What is normal air pressure in N/cm^2?

[rogerk8]

Hi!

I have a very hard time grasping the pressure concept.

Look at normal air pressure (1atm)

$$1atm≈10^5Pa=10^5N/m^2≈10^4kg/m^2=1kg/cm^2$$

In other words, one whole kg per square centimeter!

How can this be?

I don't feel the air pressing against my body with one whole kg at each square cm.

Roger

 

[russ_watters]

Pressure isn't felt, only pressure difference. So since your body is in equilibrium with the atmosphere, there is nothing to feel.

On the other hand, if you surface from a scuba dive without exhaling or equilizing your inner ears, you will definitely feel it. You can also feel it less dramatically when traveling in a plane or driving up/down a mountain.

 

[rogerk8]

Thank you for your reply!

Could you please explain that again?

What do you mean by equlibrium with the atmosphere?

I simply do not understand this.

Everything around us including my body should be affected by 1kg/cm² is my thought.

The other part of your explanation sounds understandable though.

Roger

 

[SteamKing]

The human body has evolved in this atmospheric pressure from the get go. In fact, your lungs use pressure difference to help you breathe. When you get a collapsed lung, what that means is that the pressure difference has been lost and you are not able to use the collapsed lung to help you breathe until you get treatment.

Now, if you got into a situation where one side of your body experienced a pressure of 5 kg/cm^2 while the other side was still at 1 kg/cm^2, then you would feet that, because the pressure differential might knock you off your feet. This is one reason why explosions, hurricanes, and tornadoes are so destructive: the increase in air pressure due to the shock wave or the wind is strong enough to damage or destroy many structures.

 

[rogerk8]

This is interesting. Please excuse my stupid questions (and statements).

Inhaling would mean that the pressure in my lungs would have to be lower for the air to flow into my lungs, exhaling would mean that the air have to be of higher pressure to get out.

I get this.

What I still don't get is the "enormous" 1kg/cm^2 that should affect everything around us.

What would happen if the pressure around us was 100kg/cm^2?

With your reasoning this would not matter and I don't buy that.

The existence of breathable air have to have some pressure otherwhise it is called vacuum.

But 1kg/cm^2...

No, this pressure business is extremely hard for me to understand.

Roger

 

[russ_watters]

Could you please explain that again?

What do you mean by equlibrium with the atmosphere?

I simply do not understand this.

Everything around us including my body should be affected by 1kg/cm² is my thought.

"Equilibrium" means the pressure inside is exactly equal to the pressure outside.

What would happen if the pressure around us was 100kg/cm^2?

With your reasoning this would not matter and I don't buy that.

Do you buy that scuba divers don't all die?

 

[rogerk8]

"Equilibrium" means the pressure inside is exactly equal to the pressure outside.

Thank you for this explanation.

Do you buy that scuba divers don't all die?

Point taken

Roger

 

[AlephZero]

What I still don't get is the "enormous" 1kg/cm^2 that should affect everything around us.

Part of the reason is that 1kg/cm^2 is not an "enormous" pressure. For example the crushing strength of bone is about 1800 kg/cm^2.

Even your blood pressure is about 0.3 kg/cm^2 higher than atmospheric pressure, but that doesn't make your blood vessels spontaneously explode (except in horror movies, of course).

 

[SteamKing]

What I still don't get is the "enormous" 1kg/cm^2 that should affect everything around us.

What would happen if the pressure around us was 100kg/cm^2?

With your reasoning this would not matter and I don't buy that.

When man first began to make extended deep sea dives (William Beebe in the 1920s and 1930s), new species of fish were discovered which had never been seen before. These fish lived their entire lives in the deep ocean where there is perpetual darkness and, of course, high pressure due to the depth of water. While these fish were perfectly happy to live at depths which would crush fish living in shallower waters, when they were brought out of their highly pressurized environments, they quickly died. The point is that life adapts to extreme environments where it is able, even if we land dwellers find it hard to comprehend.

Since Beebe's day, we have discovered sea worms which live only near volcanic vents deep in the ocean, which emit clouds of heavy metal with temperatures high enough to boil the surrounding water. Bacteria have also been found which can thrive quite happily in highly acidic environments which would cause extreme injury, if not death, to other living things.

http://en.wikipedia.org/wiki/William_Beebe

BTW, in the ocean, each 10 m of depth adds about 1 atmosphere of pressure.

 

[rogerk8]

I am far from understanding this.

With your reasoning ordinary air pressure might as well be 100kg/cm^2.

The only difference as I understand it is that the scuba divers then would have to dive 1km for a doubling of pressure.

Considering
$$p \propto \frac{N}{V}$$
while T is constant and N is the number of molecules (regardless of k or R).

Our lungs suck in air by increasing V (if N is constant) and thus lowering the pressure, we then exhale by lowering V thus increasing pressure (once again if N is constant).

But to me N/V is approximatelly constant because inhaling would mean that the number of molecules also increases and exhaling would mean that the number of molecules in the same way decreases.

On the other hand, this flow of molecules must take a certain time due to the width of our throat. So maybe the first reasoning is correct in as much as that it takes time for the molecules to actually generate equilibrium.

Is it perhaps so that the sucking comes first, then the N?

By the way, how does trees do when they suck up water?

There has to be some circulation of water (and nutritions) to keep the low sucking pressure. Wintertime there are no leafs so the area is lower which might mean less water evaporated. But there should still be some circulation, right?

Roger
PS
How do I measure normal air pressure? And I don't mean a barometer (which only measures the difference anyway). I will try to study this on Wikipedia after I have posted this

 

[klimatos]

One way to think of air pressure is the simple product of the number of molecular impacts per unit area per second and the mean molecular impulse per molecule. The dimensions of pressure are force per unit area. The current practice is to use force (impulse) in Newtons, area in square meters and pressure in pascals.

The pressure of your bodily fluids on the inside of your skin is balanced by the pressure of the atmosphere on the outside of your skin at normal temperature and pressure (NTP). This pressure of 101,325 pascals may seem enormous to you, but it is entirely normal (hence NTP).

By the way, what is so wrong with using pressure differences to measure ambient pressure. As long as the result is accurate, why should the technique used to get those results matter?

I suspect that you are new to science, in general, and to scientific ways of looking at things. In time, you will no longer be surprised at either very large numbers (3.01 x 10²⁷ molecular impacts per square meter per second) or very small ones (3.37 x 10^-23 Newtons per impact).

 

[SteamKing]

I am far from understanding this.

With your reasoning ordinary air pressure might as well be 100kg/cm^2.

The only difference as I understand it is that the scuba divers then would have to dive 1km for a doubling of pressure.

Considering
$$p \propto \frac{N}{V}$$
while T is constant and N is the number of molecules (regardless of k or R).

Our lungs suck in air by increasing V (if N is constant) and thus lowering the pressure, we then exhale by lowering V thus increasing pressure (once again if N is constant).

But to me N/V is approximatelly constant because inhaling would mean that the number of molecules also increases and exhaling would mean that the number of molecules in the same way decreases.

On the other hand, this flow of molecules must take a certain time due to the width of our throat. So maybe the first reasoning is correct in as much as that it takes time for the molecules to actually generate equilibrium.

Is it perhaps so that the sucking comes first, then the N?

By the way, how does trees do when they suck up water?

There has to be some circulation of water (and nutritions) to keep the low sucking pressure. Wintertime there are no leafs so the area is lower which might mean less water evaporated. But there should still be some circulation, right?

Roger
PS
How do I measure normal air pressure? And I don't mean a barometer (which only measures the difference anyway). I will try to study this on Wikipedia after I have posted this

At a depth of 1 km in the ocean, the hydrostatic pressure is approx. 100 atmospheres. I don't know what you mean by scuba divers having to dive 1 km to double pressure. No one dives to such depths, even scuba divers: the pressure is simply too great.

Obviously, you are having a hard time grasping that terrestrial organisms have evolved differently than creatures which dwell in the sea. These deep sea dwelling creatures have adapted themselves to the pressures of their environments, just like land animals have adapted to theirs. Neither organism is suited to live in the other's environment.

As far as respiration in trees is concerned, this is another topic where you are dealing with scanty and incomplete information. Trees which shed their leaves in winter lie dormant until spring. Their sap is concentrated in the lower trunk, and since they have shed their leaves, it is not possible for photosynthesis to take place. The tree is dormant during the winter, just like bears hibernate at the same time to reduce their energy consumption when food is scarce.

No barometers don't measure pressure differences: you are thinking of a manometer; there's a difference. A barometer is made so that the upper part is a vacuum: the height of the column of mercury indicates the actual atmospheric pressure.

 

[rogerk8]

At a depth of 1 km in the ocean, the hydrostatic pressure is approx. 100 atmospheres. I don't know what you mean by scuba divers having to dive 1 km to double pressure. No one dives to such depths, even scuba divers: the pressure is simply too great.

I simply used my hypothetic statement that if absolute air pressure was a 100 times higher than normal (i.e 100kg/m^2) the scuba divers would have to dive down to 1km to double that pressure. Which is a consequence of your nice teaching that pressure doubles for every 10m if absolute pressure is 1kg/cm^2 (1atm). On the other hand, an "overpressure" of as little as 1kg/cm^2 sounds high enough so at a depth of 10m each square cm of my body would feel an additional 1kg/cm^2. But this is regardless of absolute air pressure outside of the water, right?

I think I get this. Yet I find a whole kg per cm^2 simply too much to comprehend. It is only when we view it relatively like above that it makes sence.

Obviously, you are having a hard time grasping that terrestrial organisms have evolved differently than creatures which dwell in the sea. These deep sea dwelling creatures have adapted themselves to the pressures of their environments, just like land animals have adapted to theirs. Neither organism is suited to live in the other's environment.

It is amazing how they can survive down there at those pressures. The deepest part of our seas is some 10km, right?

$$p=\rho g h$$

then tells us that pressure there is some 1000kg/cm^2!

As far as respiration in trees is concerned, this is another topic where you are dealing with scanty and incomplete information. Trees which shed their leaves in winter lie dormant until spring. Their sap is concentrated in the lower trunk, and since they have shed their leaves, it is not possible for photosynthesis to take place. The tree is dormant during the winter, just like bears hibernate at the same time to reduce their energy consumption when food is scarce.

Thanks for this basic explanation! I had to look "sap" and "trunk" up in my dictionary by the way

No barometers don't measure pressure differences: you are thinking of a manometer; there's a difference. A barometer is made so that the upper part is a vacuum: the height of the column of mercury indicates the actual atmospheric pressure.

I saw a nice picture and a formula for a U-shaped tube the other day on what I think was Wikipedia but now I can't find it.

It did however tell something about that the shift of height had to do with the pressure difference.

Maybe it was like this:

$$h=\frac{P_1-P_0}{\rho g}$$

So if we pour water into the U-tube we have the same pressure at either hole and h=0. But if we put a cork on one side we just have to wait for the absolute air pressure to change (and compress/expand the corked side).

Which is another interesting question. Why does absolute air pressure change in the first place?

I spoke with a friend of mine this morning and he told me his logical thought that bad weather should mean a higher pressure and good weather should mean a lower pressure. This simply due to the fact that there are clouds above us full of water before raining so pressure should be higher. I told him "No, bad weather simply means lower pressure" because that's what I remember my grandpa told me about his precious barometer readings. But I probably remember wrongly.

Finally, I read most of the Wikipedia article about pressure (there's much to learn, to say the least).

Except for the boring part about different units, it was a very interesting article that made me realize some important stuff like the pressure in water is due to the depth and not the volume. This for instance mean that a water power plant dam with say 20m high walls does not put more pressure on the wall as if I put 20m of water into a 1dm wide tube, right?

The interesting part about this is however that it is quite easy to relate to liquids like above but not so easy for gases.

Roger

 

[SteamKing]

It is amazing how they can survive down there at those pressures. The deepest part of our seas is some 10km, right?

The deepest part of the Mariana Trench is almost 11 km deep:

http://en.wikipedia.org/wiki/Challenger_Deep

The pressure there is approx. 1250 kg/cm^2

I saw a nice picture and a formula for a U-shaped tube the other day on what I think was Wikipedia but now I can't find it.

It was probably a picture of a manometer you were looking at. There is one in this article:

http://en.wikipedia.org/wiki/Pressure_measurement

Which is another interesting question. Why does absolute air pressure change in the first place?

I spoke with a friend of mine this morning and he told me his logical thought that bad weather should mean a higher pressure and good weather should mean a lower pressure. This simply due to the fact that there are clouds above us full of water before raining so pressure should be higher. I told him "No, bad weather simply means lower pressure" because that's what I remember my grandpa told me about his precious barometer readings. But I probably remember wrongly.

The air in the atmosphere is heated by the sun during the day, and it cools at night. The air at the poles does not receive as much heat from the sun as the air at lower latitudes. The change in temperature of the air leads to changes in density, which in turn affects the local barometric pressure.

Storms like tornadoes and hurricanes and typhoons are low pressure phenomena. The barometric pressure in the center of such cyclonic storms is much lower than the surrounding atmosphere.

Typically, bad weather is associated with regions of low pressure, while good weather is associated with regions of high pressure. The pressure differentials are so slight that normally, people are not aware of them, but they can be detected with the aid of instruments like a barometer.

In the days before weather forecasts and weather satellites, people living in tropical climates typically kept a barometer handy to tell them if a storm like a hurricane was approaching. By watching the barometer readings and how fast they dropped, it was a good indication that a storm was on the way.

This for instance mean that a water power plant dam with say 20m high walls does not put more pressure on the wall as if I put 20m of water into a 1dm wide tube, right?

Yes, hydrostatic pressure depends only on the depth of fluid and the density of the fluid.

 

[Khashishi]

You don't believe in the power of air pressure? Watch some things implode when you suck out the air.

How do you explain that?

You ever seen a mercury barometer? You have mercury in a tube, with the air pressure pushing against the force of gravity. That's because the mercury has a vacuum on one side and the air on the other side, so it has the full force of the air pressure with no balancing air pressure on the other side. Now, mercury is a very dense metal, yet the air will push a whole 0.76m of it up a column.

 

[rogerk8]

You don't believe in the power of air pressure? Watch some things implode when you suck out the air.

How do you explain that?

You ever seen a mercury barometer? You have mercury in a tube, with the air pressure pushing against the force of gravity. That's because the mercury has a vacuum on one side and the air on the other side, so it has the full force of the air pressure with no balancing air pressure on the other side. Now, mercury is a very dense metal, yet the air will push a whole 0.76m of it up a column.

I have never questioned the actual power of air pressure, I am just wondering how it can be so large as 1kg/cm^2.

Your amazing demonstration even confirms my doubts. Because where does it come from?

Using the Barometic Formula

$$p=p_0-\rho gh...[1]$$

and counting backwards to where p=0 it would mean that the hight of the atmosphere is

$$h=\frac{p_0}{\rho g}≈\frac{10^5}{1*10}=10^4m=10km$$

I have heard something about airoplanes almost reaching those altitudes so it sounds right to me.

So, while $\rho gh$ seams right when it comes to gauge pressures in water, this equation seam to fit for gases too.

But what about

$$p=n_mRT=nkT...[2]$$

Does this equation perhaps tell the pressure in closed systems only?

While I'm at it, Physics Handbook tells me about another completelly different "Barometic Formula"

$$p=p_0e^{-\frac{mgh}{kT}}=p_0e^{-\frac{E_p}{E_k}}...[3]$$

where m is the molecular mass and Ek and Ep is my way of simplifying it.

But what does this mean? I am totally lost here. Two equations for the Barometer...

And to make things complete (Bernoulli's Theorem)

$$p+1/2\rho v^2+\rho gh=p+E_k'+E_p'=constant...[4]$$

which may be viewed as if Ek' of the fluid increases, Ep' will have to decrease. But viewing this like in a waterfall Ep' is obviously constant so if Ek' increases somehow, p will have to decrease. Whatever p is in this case.

No, I don't get this either

Finally, it is fascinating that pressure is a scalar and thus omnidirectional in my preliminary world of understanding.

Roger
PS
Reading the Wikipedia article about Barometer makes me think that you first fill that >85cm tube with Mercury and then turn it around in the reservoir yielding a vacuum in the top of the tube. The hight of the Mercury column then equals normal air pressure according to [1] above?

 

[klimatos]

I spoke with a friend of mine this morning and he told me his logical thought that bad weather should mean a higher pressure and good weather should mean a lower pressure. This simply due to the fact that there are clouds above us full of water before raining so pressure should be higher. I told him "No, bad weather simply means lower pressure" because that's what I remember my grandpa told me about his precious barometer readings. But I probably remember wrongly.

Your friend's belief that atmospheric pressures somehow measure the weight of everything above the barometer is shared by many people. It is incorrect, of course.

Rogers and Yau, "A Short Course in Cloud Physics" on page 235 give the mean precipitation content of an isolated thunderhead as on the order of 10⁹ kilograms. Yet, when a thunderhead passes overhead, the pressure usually drops. This is because such clouds are fed by enormous updrafts, and updrafts on the surface drop the ambient surface pressure.

A barometer simply measure the impulses transferred to its sensing surface by the impacts of the air molecules on its surface. Fewer impacts mean lower pressures.

 

[SteamKing]

The thing to remember about the atmosphere is that with respect to altitude, the amount of air varies the higher you go.

http://en.wikipedia.org/wiki/Atmosphere_of_Earth

The troposphere, the lowest layer of the atmosphere, contains approx. 80% of the mass of the atmosphere, and this layer is about 12 km deep. About 50% of the atmosphere is contained below an altitude of 5.6 km. The pressure at altitude is given by Eq. 3 in Post #16, and you can see that this variation in pressure with altitude is not linear. The temperature also varies with altitude, but its variation is much more complex, as you can see from the graphs in the attached article.

 

[Sam Ku.D]

How can this be?

I don't feel the air pressing against my body with one whole kg at each square cm.

its because you have almost equal pressure acting inside of your body! (from your calculation its 1kg per cm )

 

[rogerk8]

The thing to remember about the atmosphere is that with respect to altitude, the amount of air varies the higher you go.

http://en.wikipedia.org/wiki/Atmosphere_of_Earth

The troposphere, the lowest layer of the atmosphere, contains approx. 80% of the mass of the atmosphere, and this layer is about 12 km deep. About 50% of the atmosphere is contained below an altitude of 5.6 km. The pressure at altitude is given by Eq. 3 in Post #16, and you can see that this variation in pressure with altitude is not linear. The temperature also varies with altitude, but its variation is much more complex, as you can see from the graphs in the attached article.

Thank you for this information!

Just to show you that I have understood something:

4) Thermosphere (80km-Karman Line@100km)
3) Mesosphere (50km-80km)
2) Stratosphere (10km-50km)
1) Toposhere (<10km)

Where the Karman Line represents where it still is possible to fly flat but at the orbital velocity where the centrifugal force equals the gravitational force, right?

While I was aiming for vacuum at 10km it is a surprising fact that you actually can fly at 100km!

Viewing https://en.wikipedia.org/wiki/Atmosphere_of_Earth#Pressure_and_thickness it however tells me that neither pressure nor density is linear above some 5km. It is here where I sense that equation no.3 above, as you already have explained, holds.

So up to some 5km it is possible to use equation no.1 to determine total air pressure, p. It is thus almost not possible to determine the air pressure at our highest mountain, Mount Everest.

By the way, the different "spheres" seems to be determined by sharp temperature gradients, right?

While I was surfing around in the above article I also got to read about the Karman Line. It was interesting to note that for a moving object in a fluid (such as air) it seems like

$$p=p_0+1/2\rho v^2=p_0+p_k...[5]$$

The gauge-part of the pressure is what adds to our normal pressure when a vessel is moving.

The extra force felt by the movement must then be

$$F=p_k*A...[6]$$

Where the area A is perpendicular to the movement.

I played around with this thought by moving my hand very fast in the air. And yes, I felt some force against it even though I did not feel the square-dependency part

Roger

 

[bp_psy]

A really easy way to see what a pressure difference means is to put your hand underwater. About 30-50cm deep is enough to sense the difference and it is only ~0.05atm.

 

[rogerk8]

In my world however, normal air pressure should be around 1grams/cm^2. That would sound reasonable to me just by poking at something. Not 1kg/cm^2...

Everything points in the direction that I am wrong. The implosion above most of all. As well as simple use of the Barometic Formula [1].

Yet, I find the normal air pressure too high to comprehend.

And today I learned that the atmosphere was 10 times higher in altitude than I thought/calculated

I really do not understand much.

Roger

 

[russ_watters]

In my world however, normal air pressure should be around 1grams/cm^2. That would sound reasonable to me just by poking at something. Not 1kg/cm^2...

Everything points in the direction that I am wrong. The implosion above most of all. As well as simple use of the Barometic Formula [1].

Yet, I find the normal air pressure too high to comprehend.

Haven't you ever been swimming? You can swim down a few meters and feel a significant pressure difference - it can even be painful in your ears. But once you equalize the pressure (by holding your nose and blowing), the pain goes away. It is all about pressure difference, not pressure itself.

It's the same as wondering why if the pressure inside your house is so high is it possible to open your front door.

 

[SteamKing]

There are various altitude records set for different types of flying craft (balloon, jet aircraft, rocket, etc.)
Most of these records are still well within the stratosphere and well below the Karman line:

http://en.wikipedia.org/wiki/Flight_altitude_record

 

[rogerk8]

I won't disturb you with anymore stupid questions about the normal air pressure.

I got it now, thank you all!

One last thought though

If normal air pressure was 100 times higher, my lungs would have to push my chest with 100 times higher force. And this is the "only" difference. I hope this is correct?

The fishes down at 5km(?) experience a [gauge] pressure of some

$$1000*5000kg/m^2=500kg/cm^2$$

Which is 500 times normal air pressure. They must be very muscular

Take care!

Roger

 

[russ_watters]

If normal air pressure was 100 times higher, my lungs would have to push my chest with 100 times higher force. And this is the "only" difference. I hope this is correct?

No. This is getting frustrating because you are not listening: pressure DIFFERENCE.

 

[rogerk8]

I'm sorry but it is frustrating for me too because I simply do not understand.

Please explain to me like I was a child why the enourmously higher pressure at 5km down the sea doesn't even matter for the fishes and it what way it does.

Because if it's only about pressure difference why even point out the high pressure that surrounds the deep water fishes?

I can for sure understand that if the pressure on either side of for instance a curtain is the same, the curtain will remain still (untill I poke it). So yes, the pressure difference here is zero.

But what about above?

Roger

 

[dauto]

I'm sorry but it is frustrating for me too because I simply do not understand.

Please explain to me like I was a child why the enourmously higher pressure at 5km down the sea doesn't even matter for the fishes and it what way it does.

Because if it's only about pressure difference why even point out the high pressure that surrounds the deep water fishes?

I can for sure understand that if the pressure on either side of for instance a curtain is the same, the curtain will remain still (untill I poke it). So yes, the pressure difference here is zero.

But what about above?

Roger

The fish, like the curtain, remains still. the pressure doesn't crush the fish because the inside of the fish is also at high pressure. Pressure, nevertheless, is very important because of subtle physiological effects. Notably, the amount of air dissolved in your blood varies with pressure. Just a few extra kg/cm2 will dissolve enough nitrogen in your blood to get you drunk. That's why scuba divers breath helium (plus oxygen, obviously) instead of nitrogen.

 

[russ_watters]

I'm sorry but it is frustrating for me too because I simply do not understand.

Please explain to me like I was a child why the enourmously higher pressure at 5km down the sea doesn't even matter for the fishes and it what way it does.
[snip]
I can for sure understand that if the pressure on either side of for instance a curtain is the same, the curtain will remain still (untill I poke it). So yes, the pressure difference here is zero.

It seems like you understand it fine, you just keep forgetting or just don't want to believe it. I don't know what to tell you other than to keep rereading the thread and keep saying it to yourself until you stop forgetting and/or start believing it.

Or go for a swim and test it. Or try to suck air through a really tall straw.

Ultimately there isn't all that much we can do if you just choose not to accept it. We can teach you, but learning is completely up to you.

Because if it's only about pressure difference why even point out the high pressure that surrounds the deep water fishes?

You are the one who pointed it out!

 

[rogerk8]

It seems like you understand it fine, you just keep forgetting or just don't want to believe it. I don't know what to tell you other than to keep rereading the thread and keep saying it to yourself until you stop forgetting and/or start believing it.

Or go for a swim and test it. Or try to suck air through a really tall straw.

Ultimately there isn't all that much we can do if you just choose not to accept it. We can teach you, but learning is completely up to you.

You are the one who pointed it out!

I'm getting to become a believer now.

If I'm wrong at this one, you might as well close the thread

Imagine two me's. One has adapted to 1atm the other to 500 atm. We both have lungs. The one deep below water can somehow extract oxygen with his lungs too. We both need to differentially change the pressure to breath. A childish estimate for us both would be some +/-0,01atm (which equals +/-7,6mm Hg, a unit I have come to like due to Kashishi above). This change of pressure is thus the same for us both in spite of totally different ambient pressure.

So far, so good (I think).

Looking at the actual breathing it is getting harder.

$$p=n_mRT=nkT≈nE_k$$

were the last rough equality struck me not until yesterday.

Anyway, while n=N/V both N and V may change. Taking a "fast" breath I recon N is constant while the sucking is created due to increasement of volume and after a short while assimilated to 1atm (i.e zero difference) due to particle flow and thus pressure equalisation. Hope I'm right here.

I don't know why this pressure business is so hard to understand but I think it's because it's a scalar and thus omnidirectional.

Even your simple straw-example gets me puzzled. A tall straw would mean a "high" weight of air that I need to suck before I can get me any air. But what has this to do with pressure? The straw has a tiny area and if the pressure is constant regarding my inhale performance, the force on the (weight of the) air inside the straw will be low, thus making it hard to breath. Because in the same time I could take a M5 nut and breath through it's hole without problem.

Sorry, I'm so stupid!

Finally I just want to state the most interesting part of what I've learned (along with the pressure difference part):

1) Water pressure increases 1 atm per 10m (thank you SteamKing)
2) Neither pressure nor density is linear above some 5km (temperature varies too).
2) The atmosphere is as high as 100km which is a full magnitude higher than I thought and linearly calculated
3) Scuba divers suffer from drunkness after only a few atm and thus breaths Helium instead of Nitrogen (thank you dauto).

Roger

 

[russ_watters]

Even your simple straw-example gets me puzzled. A tall straw would mean a "high" weight of air that I need to suck before I can get me any air. But what has this to do with pressure? The straw has a tiny area and if the pressure is constant regarding my inhale performance, the force on the (weight of the) air inside the straw will be low, thus making it hard to breath. Because in the same time I could take a M5 nut and breath through it's hole without problem.

Sorry, I worded that wrong. What I meant to say was suck water through a really tall straw. It is a good way to demonstrate just how weak our lungs are/how little pressure difference they can deal with.

The opposite side of that coin is a tall snorkel under water. Since the air inside the snorkel is at atmospheric pressure, the pressure in your lungs will be a little lower than the pressure outside your lungs (from the water pressing in). You don't have to go very deep for it to become impossible to breathe through a snorkel.

 

[dauto]

Finally I just want to state the most interesting part of what I've learned (along with the pressure difference part):

1) Water pressure increases 1 atm per 10m (thank you SteamKing)
2) Neither pressure nor density is linear above some 5km (temperature varies too).
2) The atmosphere is as high as 100km which is a full magnitude higher than I thought and linearly calculated
3) Scuba divers suffer from drunkness after only a few atm and thus breaths Helium instead of Nitrogen (thank you dauto).

Roger

On point 2). Atmospheric pressure and density aren't linear below 5km either. An exponential decay is a much better description.

 

[rogerk8]

On point 2). Atmospheric pressure and density aren't linear below 5km either. An exponential decay is a much better description.

Sorry, but viewing this picture https://en.wikipedia.org/wiki/Atmosphere_of_Earth#Pressure_and_thickness tells me that you are wrong.

Both pressure, density and even temperature (from some 280K@sealevel to 220K@10km) seams quite linear to me. Point 2) height should even be changed to 10km.

Considering the common expression for pressure in a "closed" system above and the two me's I have come to the conlusion that the differential pressure for our lungs are around +/-0,1atm and thus a full magnitude higher than my preliminary guess.

The way I have calculated this is:

$$p\propto 1/V$$

Esimating our lungs volume to be some 5L, the rest breathing being some +/-0,5L and the normal pressure being 1 atm we roughly get +/-0,1atm.

Now, my deep water me experience 500atm and we both need the same amount of molecules. This wile the much higher pressure ensures higher molecular density (due to V and for simplicity, T being the same). This in turn means that the deep water me needs 500 times lower relative pressure for the same amount of molecules. And we wind up with the same differential breathing pressure. Hope this is right.

One last puzzling but extremely trivial example. This is however an example where pressure is not omnidirectional which I think is important to point out.

Consider a one ft plastic tube of say 1cm^2 cross sectional area, A. We thus have a pea-tube for shooting peas. Pressing our exhale into the tube and omitting the leakage around the pea, the (omnidirectional) pressure is suddenly turned into a force (of direction) by p*A.

Rediculous example, yet interesting somehow

Roger
PS
What is actually the use for

$$p=nkT≈nE_k[J/m^3]$$

when it comes to plasma physics?

What does p help us understand/enable?

 

[dauto]

Sorry, but viewing this picture https://en.wikipedia.org/wiki/Atmosphere_of_Earth#Pressure_and_thickness tells me that you are wrong.

Both pressure, density and even temperature (from some 280K@sealevel to 220K@10km) seams quite linear to me. Point 2) height should even be changed to 10km.

Sorry, but actually solving the equations should tell you that I'm right. Do not rely on simple visual inspection of a sketch.

 

[rogerk8]

Does not the skin of the fish deep down in water suffer from the high pressure?

Let's say that the fish lives down at 5km with 500atm of pressure.

This pressure is both outside of its body and inside of its body so it summarizes to zero.

But how about its skin?

It has got to be affected by some force like p*A?

Or?

Roger