Does the atmosphere cool with altitude due to gravity?

[kyle Bacon]

I read an article a day or so ago titled Hydrostatic Lapse, which makes the case for a phenomenon that I thought was well and truly confirmed; that gravity is responsible for the cooling of air with altitude. However I discover in the sequel article The Gemini Cycle that this phenomenon is supposedly in conflict with the second law of thermodynamics.

After a bit of research, I find that Vanquish Opprobrium (author of those articles) isn't the only one who thinks the two are incompatible. Robert G brown from Duke University also makes the case for incompatibility in Refutation of Stable Thermal Equilibrium Lapse rate.

Although both agree that they are incompatible and assert that in very similar ways. Robert G Brown comes to the conclusion that a hydrostatic gas is isothermal in nature, where as Vanquish Opprobrium comes to the conclusion that the second law is likely false.

Vanquish Opprobrium is probably a crackpot, hence the nom de guerre. Yet hydrostatic lapse seems so intuitive. If an atmosphere's gas particles didn't slow at higher elevations, they would escape Earth's gravity field and we would be left with no atmosphere.

Surely they are both wrong?

 

[rootone]

They are.

 

[davenn]

hi kyle
welcome to PF

atmospheric temperature profile is very well known and if it had anything to do with gravity, a profile like this wouldn't be possible...
note the dramatic positive and negative temperature swings with altitude !

source ...
http://www.srh.noaa.gov/jetstream/atmos/atmprofile.htm cheers
Dave

 

[kyle Bacon]

Dave,

Does that diagram hold true at night?

 

[rootone]

Night/Day makes little difference to avarage atmosphere temperature.
It does have noticable effect, but nothing very dramatic.

 

[moatilliatta]

They are.

As in both wrong?

Neither of those bloggers makes the case for an isothermal or uniformly decreasing atmosphere. They are talking isolated hydrostatic gases.

 

[kyle Bacon]

Neither of those bloggers makes the case for an isothermal or uniformly decreasing atmosphere.

Well, I remember hearing that if Nazi Germany built the Volkshalle (gigantic dome structure) that it would of rained inside. Can that dome be classed as hydrostatic? Why would it rain inside?

 

[Darryl]

Does the atmosphere cool with altitude due to gravity?

The atmosphere is cooler with altitude because gravity is less, and pressure is therefore less, it is colder because of the altitude, I don't think 'it cools'.
The reason why it is colder is because there is less of it. This is because of the 'adiabatic process', the same amount of heat over a large area is a lower temperature. This the same amount of heat (I think) but in a large area.

But if you got a sealed box of 30 Deg air and went to high altitude on a balloon that air would not get cooler because of the effect of gravity being less, even if you took it into outer space with no gravity, it would still be 30 Deg, but if you doubled the volume of the box (with the same amount of air) the temperature would be 1/4 of it's original.

Well this is my take on this anyway, I could well be wrong (I guess that is a possibility!) :)

 

[moatilliatta]

Well, I remember hearing that if Nazi Germany built the Volkshalle (gigantic dome structure) that it would of rained inside. Can that dome be classed as hydrostatic? Why would it rain inside?

It is a difficult thing to explain the temperature of the atmosphere at different altitudes due to immense number of factors that come into play. While explaining how the air of a massive hall might cool with elevation is simpler than dealing with the atmosphere as a whole, it is still a very complicated question. You have the breaths of people to contend with in addition to the thermal conductivity of the walls and dome.

To get a clear cut answer requires a clear cut question. A similar "question" was asked 5 year ago in this thread

 

[davenn]

Does the atmosphere cool with altitude due to gravity?

The atmosphere is cooler with altitude because gravity is less, and pressure is therefore less, it is colder because of the altitude, I don't think 'it cools'.
The reason why it is colder is because there is less of it. This is because of the 'adiabatic process', the same amount of heat over a large area is a lower temperature. This the same amount of heat (I think) but in a large area.

But if you got a sealed box of 30 Deg air and went to high altitude on a balloon that air would not get cooler because of the effect of gravity being less, even if you took it into outer space with no gravity, it would still be 30 Deg, but if you doubled the volume of the box (with the same amount of air) the temperature would be 1/4 of it's original.

Well this is my take on this anyway, I could well be wrong (I guess that is a possibility!) :)

no it doesn't. and yes you are wrong

did you not see my post and graphic above ?

Dave

 

[moatilliatta]

Has anyone reviewed that math proof in the hydrostatic lapse article?

Instead of going through and checking the proof each step at a time, I just inserted a formula for the change in pressure with altitude in excel and then had the discreet adiabatic formula reference its result. This should determine the temperature of a parcel of air that has risen.

In another cell I inserted the typical dry adiabatic lapse formula, that should also determine the temperature of the parcel of air.

For those two formulas to produce the same result, I had to input the ALR as the hydrostatic atmosphere's lapse rate. So that proof might be right...or not

 

[kyle Bacon]

For those two formulas to produce the same result, I had to input the ALR as the hydrostatic atmosphere's lapse rate.

Can you link the spreadsheet?

 

[moatilliatta]

Can you link the spreadsheet?

https://docs.google.com/spreadsheets/d/1GBcf9imIkRdGd54UbzpvH4R2kQwndfQ6gQD_7GNpMoI/edit#gid=0 linked

 

[kyle Bacon]

What is the scientific consensus on a static gas in an insulated box?

Will its temperature vary or be uniform throughout?

 

[infinitexzer0]

Well, I remember hearing that if Nazi Germany built the Volkshalle (gigantic dome structure) that it would of rained inside. Can that dome be classed as hydrostatic? Why would it rain inside?

I'm not sure if it would be hydrostatic, but it would be possible theoretically to rain inside if there was room enough for the water cycle to take place. Givin enough humidity and whatnot. I'm obviously not a genius in this field but from my own knowledge it should be able to maybe someone else could give an in depth answer

 

[moatilliatta]

I'm not sure if it would be hydrostatic, but it would be possible theoretically to rain inside if there was room enough for the water cycle to take place. Givin enough humidity and whatnot. I'm obviously not a genius in this field but from my own knowledge it should be able to maybe someone else could give an in depth answer

I would expect the warm breaths of cold hearted Nazis to rise, given what I know about convection. As those breaths rise they will be moving from higher pressure low elevations to lower pressure high elevations, resulting in adiabatic cooling of the moisture laden breaths. If the dome is the right height, the breaths moisture will eventually condense due to cooling. Could be wrong on this, fell free to correct or elaborate.

 

[klimatos]

The standard climatological explanation for the tropospheric lapse rate is distance from the Earth's surface. The troposphere receives some 74% of its thermal energy from the Earth's surface--mostly in the form of longwave thermal radiation. Hence, the farther you get from the surface, the cooler the air is. Above the troposphere, a number of other factors come into play, and the lapse rate changes in response.

 

[moatilliatta]

Hence, the farther you get from the surface, the cooler the air is. Above the troposphere, a number of other factors come into play, and the lapse rate changes in response.

Seems too simple an explanation of a complicated topic. This logic works well when explaining why you should move away from a camp-fire when you want to cool down, but works less well when dealing with the earth. The inverse square law may not have much of an effect when you consider the curvature of the Earth is so faint, that many in the past thought it was flat.

The big ? for me is if a hydrostatic gas will cool altitude because of gravity. This should have an answer that doesn't require a book!

 

[klimatos]

Seems too simple an explanation of a complicated topic. This logic works well when explaining why you should move away from a camp-fire when you want to cool down, but works less well when dealing with the earth. The inverse square law may not have much of an effect when you consider the curvature of the Earth is so faint, that many in the past thought it was flat.

The big ? for me is if a hydrostatic gas will cool altitude because of gravity. This should have an answer that doesn't require a book!

Simple or not, this is the explanation proposed in nearly all textbooks on the atmospheric sciences. Moreover, it is supported by the weight of scientific evidence. Yes, it does get more complicated in advanced studies, but the basic premise remains valid. The inverse square law doesn’t really enter into it, since virtually all surface radiation is absorbed by the lower 100 meters of the atmosphere. It is more a matter of absorption and radiation, absorption and radiation, with each layer radiating more to the layer above it than to the layer below it. Eventually, the outgoing global terrestrial radiation is equal to the incoming global absorption of insolation and the Earth’s “heat budget” is maintained. * * * * *Your question on gravitational cooling of the atmosphere gets a qualified “sometimes” answer from kinetic gas theory and statistical thermodynamics. When a mass of air is at rest (no winds or currents), the number of molecules with an upward component of translational motion is equal to the number with a downward component (about 1.21 X 10²⁵ per cubic meter each way at 25°C and 10⁵ Pascals). The speeds of all upward molecular movements are decelerated by gravity; and the speeds of all downward molecular movements are accelerated by gravity. These changes in molecular speeds are reflected in the mean kinetic energies of translation and hence in the temperatures. In still air, these ongoing temperature changes cancel one another out. When there is net upward or downward air movement, the proportions change—with more molecules moving in the flow direction than in the opposite direction. If the net air mass movement is upward, more molecules will be cooled than warmed. If the net air mass movement is downward, more molecules will be warmed than cooled. This explains the adiabatic cooling and heating of rising and sinking air masses. You might think that these convectional flows would average out and that there would be no net heating or cooling because of them. This is true except for water. Atmospheric water rises as a gas (water vapor), but sinks as a solid or liquid (rain, snow, hail, etc.). Hence there is slightly more adiabatic cooling of the atmosphere then there is adiabatic warming. How much more is a good question. * * * * *I am a bit uncomfortable with your reference to the atmosphere as a “hydrostatic gas”. The hydrostatic equation requires a number of conditions to be valid—conditions that are not met by the free atmosphere. Most importantly, however, it requires that a condition of equilibrium exist in order to be valid. The global atmosphere is never in a condition of equilibrium or even close to it. Hence, the hydrostatic equation does not really apply to it.

 

[kyle Bacon]

The speeds of all upward molecular movements are decelerated by gravity; and the speeds of all downward molecular movements are accelerated by gravity. These changes in molecular speeds are reflected in the mean kinetic energies of translation and hence in the temperatures. In still air, these ongoing temperature changes cancel one another out.

Ok, so a massive insulated box full of still air will be isothermal?

 

[moatilliatta]

Simple or not, this is the explanation proposed in nearly all textbooks on the atmospheric sciences. Moreover, it is supported by the weight of scientific evidence. Yes, it does get more complicated in advanced studies, but the basic premise remains valid.

I thought that atmospheric lapse was mainly a result of adiabatic heating and cooling resulting from convection. My bad

The speeds of all upward molecular movements are decelerated by gravity; and the speeds of all downward molecular movements are accelerated by gravity. These changes in molecular speeds are reflected in the mean kinetic energies of translation and hence in the temperatures. In still air, these ongoing temperature changes cancel one another out.

So a Hydrostatic Gas is Isothermal. Thanks for answering the question.

Could you please also explain why https://medium.com/criminal-clouds/the-isothermal-elevator-part-3-of-3-7d828edb3c1a wouldn't work?

I am a bit uncomfortable with your reference to the atmosphere as a “hydrostatic gas”. The hydrostatic equation requires a number of conditions to be valid—conditions that are not met by the free atmosphere. Most importantly, however, it requires that a condition of equilibrium exist in order to be valid. The global atmosphere is never in a condition of equilibrium or even close to it. Hence, the hydrostatic equation does not really apply to it.

I never said the atmosphere is hydrostatic.

Surely they are both wrong?

Robert G Brown and Vanquish Opprobrium focus on hydrostatic gases. To answer Kyle's second question you need to deal in terms hydrostatic gases.

 

[kyle Bacon]

So a Hydrostatic Gas is Isothermal. Thanks for answering the question.

Could you please also explain why https://medium.com/criminal-clouds/the-isothermal-elevator-part-3-of-3-7d828edb3c1a wouldn't work?

.

I'm thinking that the Isothermal explanation has to be the right one. Everyone thinks that hydrostatic lapse is in conflict with the second law, except it seem the simple wiki for the second law which states:

"over time, differences in temperature, pressure, and density tend to even out in a horizontal plane, but not in a vertical plane due to the force of gravity. For example, density and pressure do not even out in a vertical plane, and nor does temperature because gravity acts on individual molecules, and this means molecular kinetic energy interchanges with gravitational potential energy in free path motion between collisions."

Good reason to never trust Wikipedia maybe.

The last week I've been daydreaming something that looks very similar to the below, trying to get my head around all this:

I think its time to return to daydreaming about a Ducati I've been saving for, yet I don't think I will able to until this is all put to bed.

For the sake of my sanity, can someone please why this isothermal elevator doesn't work!

 

[russ_watters]

Simple or not, this is the explanation proposed in nearly all textbooks on the atmospheric sciences. Moreover, it is supported by the weight of scientific evidence.

Please provide a reference for this claim. An obvious source, like wikipedia, says:

In the lower regions of the atmosphere (up to altitudes of approximately 12,000 metres (39,000 ft), temperature decreases with altitude at a fairly uniform rate. Because the atmosphere is warmed by convection from Earth's surface, this lapse or reduction in temperature is normal with increasing distance from the conductive source.

Although the actual atmospheric lapse rate varies, under normal atmospheric conditions the average atmospheric lapse rate results in a temperature decrease of 6.4C°/km (3.5F° or 1.95C°/1,000 ft) of altitude above ground level.

The measurable lapse rate is affected by the moisture content of the air (humidity). A dry lapse rate of 10C°/km (5.5F° or 3.05C°/1,000 ft) is often used to calculate temperature changes in air not at 100% relative humidity. A wet lapse rate of 5.5C°/km (3F° or 1.68C°/1,000 ft) is used to calculate the temperature changes in air that is saturated (i.e., air at 100% relative humidity). Although actual lapse rates do not strictly follow these guidelines, they present a model sufficiently accurate to predict temperate changes associated with updrafts and downdrafts.

https://en.wikipedia.org/wiki/Lapse_rate

Indeed, the "lapse rate" is often called the "adiabatic lapse rate".

The wiki makes mention of the idea of radiation playing a role (only above the tropopause?), but the reference is a dead, self-published source that is plastered with warnings about it being an insufficient source.

 

[votingmachine]

Could you please also explain why https://medium.com/criminal-clouds/the-isothermal-elevator-part-3-of-3-7d828edb3c1a wouldn't work?

I think I would have to look closer, but at a glance it looks like they leave out a lot of work. You can fill up a tank with sea level air, and fly it in a plane to high altitude, and extract the energy from the pressure difference, but you use a lot of energy in raising the container.

I would guess that including the process of pumping a vacuum in the cylinder would stop the process from being free. Same as it takes energy to fly that tank to a high altitude, you cannot get it very high without an energy expenditure.

Further, you know how the atmosphere supports a column of mercury 760 mm high, or a column of water about 30 feet high? There is a limit to how high any particular mass can be lifted. Even if the mass was just the mass of the air and the tank was weightless, there is still a thing being lifted. And that air would not actually be lifted all the way to the top by a vacuum. The perfect vacuum would only lift a (weightless) tank of air at the lower mass of the upper altitude air, to the top. It would not lift a (weightless) tank of air at the lower pressure, because it weighs more.

To me it looks like they are leaving out the fact that the tank will never get lifted that high by a vacuum, and that it takes energy to make that vacuum. I rather suspect that it is a huge waste of energy overall.

If that is all completely wrong, feel free to tell me. I formed that opinion without a detailed reading.

 

[moatilliatta]

I think I would have to look closer, but at a glance it looks like they leave out a lot of work. You can fill up a tank with sea level air, and fly it in a plane to high altitude, and extract the energy from the pressure difference, but you use a lot of energy in raising the container.

Best to have a closer read votingmachine,

The gas containing vessel is lifted by a pulley inside an elevator shaft. I'm guessing that the elevator shaft is a vacuum so buoyancy doesn't have to be factored in, not to actually conduct the work. Because of this, I'm actually quite confident on the vessel lifting calculations being correct. The calculations are just a combination of the ideal gas law and gravitational potential energy formulas. They are so straightforward I expect high-school students to be able to do the calculations.

The isothermal expansion formulas are far more complex. I'm thinking that they must be out by a factor of 2. If the expansions only produced half the energy then there would be no net energy production. This is what I would expect from anything that has been idealised, so that friction and other inefficiencies are not factored in.

 

[votingmachine]

My bad. On closer reading I see the raising and lowering hoist. I thought the picture showed a piston. And I agree that the bouyancy is probably why they specify a vacuum.

I don't see any math errors. It looks to me like that is a process for extracting energy from a higher pressure gas to a lower pressure gas. It does seem a bit surprising that there is an energy net. But it is not impossible, not a violation of perpetual motion. I think the temperature issue would be the reason it breaks down. I think it is more like one of those "drinking bird heat engines". It postulates that you will expand the ground temperature air into the low pressure high altitude air, in an isothermal expansion. You can do that once, but the next time, the air is a little cooler. eventually you extract enough heat from the high altitude air, that there is no more isothermal expansion.

It is possible that this is another basis for the temperature drop in the air with altitude. I can't quite reason that out, but this process for extracting work if the temperatures are the same suggests that the work would instantly be done by the unrestrained atmosphere.

If you put in a T1 and a T2 for the top and bottom temperatures the work changes. From a quick google ... the summit of Mt Everest runs between -20 C and -35 C. Call that 250 and 235 K. If you used Hawaii, Mauna Kea is 13,000 feet high and the summit looks to be about 40F (275K) when the sea level is about 80F (300K). Summit pressure is about 0.6 atmospheric.

The work on the up and down cycles also changes a slight bit. The colder air at the top is more atoms at that pressure ... a bit more mass. And the air you raise up is warmer ... a bit less mass. That is beneficial.

Interesting read, and other than the temperature bit, the math doesn't appear to be in error. I still put it as similar to the "drinking bird heat engine". And since the requirement is for a pretty substantial amount of infrastructure that has to be extraordinarily efficient ... I can't see anyone really testing this out. Perhaps taking a bottle up and down Mauna Kea would be a great research grant to try and get. I volunteer to surf out and get the cleanest ocean air for you. You can walk to the summit and back down, then I will get the good stuff from a wave. Heck, put TWO bottles in the grant. I'll fill 'em both.

 

[moatilliatta]

I don't see any math errors. It looks to me like that is a process for extracting energy from a higher pressure gas to a lower pressure gas. It does seem a bit surprising that there is an energy net. But it is not impossible, not a violation of perpetual motion. I think the temperature issue would be the reason it breaks down. I think it is more like one of those "drinking bird heat engines". It postulates that you will expand the ground temperature air into the low pressure high altitude air, in an isothermal expansion. You can do that once, but the next time, the air is a little cooler. eventually you extract enough heat from the high altitude air, that there is no more isothermal expansion.

From a practical sense I think you have a point. I don't think this thing is going to be putting any coal mines out of business. However everything in the isothermal elevator's idealistic scenario is thermally connected and true isothermal expansion is quasi-static. Everything in that system has same temperature, because the contraption isn't robbing heat from one location fast enough for it to cool without everything cooling.

t is possible that this is another basis for the temperature drop in the air with altitude. I can't quite reason that out, but this process for extracting work if the temperatures are the same suggests that the work would instantly be done by the unrestrained atmosphere.

If I were a betting man I would be backing "hydrostatic gases are isothermal and the math is flawed" all other options are rather controversial. If I were a betting man I would also be getting more nervous by the hour. I can hear crickets over senior forum members

Perhaps taking a bottle up and down Mauna Kea would be a great research grant to try and get. I volunteer to surf out and get the cleanest ocean air for you. You can walk to the summit and back down, then I will get the good stuff from a wave. Heck, put TWO bottles in the grant. I'll fill 'em both.

I probably could do with the exercise...

 

[DrStupid]

Could you please also explain why https://medium.com/criminal-clouds/the-isothermal-elevator-part-3-of-3-7d828edb3c1a wouldn't work?

The work of the total process is zero. The calculation in the link results in a net energy production because W₂ and W₄ do not include the volumetric work done by pushing the air out of the vessel or into the vessel. They just cover the expansion or compression and are therefore be valid in an evacuated environment only only.

 

[votingmachine]

The work of the total process is zero. The calculation in the link results in a net energy production because W₂ and W₄ do not include the volumetric work done by pushing the air out of the vessel or into the vessel. They just cover the expansion or compression and are therefore be valid in an evacuated environment only only.

I'm not following that work. The air moves from high pressure to low pressure. The work is taken when the environmental, sea-level atmospheric pressure pushes air into the (high altitude, low pressure) vessel, and again when the (sea level pressure) vessel is allowed to push air out, into the high altitude, low pressure environment.

Where/what is the volume work?

 

[DrStupid]

Where/what is the volume work?

For a better understanding let's divide step 2 into two separate sub-steps:
In step 2a the air in the vessel expands from bottom pressure to top pressure. This process releases energy. But now there is an excess volume that does not fit into the vessel and needs to be released into the environment. As there already is air that needs to be pushed away, this process consumes energy. Therefore only a part of the energy released during the expansion remains inside the system. The other part is gone with the volumetric work done against the environmental pressure. Step 4 can be divided in a similar way.

Here is my calculation:

With a bottom pressure p₁ the top pressure is

$p_2 = p_1 \cdot \exp \left( { - \frac{{M \cdot g \cdot h}}{{R \cdot T}}} \right): = p_1 \cdot \exp \left( { - k} \right)$

All processes are assumed to be isothermal and reversible.
The vessel has the volume V₁ and its empty weight shall be compensated by a counter weight. Thus moving the empty chamber requires no work.

Step 1: Lifting

$W_1 = - m \cdot g \cdot h = - n \cdot M \cdot g \cdot h = - p_1 \cdot V_1 \cdot \frac{{M \cdot g \cdot h}}{{R \cdot T}} = - p_1 \cdot V_1 \cdot k$

Step 2a: Expansion

The resulting volume is

$V_2 = V_1 \cdot \frac{{p_1 }}{{p_2 }} = V_1 \cdot \exp \left( k \right)$

and during this process the pressure decreases according to

$p\left( V \right) = p_1 \cdot \frac{{V_1 }}{V}$

The corresponding work is

$W_{2a} = \int\limits_{V_1 }^{V_2 } {p\left( V \right) \cdot dV} = p_1 \cdot V_1 \int\limits_{V_1 }^{V_1 \cdot \exp \left( k \right)} {\frac{{dV}}{V}} = p_1 \cdot V_1 \cdot k$

Step 2b: Pushing the excess volume out

$W_{2b} = - p_2 \cdot \left( {V_2 - V_1 } \right) = - p_1 \cdot V_1 \cdot \left[ {1 - \exp \left( { - k} \right)} \right]$

Step 3: Lowering

$W_1 = p_2 \cdot V_1 \cdot k = p_1 \cdot V_1 \cdot k \cdot \exp \left( { - k} \right)$

Step 4a: Compression

The resulting volume is

$V_3 = V_1 \cdot \frac{{p_2 }}{{p_1 }} = V_1 \cdot \exp \left( { - k} \right)$

And during this process the pressure decreases according to

$p\left( V \right) = p_2 \cdot \frac{{V_1 }}{V} = p_1 \cdot \frac{{V_1 }}{V} \cdot \exp \left( { - k} \right)$

The corresponding work is

$W_{4a} = \int\limits_{V_1 }^{V_3 } {p\left( V \right) \cdot dV} = p_1 \cdot V_1 \cdot \exp \left( { - k} \right)\int\limits_{V_1 }^{V_1 \cdot \exp \left( { - k} \right)} {\frac{{dV}}{V}} = - p_1 \cdot V_1 \cdot k \cdot \exp \left( { - k} \right)$

Step 4b: Getting air into the empty volume

$W_{4b} = p_1 \cdot \left( {V_1 - V_3 } \right) = p_1 \cdot V_1 \cdot \left[ {1 - \exp \left( { - k} \right)} \right]$

Total process:

Due to

$W_1 + W_{2a} = 0$
$W_{2b} + W_{4b} = 0$
$W_3 + W_{4a} = 0$

There is no resulting net energy production.

 

[moatilliatta]

The work of the total process is zero. The calculation in the link results in a net energy production because W₂ and W₄ do not include the volumetric work done by pushing the air out of the vessel or into the vessel. They just cover the expansion or compression and are therefore be valid in an evacuated environment only only.

Hi DrStupid,

I'm very glad someone is having a go at this. I'm not sure I'm reading you correctly, but I am pretty sure the vessel does not expand. I think its quite clear that the energy is harnessed from air moving through expanders that are stuck to side of the elevator shaft. There is no volume to be expelled.

I could easily be misunderstanding you so please elaborate.

 

[votingmachine]

I think that is not quite the set up they describe. The vessel is a fixed volume. The vessel is attached at the bottom for "filling", then detached, and attached again at the top for "emptying". I follow the math, but the intuitive sentence for step 2b and 4b elude me.

I still half-think there is a heat engine going on. As a mental image, picture a box with a gossamer divider. Say the top pressure is 0.5 atmospheres and the bottom is 1.0. The divider is at the end of the box and the air inside is at 0.5. At the bottom, the expanding 1.0 atmosphere air moves the divider to 1/2 the volume, and fills the other side. That makes the box full of 1 atmosphere air. Then at the top the divider slides to the end again, as the air expands.

At the bottom, the isothermal expansion absorbs heat, to keep the temperature constant as 1.0 atmosphere air expands, which would ordinarily cool. At the top the expanding gas again absorbs heat. To have an isothermal expansion, heat has to be added.

I'm not sure that the answer cannot be non-zero. It seems like it ought to be zero. But if it is not I think the heat flow explains it.

 

[mic*]

It IS a heat engine in a guise where the external heat added that drives a delta-P (to do work) is subsituted for gravity driving the delta-P.

So the true question is can we harness gravity like this? I am questioning all the idealised elements of the elevator.

I suspect the 2nd law of thermodynamics isn't really in jeopardy.

 

[willem2]

I think that is not quite the set up they describe. The vessel is a fixed volume. The vessel is attached at the bottom for "filling", then detached, and attached again at the top for "emptying". I follow the math, but the intuitive sentence for step 2b and 4b elude me.

You need to push the outside air away. The usual formula for the work gained from isothermal expansion, assumes that the gas is in a cylinder pushing against a piston with vacuum on the other side. If there's air on the other side, you need to do work to push it away.
Note however that the work on the outside air at the top and at the bottom cancel, so you could just ignore this.

At the bottom, the isothermal expansion absorbs heat, to keep the temperature constant as 1.0 atmosphere air expands, which would ordinarily cool. At the top the expanding gas again absorbs heat. To have an isothermal expansion, heat has to be added.

So you could have a process that takes or produces no work and cools the whole elevator?
The expansion at the top produces work and removes heat, and the compression at the bottom takes work and removes heat. This is only OK as long as there is no temperature difference, since otherwise the 2nd law would be violated. In the real atmosphere, the elevator would take work, cool the top and heat the bottom more.

I'm not sure that the answer cannot be non-zero. It seems like it ought to be zero. But if it is not I think the heat flow explains it.

The problem with the calculations on the Isothermal Elevator page is in

$$W_{exp} = \Delta nRT ln (\frac {P_{final}}{P_{start}})$$

used in the calculation for the expansion or compression of gas.

I don't know where he gets the Δn from.

 

[moatilliatta]

The problem with the calculations on the Isothermal Elevator page is in

Wexp=ΔnRTln(PfinalPstart)​

W_{exp} = \Delta nRT ln (\frac {P_{final}}{P_{start}})

used in the calculation for the expansion or compression of gas.

I don't know where he gets the Δn from.

Yes,you must be right. So if that is the wrong formula, what is the right formula for the process?