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 N 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 N 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.