What is Air Pressure? Understanding the Weight of the Atmosphere

[russ_watters]

Take a barometer out of its case during a windstorm and watch how the needle moves wildly. Do you think that this truly represents the "weight of the overlying air"?

This and most of the "incorrect assumptions" you listed above mix a concept with a model:

The barometric formula relating pressure to the weight of the overlying air is invalid in many respects.

While a certain formula may require such a simplifying assumption (I don't think it is correct to call them "incorrect assumptions"...but then I am an engineer...), the concept is unaffected by variations in temperature.

When the wind blows across a building with open windows, barometers in the different rooms will show different pressures. And these pressures will be different from that registered by a barometer in the basket of a nearby drifting balloon at the same elevation.

If the barometer is measuring velocity pressure, then it isn't just measuring atmospheric pressure. That's not a flaw in the description of "atmospheric pressure", it's a flaw in our ability to measure it.

Don't let the complexities of a moving atmosphere distract you from the reality that whether you model it as a pile of solid balls or a bunch of bouncing balls, the pressure is the same as long as you have the same number/weight of balls.

The conic section issue is interesting, though. I've never heard it and it sounds reasonable, but how big of a difference does it really make? We're not talking about a very large distance, since most of the atmosphere is packed close to the earth. Similar issue with g. I guess if I were being pedantic with the conic section one though, I'd say that saying "above" implies it must be conic/pyramidal, though I do admit I've never thought of it.

 

[sophiecentaur]

@klimatos
Your comments are mostly just additions to the basic model. As Russ says, the effect of wind is not relevant to the average situation.

Taking up your 'conic' objection and applying the actual figures. Effectively, 'space' takes over at a height of about 100km, beyond which there is very little mass of atmosphere. As a rule of thumb, Atmospheric `pressure doubles every 5000m of altitude. That means that half of the atmosphere's weight is contributed by the lowest 5km. By the time you are 20km up, the AP will be 1/16 of that at sea level; most of the atmosphere will be below you..
Lets examine this frustum (truncated cone) to which you refer and compare it with a 1m radius cylinder. The Earth's radius is about 6000km so, geometry tells us that the radius of a conic section at 20km will be 6020/6000 of that at the surface. Your 1m cylinder (at sea level) has now stretched to a radius if 1.003m. That's a difference in area of about 0.6% extra. Perhaps not worth making a big shout about.
Furthermore, the factor by which the area (and hence the difference in volume of an elemental slice) will be increased is exactly the same as the factor by which g will decrease, due to the inverse square law. So the effect would be exactly the same as for the simpler cylindrical model. There would be more molecules up there but they would all weigh a bit less each. (I rather like that argument actually - preen preen). Your 'dimensional' objection doesn't really seem relevant.

You have an objection based on the use of (and possibly failure of) the gas laws and their assumptions. We all know that real molecules have finite radius and that the mean free path of an air molecule is not many mm at AP, so they really can be expected to collide and pass on / exchange their momentum. The upper ones are kept up there by bouncing against the lower ones. Interestingly, if you take an atmosphere of ideal molecules, which never hit each other, you would have a situation in which the only thing keeping molecules up there would be that they would all, at some stage, hit the ground and bounce off, with a range of velocities, determined by the thermal motion of the molecules on the solid ground. They would rise through the atmosphere in a parabolic trajectory and they will lose Kinetic Energy and gain Gravitational Potential Energy, eventually falling to Earth again for another bounce. In a simple model, this would imply that the temperature of the atmosphere would steadily reduce as you go higher because the average KE would reduce. But, effectively, each molecule, by striking the ground once every so often, would be just like the ball which my man on the scales keeps hitting upwards. The average force will just be due to its weight. So, with or without the gas laws, the 'weight' argument must still apply.

But, in the end, what is the problem with relating AP to the weight of the atmosphere? Numerically, it's the same, irrespective of the varying constitution of it with height. I think the main problem that was expressed in this thread was the question of how a nebulous thing like a gas can have the same effect as a solid piston. I think that;s something one just has to come to terms with - because it can clearly be shown to be true.

 

[Feeble Wonk]

From one non-physicist to another... perhaps the most basic way of thinking about this is that the atmosphere is simply an accumulation of gas molecules that respond to the forces placed on them. Gravity acts on them because they are massive particles, thus pulling them toward the Earth's surface. The electromagnetic forces applied by the negatively charged electron shells repel the individual molecules away from each other. Air pressure builds as these forces compete against each other. Therefore, as more air molecules are accumulated in a given area of atmosphere, the air pressure rises, and will attempt to reduce the pressure by moving the air mass toward adjacent areas that lesser accumulations of air molecules. Obviously, any additional factors that effect molecular action/force (such as temperature) will have their effect as well. It all is simply the result of the system, as a whole, trying to achieve higher entropy.

 

[sophiecentaur]

From one non-physicist to another... perhaps the most basic way of thinking about this is that the atmosphere is simply an accumulation of gas molecules that respond to the forces placed on them. Gravity acts on them because they are massive particles, thus pulling them toward the Earth's surface. The electromagnetic forces applied by the negatively charged electron shells repel the individual molecules away from each other. Air pressure builds as these forces compete against each other. Therefore, as more air molecules are accumulated in a given area of atmosphere, the air pressure rises, and will attempt to reduce the pressure by moving the air mass toward adjacent areas that lesser accumulations of air molecules. Obviously, any additional factors that effect molecular action/force (such as temperature) will have their effect as well. It all is simply the result of the system, as a whole, trying to achieve higher entropy.

Pretty well summed up - the molecules are kept apart by just the same forces that keep the molecules of a solid apart - it's just that they are more sporadically applied.

 

[klimatos]

This and most of the "incorrect assumptions" you listed above mix a concept with a model: While a certain formula may require such a simplifying assumption (I don't think it is correct to call them "incorrect assumptions"...but then I am an engineer...), the concept is unaffected by variations in temperature.

We may have a semantic problem here. My position is that if they do not match the observed characteristics of the real atmosphere, then they are incorrect.

[/QUOTE] If the barometer is measuring velocity pressure, then it isn't just measuring atmospheric pressure. That's not a flaw in the description of "atmospheric pressure", it's a flaw in our ability to measure it.[/QUOTE]

A barometer is just another form of manometer. It has no magical properties. It cannot separate out the "weight of the atmosphere" from the myriad other factors that influence how often and how hard the air molecules impact on its sensing surface. It measures the pressure of the ambient air. That is all that it does.

[/QUOTE]Don't let the complexities of a moving atmosphere distract you from the reality that whether you model it as a pile of solid balls or a bunch of bouncing balls, the pressure is the same as long as you have the same number/weight of balls.[/QUOTE]

I'm sure you wrote that without thinking it through. Change the "bouncing" speed and the pressure changes. The mass remains the same.

[/QUOTE]The conic section issue is interesting, though. I've never heard it and it sounds reasonable, but how big of a difference does it really make? We're not talking about a very large distance, since most of the atmosphere is packed close to the earth. Similar issue with g. I guess if I were being pedantic with the conic section one though, I'd say that saying "above" implies it must be conic/pyramidal, though I do admit I've never thought of it.[/QUOTE]

It doesn't make a big difference. But it implies sloppy thinking.

 

[klimatos]

I have another objection to the weight-force hypothesis. For the sake of argument, (I don't really believe it.) let us assume that in a still atmosphere the ground pressure did actually represent the "weight of the overlying air".

Now let us postulate a wind aloft. In keeping with Bernoulli's principle, this wind would drop the pressure on all surrounding parcels of air. The pressure drop would be ultimately measured at the surface where the air was still. The mass of the air in a column of air is still the same as before the wind started blowing; but the pressure is less.

A second and opposite wind at another elevation does not cancel out the pressure drop of the first wind, but simply adds to it.

Since winds are blowing at some elevation virtually everywhere, the logical conclusion is that the hydrostatic equation actually underestimates the weight of the overlying air.

 

[sophiecentaur]

I have another objection to the weight-force hypothesis. For the sake of argument, (I don't really believe it.) let us assume that in a still atmosphere the ground pressure did actually represent the "weight of the overlying air".

Now let us postulate a wind aloft. In keeping with Bernoulli's principle, this wind would drop the pressure on all surrounding parcels of air. The pressure drop would be ultimately measured at the surface where the air was still. The mass of the air in a column of air is still the same as before the wind started blowing; but the pressure is less.

A second and opposite wind at another elevation does not cancel out the pressure drop of the first wind, but simply adds to it.

Since winds are blowing at some elevation virtually everywhere, the logical conclusion is that the hydrostatic equation actually underestimates the weight of the overlying air.

But no one is saying that the atmosphere is still. the fact is that the pressure variations which we experience due to convection, the Earth's rotation and other thermal effects doesn't actually perturb the general atmospheric pressure very much (a range of about 6%, I think). Right inside a tornado it may get a bit extreme but that is due to very local effects. The effects are all accounted for.
I don't actually see why you object to the term 'weight' so much. If it weren't for the weight of the molecules they would just drift off into space. Yes, the basic weight contribution is subject to modification by other effects; it's a complicated and very energetic system but why take issue with a basic fact that the actual average pressure relates well to the mass of gas in the atmosphere. Can you suggest another explanation or description?

I think your argument about air flow and pressure needs justifying quantitatively because any change in pressure would surely result in a reduction in volume (all things being equal) - and that would alter the density of a moving volume and also the surrounding regions. Why should that alter the value (on average) due to the 'weight'? And how much would you calculate the difference to be?

 

[klimatos]

I don't actually see why you object to the term 'weight' so much. If it weren't for the weight of the molecules they would just drift off into space. Yes, the basic weight contribution is subject to modification by other effects; it's a complicated and very energetic system but why take issue with a basic fact that the actual average pressure relates well to the mass of gas in the atmosphere. Can you suggest another explanation or description?

I object because I do a lot of work with statistical mechanics and kinetic gas theory. Down at the molecular level, the idea that pressure represents the weight of the overlying molecules is absurd. Let me give you an example.

In any given instant in time, an area of sensing surface equal to the presumed molecular cross-sectional area will have no molecular impacts. The pressure is zero. Does this mean that the volume of air above that area has no mass. Of course not.

Then, along comes an air molecule and impacts on that surface. The impulse generated depends on that molecule's mass and speed normal to the surface. Let's postulate average mass and average speed and an air temperature of 25°C. The impulse generated will be equivalent to some 186,000 hectoPascals. Does that represent the weight of the overlying air. It does not.

Consequently, at what "magic" combination of duration and area does pressure stop being simply the product of frequency of molecular impact and mean impulse and also become the weight of the overlying air?

Why can't we just accept the fact that atmospheric pressures measure the impacts of air molecules on a surface? Why do we have to invoke the "weight of the overlying air"? What is gained by this?

 

[klimatos]

I see from the comments posted that I have not made myself clear. Since there are many of you and one of me, the fault is obviously mine. Let me try to clarify.

1) I have no trouble in accepting that the pressure of 1013.25 hPa closely measures the mass/weight of the atmosphere. There is abundant evidence to support this hypothesis.

2) I do have trouble believing that a barometer always measures the weight of the overlying air at that time and place. I think that it simply measures the local ambient air pressure; and that this pressure does not necessarily reflect that weight.

3) I do not see the either the mechanical or the statistical linkage between molecular mass at elevation z and air pressure on the ground.

4) Any weight-force at elevation z would be diminished by the inverse square law before it reached the surface.

5) A thunderclap is far more forceful than such a weight-force, and it is my understanding that no thunderclap has ever been heard at a distance of more than twenty kilometers. That force does not simply diminish below the hearing threshold, it utterly disappears into the entropy of the air. I can see no reason why the same cannot be said for the weight-force.

6) I am quite familiar with the barometric formula and its derivation from the hydrostatic equation. This derivation proves nothing, since the hydrostatic equation assumes that pressure equals weight. Proving your assumptions is bad form in all disciplines.

7) I am open to any argument that does not start with the assumption that a barometer measures the weight of the overlying air at that time and that place--regardless of weather conditions on the ground or aloft.

 

[sophiecentaur]

Two brief responses.
Where does the inverse square law apply? If you insist on dissipating the wright of a molecule then you have to accept that the shares from many molecules will arrive at each spot on the ground.
Also, the pressure from a lightning flash is an impulse which will be dispersed / absorbed. The weight effect is 'dc'. There is no comparison.

 

[russ_watters]

We may have a semantic problem here. My position is that if they do not match the observed characteristics of the real atmosphere, then they are incorrect.

I think it was more basic than that: I think you confused a concept with an equation and if you're using the equation I'm thinking of, it's an ad-hoc model of pressure vs altitude that doesn't make any attempt to explain why ground level atmospheric pressure is what it is -- which is the entire question being asked! Is this what you were referring to?: http://hyperphysics.phy-astr.gsu.edu/hbase/kinetic/barfor.html

But on another level, you are right about the difference between a physicist and an engineer. I see 99% correct - particularly when answering a basic question - as being close enough. I don't think it's useful to confuse a novice with things that make very little difference.

A barometer is just another form of manometer. It has no magical properties. It cannot separate out the "weight of the atmosphere" from the myriad other factors that influence how often and how hard the air molecules impact on its sensing surface.

Agreed!

It measures the pressure of the ambient air. That is all that it does.

What pressure? If it can't distinguish between velocity pressure and static pressure, then it isn't measuring "the pressure of the ambient air" - ie, atmospheric pressure, which is static pressure only.

I'm sure you wrote that without thinking it through. Change the "bouncing" speed and the pressure changes. The mass remains the same.

Nope. You are incorrect - its true for a closed container, but not for the atmosphere. If you change the bouncing speed, the force of each collision increases, but the frequency of collisions goes down and the net primary effect is absolutely no change in pressure. The secondary effect of increasing the temperature and therefore "bouncing speed", though, would be to expand the atmosphere and due to the reduced g, you'd have a tiny reduction in atmospheric pressure.

 

[russ_watters]

I have another objection to the weight-force hypothesis. For the sake of argument, (I don't really believe it.) let us assume that in a still atmosphere the ground pressure did actually represent the "weight of the overlying air".

Now let us postulate a wind aloft. In keeping with Bernoulli's principle, this wind would drop the pressure on all surrounding parcels of air. The pressure drop would be ultimately measured at the surface where the air was still. The mass of the air in a column of air is still the same as before the wind started blowing; but the pressure is less.

A second and opposite wind at another elevation does not cancel out the pressure drop of the first wind, but simply adds to it.

Since winds are blowing at some elevation virtually everywhere, the logical conclusion is that the hydrostatic equation actually underestimates the weight of the overlying air.

Now you're oversimplyfing - to create an effect that doesn't exist. Jets streams still have the same underlying cause as other winds: rising currents of warm air (and complimentary dropping currents of cold air somewhere else). So they aren't just complementary winds moving in opposite horizontal directions. The net effect is all that cancels itself out.

Moreover, even if there were no complimentary rise in atmospheric pressure somewhere else due to a falling, cold mass of air, you're again quibbling over an effect that if it existed would account for a tiny fraction of a percent of atmospheric pressure.

Now there is an effect that probably makes a difference (again, by a tiny fraction of a percent): The atmosphere is one big steam engine and if you count the water going up as being part of the atmosphere but don't count the water going down, that would cause a discrepancy.

 

[russ_watters]

I object because I do a lot of work with statistical mechanics and kinetic gas theory. Down at the molecular level, the idea that pressure represents the weight of the overlying molecules is absurd.

I think that may well be the problem: if you zoom into the molecular level, you're now only looking at the interactions of the molecules and ignoring the container they are in that causes them to be constrained in the first place! For a car tire, the "container" is the tire. For the atmosphere, the "container" is the molecules above you. You are almost literally missing the forest for the trees.

Let me try it another way: if the concept is "absurd", is it just a crazy coincidence that it comes damn close to giving exactly the right answer?

Consequently, at what "magic" combination of duration and area does pressure stop being simply the product of frequency of molecular impact and mean impulse and also become the weight of the overlying air?

That sentence is not grammaticality correct and I'm not being pedantic, but it matters here: it doesn't "stop" being the frequency/impulse of molecular impacts. But that's an effect without a cause. It can tell you what atmospheric pressure is (given a pre-determined local set of conditions), but not why.

Why can't we just accept the fact that atmospheric pressures measure the impacts of air molecules on a surface? Why do we have to invoke the "weight of the overlying air"? What is gained by this?

We gain "why?".

 

[russ_watters]

Klimatos, I think maybe going back to the op would focus us: the op was asking about the atmosphere vs a closed container. Zooming into the molecular level just doesn't address the question.

Edit: actually its worse than that. Given a certain average position, an air molecule imparts more impulse in a downward direction than an upward collision. The only way to deal with that is to insert gpe into the model. Integrating that over the height of the atmosphere yields hydrostatic pressure!

Or looking at it from the other direction: a molecular model that doesn't account for the weight of the molecules yields an atmosphere that deflates like a popped balloon.

 

[klimatos]

Russ,

Again I see I am not making myself clear. I do not object to the concept that the mean sea level pressure reflects the mass of the global atmosphere. I object to the notion that a barometer measures the weight of a column of overlying air. I can see nothing in the mechanism of a barometer that can distinguish between an impulse generated directly overhead from an impulse generated off at an angle. I can see nothing in the mechanism of a barometer that can distinguish static pressure from dynamic pressure.

Consequently, I believe that barometers simply measure the frequency and force of the air molecules that impact on their sensing surface. In other words, they measure ambient pressure (static plus dynamic plus whatever else) and not simply the weight of the overlying air.

 

[klimatos]

1] Yes, the basic weight contribution is subject to modification by other effects; it's a complicated and very energetic system but why take issue with a basic fact that the actual average pressure relates well to the mass of gas in the atmosphere. Can you suggest another explanation or description?

2]I think your argument about air flow and pressure needs justifying quantitatively because any change in pressure would surely result in a reduction in volume (all things being equal) - and that would alter the density of a moving volume and also the surrounding regions. Why should that alter the value (on average) due to the 'weight'? And how much would you calculate the difference to be?

Sophie,

1] I don't take issue with the idea that the average sea level pressure generally reflects the mass of the global atmosphere. I don't believe that I said that. I take issue with the notion that a barometer measures the weight of an overlying column of atmosphere.

2] Bernoulli's Equation makes the pressure drop proportional to the density of the fluid and the square of the velocity. Although only truly valid for incompressible fluids, it works well enough for air at the atmospheric range of wind speeds. This drop in pressure is sufficient to pull a heavy sheet of plywood out of the bed of a pickup truck at moderate road speeds. It can lift several inches of liquid through the stem of an atomizer. It can pull the roof off of a house in a windstorm. It is certainly strong enough to drop the pressure on a barometer. I have seen it do so on innumerable occasions. The pressure drop depends upon whether or not the barometer was dampened by enclosure in a near-airtight case or standing free.

Standing free, the highest pressure drop that I have seen was 33 hPa.

 

[klimatos]

Sophie,

1] I don't take issue with the idea that the average sea level pressure generally reflects the mass of the global atmosphere. I don't believe that I said that. I take issue with the notion that a barometer measures the weight of an overlying column of atmosphere.

Mea Culpa! Pardon me while I scrape the egg off of my face. In going over my previous posts, I find that I did say that (post #36). Allow me a moment for cool contemplation so as to resist my natural tendency to jump into an ongoing argument with both feet.

Years ago, when I had just returned from overseas, I bought bicycles for my three children. They were foreign-made and came unassembled. The instructions had obviously been literally translated from the original language. Instruction #1 was, "Approach assembly in contemplative frame of mind."

That was good advice then, and is good advice now.

Therefore: I believe that the mean sea level pressure closely approximates the mass of the overlying global atmosphere, but that it underestimates it somewhat because of the effect of winds. There.

Does anyone know of a good paper on calculating the mass of the atmosphere that does not start with the hydrostatic equation, but that arrives at essentially the same evaluation?

 

[sophiecentaur]

@klimatos
I'm not sure where this leaves us now. Are we waiting for another word from you, after contemplation?
Meanwhile, here is a thought model / experiment.
Imagine taking the atmosphere and dividing it up into a number of 'bubbles' with massless and totally flexible envelopes. Each bubble would have 'weight' in its own right. Its volume would depend upon the temperature and the pressure exerted on it by the adjacent bubbles. In fact, make the model into a vertical cylinder (/cone, if you like) with massless, frictionless etc. horizontal divisions, breaking the column up into cells. The pressure difference between the upper and lower faces of one of the cells would just be due to the weight of the air contained in that section. The pressure on the base of the bottom cell must be the total weight of the cells above divided by the area - just the same as if it were a pile of coins.
The same situation would exist if you were to remove the division between pairs of adjacent cells. The molecules in each half would be providing the same pressure against each other as when the intervening wall was present. This is the simple, hydrostatic approach. Because we are dealing with gases, the volumes of the segments will be affected by temperature.
Starting with equilibrium, whatever you do to change the molecular activity within each cell, (which you say will affect the pressure and invalidate the 'weight' argument), the volume will adjust until equilibrium is again reached. Although the volume may change, you still have the same mass up there so the force (hence the pressure) will be the same. The only thing that could change the result would be that the changing size of the cells can cause a small inverse square law change in the weight of each of the cells as their altitude changes.

We all know that the local pressure on Earth is affected by the different masses of air moving about the globe, causing the weather but in the end, the mean pressure over the Earth's surface can only be due to the total weight of all the cells up there, modified, slightly by the actual height / depth of the atmosphere (a result of an overall change in temperature).

 

[klimatos]

Sophie,

I have two problems with your thought experiment.

1) It assumes closed containers. All of my 55 years of studies have been of the free atmosphere--unbounded, unlimited, and unconstrained. I have never done research on closed containers. I know nothing about them except what I have read in physics publications. There is a big difference between the behavior of air in the free atmosphere and air in closed containers. In the containers, the standard gas laws apply. In the free atmosphere most of them do not apply. There are two basic reasons for this: One, you cannot hold A steady while varying B. Two, conditions of equilibrium do not exist.

2) The assumption of equilibrium is my second problem. I have never seen a significant portion of the free atmosphere in a state of equilibrium. It is always being heated or cooling, being humidified or undergoing condensation, and 99% of the time a wind is blowing at some elevation or other. In other words, there is weather going on. By definition, when weather is occurring, equilibrium cannot exist.

I repeat what I have said before: I have no problem in believing that the mean atmospheric pressure approximates (not equals) the mass of the atmosphere. I simply cannot bring myself to believe that a barometer always measures the weight of an overlying column of air.

If you respond by saying that sometimes is does and sometimes it doesn't, then I have to ask how we can we tell when it is and how can we tell when it is not.

 

[russ_watters]

Again I see I am not making myself clear. I do not object to the concept that the mean sea level pressure reflects the mass of the global atmosphere.

Oh, ok - well then I don't think we really have much to disagree on! That's the primary point of the thread!

I object to the notion that a barometer measures the weight of a column of overlying air.

I haven't seen anyone claim that does. There are many types of barometers and all have at least some flaw. Though people in the thread have used the word "barometer", I'm reasonably certain they aren't talking about real instruments with flaws, but rather using the word as a proxy for "a device that measures atmospheric pressure". I think you're being too pedantic here.

[then I saw this...]

Mea Culpa! Pardon me while I scrape the egg off of my face. In going over my previous posts, I find that I did say that (post #36). Allow me a moment for cool contemplation so as to resist my natural tendency to jump into an ongoing argument with both feet.

Meh. Internet forums make people argumentative and stubborn. Never happens to me though.

No prob.

Therefore: I believe that the mean sea level pressure closely approximates the mass of the overlying global atmosphere, but that it underestimates it somewhat because of the effect of winds. There.

I disagree as said in a previous post (I think your concept violates conservation of energy), but we're getting pretty far down in the weeds now.

 

[russ_watters]

I simply cannot bring myself to believe that a barometer always measures the weight of an overlying column of air.

Well, scales don't always measure the weigh or mass of objects either, but we still use them for exactly that purpose. When the flaws are known/understood, they can be mitigated. Ironically, the flaw you pointed out in barometers is exactly the same as the primary flaw in scales. That flaw (for both) can be mitigated with an enclosure that blocks airflow around the device.

 

[sophiecentaur]

@klimatos
The only way that the gas law could not apply to a region of gas 'out in the middle' of another larger mass, would be if it could, somehow 'know' where it was. How could that be?

 

[klimatos]

No prob. I disagree as said in a previous post (I think your concept violates conservation of energy), but we're getting pretty far down in the weeds now.

Thanks, Russ

I'm beginning to feel that all of us could probably spend our time more profitably on some other topic, but I'm going to make one last stab at justifying my position that the mean atmospheric pressure underestimates the mass of the atmosphere.

Take a laboratory container of still water. The pressure on the bottom of the container accurately measures the weight of the overlying water.

Now take a rushing river. The pressure on the bottom of the river does not accurately measure the weight of the overlying water because of the Bernoulli drop in pressure due to the water's density and the square of the velocity. The faster the water flows, the greater the drop in pressure.

Like the river, the atmosphere is a dynamic system. Mutability might well be its defining characteristic. It is always in movement at one elevation or another. Therefore, by the same principles that we applied to the river, the mean sea level pressure must underestimate the weight of the overlying atmosphere.

I get the distinct impression that most of my responders are most comfortable with laboratory observations and static conditions. I also get the impression that they are not comfortable with kinetic gas theory and statistical mechanics. In this, I may be doing them an injustice. It just appears the most of the responses are based on statics and closed systems. Neither concept can be applied to the free atmosphere.

 

[DaveC426913]

Like the river, the atmosphere is a dynamic system. Mutability might well be its defining characteristic. It is always in movement at one elevation or another. Therefore, by the same principles that we applied to the river, the mean sea level pressure must underestimate the weight of the overlying atmosphere.

A few assumptions here.
1] How much is the air moving at any given time? How much does that affect the air pressure? Certainly, any attempt at an accurate measurement of air pressure will be confounded with the presence of nearby winds. An accurate scale would bob up and down a little. Measure it when the air is still.

2] Does Bernoulli's Law apply to an entire body of air/fluid, despite it flowing at varying rates (including zero and negative), throughout its thickness? I suspect it does not apply so simply. I am not convinced that your river is a valid analogy to the atmosphere in termes of Bernoulli's Law.

 

[klimatos]

@klimatos
The only way that the gas law could not apply to a region of gas 'out in the middle' of another larger mass, would be if it could, somehow 'know' where it was. How could that be?

Sophie,

I believe that you misspoke when you referred to "the gas law". As I'm sure you know, there are many gas laws, from Amonton's Law to Van der Waal's Law. Perhaps you are referring to the Ideal Gas Equation of State: PV=RT. That equation is only valid under conditions of equilibrium, and equilibrium is virtually never present in the free atmosphere.

Neither Boyle's Law nor Charles' Law can be applied to processes in the free atmosphere because each requires that the parcel of air under observation have a measurable volume (not possible in the free atmosphere) and that either temperature or pressure be kept constant (not possible in the free atmosphere). The same basic objections apply to a number of other gas laws.

On the other hand, Avogadro's Law works on the free atmosphere, as does Bernoulli's principle, Dalton's Law of Partial Pressures, Graham's Law of Diffusion, the Maxwell-Boltzmann Distribution Function, and Van der Waal's Law.

 

[klimatos]

A few assumptions here.
1] How much is the air moving at any given time? How much does that affect the air pressure? Certainly, any attempt at an accurate measurement of air pressure will be confounded with the presence of nearby winds. An accurate scale would bob up and down a little. Measure it when the air is still.

2] Does Bernoulli's Law apply to an entire body of air/fluid, despite it flowing at varying rates (including zero and negative), throughout its thickness? I suspect it does not apply so simply. I am not convinced that your river is a valid analogy to the atmosphere in termes of Bernoulli's Law.

Dave,

For air with a temperature of 25°C at 50%RH and an ambient pressure of 1000 hPa, the drop in pressure on a parallel surface (assuming no turbulence) will be 56 Pascals at a wind velocity of 10 meters per second. It will be a pressure drop of 4,248 Pascals at a wind velocity of 100 meters per second.

You say measure it when the air is still. Just because the air is still at sea level does not mean that the air is still at all elevations. Winds aloft still induce drops in pressure.

Bernoulli's Law applies to all parts of a moving fluid. As you say, the application is not a simple one. However, the principle still applies. I will bow to experts in fluid mechanics as to the details.

You say you remain unconvinced by my river analogy. What part of it don't you like?

 

[Drakkith]

If we remove the top 50 miles of the atmosphere, it will be mostly replaced as the air lower in height moves up thanks to the lack of air above it "pushing" it down correct? Does that have any relation to weight? Looking from an overall picture it looks like the air above it is weighing it down.

 

[DaveC426913]

You say you remain unconvinced by my river analogy. What part of it don't you like?

For all of the above reasons. Air is much more turbulent and at the same time, less viscous than water, so the effects would essentially be lost in the noise. I am not convinced that Bernoulli's Law is sufficient to draw a proportionality correlation between the complex (read: noisy) movement of the air and some specific drop in pressure in a real world situation.

You say we would be "underestimating" the mass of atmo - i.e. consistently on the low side. I'm not convinced that margin would consistently err on the low side such that it would count as a consistent underestimation. It seems sufficient to me to conclude that, due to a shifting atmo, we would expect a margin of error +/-.

 

[sophiecentaur]

I think klimatos is actually shifting his ground from a position of 'nothing to do with weight' to one of modification of the basic principle by a dynamic situation.
Whether, in the overall scenario of energy balance (heat in - heat out) there is any overall modification to the 'weight' figure, is questionable. I have a feeling that the presence of winds is no different, in principle to the 'internal energy' corresponding to average ambient temperatures.

All this stuff about needing a fixed mass of a gas in order to derive the gas LawS is only to do with the derivation and where the gas happens to be. Would no Gas Laws apply inside a gas nebula?

 

[klimatos]

For air with a temperature of 25°C at 50%RH and an ambient pressure of 1000 hPa, the drop in pressure on a parallel surface (assuming no turbulence) will be 56 Pascals at a wind velocity of 10 meters per second. It will be a pressure drop of 4,248 Pascals at a wind velocity of 100 meters per second.

CORRECTION! I grabbed the wrong column from my Excel spreadsheet. (That will teach me to use explicit headings!)

At a wind speed of 10 meters per second, the pressure drop on a parallel surface will be 14 Pascals. At a wind speed of 100 meters per second, the pressure drop will be 1447 Pascals.

If you wish to check my calculations (not a bad idea), Bernoulli gives the drop in pressure as equal to one-half the product of the mass density in kilograms times the square of the wind velocity in meters per second. Under the stated conditions, I used a mass density of 1.15792178 kilograms per cubic meter.

The erroneous values were for the pressure drop on the leeward side of an obstruction (assuming no turbulence).

Technically speaking, Bernoulli's Equation only applies to incompressible fluids. However, wiki articles on wind force state that it is valid for wind speeds up to 0.3 Mach.