Re: A physics question Message #20 Posted by Rich Messeder (US) on 16 June 2012, 7:02 a.m., in response to message #19 by bill platt
Thank you! Yes, I think that I should like to teach, for I regard teaching highly, considering the impact of teaching on shaping students' lives.
I had a few minutes so took the discussion a bit further for those who might want to know whence comes weight in the discussion.
Derive the pressure at the bottom of a column of air from the mass of the air in the column.
Ideal Gas Law (r = rho, density):
R = ideal gas constant
k = Boltzmann's constant
m = molecular mass of air
N = # of molecules in a sample of air
n = # of moles in a sample of air
V = volume of a sample of air
r = density of air
P = pressure of air
T = temperature of air (K)
h = altitude above Earth's surface
g = acceleration of gravity
PV = nRT = NkT
P = NkT/V = rkT/m
r = mP/kT
P0 = psia @ sea level
Imagine a slab of area A and thickness dh. In hydrostatic equilibrium, the change in pressure over an infinitesimal change in altitude must oppose the gravitational force on the air in that infinitesimal layer. The change in pressure, dP, from the bottom to the top of the slab, dh, for some air density rho is (using calculus notation and implicit multiplication). Take care to distinguish twixt an expression with variables in it, where, for example, m means mass, and an expression that has units, where, for example, m means meters.
[P(h) - P(h+dh)] A = r g A dh.
r A dh is: the density times the volume of the slab = mass
This leads us to [P(h+dh) - P(h)] / dh = - r g, where [P(h+dh) - P(h)] / dh is the definition of dP/dh.
Substituting for r, we find
dP/dh = - r g = -mgP/kT
This leads us to an expression for pressure as a function of height, assuming some fixed temperature (not too bad an assumption when T is measured in K)
P(h) = P0 exp (-mgh/kT) where mgh is potential energy, and kT is thermal energy (units cancel).
Going back to
dP/dh = - r g, we can rearrange as dP = - r g dh
If we integrate both sides (wrt P on the left and wrt h on the right) from some height h to infinity, we have (pardon trying to do calculus in a text base)
- integ (P(h), P(inf), dP) = integ (h, inf, g r dh).
Because P(inf) = 0 (ideally), P(h) must equal integ (h, inf, g r dh), but this is just the weight of the air in the vertical column of unit cross-sectional area lying above that level. Unit analysis yields
g r h = g m/V h: m/s^2 * kg/m^3 * m = kg/(m s^2). This is the Pascal, the SI unit of pressure, which we commonly know in the US as pound/sq in, or psi, and it is not in units of "weight," which is W = m g, but rather in units of "weight"/area, i.e., N/m^2. I put "weight" in quotes because in US units we use weight specifically for the product m g, but in SI there is the concept of force and mass. Socially, this distinction is still confused, IMO, in a general sort of way because if I am asked how much I "weigh" I can respond with, 74 kg, which is not a measure of /weight/, but of /mass/, or I can respond with 160 lbs, which /is/ a measure of weight (force), and no one wants me to muddy the water with slugs ;-) In SI units, I really "weigh" 726 N, but that just makes me sound fat, and who needs that? ;-) My torque wrench has scales in ft-lbs and N-m, but I expect my tools to be technically correct.
Note that an important part of this discussion is my assumption that the volume enclosed by the closed end of the cylinder and the piston is a perfect vacuum. I think this agrees with the original condition of "with the piston completely at the bottom."
I think that one finds the most interesting discussions on this forum...the original 15C manual and the Advanced manual covered material, and presented it in a manner, that one simply does not see today, and I think that anyone who enjoys that sort of material has an interesting mind.
Yours, &c.
~R~
Edited: 16 June 2012, 7:03 a.m.
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