How Does Absolute Pressure Differ from Gauge Pressure?

[sgstudent]

The formula for absolute pressure is absolute P=gauge P+atmospheric pressure. So when the gauge is at 0 it's actually 1 atm. So using the gas law if I were to half the volume the pressure would increase to 2 atm. So would the gauge read 1 atm when this happens?

So how does the mechanism allow it to read the absolute pressure minus the atmospheric pressure only?

Thanks :)

 

[russ_watters]

You understanding of the math is correct.

The thing to get about how pressure gauges work is that all of them measure differential pressure between two points.

They tend to just be a membrane or similar apparatus that has the atmosphere on one side and what you are measuring on the other.

 

[sgstudent]

Hi thanks for the reply :)

Actually its not just a pressure gauge. There's a graduated cylinder with a piston, manometer and temperature sensor. So I thought when we pump in air from the outside the pressure inside would be 1 atm but the gauge would read 0. So now when we half the volume, the pressure doubles. So the absolute pressure should be 2 atm. So using the formula absolute P=gauge P+1atm. Gauge pressure should be 1atm.

However I don't really get why the 1 atm remains constant here. Aren't we just manipulating the 1 atm we already had in there? So I don't get the formula in this sense.

Thanks :)

 

[sophiecentaur]

Whatever system is used, it's based on the difference between two pressures. I think your concern is about comparing an external pressure with a reference pressure (inside some reservoir). I think it has to be true that altering the (finite) volume of the inside reservoir will change its pressure (i.e. the internal pressure will no longer be 1atm) but that effect can be eliminated by calibrating the scale on the device.

 

[Aero_UoP]

So I don't get the formula in this sense.

Thanks :)

P(abs) = P(atm) + P(gauge) IF P(abs) > P(atm)

P(abs) = P(atm) - P(gause) IF P(abs) < P(amt)

I hope this helps ;)