4.1 x 10^-26 cubic meters. The cube root gives us an average distance between molecules of 3.4 x 10^-9 meters, or 3.4 nm. The size of the nitrogen gas molecule is on the order of tens of picometers, so 100 times smaller. The size of a water molecule is on the order of hundreds of picometers, so 10 times smaller.
Problem 1.16: The exponential atmosphere.
(a) Consider a horizontal slab of air whose thickness (height) is dz. if this slab is at rest, the pressure holding it up from below must balance both the pressure from above and the weight of the slab. Use this fact to find an expression for dP/dz, the variation of pressure with altitude, in terms of the density of air.
It’s embarrassing how rusty my math has gotten. I had to look back at my old work to recall how to start solving this problem. Density is mass per unit volume, and usually denoted by \rho (the Greek letter rho). If you assume a fictitious “air molecule” with mass m, then for a slab of air the density would be given by \rho = \frac{Nm}{V}. Since I’m assuming that the air is uniform horizontally and only varies in the z direction, let the area of the top of the slab go to zero, which gives \rho = \frac {Nm}{dz}. Then the mass of the air in the slab is just \rho dz.
If the pressure below the slab is equal to the pressure above plus the weight of the slab, then we have
P(z) = P(z+dz) + (\rho dz)gwhat can be rewritten
P(z) - P(z+dz) = \rho g dzIf the change in pressure dP = P(z + dz) - P(z), then the left side of the equation above is -dP. So I have
-dP = \rho g dzwhich gives me
\frac {dP}{dz} = -\rho g.
(b) Use the ideal gas law to write the density of air in terms of pressure, temperature, and the average mass of air molecules. Show, then, that the pressure obeys the differential equation \frac {dP}{dz}= -\frac {mg}{kT}P, called the barometric equation.
Let’s consider my fictitious air molecule from the previous part to have a mass equal to the average molecular mass of air. (If I were doing this for real, I’d need to make use of the fact that air is 78% molecular nitrogen, 21% molecular oxygen, and 1% argon. But that seems tedious.) Going back to the definition of density, I can rewrite my last result
\frac {dP}{dz} = -\rho g = -\frac {Nm}{V}g.
Then I can use the ideal gas law PV=NkT \rightarrow \frac {N}{V}=\frac {P}{kT} to show
\frac {dP}{dz} = -\frac {P}{kT}mg = -\frac {mg}{kT}P.
(c) Assuming that the temperature of the atmosphere is independent of height (not a great assumption but not terrible either), solve the barometric equation to obtain the pressure as a function of height: P(z) = P(0){e}^{{-mgz}/{kT}}. Show also that the density obeys a similar equation.
Finally, it’s time to pull out my hideously rusty calculus skills. I only really remember how to solve this for a separable differential equation. Fortunately, it looks like this is one; it fits the form \frac {dy}{dt}=g(t)h(y). So I can take my differential equation from part b and do this
\frac {dP}{dz} = -\frac {mg}{kT}P \rightarrow \frac {dP}{P} = -\frac {mg}{kT} dz.
Now I integrate both sides of the separated equation
\int {\frac {dP}{P}} = \int{-\frac {mg}{kT} dz} \Rightarrow \ln {P} = -\frac {mgz}{kT}+C.
To solve for pressure I exponentiate both sides
P(z) = {e}^{{-mgz}/{kT}+C}={e}^{C}{e}^{{-mgz}/{kT}}.
Notices that when z=0, P(z)=P(0)=e^C. Thus I can rewrite the constant term and get the sought equation,
P(z) = P(0){e}^{{-mgz}/{kT}}.
For density, I recall that \rho = \frac {Nm}{V}, which lets me rewrite the ideal gas law as P=\frac {NkT}{V}=\frac {\rho kT}{m}, and thus \rho = \frac {Pm}{kT}. Plugging that into my equation for pressure, I get
\rho (z) = \frac {m}{kT} P(z) = \frac {m}{kT}P(0){e}^{{-mgz}/{kT}}.
From the ideal gas law above, though, we can say that \frac {P(0)m}{kT}=\rho (0), so I can rewrite the above as
\rho (z) = \rho (0){e}^{{-mgz}/{kT}}.
Problem 1.22: If you poke a hole in a container full of gas, that gas will start leaking out. In this problem you will make a rough estimate of the rate at which gas escapes through a hole. (This process is called effusion, at least when the hole is sufficiently small.)
(a) Consider a small portion (area = A) of the inside wall of a container full of gas. Show that the number of molecules colliding with this surface in a time interval \Delta t is PA\Delta t/(2m\bar { { v }_{ x } }), where P is the pressure, m is the average molecular mass, and \bar { { v }_{ x } } is the average x velocity of those molecules that collide with the wall.
First, we need to consider only those molecules that are able to reach the area A in the time allowed. Those would be only those molecules in the volume given by A\bar {{v}_{x}}\Delta t. The number of molecules in this volume would be (\frac {N}{V})A\bar {{v}_{x}}\Delta t, which I can use the ideal gas equation to rewrite as \frac {P}{kT}A\bar {{v}_{x}}\Delta t. That’s the number of molecules in the volume we’re concerned with, but I wouldn’t expect all of them to be moving toward the wall of the container. At any given time I’d expect half of them, on average, to me moving toward the wall of the container, and the other half to be moving away. So let’s express the molecules in the relevant volume that are actually going to hit the wall as \frac {1}{2}\frac {P}{kT}A\bar {{v}_{x}}\Delta t.
A result from earlier in the chapter had that kT=m\bar {{v}_{x}^{2}}, so I’ll invoke that now to rewrite the previous expression as \frac {1}{2}\frac {P}{m\bar {{v}_{x}^{2}}}A\bar {{v}_{x}}\Delta t. Then I cancel to get the sought expression, PA\Delta t/(2m\bar { { v }_{ x } }).
There’s more to this problem, including developing a differential equation for the number of molecules that escape from the container as a function of time once you punch a hole in it. But I had to interrupt Thermo Thursday this week for a student meeting, so I’m not going to get to them here.