arkie87

I have to agree with Wedge here. Residence time is not nearly as important as fractional changes per time. Yes, residence time is related to fractional changes, but what is directly used by modelers is fractional changes per time, not residence time. Specific heat (and even latent heat) from water/water-vapor is not a negative feedback in the way you described. All it effects is thermal mass i.e. how quickly the system reacts to changes in forcing, but not the steady-state solution. However, water vapor in air can cause negative feedback by forming clouds, which reflect sunlight back into space; it is my understanding that clouds are responsible for the majority of the earth's non-zero albedo (correct me if i am wrong; and yes, ice caps reflect sunlight too, but the solar flux in the poles is much weaker and the fraction of land covered in ice/snow is only a small fraction of the total, whereas clouds can form all over the planet) Because it should be negligible, since the fraction of energy from the Sun in IR is much less than in visible... - - - Updated - - - Not sure why he is arguing this point. Greenhouse gas effect and the way greenhouses work are obviously slightly different. Greenhouse effect is just an analogy. It's like arguing that "color" in quantum mechanics is misleading, since it has nothing to do with color.... In his defense, there is liquid water on the ground, so the heat capacity of the air might not change by adding water vapor into it (directly), but water itself can absorb energy (and the oceans have much mroe thermal mass than the atmospehre itself). Furthermore, while you need liquid water to dampen warming via evaporation, you need water vapor to damping cooling via dew point and condensation. Agree with Peadar here. Arguing that there is ice, so albedo is lower does not explain why there is ice to begin with... Well, it is my understanding (from a previous thread) that global warming via CO2 actually requires a positive feedback loop with water vapor to produce a scary amount of warming i.e. more CO2 raises temperature, which raises solubility level of water vapor levels in atmosphere (so the water vapor wont necessarily be fully removed by natural water cycles and negative feedbacks), which causes CO2 to become insoluable in ocean and enter atmosphere, which warms atmosphere, etc...

No, surface must still be warmer than atmosphere (assuming non-infinte heat transfer coefficient from surface to lower atmosphere) since it is receiving heat flux from sun, and wouldnt be able to reject any heat to lower atmosphere if it was at the same temperature as it. If my logic wasnt flawed, it would have pointed out something very important... Well, i was imagining a thermocouple (since thats what we use in our lab). Since the thermocouple is so thin, heat will easily conduct between the two sides, so if one side is facing 3K and one side is facing 300K, the measured temperature will not be accurate at all. You need to radiatively insulate the thermocouple from the cold side Actually, i figured out how to prove (not so rigorously) that outgoing radiative flux from an opaque volume of gas is equal to a blackbody. Imagine an infinite plane of isothermal gas separated by a distance (so no convection or conduction) from a infinite planar surface at the same temperature of the gas (with a heater to maintain the temperature if needed). The system is in equilibrium, so no temperature changes would take place. Since the surface is emitting like a blackbody, and all the radiation is being absorbed by the gas, the amount radiated by the gas to surface must be equal. Therefore, gas must behave like a blackbody as well, despite the fact that radiation in gasses is volumetric and not a surface phenomena.

Scientists often makes approximations. For instance, the boundary layer thickness over a wing. True, boundary layer technically extends out to infinity, but scientists define the minimum y-distance that U(y)=0.99*U_0. Making this approximation allows for useful results, such as extremely simple correlations for friction factor, heat transfer coefficient, etc... If we assume something is essentially opaque when 99% of the light passing through it is absorbed, then i am saying we should have ten times that thickness, such that even if we discritize the volume into ten parts, each part is, by itself, still opaque. I dont see whats the big deal. I apologize for not understanding that you were trying to model radiation using thermal conductivity, but in my defense, you never specified that. All you would have had to say is: "if we create a simplified model to simulate radiation absorption and transmission via thermal conductivity" or something like that... No need to be rude and condescending. I already explained why i even bothered to mention that. If you are going to continue to be rude, i will kindly request you stop responding to my threads. If you have ever submitted a paper to a science journal, you know that there is no assumption of competency. Expert reviewers will read what you say literally, and not give you the benefit of the doubt, even if it is obvious that you meant something else or made a mistake. Just correct the mistake and move on... I am well aware of the difference... What does the following post mean: Does this not mean that thermal conductivity of a greenhouse atmosphere is higher than a non-greenhouse one? Edit: after review, it appears there was some ambiguity; you meant "... thermal conductivity of the atmosphere with greenhouse gasses is lower ..." . I apologize for the confusion. Yes, that was the problem. I moved too quickly from my "naked kerbal" example to a planet, without taking into account the fact that the surface is hit with a heat flux and must be hotter than an isothermal atmosphere in order to transfer net IR heat to it (even if the atmosphere itself is well mixed). For the future, I think it is a good idea whenever trying to help someone who is conducting a thought experiment to explain the flaw in the thought experiment by directly using the assumptions and logic of the thought experiment rather than what happens in the real world. It is equivalent to me claiming: a=1 b=0 And then: a+b = 0 And then you say: actually, a+b = 1 + 0 = 1 not 0, or, correct my assumption that b=0, by saying: actually b= - 1, and then a+b = 0; This would have fostered a direct discussion of the thought experiment, and pointed out the flaws in calculations, assumptions, or logic. What you did is ignore the thought experiment and just explain how greenhouse works in general, which is closer to something like saying: c=1 d=-1 And then saying: c+d = 0 Yes, it is true, but I am talking about a case where a=1, and b=0, so either claim b=/=0 or point out that a+b=/=0.

Didnt mean to come off arrogant. Just meant to suggest you dont need to explain basic things to me. Intended both temperatures to be equal to each other. I dont understand what you are talking about here. There is a certain atmospheric thickness (given a certain ppm... or in more appropriate units: atm-cm i.e. distance*pressure) at which just about 100% of a given wavelength will be absorbed. It is exponential and is related to absorption coefficient, which has units of 1/length... If the absorption coefficient is 1 [1/cm], then 1 [cm] thick of gas will transmit exp(-1) percent of the radiation going through it. It will never equal zero, but will approach it... So if you have, lets say, 10x the absorption coefficient length such that you end up with exp(-10) being transmitted, which is non-zero, but very small... After reading the rest of your post, it appears you think greenhouse effect is related to thermal conductivity of the air, which is why you think my response makes no sense... I think you need to read up on greenhouse effect. Thermal conductivity???? Are you serious? I repeat: thermal conductivity? Are you serious? I guess you are serious. Greenhouse effect has nothing to do with thermal conductivity... https://en.wikipedia.org/wiki/Thermal_conductivity In addition, your logic itself is backwards. If increasing greenhouse gas concentration increased thermal conductivity, and the greenhouse effects act through thermal conduction (even though in reality, it actually acts through radiation), then greenhouse gasses should make the earth cooler, since less of a temperature gradient would be needed to release the same amount of heat for a higher thermal conductivity value. Obviously, it doesnt work this way, but im trying to point out the flaw in his (misinformed) logic. No... I'm not setting T_0 = T_end... and no, i'm not assuming anything about the sun... my argument holds true if the sun is shining in visible light (and indeed, this is what i actually intended in my example)... maybe i should make a graphic to illustrate my point, or something.... - - - Updated - - - Nevermind, i figured out the answer, everybody. Thank you those that were trying to help me, and i'm sorry we didnt communicate well...

I appreciate your effort to understand me It is a thought experiment to understand exactly how greenhouse effect works "Well-mixed" means there is lots of vertical wind such that the atmosphere is isothermal (at uniform temperature). It is not important how its possible to have this in real life. It is a thought experiment. It is a simplified discussion for the purpose of the thought experiment. I am talking about a hypothetical (made up) case where temperature is uniform. The atmosphere, in my though experiment, does not have a temperature gradient. It is all at, say, 15C... surface, lower atmosphere, upper atmosphere, etc... I know this isnt realistic and/or the reality, but I am conducting a thought experiment right now... Well, i appreciate your effort Thanks! (100^4-3^4) =/= 97^4 (100^4-3^4) ~= 100^4 Are you starting from the top going down or vice versa? It's hard to follow what you are saying... So it seems like you agree with me. A thermal gradient is necessary for the greenhouse effect to occur. Without it, there wouldnt be any... The T^4 dependence would cause a lot more error than that. Assuming water is 30C and space is 3K, the equation is: (273.15+30)^4 + (3)^4 = 2*(273.15+T)^4 Solving for T, we get: T=-18.23C Not 27C; huge error. It has everything to do with it. If there is a large thermal resistance between measurement point of thermometer and space-facing surface, the top portion will be colder and will reach equilibrium with space, while the bottom (measurement point) of thermometer will reach equilibrium with ocean.

WOAH! Sorry. Huge typo. Apparently compulsively re-reading my posts a few times isnt enough!! I have corrected it in the post. Please reread now... That is a very interesting graph; how is it possible that higher up it gets hotter? Solar particles? Ozone? However, this graph has nothing to do with my question... my question is about an isothermal atmosphere and surface. Do you have a mathematical source that proves this to be the case that it perfectly balances and that an isothermal semitransparent media always emits just like a blackbody perpendicular to its containment sphere? If not, do you have a better/more-detailed logical argument as to why all the components still sum to one; i would have thought it might be between 0.5 to 1. This is good to know though... I understand. We are agreeing. At some points in this thread we were arguing about temperature swings (like in deserts), and were discussing thermal mass/atmosphere. At other points, we were discussing average, and were discussing greenhouse gasses and effects. The thermometer presumably has two sides. So whereas on the side facing the sea, it absorbs and emits an equal amount of radiation, yielding zero net heat flow, on the other side, it emits but does not receive, thereby making the other side cooler. So unless there is a large thermal resistance between them, the temperature sensor will measure a lower temperature. The smaller the thermometer, the smaller the thermal resistance between the side facing the sea and the side facing space; therefore, all things being equal, the error is magnified as the sensor gets smaller... - - - Updated - - - First of all, i never said ANYTHING about the Kerbal being naked I dont know if your reading comprehension is off, but you are definitely not answering the question i asked. I understand how heat transfer works (perhaps better than most, as I have a degree in it), and I think the entire discussion up until now should indicate that. Your analogy is a good one for explaining how the greenhouse effects works, but you should carefully read my question, which i will repeat here: Imagine this example: an isothermal blackbody radiating surface surrounded by an isothermal (i.e. really well-mixed) pure greenhouse atmosphere. Imagine the atmosphere is 10x thicker than it needs to be to block all outgoing surface IR radiation. Under these conditions, there should be no greenhouse effect, since the upper most layer is emitting the same amount of heat to space as the surface would be if there was no atmosphere to begin with. The fact that it is absorbed and re-emitted millions of times before finally escaping the planet doesnt matter, since the NET effect is radiation from the surface/atmosphere leaving the planet. So actually, for the greenhouse effect to exist, it requires that the upper atmosphere be cooler than the surface (not that it "can nevertheless result in higher surface temperatures")

This does not at all address the points and questions i have raised in the quoted text. I asked what is the derivative of f(x)=x^2*cos(exp(x)) with respect to x, and you answer that the derivative of f(x) with respect to y is 0... true, but it is a lot simpler to calculate and doesnt answer my question...

I dont think that part is true. It is just a thermal resistance network. Temperature is like voltage and heat is like current. In order for the atmosphere to release the same current (heat) as it receives from the sun, it must be at a lower voltage (temperature) than the surface, since current (heat) is flowing from the surface to the atmosphere, and there is a/many (thermal) resistance(s) between the surface and the atmosphere. Imagine this example: an isothermal blackbody radiating surface surrounded by an isothermal (i.e. really well mixed) pure greenhouse atmosphere. Imagine the atmosphere is 10x thicker than it needs to be to block all outgoing surface IR radiation. Under these conditions, there should be no greenhouse effect, since the upper most layer is emitting heat to space (and back to the planet) at the same temperature as the surface (and it doesnt cool since it is isothermal/well-mixed). Here, I am making a simplification/assumption that the average heat flux escaping the gas is equivalent to that of a blackbody radiator at the same temperature despite the fact that the gas emits equally in all directions, not just vertically or into space. However, since layers below the edge near the edge also emit radiation in all directions, some of which end up escaping to space, it is possible that the net effect is still to emit, on average, the same heat flux a blackbody radiator would. If this is not the case, that is fine, but it would only have the effect of reducing the effective emisivity of the radiating layer to space; thus, a slight qualifier is needed: if the surface emisivity were equal to this effective emisivity, there would be no greenhosue effect. I think this is a very good example to discuss since it really describes how the greenhouse effect works (and DOESNT work)... Why do the triggers matter; if the positive feedback exists, it doesnt matter what triggers it? each individual CO2 molecule might only last 30 years before it is absorbed by the ocean or into a plant, but that doesnt matter if another one (or two!) take its place. In water cycle, residence time of each molecule might be much shorter but there is usually always another molecule to take its place, since humidity is always rising up and down. I dont know what you were talking about, but i was talking about the temperature swings from day to night only when referring to atmosphere; and only the fact that the earth is warmer than the -33C when referring to greenhouse gasses. It would have to be facing the sea and insulated from the lack of radiation from cold space. - - - Updated - - - Ok, this is what i thought you were suggesting. Please cite a reference that shows IR intensity from sun is non-negligible... I have one here that shows it is negligible, so this cannot be the mechanism: I suspect you might argue that the IR radiation the water vapor in the atmosphere is absorbing isnt coming from the sun, but from our own atmosphere. This is backwards, as the net effect is always to transfer heat from hot to cold. Since the upper atmosphere is colder than the lower atmosphere and surface, it is not possible that the upper atmosphere is warming the lower atmosphere i.e. net heat flow is from upper (cold) to lower (hot). If it were, then you have discovered a way to get free energy! - - - Updated - - - Sorry for my lazy ambiguous writing. Yes, i meant lack of.

You say water vapor is the reason deserts are hotter, and then cite how water is a strong absorber in the IR range. The fact that it is lacking in the deserts should make deserts colder during the day, since the radiated IR heat isnt trapped... If by water vapor, you meant cloud coverage, which can reflect a large portion of visible light, then cities should get just as hot as deserts on cloudless days. Clearly, it is more complicated than just one factor.

On Venus, CO2 concentration is 965,000 ppm wheres on earth it is ~400 ppm; one would expect IR radiation to bounce many more times on Venus, though that effect might not matter after a certain point, where the atmosphere is essentially opaque. However, one thing that has confused me for a bit is this: Consider a blackbody radiator in the shape of a Kerbal separated by a volume of isothermal greenhouse gas (well mixed) from a IR camera. As the pressure of the gas is increased, the image of the Kerbal will slowly disappear; however, the image will not be dark. It will change from a picture of a Kerbal to a uniform bright light, since the gas absorbs and emits the light. Thus, while the light from the Kerbal will not reach the other side of the detector, plenty of light will reach the IR camera (it will just be diffuse). Thus, there appears to be a limit on the maximum opacity, since the outer-most layer will always radiate to space (even if the original light waves were absorbed and re-emitted). However, i suppose what causes greenhouse effect is when the outermost layer is very cold (i.e. when the outer most layer is thermally insulated from the surface), such that the amount of radiation leaving the planet is less than would leave if the surface could radiate to space directly. Does that sound about right? See response below regarding positive feedback loops involving water. Also, positive feedback is a very dangerous thing. Even small positive feedback, in the absence of any negative ones, can result in exponential growth until saturation. What doesnt make sense to me is how this positive feedback can be strong enough to raise the earth temperature by only a few degrees but not skyrocket the planet towards Venus conditions. And the fact that more heat is radiated at elevated temperatures doesnt help, since the heat is still trapped by the CO2 (it would only help if the temperature increases enough to change the outgoing wavelengths significantly such that CO2 no longer absorbs them). In addition, how does one explain this positive feedback, given that millions of years ago, the CO2 concentration in the planet was ~3000 ppm, but the temperature was only ~10 C warmer (i.e. not Venus)? The water vapor cycle is very stable and has many negative feedbacks to take water out of atmosphere (residence time of water molecule of short). Any positive feedbacks of CO2 that act through water vapor might be negated by negative feedbacks of water cycle itself. Of course not (it doesnt affect average temperature). But it DOES change the temperature difference from night and day! Sounds reasonable! I'm not arguing an atmosphere (non-greenhouse) will affect average temperature, just temperature difference between day and night. Atmosphere might not be a significant thermal mass compared to the oceans or the whole mass of the earth, but on the time scale of a day, the heat from the sun does not have a change to penetrate through the whole ocean or the earth. With wind and natural convection, the thermal mass of the atmosphere is the one that is mostly used.

This would be true even without greenhouse effect, since higher temperature black bodies emit higher intensity light at ALL wavelengths, not just the peak. For that reason, the graph you show is slightly misleading. Though i suppose distance can reduce intensity...

Very good point. Mun... uh... i mean Moon day is 28 times longer than Earth's. Being there doesnt mean its significant at all. Without greenhouse gasses, air (O2 and N2) wouldnt have any significant greenhouse effect. Greenhouse effect (raising average temperature) is different than atmosphere effect (dampening temperature swings between day and night).

Not sure what question in the OP you are answering? The post isnt about greenhouse effect in general; it was to clarify the arguments made by the source in the OP. - - - Updated - - - I dont see how this example is any simpler than just directly explaining what is actually happening... I also dont see what question you are answering. It seems like you read the thread title and are trying to explain greenhouse gas effect in general, not having read the OP... - - - Updated - - - Good to know... what is a better site to use? Why? I would argue the atmosphere would give the surface more thermal mass and convective cooling, and would dampen temperature swings somewhat... - - - Updated - - - Thanks for the graphs. Do these graphs show the valleys and spectra deepening? Do you have other graphs that do? Also, I assume, that: (1) when looking up at the atmosphere, we are looking at the intersection of the IR frequencies from the surface and what is absorbed, reflected, and/or scattered by the atmosphere and what is emitted by the atmosphere (since nothing is coming from space)? (2) when looking down at the land, we see what is emitted from the surface and transmitted through the atmosphere, a fraction of what is emitted from surface and absorbed and then re-emited by the atmosphere, and what is emitted by the atmosphere on its own?

Always a pleasure AngelLestat Ok, i would agree if this were an article in a newspaper. But it is on skeptical science, which is supposed to be detailed enough to counter skeptics' arguments. Giving simplified arguments will just add fuel to skeptics who think global warming is a conspiracy or based on "bad" science. Not sure what you mean here. Add 1 bar to current atmosphere (i.e. with O2, N2, CO2 etc... in current proportions?) Then, yes, adding more mass of non-greenhosue gasses wont raise average temperature, but will dampen temperature swings in a given location. It is my understanding that they are referring only to long-wave radiation. But i guess, you can read "long-wave radiation" as referring only to outgoing radiation and not incoming i.e. specifying long-wave is redundant since all outgoing is long-wave.

(Before I start, I will repeat my request from my previous thread: please be respectful in this forum; given the subject matter, I imagine it is quite easy to get frustrated and condescending and would like to request that those that do not feel they can be respectful kindly not enter the discussion; I am coming here with questions and want to learn). So, I am researching the greenhouse gas effect and was reading this article on Skeptical Science and have some questions which i hope this community could (respectfully) answer: This seems to me to be a flawed argument. One could argue the reason the moon gets so hot during the day and cold during the night is not because it has no greenhouse effect, but because it has no atmosphere. The atmosphere, through convection, dampens temperature swings, and if there is water content in the air, it can rapidly slow cooling if dew point is reached. In dry climates, such as desert, the temperature swings are much larger, since there is no water vapor to slow cooling at night or foliage to absorb the sun's rays. It doesnt necessarily have anything to do with a greenhouse effect i.e. preventing IR radiation from reaching space, but rather, in dampening temperature swings. The fact that earth is 33C warmer than it would be (on average) without an atmosphere IS evidence of the greenhouse effect. Why shouldnt this be the case? It is actually a requirement of the Planck distribution that at higher temperatures, the intensity at every wavelength is higher (not just the that the peak intensity is higher) To me, this seems more important than the first part (not sure why it is in parentheses)? Are they saying that they observed the wavelengths emitted in 1970 and now, and can see noticeable attenuation in the wavelengths absorbed by greenhouse gasses (presumably CO2, methane etc... but not water)?

It is pretty widespread the (false) idea that things are hotter because they are under pressure: "Since Jupiter is a gas giant, it has no solid surface, so it has no surface temperature. But measurements taken from the top of Jupiter’s clouds indicate a temperature of approximately -145°C. Closer to the center, the planet’s temperature increases due to atmospheric pressure." from Universe Today I even remember hearing that as a kid, even though it is wrong.

I suppose if the piston had mass, the inertia would allow the piston to over-compress, until the pressure stopped the momentum, which would then lead to expansion, and over-expansion, etc... Which seems very similar to Kelvin-Helmholtz contraction to me... This is isothermal compression. I dont know enough about Kelvin-Helmholtz contraction, but this sounds an awful lot like it... If the first or last have anything to do with his response, i quit

If you could find a gas with Cp~=Cv; i.e. gamma = 1, then the gas would not heat up upon compression. The smallest gamma i could find is for n-heptane with a value of 1.05 Isentropic processes follow: P*V^gamma = constant. If gamma is 1, then P*V=constant, so T cannot change. To reiterate, the fact that T increases is due to the process that is used to get there; not from the compression itself. If you had a 1 m3 volume of gas at 300 K and 1 bar pressure, and increased the pressure by a factor of 2 isentropically: For air: gamma = 1.4, so T would increase to 365.7 K. Because of the increase in temperature, volume would only decrease to 0.6095 m3 (notice the ratio of 365.7/300 = 0.6095/0.5). For n-heptane: gamma = 1.05, so T would increase to 310.1 K, and V would decrease to 0.5168 For a gas with gamma = 1, T would remain at 300 K and V would decrease to 0.5

So, i dont understand what you are arguing then. The discussion was about Venus being hot because it was compressed. There is no Kelvin-Helmholtz mechanism, and there is no reason the atmosphere would have undergone isentropic compression recently. And arguing that "technically, the process doesnt happen instantly, so that increasing pressure always results in an instantaneous (albeit fleeting) increase in temperature is a pointless point to argue for the sake of being technically right... if that is what you are arguing, then i have no desire to continue this conversation. Actually, what i said that if the volume goes down, either temperature or pressure or both must change to satisfy the ideal gas law. There are three variables (pressure, temperature, and volume), and if you specify a change in one, with the ideal gas law, you still have one degree of freedom (2 equations with three unknowns), so there are infinite solutions. The third equation, which decides which of the infinite solution actually occurs, comes from the process used to get there (isothermal, isentropic, isobaric, isochoric etc...) This case you are describing is impossible (assuming there is no friction etc... or other monkey business). In real life,if you apply a force to a piston previously at equilibrium, the gas will not spontaneously heat up in order to maintain at constant volume; it will compress. It is possible, however, to increase the force at a steady rate while adding heat to the gas to maintain equilibrium. However, applying additional force will not, on its own, increase the temperature to oppose motion while maintaining its original volume. Heat must be added. Temperature doesnt (directly) generate a force; pressure does. You can have two gasses at different temperature but at the same pressure and the forces will be the same; two gasses at the same temperature will not generate the same force unless they are also at the same pressure. And as i said above, the temperature cannot increase unless volume is reduced. Nothing. If a piston is initially at equilibrium and you apply force to it, it will reduce in volume until the pressure increases enough to oppose your additional applied force. It will not spontaneously heat up without reducing in volume. If you think this is possible, then there is a serious misunderstanding of the physics here...

Change one of the pressures to temperature?

Thats cool! Try: 200, 0, 0, 0 -1 50 142 0 0 140 1 -50 0 0 280 I think i know why that happens...

More massive objects attract other objects with a larger force, true. But more massive objects have... well... more mass. Thus, it requires more force to cause the same acceleration, so the acceleration is still constant. Hope that helped

I assume by expand, you meant contract, since you are applying pressure. The fact that temperature goes up in your real world experience has nothing to do with ideal gas law, it has to do with isentropic compression (since that is what it is). When pressure is applied, temperature does not necessarily increase. Whether or not temperature increases depends on the compression mechanism. There are two types: isentropic and isothermal. Isentropic compression requires the process to be adiabatic (no heat lost) so the process must either be quick or the volume of gas well-insulated. Isothermal compression is either slow or not well insulated, and, as the name implies, is performed at constant temperature. In both processes, pressure goes up and volume goes down. In isentropic process, no heat is lost, so the temperature goes up as well; in isothermal case, some heat is lost so temperature remains the same. You can think of isothermal compression as such a slow compression that as you compress the gas slightly, its increase in temperature causes it to be cooled faster than compressing it causes it to heat up. In the case of Kelvin-Helmholtz mechanism, it is actually relatively slow process, and the planet radiates the excess energy away, so it is actually closer to isothermal compression, since all the heat/temperature gained from compression, is released as heat instantly, since the compression is so slow....

Can agree with that! - - - Updated - - - Thank you. That's all I needed. Charles Law does not state that increasing pressure will increase temperature. It states that the ratio of temperature to volume will remain constant if pressure is held constant. More directly. Charles' Law states: "When the pressure on a sample of a dry gas is held constant, the Kelvin temperature and the volume will be directly related." Note that pressure must be held constant! P*V=n*R*T If T goes up, EITHER P or V can increase to compensate. If T is doubled, doubling V will prevent P from changing! Similarly, if P increases, V can go down such that T can be held constant! Increasing pressure does NOT necessitate an increase in temperature!!!!!!!! Compression is a change in volume, not pressure (though sometimes pressure changes as well). You can compress a gas by cooling it at constant pressure (isobaric process)! Compression can also cause heating if done quickly/adiabatically.

I think the confusion is from ideal gas law. P = rho*R*T only shows that increasing temperature leads to an increase in pressure when density (or volume and mass) is constant. For planets, volume is not constant, and instead of pressure increasing when temperature changes, density and/or volume changes to compensate. - - - Updated - - - I'm not talking about every case. I am just talking about that graph, since that graph has been used to show greenhouse effect. My only argument is that even if greenhouse effect magically didnt exist, the graph would look the same due to release of CO2 from ocean, due to solubility. Therefore, it is a bad graph to use to that end, since it doesnt isolate the greenhouse effect (or even necessarily show it). Please address that point... Or are you agreeing, saying that the reasons CO2 has gone up and down in the past has been so variable, that one cannot use it to show greenhouse effect either?