Pressure Experiment - P=P0±mg/A

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Homework Help Overview

The discussion revolves around an experiment involving the pressure of a fixed mass of air in an upside-down syringe, where weights are added to decrease the gas pressure and increase its length. The participants are exploring the relationship between pressure, force, and area, referencing the gas law equation pV = constant.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the formula for force and its implications for calculating pressure. There is confusion regarding the correct expression for pressure in terms of P0, A, M, and g, with differing interpretations of the relationship between internal and atmospheric pressure.

Discussion Status

Some participants are questioning the validity of the derived pressure expression, while others are clarifying the forces at play in the experiment. There is an ongoing exploration of the assumptions regarding pressure and the behavior of the plunger in response to added weights.

Contextual Notes

Participants are grappling with the implications of their findings in light of the experimental setup and the definitions of pressure involved. The original poster expresses frustration over the conflicting interpretations of the pressure formula.

charlottexo
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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
 
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charlottexo said:
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:
Charlotte xx

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

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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