# Atmospheric pressure- where do we get the mass?

1. Nov 26, 2018

### 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?

2. Nov 26, 2018

### Staff: Mentor

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 $m = 15\ lb_m$

3. Nov 26, 2018

### DaveC426913

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

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.

4. Nov 26, 2018

### 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?

5. Nov 26, 2018

### opus

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?

6. Nov 26, 2018

### Staff: Mentor

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.

7. Nov 26, 2018

### opus

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.

8. Nov 26, 2018

### CWatters

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.

9. Nov 26, 2018

### Staff: Mentor

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.

10. Nov 26, 2018

### opus

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

11. Nov 26, 2018

### opus

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.

12. Nov 26, 2018

### DaveC426913

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

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.

13. Nov 26, 2018

### 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.

14. Nov 26, 2018

### DaveC426913

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

15. Nov 26, 2018

### opus

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

16. Nov 27, 2018

### davenn

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 used

Dave

17. Nov 27, 2018

### opus

Thanks Dave!

18. Nov 28, 2018

### 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.

19. Nov 28, 2018

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.