I heard a climatologist on a talk show saying that one of the widely known arguments used by climate scientists to exemplify what a runaway greenhouse effect could cause to Earth's temperature and climate (a comparison with Venus' atmosphere) is basically a farce.

He argued that Venus' high surface temperature (which is over 400ºC) is caused mainly by its high surface pressure (which is over 90x greater than Earth's) and that the so-called "greenhouse effect contribution" to this temperature was "negligible" or simply "non-existent". He mentioned the ideal gas law in order to "prove" his argument, saying that temperature grows proportionally with pressure (which, of course, is true for an ideal gas). And this pretty much settled his train of thought.

This made me wonder and I'm trying to come up with a reasoning that debunks his argument. At first, I thought one could not simply apply the ideal gas law to Venus' atmosphere, since the molar density at the surface is too high and that one should use more terms of the virial expansion to account for the behavior of the pressure near the surface. However, even if that is true, his argument remains undisputed, since the leading term of the expansion is still the "ideal gas" term.

Then I realized his oversight may have been the cause-effect relationship he tried to establish between pressure and temperature. As a matter of fact, an equation of state doesn't tell us anything about causality: it only states the relationship between different state variables that describe a system. This takes us back to my question: Is it the really the high pressure of Venus' at surface the agent responsible for its high surface temperature, or is it the other way around or maybe even a mix of the two?

To better clarify my thinking, imagine this: if we take Venus and place it out of the Solar system where its exposure to solar radiation would be negligible, I think we can agree that its temperature would decrease drastically. Now, I know most of its atmosphere would solidify in such conditions, but the thing is: would the pressure on the same layer of matter where the surface of the planet was previously remain the same? No. So, does this tells us that it is the temperature of the planet that determines its pressure?

On the other hand, I am also aware that the pressure at the surface is simply a function of how much matter (gas) there is above it (no clear connection with temperature here).

I've reached this crossroads and can't seem to establish any clear causality relationship between pressure and temperature. Am I going in the right direction or is it that I am just plain wrong and the climatologist correct when he made his assertion?

  • $\begingroup$ Not enough time to answer, so a couple of quick comments. First off, the phrase "don't believe everything you see on the internet" applies to the Nth degree to what you hear on right wing talk radio. Was this really a climatologist, or yet another quack who claims to be a climatologist? Denying the existence of the greenhouse effect is pure quackery. That the greenhouse effect must exist is basic atmospheric science. Moreover, without this effect, the Earth's surface temperature would be 255 kelvin. Life on Earth depends on the existence greenhouse effect. $\endgroup$ – David Hammen Nov 4 '14 at 16:16
  • $\begingroup$ See also Earth Science Stack Exchange. $\endgroup$ – gerrit Nov 4 '14 at 16:34
  • $\begingroup$ I know it's kind of hard to swallow that the guy is a real climatologist, but it's true. He is a professor at a renowned university in Brazil. His resumé is mediocre though and I'm pretty sure his colleagues don't really agree with him. And, yes, I was reading an atmopheric physics book and realized the only way to account for this extra 30K in global temperature is to include the greenhouse effect in the picture. $\endgroup$ – Caio Nov 4 '14 at 16:47

Well, to clarify some things first In atmospheric science, or more correct: If you do the math... your only to free Variables are Density and Temperature. The equation of state which gives you the pressure, is a material property.

The equations for the atmospheric variables are interconnected at any moment, it is nonsense to say P causes T or T causes P. You could only make sense of this, if you would have a concrete change in T due to some radiative effect, or change P due to contraction.

The point here is, that your 'climatologist' doesn't recognize, is that the energy content of an atmospheric layer is always in flux. If the $F_{up} = F_{down}$ then this is $= \sigma A T^4$.

However to reach a high state of equilibrium flux (meaning: to reach a high T!) you have to have a mechanism to contain this flux somehow, and that is achieve by pulling a blanket over your atmosphere (a.k.a greenhouse effect). This makes some initial flux never be able to escape.

So to summarize: The Equation of state is true at any given moment, but doesn't explain how all the heat got there, or stayed there. You have to look at the thermal history of Venus or the body of interest, to understand how the heat got there. (and thermal history means to solve the above mentioned equations, involving radiative windows etc. that this guy ignore.)

Tell me, if you want me to clarify things up.

  • $\begingroup$ Thanks for clearing up my confusion. I just have one further question: You said the equation of state which yields the pressure is a material property. And that the only free variables are the density and temperature. Does that mean that pressure and temperature are not related by an equation of state and the pressure is only dependent on properties of the material? A "constant"? What kind of equation of state would that be? Or is it that the pressure depends on material properties through the density and temperature like in the ideal gas law? Sorry, but I didn't get that. $\endgroup$ – Caio Nov 4 '14 at 17:08
  • $\begingroup$ What I meant is that $P=P(\rho,T)$, the pressure is some functional dependency of $\rho$ and $T$. This function will stay the same for the whole $\rho - T$-space that doesn't violate the assumptions used to derive $P(\rho,T)$. A concrete example would be the validity of the ideal gas law, as long as you have elastic collisions and long mean free paths. This breaks down with e.e. degeneracy at high densities. $\endgroup$ – AtmosphericPrisonEscape Nov 4 '14 at 21:18
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    $\begingroup$ I'd want to modify David's thought experiment for clarification: Take two identical planets, one with a greenhouse (G) layer high up in the atmosphere, one without (W). G will retain its energy content. W will radiate away its energy content, loose temperature. This will increase density, again increasing P. But because of this feedback loop, it's depending on how much energy is actually lost, this can lead to a netto increase or decrease in P, that's why I take the climatologist's argument as invalid. $\endgroup$ – AtmosphericPrisonEscape Nov 4 '14 at 21:36
  • $\begingroup$ @AtmosphericPrisonEscape in a moving system, yes. If you start with an atmosphere that's spread out and let it fall closer to the surface of the planet, that in-falling would create heat, but you're not thinking things through. Atmospheres and how spread out they are has to do with an equilibrium of temperature, pressure and gravity, much like stars, if you remove the greenhouse gas, the planet would slowly cool, and slowly contract, as it loses heat the contracting would correspond to that. There's no event in the theoretical removal of greenhouse gas that would lead to heating. $\endgroup$ – userLTK Jul 5 '18 at 7:06

(I have only read the first paragraph of the question)

The pressure of Venus's atmosphere is about 90x greater than that on earth. It also happens to be about 90x more dense that that on earth. Coincidence? No. The density is the reason the pressure is so high.

If you were to descend 1 km into one of our oceans, the pressure would be comparable to that of the Venusian atmosphere. And yet the ocean isn't several hundreds of degrees is it? The reason for the high temperature on Venus is the greenhouse effect.

  • $\begingroup$ Wrong! The temperature difference betweens 1 km down in the ocean and Venus is due to the adiabatic temperature gradient. Imagine the adiabatic temperature gradient as the speed and thus temperature being added to a molecule as it falls down in the gravity field. $\endgroup$ – David Jonsson Aug 13 '19 at 10:13
  • $\begingroup$ @DavidJonsson Which part of my answer are you claiming is wrong? My last paragraph simply makes the point that pressure does not dictate temperature by giving an example of two systems that have the same pressure but different temperatures. $\endgroup$ – lemon Aug 13 '19 at 10:37

I look at the question this way, and it's very simple.

What happens when you pump up your bycycle tyre? ... it gets hot, because work is done on the air to compress it into the tyre. You leave it when finished. It cools down. The air is still at pressure (OK a little less on cooling). Why does it cool - because work is NO longer being done on the air. But it is still at high pressure. The work done on the atmosphere by it's compression due to mass/gravity is a ONE time only event and NOT kept there by that process. In the same way your bicycle tyre does not stay forever hot.


Your Brazilian Professor is partly right: the high temps at the surface of Venus are partly due to the high pressures at its surface. But that's not sufficient to explain the total temp. Other factors are that Venus is a lot closer to the sun than Earth is, and of course the greenhouse effect.

As a thought experiment, if you took two planets that were identical in every way except that one had a thicker atmosphere than the other, the one with the thicker atmosphere would have higher surface temps than its twin, even if global warming were not a factor. This is because higher pressures produce higher temps, and explains why it's hotter at the bottom of Mt. Everest than at the top.

I think part of the problem is that most people don't understand how the greenhouse effect really works. Here's a brief explanation: Gasses that are deemed "greenhouse gasses" are transparent at the peak wavelengths that the Sun emits, so that energy travels through the atmosphere relatively unscathed and heats up the surface of the Earth. The Earth gets rid of that energy by re-radiating it back into space as black-body radiation. But that re-radiation is at much longer wavelengths (well into the infrared), at which greenhouse gasses are not transparent. (Non-greenhouse gasses are transparent at both the incoming and outgoing wavelengths). This is not a matter of speculation: you can take a container full of methane or carbon-dioxide and measure its transparency as a function of wavelength in any well-equipped lab in the world. We know exactly how much extra energy is being trapped inside the atmosphere due to these gasses. The only dispute (among those versed in the science) is what the effects will be on climate, ocean currents, sea level, etc., and how fast those changes will occur.

  • $\begingroup$ I do understand the greenhouse effect explanation since I've just done the math here. I was able to obtain an average global temperature of 286K considering a high short wavelength transmittance (0.9) and low long wave transmittance (0.2) as one would expect for greenhouse gas atmosphere. This part is clear now. The high temps due to high pressures part is not clear yet though.I must agree your thought experiment argument is very compelling. However, don't things work the way lemon stated above: ins't there a change in density that offsets the effect the pressure would have on temperature? $\endgroup$ – Caio Nov 4 '14 at 18:48
  • $\begingroup$ I guess I don't understand lemon's argument. The pressure at the bottom of the atmosphere is due to the weight of the air above it, and that weight compresses the air to the observed density. Lemon is correct that descending into the ocean would increase the pressure and not the temp. But that's because the ocean is not a gas. If you filled a balloon with air and sank it 1 km into the ocean, it would indeed increase in pressure, density and temperature according to the ideal gas law (assuming the walls of the balloon were insulated to prevent cooling by contact with the water). $\endgroup$ – David Rose Nov 4 '14 at 20:37

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