I dont know about outside, but in my eight storey buiding without A/C but with good insulation and heating the same in all apartments, it is definately warmer on the seventh floor than the ground floor. It is true of the rooms and particularly of the stairwell. I can only assume that it is a result of warm air rising through the building.
From what I have read about construction work, it’s often much more windy at the top of skyscrapers than at ground level. That would account for some cooling … wind chill effect.
PV = nRT applies to gases, period. The model deviates a bit in dynamic conditions (it’s a steady-state model), but it most definitely applies to gas in the atmosphere (i.e., air). Because we have gravity on this planet our atmospheric pressure is higher at lower altitude, so P goes down as one goes up. T goes down accordingly. Very, very simple model.
The statement: "In fact, an object sitting at earth’s orbit, and rotating, will have a mean temperature comparable to the earth’s - even though the pressure is zero. " doesn’t seem to have anything to do with gases, as near as I can tell.
“PV = nRT explains what happens to closed containers as you play with the four variables”, isn’t correct, as there is no assumption about closed containers. n moles of a gas will behave as described. Whether or not there is a container doesn’t affect the basic gas laws.
Connie,
Your friend is correct. If it’s winter in Chicago and you wish you were lying on the beach in Rio, just do the following…
[list=1][li]Head for the Sears Tower or any other convenient large building – with beach chair, suntail oil, bottle of rum, etc. [/li][li]Especially the bottle of rum.[/li][li]Proceed to the top floor.[/li][li]Do whatever you need to in order to gain access to the roof (good luck in these post 9-11 times)[/li][li]Locate one of the giant exhaust fans you’ll find there and set up your chair under or near it.[/li][li]Close your eyes and imagine real hard that you’re at the beach. The rum can be a useful aid here.[/li][li]Click your heels together three times and say, “There’s no place like Rio, There’s no place like Rio”[/li][/list=1] Of course you should realize the unfortunate truth that Rio, like so many other things, probably wouldn’t seem as “hot” to you (or me) today as when we were twenty-two years younger 
Absolutely! But it is also insufficient in this case. I said a closed container, because then you have a definite volume. What is the volume to which you are applying the ideal gas law? Clearly not the whole atmosphere. In fact, it only applies to volumes small enough so that the terms “pressure” and “temperature” are defined. So you pick small enough parcels of atmosphere of similar volume in a column within the atmosphere. Now, the pressure drops with altitude. This does not necessarily imply that the temperature drops! Why? Well, the one thing you know for sure is that the number of moles in the parcels drops with altitude. You can only determine what happens to the temperature when you have a model for the change in the number of moles.
No matter how you slice it, the ideal gas law is an equation with four unknowns. If you are measuring one, the pressure in your case, then you can only find a second, the temperature, when you have some mechanism for eliminating the other two. (In my example, I picked the volume to be some constant, arbitrary value.) No insistence that the ideal gas law applies to the atmosphere will change this mathematical fact.
To see that it is not all just esoteric, physics mind games, go to this site: http://daac.gsfc.nasa.gov/CAMPAIGN_DOCS/ATM_CHEM/atmospheric_structure.html
A brief quote:
Now, I happen to know that the pressure within the stratosphere does not increase with altitude, but drops exponentially, so that it is a fact that the temperature increases as the pressure decreases.
The temperature rises in the stratosphere because the expansion is not adiabatic - ozone is absorbing UV light so energy is entering the system.
Ah, apologies for the double post. I typed this out earlier but I couldn’t get it to post properly. This is my first time posting so I hope you will bear with me if I step on any toes.
Adiabatic (no energy enters or leaves) expansions can easily be explained with PV = nRT. For a given parcel of air (n constant) rising through the atmosphere, pressure § decreases exponentially but volume (V) cannot increase exponentially because of the weight of the air above. Hence, temperature (T) must decrease to compensate.
More rigorously from thermodynamics, dU = -PdV for adiabatic expansions (q = 0), so dH = VdP. For an ideal gas dH also equals Cp * dT, so combining the two equations (and substituting V = nRT/P) gives Cp/T * dT = nR/P * dP. Since Cp, T, n, R, and P are all positive, then the sign of dP must equal the sign of dT. If the change in pressure is positive, then the change in temperature must be positive.
This holds true for all adiabatic expansions.
The problem with the stratosphere is that the adiabatic conditions don’t hold. Ozone absorbing UV light pumps energy into the system, so the expansion is no longer adiabatic and the equation doesn’t hold.
The problem with the interior of a building is that the roof holds up the weight of the air above it. The pressure inside the building does not decrease exponentially (unless someone opens a window) and hence, there is no drop in pressure from the ground floor to the top floor.
Welcome aboard Dolphin, nice first post.
I agree with almost everything you said. You are absolutely correct in asserting that you must assume an adiabatic expansion in order to use the ideal gas law. That was a point I was trying to make.
A couple of nits, however. The weight of the air above is not the reason the volume can not increase exponentially; the weight of the air above is the source of the pressure in the first place. So, if the pressure decreases exponentially, the volume could increase exponentially. But, of course, it doesn’t, because of the constraints of adiabatic expansion.
Also, the building is either a sealed container, or it isn’t. If it isn’t, then the pressure within the building will decrease exponentially with altitude. If it is sealed, then it is its own system, and obeying the same laws as the earth, the pressure will again decrease exponentially. It seems to me that some Chicagoans would know if high speed elevators can make your ears pop - due to the sudden pressure drop.
About the ideal gas law… True, we don’t have a reference for V, the volume, but then, we also don’t have a reference for N, the number of molecules. Why not re-arrange the equation to P = [symbol]r[/symbol]kT (where [symbol]r[/symbol] is the number of molecules per volume), and thereby cancel out our ignorance? Density is a perfectly well-defined quantity, regardless of what volume we choose.
(Note for chemists: I used the logical, physicist’s version of the Ideal Gas Equation. If you prefer, you can multiply and then divide the right-hand side by a large, arcane number whose significance is lost to most students and which isn’t really relevant here anyway. I prefer my equations simple.)
Correct me if I am wrong Chronos, but surely the density is also a function of altitude? And, it is the fact that density does not drop exponentially (uniformly) with altitude, while pressure does, that leads to the temperature dropping with altitude (below the stratosphere).
Dolfin implicity assumed N to be a constant (and so is following the same chunk of air), and explicitly that dQ vanished, so that dU = PdV. This then allows using the ideal gas law.
As far as getting a tan is concerned, the top of the building definitely would be better (although perhaps not significantly). There’s slightly less atmosphere for the UV to pass through, and if there’s a lot of pollution at ground level, then there’s less ozone to pass through as well.