Earth Atmosphere properties:

Your argument is fundamentally correct, as in the end, the atmosphere is a very thin shell laying on the rocky planet, and because it's so thin, even if it expands, the gravitation attraction on it hardly changes as a function of altitude - same as on Earth. If you are in orbit and you suddenly stop moving, like the Road Runner cartoons, you will immediately feel the full force of gravity and fall like a rock just as if you were much closer to the surface.

Hence the pressure on the surface only depends on the net mass of the atmosphere - nothing more. Its just the weight of the CO

_{2} impinging on the surface - even if it expands it still weights the same. The distribution of density as a function of height doesn't matter as far a bottom pressure is concerned. Double the thickness of a thin atmosphere, and you still have a thin atmosphere. So pressure should not change until, as Mr. Skummins suggested, you lock the CO

_{2} into a solid, reducing the partial pressure of CO

_{2} in the atmosphere.

If you heat the atmosphere, this atmosphere will mostly just expand vertically into space, as it has nowhere else to go along the spherical surface, so the height/thickness H of the atmosphere must be strongly variant with Temperature, and if the atmosphere is thin, it will be a linear variation of temperature (you have only one dimension left for expansion) - see chart above. So Temperature is equivalent to atmosphere thickness - a linear relationship.

HOWEVER, still that doesn't excuse the use of constant density, because over that very small thickness, density does vary greatly from a big constant on the surface to zero at the edge. Same for Pressure.

If you want to

**prove** there is no effect of atmosphere thickness change on the bottom pressure, unfortunately, you will have to derive an equation that takes into account Pressure and Temperature as a function of height, and conflate that with the proper altitude H which depends on those non-linear curves above.

So I set to see if I could actually derive the "Zero Change" in pressure as a function of height/temperature

As it turns out, You have to divide the atmosphere into very thin layers where you could assume that P=ρgh was correct. Basically you need to use calculus. An infinitesimally thin layer of air of "Delta Y" thickness, dy, contributes a "Delta P" dP of pressure. Density is a function of altitude y, that is ρ=ρ(y)

So the contribution to pressure is

dP(y) = ρ(y)dy

And you need to sum an infinite set of layers from altitude zero to the edge of the atmosphere (integral P = ∫dP ):

P

_{o} = ∫ ρ(y)dy, where P

_{o} is the pressure on the surface of the planet

And using Ideal Gas Law: P= ρRT (R is the specific gas constant of CO

_{2} - not the universal one you used)

P

_{o} = (g/R) ∫ [P(y)/T(y) ]dy

Now we said the temperature varies linearly with altitude so one can write:

T(y) = [T

_{o}/H]y - T

_{o} in other words, a straight line... T

_{o} is the temperature on the surface.

Now this is where I got confused, because you are adding the weight of the atmosphere from top to bottom (weight of the column of air), so the negative sign above may in fact be positive (y coordinate is downwards) - I'm still trying to re-do my equations- I had to finagle a sign - this causes major troubles later on... but I'll ignore that for now...

Now the Pressure equation becomes:

P

_{o} = (gH)/(RT

_{o}) ∫ [P(y)/(H - y)]dy (that negative sign may be a plus)

So you really can't solve this because you don't know what the variation of P is as a function of y. Or can't we? This is actually a First Order Linear Differential equation masquerading as an integral.

Let's forget about end point conditions a bit and take a derivative (walking backwards, so to speak) - throw the constants away - I just want to see the "shape" of it:

dP(y)/dy = (gH)/(RT

_{o}) P(y)/(H-y)

or

dP(y)/dy + G(y) P(y) = 0,

where G(y) = -(gH)/(RT

_{o}), for convenience when using integral tables - I'm very rusty on my calculus folks.

The general solution looks like:

P(y) = C exp [-∫ G(y)dy ]

where C is some constant (I'm being very dirty and ignoring details here folks)

and if we re-write it (skipping a few steps, you can play with particular solutions and constants at home if you like)

P(y) = C exp{ +[ (gH)/(RT

_{o}) ∫ dy/(H-y) ] }

This is the moment of truth. This has a definite solution and not only can you tell P as a function of y but you can hold y at a constant value and play with H to see what effect changing the atmosphere has on the ground pressure, P

_{o}If from tables ∫ dy/(H-y) = -ln(h-y)

P(y) = C exp {+(gH)/(RT

_{o}) [-ln(H-y)] }

An exponent of a logarithm. The plus sign just indicates that two negative signs cancelled one another and there is one more negative sign on the natural logarithm from the last step. The overall exponent function still has a negative argument.

Here you'll get infinities and zeroes when evaluation limits and constants if your negative sign in (H-y) should be a positive - I'm still trying to figure exactly what happened, but when comparing the general solution to a known solution (link below), I saw that it had basically the same form if the argument is (H+y), so I used the graphing calculator to play with the equation, until I made it look like the actual plots

Noting the mathematical identity, b

^{x} = exp[x ln(b)] ...this is a nasty trick

P(y) = C [H-y]

^{-(gH)/(RTo)}Which has a very simple form, and looks a lot like known atmospheric plots:

My plots of P(y) = C [H+y]^{-(gH)/(RTo)}

Now look at the peak of that plot at X=0 (X is actually my altitude y in the equations above). This is not obvious, but if you change the value of the atmosphere thickness H, that peak will remain pinned at y=1 (Pressure on the surface P

_{o}). That is changing the thickness of the atmosphere H does not affect the pressure on the surface.

The pressure will asymptotically go to zero, but you have to define where H actually is. At 0.005 P

_{o} perhaps?

And so there it is. The pressure varies inversely proportional to altitude, but both pressure density is heavily biased toward the surface as the gases as given by "1/y" relationships, compressed under their own weight, and the atmosphere expands with temperature - linearly if very thin. The temperature is linear with altitude. The pressure at the surface is constant -until the gases precipitate or freeze out of the atmosphere.

Class dismissed, don't forget your homework and we have a quiz on Friday