# Pressure Experiment - P=P0±mg/A

1. Mar 28, 2013

### charlottexo

Hey guys, we did an experiment that involved pressure of a fixed mass of air in an upside down syringe. We hung weights off the bottom and decreased the pressure of the gas, and increased the length of it. Quite a basic, gas law experiment based on the fact that pV = constant.

Anyway, when we added our masses, we had quite a strange formula to work out the force which was given to us as F=P0*A - mg. P0 was 1.01x10^5 and A was the cross sectional area of the syringe.

So we calculated force for different lengths of the syringe and plotted Log(L) on the y-axis and Log(F) on the x-axis (I'm pretty sure that is the right way around to do it, right guys?).

Now, that is all fine and dandy but this question is asking me:

Write an expression for pressure of the air (p) in terms of P0, A, M and g. Pressure is F/A obviously, so using our original equation for force, F=P0*A-mg:

P = P0 - mg/A, right? However, apparently it is P=P0+mg/A. Can anyone please shed some light on why this is? Because I'm going insane right now. :(

Charlotte xx

Last edited: Mar 28, 2013
2. Mar 29, 2013

### Sunil Simha

Is the question referring to the experiment or is it from someplace else?

3. Mar 29, 2013

### rude man

You have 3 forces to consider:

1. Gravitational due to the dangling masses pulling down on the plunger;
2. internal pressure inside the syringe pushing down on the plunger;
3. atmospheric pressure pushing up on the plunger.

These forces must add to zero since the plunger doesn't move at each step of your experiment.

So what is the net force counterbalancing the internal pressure pushing down on the plunger?

4. Mar 29, 2013

### rude man

Well, it can't be p = p0 + mg/A since that would say that the internal pressure exceeds atmospheric pressure which, when the masses are added, would make the plunger move downwards, wouldn't it?

Since p = p0 - mg/A was given to you, why are you questioning it?

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