What is Air Pressure? Understanding the Weight of the Atmosphere

[Graeme M]

Reading a website and I noticed that the author claimed that air pressure is essentially equal to the weight of the atmosphere. I googled that fact and sure enough, everywhere I go there is that statement. But...

Caveat. I have no maths or science background, so won't cope with too many equations in answering my question...

I assume the concept pressure=weight is a sort of generalisation. Because it does not seem obvious to me that this could be so.

Here is a thought experiment. I have a safe of solid steel 1 metre on all sides with an airtight seal. With the safe open and a barometer inside, the air pressure inside will be whatever the extant air pressure is at my location. If I close my safe, there is no longer a column of air acting upon my barometer. True, the atmosphere may be weighing heavily upon my safe's surface, but I see no way for that to affect my barometer. The parcel of air available to exert some kind of vertical pressure is little more than a few cm high. It won't weigh much...

What do I not understand?

 

[tiny-tim]

Welcome to PF!

Hi Graeme! Welcome to PF!

the author claimed that air pressure is essentially equal to the weight of the atmosphere. I googled that fact and sure enough, everywhere I go there is that statement. But...
…
… If I close my safe, there is no longer a column of air acting upon my barometer. True, the atmosphere may be weighing heavily upon my safe's surface, but I see no way for that to affect my barometer. The parcel of air available to exert some kind of vertical pressure is little more than a few cm high. It won't weigh much...

Yes, you're right to be suspicious …

technically, the air pressure is essentially equal to the weight of the atmosphere plus any external forces (other than weight) …​

we usually take a column of air that extends to "outer space", where the external force is zero,

but in your example the column extends only to the roof of the safe, which is exerting a downward force on the air.

(in practice, of course, the air pressure inside the safe is simply whatever it was when you closed the door, and from good ol' https://www.physicsforums.com/library.php?do=view_item&itemid=373" from the roof is equal and opposite to the force of the air on the roof )

 

[Graeme M]

Thanks tim. Hope you can bear with me on this, just a few more questions to clarify that. I'll have to read up on what the third law says (and the first and second too clearly!).

I would have imagined that weight is the force exerted on a surface by an object that is accelerated by gravity. That is (and I am sorry if I don't know the correct physics terms/concepts to describe what I am saying), any object with mass will, in the presence of a gravitational field, be accelerated by that field, presumably only in a direction downwards perpendicular to the surface of the gravitational object. That object will not 'weigh' anything while being so accelerated. It will only exhibit 'weight' when it is prevented from accelerating. So weight is the effect of an object being prevented from responding to a gravitational field. All objects on Earth's surface will fall to the surface, where weight can be measured.

However, the atmosphere is not falling. As I understand it, the atmosphere is composed of free moving molecules which are in effect unconstrained.

If those molecules are not 'connected' and not falling, then they can exert no force from the effects of gravity on any surface ?

 

[russ_watters]

Are you sure those websites said "air pressure" and not "atmospheric pressure"?

 

[Graeme M]

Ummm... I am unaware of the distinction. I mean the pressure that one experiences at the surface of the Earth and which is measured by a barometer.

I just read tiny_tim's reaction force link. This is exactly my point. The magic is contained in the statement "between two bodies in contact". Which is precisely what I said above. There must be a surface connection for weight to be exhibited, or for a force to be realized. The atmosphere is not composed of a single object or body - it is free moving molecules which do not seem to compose a 'body'. Whilst gravity may affect each molecule, it would seem to do so independently. Solid objects are composed of molecules in some form of bond (is that E/M forces or chemical bonds?) and thus the effect of gravity on each molecule is effectively communicated across all molecules in that object. Not so an atmosphere I'd have thought?

 

[tiny-tim]

Hi Graeme!

However, the atmosphere is not falling. As I understand it, the atmosphere is composed of free moving molecules which are in effect unconstrained.

If those molecules are not 'connected' and not falling, then they can exert no force from the effects of gravity on any surface ?

The molecules are moving freely, but only until they hit something.

Usually, they just hit each other, but when they hit a solid surface they bounce off it, and that change in momentum exerts (indeed, is) a force on the surface.

(that's https://www.physicsforums.com/library.php?do=view_item&itemid=26" … force = rate of change of momentum)

(and Newton's third law applies as between the surface and the molecules that hit it)

 

[Graeme M]

Yes, but it will only be those molecules striking the surface that can impart a force. The molecules 200 feet further up, or 5 km further up, are not striking that surface. Or are you suggesting some kind of instantaneous cascading collision force through the entire atmospheric column at any instant?

 

[tiny-tim]

Or are you suggesting some kind of instantaneous cascading collision force through the entire atmospheric column at any instant?

Yes , except that it doesn't need to be instantaneous, does it?

 

[Graeme M]

Hmmm... yes, it would be largely wouldn't it?

I guess I am talking from ignorance here, but a solid object is solid at most instants isn't it? I mean if we slice down time to ever smaller slices then our solid object remains solid I assume? I am getting into very fuzzy territory now as I have no idea what actually makes a solid object solid, but it is the state of solidness that allows an object to be accelerated by gravity isn't it?

 

[Graeme M]

But getting back to my question, it seems to me that a solid object is a bound object and it is that binding that allows it to exhibit 'weight'. So I can easily conceptualise such an object bearing on a surface. I can't see how individual free moving molecules can transfer the effect of 'weight' so effectively throughout the whole vertical column.

 

[tiny-tim]

But getting back to my question, it seems to me that a solid object is a bound object and it is that binding that allows it to exhibit 'weight'. So I can easily conceptualise such an object bearing on a surface. I can't see how individual free moving molecules can transfer the effect of 'weight' so effectively throughout the whole vertical column.

You can think of a solid as being made of very tight springs.

A force at one side of a solid is transmitted at the speed of sound towards the other side.

Same thing for a fluid (liquid or gas), except that the transmission is by collision instead of by the electromagnetic forces that make up the "springiness" of a solid.

 

[Graeme M]

Forgive me for taking so much time with what must be very basic to many of you, and thanks for your replies so far, it is certainly helping me develop a decent mental model.

I must admit I've never heard of forces in solids being transmitted at the speed of sound! But there you go, you learn something new all the time.

So, you are saying that the molecules in the column of the atmosphere are all colliding with the molecules below them as they are pulled earthwards by gravity and this force is cascading through the column, and at each 'layer' this force is accumulating to create the pressure at the surface? Do air molecules only move downwards, or are they free to move in other directions? if other directions, would there equally be a force in another direction. For example, if I heated the air, the molecules would be excited and I would imagine them to move in all sorts of directions, including up. That force in a sideways or upward direction would be 'pressure' but it wouldn't be 'weight' would it?

 

[yoshtov]

So, you are saying that the molecules in the column of the atmosphere are all colliding with the molecules below them as they are pulled earthwards by gravity and this force is cascading through the column, and at each 'layer' this force is accumulating to create the pressure at the surface? Do air molecules only move downwards, or are they free to move in other directions?

They can move in other directions, absolutely! In general, you don't need to worry about differences in air pressure with respect to height, except when you start considering near-airtight structures that cover a significant vertical part of the atmosphere. There is almost no difference in air pressure as you move up by, say, 1 meter or 2.

if other directions, would there equally be a force in another direction. For example, if I heated the air, the molecules would be excited and I would imagine them to move in all sorts of directions, including up. That force in a sideways or upward direction would be 'pressure' but it wouldn't be 'weight' would it?

Absolutely! Consider the equation PV=nRT, which says that Pressure x Volume = number of molecules of gas (in moles) x R (a constant) x Temperature. Notice that nowhere does it make mention of mass of the gas involved. If you heated a volume of gas, the gas would exert more pressure on the walls of the container, but since the pressure would be equal across all of the walls of the container, the forces would cancel, and the mass would remain unchanged.

I think that the model of pressure being equal to the weight of the atmosphere has its benefits, but it must be properly illustrated, or it just leads to more confusion. Here's my take:

Air has mass. In the absence of any sort of gravitational field, the air has mass, but no weight. If you subject the air to a gravitational field, now the air has weight. It is attracted to the object that is producing this field. Assuming that there are a lot of molecules of gas, they will all try to get as close as possible, but will stack up on each other. Every molecule of gas wants to be as close to the source as possible, and so the molecules will "push" and "shove" (physics talk: collide) in an attempt to get closer. As you get closer and closer to the graviational source, the collisions become more aggressive. These collisions are what we call "air pressure." The cumulative effect of many, many collisions can produce a substantial force

 

[EnglishScient]

Air pressure is the result of gravity pulling down on the atmosphere and stopping the air from drifting off into space. So at sea level the air pressure is about 14.7 pounds per square inch, that's the weight of a 1 square inch wide column of air from sea level to the top of the atmosphere.

When you open your safe door the air inside the safe is at a pressure of 14.7 pounds per square inch, when you close the door you are trapping the same volume of air at the same pressure so the pressure remains the same inside.

Even if your safe was sent into space the pressure inside would remain the same as long as the safe didnt leak and the temperature stayed the same.

In the vacuum of space the air inside the safe would be pushing against the inside of the safe walls with a pressure of 14.7 pounds per square inch and since the vacuum outside is not pushing against the outside of the safe, the walls of the safe are stressed and the safe is trying to inflate like a balloon. Of course NASA has to make sure that spacecraft have strong enough skin to withstand this.

 

[russ_watters]

Ummm... I am unaware of the distinction. I mean the pressure that one experiences at the surface of the Earth and which is measured by a barometer.

There is: atmospheric pressure is the pressure of the atmospere. Air pressure is of any volume of air: it can be pressure in a tank.

 

[Paul Alexande]

Graeme M
You have hit the jackpot with this one!
No-one can explain how air pressure works!
Congratulations, in the spirit of pure thinking.

 

[sophiecentaur]

Graeme M
You have hit the jackpot with this one!
No-one can explain how air pressure works!
Congratulations, in the spirit of pure thinking.

"no one can explain"?
That's a bit sweeping, isn't it?
The pressure in a gas is due to the motion of the molecules. If the gas is in contact with the walls of a container then the pressure is due to the molecules bouncing against the container sides. Each molecule hitting the sides has it momentum, normal to the side, reversed. That involves an equal and opposite impulse against the side. The rate of the molecules hitting the side and their average velocity (related to the temperature) will tell you what the actual force is on every unit of area. Pressure acts in all directions equally; up down and the sideways directions. That is what pressure is and what 'causes' it. You don't need a surface to be there; the pressure is there, keeping the molecules in one region from encroaching (on average) into a nearby region of a gas, across a virtual surface between them.
The reason that there is pressure in the atmosphere is that all the molecules above us are pulled towards the Earth by gravity. The downward force on a 1square metre just above us is the same as the upward force and is due to sum of the downward forces (the weights) of all the molecules in a 1msquare column , all the way out into space. The fact that molecules are zapping about in all directions doesn't alter the above because it just averages out. The actual temperature affects the speed of the molecules, hence the pressure and, hence, how far the atmosphere extends out into space. If the whole Earth were much colder then the thickness of the atmosphere (assuming the same number of molecules) would be much less but the pressure would be much the same. ~(Very simplified model because, for a start, all the water would have condensed etc. etc. and gravity gets a bit less as you go up).
That was explained to me about 50years ago and it still is a pretty good explanation of what is happening.

 

[Paul Alexande]

Sophie (or "Centaur")
Thank you for providing a reasonable explanation.

 

[Graeme M]

Thanks again, this is good and I am making progress. But still having trouble with some conceptualisation.

I'll come back to Sophie's explanation in a moment. EnglishScientist states that my theoretical safe 'locks' in air pressure, even if my safe were in space (ie a vacuum) air pressure remains at what it was. However that ruins the earlier explanation which invoked the weight of the column acting on the safe and using Newton's third law to accumulate weight inside the safe. EnglishScientists' explanation therefore must explain air pressure as a force caused purely by molecular collisions with the barometer in the safe, surely?

Sophie's explanation suffers the same fate in my view. In fact, he/she invokes a virtual surface but also says that a surface is not needed in the atmosphere. However that leads me to contemplate the nature of weight. Weight it seems to me cannot be expressed without a surface when we are talking about objects, which I assume we are. The atmosphere is not an object is it? Therefore, it is the molecules we must consider, and these are the 'objects' being acted upon. Without a surface to bear upon, how can we have weight? Weight seems to me to be a function of the extent to which an object accelerated by gravity is prevented from reaching the velocity gravity imposes on it. But our molecules are not doing that, on the whole.

 

[Graeme M]

I also have trouble with the notion that 'collisions' impart weight. Any mass that is accelerated which collides with another mass will impart a greater force than if that first object merely rested on the second object wouldn't it? My thinking suggests that a collision is a different event to the 'contact' proposed by Newton in expressing weight. Gas molecule collisions will impart a force, but that force wouldn't be 'weight' would it?

 

[tiny-tim]

… However that ruins the earlier explanation which invoked the weight of the column acting on the safe and using Newton's third law to accumulate weight inside the safe.

No it didn't, the earlier explanation was that we have to add (1) the weight of the column of air inside the safe and (2) the force from the roof of the safe.

Weight it seems to me cannot be expressed without a surface when we are talking about objects, which I assume we are. The atmosphere is not an object is it?

I also have trouble with the notion that 'collisions' impart weight. Any mass that is accelerated which collides with another mass will impart a greater force than if that first object merely rested on the second object wouldn't it? … Gas molecule collisions will impart a force, but that force wouldn't be 'weight' would it?

You can regard the air as single body, in which case it is stationary (so no change of momentum) and it has weight, which acts as a force on the ground beneath it.

Or you can regard it as lots of separate molecules colliding with each other and with the ground. These molecules are not stationary, and it is the change of their momentum (only of the molecules next to the ground, of course) that provides the force on the ground …

this is equal to the weight calculated by the first method.

 

[Graeme M]

My apologies tiny-tim, I must have misunderstood. I thought you suggested that it was the weight of the air column on the roof of the safe that caused an opposite force to be expressed inside the safe, which then acted upon the air in the safe. In which case, if my safe were in a vacuum, there could be no such force. However, you are saying that the roof of my safe exerts a pressure on the air within? Wouldn't that mean it's just lucky that my safe knows exactly how much force to impart to give 14.7 lbs/sq inch?

 

[tiny-tim]

… Wouldn't that mean it's just lucky that my safe knows exactly how much force to impart to give 14.7 lbs/sq inch?

No, it's 14.7 because that's what it was before the safe door was sealed shut.

(assuming the temperature stays the same)

 

[sophiecentaur]

Thanks again, this is good and I am making progress. But still having trouble with some conceptualisation.

I'll come back to Sophie's explanation in a moment. EnglishScientist states that my theoretical safe 'locks' in air pressure, even if my safe were in space (ie a vacuum) air pressure remains at what it was. However that ruins the earlier explanation which invoked the weight of the column acting on the safe and using Newton's third law to accumulate weight inside the safe. EnglishScientists' explanation therefore must explain air pressure as a force caused purely by molecular collisions with the barometer in the safe, surely?

Sophie's explanation suffers the same fate in my view. In fact, she invokes a virtual surface but also says that a surface is not needed in the atmosphere. However that leads me to contemplate the nature of weight. Weight it seems to me cannot be expressed without a surface when we are talking about objects, which I assume we are. The atmosphere is not an object is it? Therefore, it is the molecules we must consider, and these are the 'objects' being acted upon. Without a surface to bear upon, how can we have weight? Weight seems to me to be a function of the extent to which an object accelerated by gravity is prevented from reaching the velocity gravity imposes on it. But our molecules are not doing that, on the whole.

OI! If I were a girlie, would I have an avatar like that? Do you ever read profiles?

But, on to pressure and in more detail. Why should there not be two ways of creating pressure? Inside a container, the pressure is caused by the force on the walls. Open the container in deep space and the molecules will just carry on in motion outwards and dissipate, bring the pressure inside the container to zero quickly. In the atmosphere, the pressure is there because of the velocities of the gas molecules hitting a 'notional' wall at any level.

Why should a molecule not have 'weight'? Weight is defined by mg (no specification about how small m can be nor how it has to be measured). This is, in fact, the force that is accelerating it downwards, whether it is 'resting on something' or falling.

In the end it comes down to this. If there were not a force keeping the molecules from falling to the ground (i.e. counteracting their weight) then they would all FALL to the ground). The fact that they are all up there must imply that there is an upwards force, keeping them there. That force must be equal to their weight. Or can you suggest another value for it?
If you have a problem with reconciling the idea of weight with the bouncing of molecules, consider this model. A man stands on some slow acting bathroom scales, holding a bat and a ball. The scales register his weight plus the weight of the ball and the bat. He then starts bouncing the ball in the air (keepy-uppy). Although, every time he hits the ball upwards, there is a large impulse, transferred onto the scale pan, most of the time the ball is not actually pressing down. The average force on the (slow acting) scales (the weight measurement) will be exactly the same as when he was just holding the ball. It's the same with the molecules, the ground has to support (indirectly) the weight of every molecule above it in the same way as the scales have to support the weight of the bouncing ball.

 

[Graeme M]

Sorry sophiecentaur, I am in the middle of a normal working day and only assigning a meager portion of what I imagine is my mind to this forum. It is all very useful but I didn't notice your avatar till after I posted. I *did* go back and change my post slightly...

Thanks for your latest post, very informative. I'll think about it and respond, though we may have got me there! :)

 

[sophiecentaur]

Sorry sophiecentaur, I am in the middle of a normal working day and only assigning a meager portion of what I imagine is my mind to this forum. It is all very useful but I didn't notice your avatar till after I posted. I *did* go back and change my post slightly...

Thanks for your latest post, very informative. I'll think about it and respond, though we may have got me there! :)

I wasn't really grumpy.

I may have put the problem in a nutshell. I must say, I never really though it through before.
It's my bedtime now!

 

[tiny-tim]

Don't feel bad, Graeme …

I used to think he was half-girl, half-horse …

turns out he's a boat!​

as a fish, i should have recognised that! ​

 

[Drakkith]

Hrmmm. It looks to me like the atmospheric pressure is due to the density and temperature of the air, not the "weight". Although the density of the air would be directly related to it's weight, as that is the reason that there is more air near the surface compared to 100 miles up.

Consider this: If I lock myself in a sealed container, the air pressure inside doesn't suddenly go away even if my container can withstand the full weight and pressure of the atmosphere. It is simply due to the density of the air inside and the temperature.

Now consider an air compressor. The compressor simply forces air into a container. The density of the air increases as the pressure does since we are shoving more air into a sealed space. A temperature change can also increase or decreases pressure, hence why you don't want a sealed container of pressurized gas thrown into a fire.

How does that look?

 

[Graeme M]

Thanks guys, I am pretty much happy with the explanation given. I will summarise, let me know if I am largely on track.

The atmosphere is composed of molecules. The molecules are in motion due to a variety of forces, eg wind, heat etc. However, by and large each molecule has mass and is being attracted to the Earth's surface. When not being pulled about by other forces, molecules fall downwards. On average, this leads to a greater density of molecules the closer to the Earth's surface we are. Also on average, at any level in the atmosphere the collisions of those molecules bears on any surface in the form of a force referred to as 'atmospheric pressure'. This varies with the density of the atmosphere (ie number of molecules). If all of those molecules immediately lost energy and fell to the ground where they could register as 'weight', the weight would be equivalent to the earlier pressure. Hence for the atmosphere, pressure broadly equates (or is exactly equivalent to?) to weight. I assume this is the same for any gas.

Close?

 

[klimatos]

So at sea level the air pressure is about 14.7 pounds per square inch, that's the weight of a 1 square inch wide column of air from sea level to the top of the atmosphere.

Since the atmosphere increases in circumference as you move away from the Earth's surface, the use of a uniform column is dimensionally incorrect. For a circular surface area, you should use a conic section. For a rectangular surface area, you should use a pyramidal section. The barometric formula is dimensionally incorrect in its use of a column.

The barometric formula relating pressure to the weight of the overlying air is invalid in many respects. It assumes a constant temperature, which is not the case. It assumes a constant gravitational constant: this assumption is invalid. It assumes a constant molecular mass--this is incorrect.

The weight-force hypothesis was developed when gas molecules were thought to be able to expand indefinitely to "fill" any space into which the gas was released. This expansion allowed molecule-to-molecule "contact" and hence transfer of weight force. The development of kinetic gas theory forced the abandonment of that concept, but the language stays on.

From a kinetic point of view, atmospheric pressure is no different from air pressure. It is the simple product of the number of molecular impacts per square meter and the mean impulse (in Newtons) transferred per impact.

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.

The near-air-tight case of a barometer serves to dampen its reactions. 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"?

 

[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.