Atmospheric pressure- where do we get the mass?

[opus]

At sea level, we experience an external force pressing down on us at any given time which is equal to about 15 pounds per square inch.
Pressure is defined as: $P = \frac{Force}{Area}$ where Force is equal to mass x acceleration.
When we say that we experience this 15 pounds per square inch of atmospheric pressure, we are using this pressure formula where Area = 1 square inch, and acceleration is equal to the gravitational constant. But where are we getting the values for mass? Is it the mass of all of the gas molecules added up that are on top of us? How would we determine that?

 

[Chestermiller]

At sea level, we experience an external force pressing down on us at any given time which is equal to about 15 pounds per square inch.
Pressure is defined as: $P = \frac{Force}{Area}$ where Force is equal to mass x acceleration.
When we say that we experience this 15 pounds per square inch of atmospheric pressure, we are using this pressure formula where Area = 1 square inch, and acceleration is equal to the gravitational constant. But where are we getting the values for mass? Is it the mass of all of the gas molecules added up that are on top of us? How would we determine that?

There is no acceleration because the system is in equilibrium. You can approximate the total mass of the air above each square inch (neglecting the variation of gravity with altitude) by applying the equilibrium force balance: $$F=m\frac{g}{g_c}=15\ lb_f$$ So $m = 15\ lb_m$

 

[DaveC426913]

But where are we getting the values for mass? Is it the mass of all of the gas molecules added up that are on top of us?

Yes. A one square inch column of air extending from the ground to space literally weighs 14.7 pounds.

How would we determine that?

A tube of mercury will tell us - i.e. a barometer.
The weight of air pushing down on a plate of mercury is sufficient to lift a column of mercury up an evacuated glass tube almost 30 inches.
That tells us the pressure is about 14.7 pounds per square inch.

 

[opus]

Thank you guys. And what does the height of 30 inches in the barometer correspond to? To my understanding, you start out with a vacuum sealed tube, open the bottom while it's submerged in mercury, and the air pressure will press down on the base of the barometer which will then push the mercury inside the tube upwards until the downward pressure in the tube equals the downward pressure of the air outside of the tube so they are balanced out. But when we measure 30 inches, what does this mean?

 

[opus]

There is no acceleration because the system is in equilibrium.

Could you explain what this means? I know that equilibrium means kind of a "balanced out" sort of manner, but what is exactly in equilibrium?

 

[Chestermiller]

Could you explain what this means? I know that equilibrium means kind of a "balanced out" sort of manner, but what is exactly in equilibrium?

The upward normal force exerted by the surface of the Earth on the column of air above is equal to the weight of the column of air (such that the column of air is in equilibrium). By Newton's 3rd law of action-reaction, the upward normal force exerted by the surface of the Earth on the column of air above is equal in magnitude and opposite in direction to the normal force exerted by the column of air on the surface of the earth.

 

[opus]

The upward normal force exerted by the surface of the Earth on the column of air above is equal to the weight of the column of air (such that the column of air is in equilibrium). By Newton's 3rd law of action-reaction, the upward normal force exerted by the surface of the Earth on the column of air above is equal in magnitude and opposite in direction to the normal force exerted by the column of air on the surface of the earth.

Ok great I'm on board with that. One final question- why is there an upward force exerted by the surface of the earth? Seems like something opposite of gravity.

 

[CWatters]

Thank you guys. And what does the height of 30 inches in the barometer correspond to? To my understanding, you start out with a vacuum sealed tube, open the bottom while it's submerged in mercury, and the air pressure will press down on the base of the barometer which will then push the mercury inside the tube upwards until the downward pressure in the tube equals the downward pressure of the air outside of the tube so they are balanced out. But when we measure 30 inches, what does this mean?

Its literally the height of the mercury column. At sea level 30 inches is the typical height of the mercury that air pressure can support. Eg If you try to make the tube of mercury taller (say 35 inches) then an empty space will appear between the top of the mercury and the sealed end of the tube because standard atmospheric pressure at sea level is unable to push the mercury up the tube any higher.

Edit: Water is less dense than mercury so atmospheric pressure will support a much higher column of water, around 33ft tall. But a barometer that tall would be a bit inconvenient so mercury is used.

 

[Chestermiller]

Ok great I'm on board with that. One final question- why is there an upward force exerted by the surface of the earth? Seems like something opposite of gravity.

The weight of the air is pushing down on the surface of the earth, and the surface of the Earth must post back with the same force. This is Newton's law of action-reaction. This force from the surface of the Earth is indeed acting in the direction opposite to gravity.

 

[opus]

The weight of the air is pushing down on the surface of the earth, and the surface of the Earth must post back with the same force. This is Newton's law of action-reaction. This force from the surface of the Earth is indeed acting in the direction opposite to gravity.

Ahhh ok. That makes total sense. I haven't gone over any of that stuff yet but I can definitely see that. Thanks!

 

[opus]

Its literally the height of the mercury column. At sea level 30 inches is the typical height of the mercury that air pressure can support. Eg If you try to make the tube of mercury taller (say 35 inches) then an empty space will appear between the top of the mercury and the sealed end of the tube because standard atmospheric pressure at sea level is unable to push the mercury up the tube any higher.

Edit: Water is less dense than mercury so atmospheric pressure will support a much higher column of water, around 33ft tall. But a barometer that tall would be a bit inconvenient so mercury is used.

Ok so we choose mercury purely out of convenience and the downward pressure exerted by the column of air pushes the mercury in the tube up. Now, when this mercury gets pushed up to 30 inches, where does 14.7 psi come into play so that we can say the atmospheric pressure at sea level is 14.7 psi? Is it that we take the amount of mercury and it's density and that leads us to 14.7 psi? I guess my hang up is how we get from 30 inches of mercury to 14.7 psi.

 

[DaveC426913]

Is it that we take the amount of mercury and it's density and that leads us to 14.7 psi? I guess my hang up is how we get from 30 inches of mercury to 14.7 psi.

If that column were one inch square, the amount of mercury in 30 inches would literally weigh 14.7 pounds.

https://www.traditionaloven.com/met...b-of-mercury-to-cubic-inch-cu-in-mercury.html

But by constricting it, to, say 0.1 square inches, it would only have to lift 1.47 pounds of mercury to reach the same height.Why 30 inches? I think that's just what the setup was when they first did it, and it's become the standard.

Same reason why a thermometer doesn't have to have some exact length for convenience. It doesn't matter.

 

[opus]

Ohhh ok. So the 14.7 pounds per square inch is the weight of the mercury at 30 inches and this corresponds to the atmospheric pressure at that altitude because the downward force of the mercury(weight) must balance with the downward force of the air column. Since the mercury's weight is 14.7 pounds for that square inch, and is balanced by the atmospheric pressure, we can deduce that the atmospheric pressure is also 14.7 pounds per square inch.

 

[DaveC426913]

Ohhh ok. So the 14.7 pounds per square inch is the weight of the mercury at 30 inches

Yup. That calculator I linked to allows you to input cubic inches of mercury (30) and spits out weight in pounds (14.7).

 

[opus]

Ok great thanks guys. That cleared everything up for me.

 

[davenn]

and acceleration is equal to the gravitational constant.

ohhh and that isn't correct ...

Gravitational Constant G ( big G)
https://en.wikipedia.org/wiki/Gravitational_constant

Earth's gravity ( which is variable) ... g ( small g) = 9.81 m/s² ... with lots of small variations in different areas depending on rock densities
But for most general calculations 9.81 m/s² can be usedDave

 

[opus]

ohhh and that isn't correct ...

Gravitational Constant G ( big G)
https://en.wikipedia.org/wiki/Gravitational_constant

Earth's gravity ( which is variable) ... g ( small g) = 9.81 m/s² ... with lots of small variations in different areas depending on rock densities
But for most general calculations 9.81 m/s² can be usedDave

Thanks Dave!

 

[DrDu]

1 comment:
The force the Earth excerts on the air is elastic force, the atmospheric pressure compresses the Earth a tiny bit and the Earth reacts with an opposing force, like a rubber ball you try to squeeze.

 

[Charles Link]

I didn't see it mentioned above: The density of mercury is 13.6 grams/cm^3. $\\$ Using 30 inches $\approx$ 760 mm= 76 cm, and using $g=980$ cm/sec^2, you get a pressure from the weight of the column of mercury in dynes/cm^2. If you do the conversion of pounds/in^2 to dynes/cm^2, you will get agreement:$\\$ (13.6)(76.0)(980)=1.013 E+6 with the last number being a result that I googled for atmospheric pressure in dynes per square centimeter.

 

[opus]

1 comment:
The force the Earth excerts on the air is elastic force, the atmospheric pressure compresses the Earth a tiny bit and the Earth reacts with an opposing force, like a rubber ball you try to squeeze.

Is there a way we can measure this? How could we measure the force that the Earth pushes back with? I understand the concept and believe that it does, but is there a way to measure it? Additionally, why don't we “feel” this atmospheric pressure? If I put 14.7 pounds over the top of my hand with a weight it wouldn't be comfortable, but I don't feel this atmospheric pressure at all.

 

[opus]

I didn't see it mentioned above: The density of mercury is 13.6 grams/cm^3. $\\$ Using 30 inches $\approx$ 760 mm= 76 cm, and using $g=980$ cm/sec^2, you get a pressure from the weight of the column of mercury in dynes/cm^2. If you do the conversion of pounds/in^2 to dynes/cm^2, you will get agreement:$\\$ (13.6)(76.0)(980)=1.013 E+6 with the last number being a result that I googled for atmospheric pressure in dynes per square centimeter.

Thanks Charles!

 

[Charles Link]

It may also be worth mentioning, that for a barometer to work properly, there must be a near vacuum in the space in the tube above the mercury.$\\$ e.g. If there were 1 atmosphere of air pressure there, the outside atmospheric pressure would not push the mercury up the tube at all.

 

[opus]

It may also be worth mentioning, that for a barometer to work properly, there must be a near vacuum in the space in the tube above the mercury.$\\$ e.g. If there were 1 atmosphere of air pressure there, the outside atmospheric pressure would not push the mercury up the tube at all.

And is the reason for this that the air in the tube has weight that is pressing down on the mercury inside? What confuses me about this is that there isn't as much air inside the tube than there is outside in the atmosphere so shouldn't the weight of the air outside the tube be much more than inside?

 

[Borek]

Charles has answered your question in the second paragraph of his post. It is not about weight of the air inside, it is about its pressure.

 

[Charles Link]

The air inside the tube, if any will not weigh very much. It's pressure is determined by $PV=nRT$. If air is first allowed in the tube that holds the mercury, it will be automatically be at a density/pressure that is at 1.0 atmosphere if steps aren't taken to make sure the barometer tube has a vacuum above the column of mercury. I think this is easily done just by putting the mercury in a U-shaped tube that extends high enough on one side, and then turning the tube upside down.

 

[CWatters]

Additionally, why don't we “feel” this atmospheric pressure? If I put 14.7 pounds over the top of my hand with a weight it wouldn't be comfortable, but I don't feel this atmospheric pressure at all.

We don't feel it because the pressure inside the body is normally the same as outside.

Sometimes when driving up or down a hill the pressure inside doesn't change as fast as outside and you can feel the pressure difference, typically in your ears.

 

[Charles Link]

@opus The air pressure is what makes a suction cup stick to a wall. If you increase the volume inside by tugging on the suction cup, it has very reduced pressure inside compared to the outside pressure.

 

[CWatters]

Is there a way we can measure this? How could we measure the force that the Earth pushes back with? I understand the concept and believe that it does, but is there a way to measure it?

Remove the air (perhaps by pumping it out from under a bell jar.) and the ground will expand slightly. Recompress it somehow measuring the force it exerts on whatever you are using to compress it.

You can feel the Earth pushing on your feet when you are standing up

 

[opus]

Thanks everyone. This now makes a great deal more sense. I never really thought of these things so it's very interesting.

We don't feel it because the pressure inside the body is normally the same as outside.

Sometimes when driving up or down a hill the pressure inside doesn't change as fast as outside and you can feel the pressure difference, typically in your ears.

Ok now here's a dumb question. If our pressure inside out body is the same as the outside, which should be about 14.7 psi or 1 atm, why don't we "lose pressure" if we got stabbed or something? Or is it that we technically would, as maybe our blood, fluids, gases in our body make up this internal pressure?

 

[russ_watters]

Ok now here's a dumb question. If our pressure inside out body is the same as the outside, which should be about 14.7 psi or 1 atm, why don't we "lose pressure" if we got stabbed or something?

How can we lose pressure if the pressure is already equal inside and out?

What do you feel in your ears when you drive up a mountain, fly up in a plane or dive deep under water?

 

[opus]

How can we lose pressure if the pressure is already equal inside and out?

What do you feel in your ears when you drive up a mountain, fly up in a plane or dive deep under water?

Good point. So then we wouldn't lose pressure because the pressure is equal on both sides. I suppose then, and this may be for another topic, what are the causes of our pressurization? A google search led me to things like cabin pressurization which isn't exactly what I'm looking for.

What do you feel in your ears when you drive up a mountain, fly up in a plane or dive deep under water?

That would be pressure from the change in ambient pressure which could be experienced from either going up (decrease in pressure) or down (increase in pressure).

 

[russ_watters]

Good point. So then we wouldn't lose pressure because the pressure is equal on both sides. I suppose then, and this may be for another topic, what are the causes of our pressurization?

You answered this question correctly in your own first post. I feel like you are looking for a complicated answer to a simple question you already know the answer to. Consider the following:

A. You set a 15 lb stack of books on a table. How much force does it apply to the table?
B. You take half the books off. How much force does the stack apply to the table now?
1. You set a 15 lb stack of books on top of a plastic bag full of water on top of a table. What is the force applied to the bag of water?
2. You take half the books off. What is the force applied to the bag of water now?
X. You have a glass of water at sea level (round off to 15 psi atmospheric pressure). What is the pressure inside the glass of water?
Y. You take the glass up to an altitude where atmospheric pressure is half sea level. What is the pressure inside the glass of water now?

None of these are trick questions and all three sets of scenarios are the same. What happens to change from the first to the second question in each set of scenarios?

 

[opus]

In the second cases, you are removing weight, or pressure, from the top. In the first two scenarios its by removing books, in the last it's removing some amount of column of air from atop the glass. I think I've been looking for a deeper, more complicated answer for no reason as you've said. Sorry about that.

 

[Charles Link]

On the gas pressure inside of us, on a previous comment above, did you ever see them talk on TV about deep sea divers need to make sure they don't come to the surface too quickly or they can get the "bends", a problem caused by gases "boiling" out of their body etc, because the internal gas pressure increased when they were in the higher pressure environment farther down, etc., and they need to release this extra pressure slowly? See https://en.wikipedia.org/wiki/Decompression_sickness

 

[opus]

On the gas pressure inside of us, on a previous comment above, did you ever see them talk on TV about deep sea divers need to make sure they don't come to the surface too quickly or they can get the "bends", a problem caused by gases "boiling" out of their body etc, because the internal gas pressure increased when they were in the higher pressure environment farther down, etc., and they need to release this extra pressure slowly? See https://en.wikipedia.org/wiki/Decompression_sickness

Yes as I understand it, the gases don't have time to be exhaled as they come out of solution too quickly and small bubbles come into formation inside the body. Are these gases the cause for our pressurization?