Very nice Peter, your reply was what I was hoping for. It is derivations, such as this, that the student must involve themselves with and apply if they want to understand the elements of physics. Pressure and temperature are not "concepts" they are measurable properties (quantities), that are in fact mathematical aggregations of properties of the gas particles themselves. The epiphanies in this development, other than the grand epiphany that we are able to develop such things in the first place, are that pressure is actually a macroscopic measure of the gas particles "bouncing" off the walls and that temperature is a macroscopic measure of the kinetic energy of those particles. Also, mathematical derivation shows how terms are (conveniently) collected, and later mathematical application of this analysis will reveal that they were collected in just the right way. Even though the physics student will not realize this immediately, they will realize it eventually. This derivat! ion of pressure and temperature was not arbitrary. It was guided by our macroscopic notions of temperature and pressure. The original derivation took quite some time (100's of years) and many iterations to get it just right. Further, this analysis can now be applied to phenomena that makes sense to us, ice melting in a glass or water, and phenomena that doesn't make sense to us (yet), like refrigeration.
Generally, a solid (high school) physics class starts with algebra and then leads into calculus. This can be in tandem with a calculus class or not. If the students are following this type of analysis, they can get applied calculus (sans the continuity arguments) quicker than you think. But they must be on top of the algebra. I found physics a very enjoyable environment in which to learn single variable calculus. It is multi variable calculus that I think suffers without a solid mathematical development of calculus. So I am not suggesting that a student should shun calculus class, only that physics and calculus blends very well. They were made for each other.
On Aug 21, 2012, at 11:15 PM, Peter Duveen <email@example.com> wrote:
> Well, Bob, I was hoping you would ask about the temperature, and how it is related to the formula . > > First let me derive the formula. > > Imagine a tiny cube of breadth d filled with a gas at pressure P and density p. > > Divide the molecules of the gas, considered equal in mass, into three groups, each of which bounces back and forth between the two parallel sides of the cube, each traveling at the mean velocity v. > > Let's focus on one molecule going between one pair of the three pairs of parallel sides of the cube. > > The mass of the molecule is m. It hits one side and bounces off, resulting in a change of momentum of 2 mv. We also have the time between collisions of the molecule on that one side, because it has to go back and forth each time. we have vt = 2d, or t = 2d/v. > > Now if we divide the change in momentum by the change in time, we get the average force exerted on the one face by this single molecule. It is mv^2/d. If this is in turn divided by the area of a face of the cube, we have the pressure exerted on one face by that molecule, or > > mv^2/d^3. > > Now if there are N molecules in the cube, we have divided them all into three groups of N/3 molecules, each group bouncing between parallel sides of the cube. Thus, the pressure on one of the faces of the cube is N/3 times the pressure of one molecule, which we have already calculated, so it is N/3 mv^2/d^3 = P. But Nm is just the entire mass of gas in the cube, and that divided by d^3 is merely the density, which we shall call "p". So we finally have 1/3 pv^2 = P, or v^2 = 3P/p. > > You will notice that we have arrived at a rather important result using simple algebra. It is important because it says the microscopic characteristic of the molecular world, namely, the velocity of molecules, is related to the macroscopic quantities of the gas, P and p, which are fairly easily measured. > > Now where does the temperature come in? We use the conventional formula for the state of an ideal gas, PV = KT, where P is the pressure, V is the volume, K is a constant of proportionality, and T is the absolute temperature, of a sample of gas. K is specific to the sample. > > Simply solve for P, and put the value in our first formula. That is, P = KT/V > > Replacing P in our first formula, we have: > > v^2 = 3 KT/Vp > > Now note that Vp is nothing more than the entire mass of the sample of gas. Moving that to the left side of the equation, we have: > > mv^2 = 3KT, or > > 1/2mv^2 = 3/2 KT > >> From this equation, we can conclude that the kinetic theory of gases implies that the absolute temperature is directly proportional to the kinetic energy of the molecules. > > I erred in saying "speed" in my previous post. I should have said "kinetic energy." > > So, using simple algebra, and starting from first principles, we have shown that the molecular theory of gases implies that the mean velocity of the molecules is given by v = sqrt 3P/p, and that the temperature is proportional to the kinetic energy of the molecules. We have thus connected our microscopic theory to important macroscopic properties--tangible properties that can be measured. > > Bob, I hope this clarifies what I was referring to in my previous reply.