Pressure Experiment - P=P0±mg/A

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SUMMARY

The discussion revolves around an experiment involving the pressure of a fixed mass of air in an upside-down syringe, where weights were added to decrease gas pressure and increase the syringe's length. The formula for force was derived as F = P0*A - mg, where P0 is atmospheric pressure (1.01x10^5 Pa) and A is the cross-sectional area. The confusion arises regarding the expression for pressure (p), with participants debating whether it should be P = P0 - mg/A or P = P0 + mg/A. The consensus is that the correct expression is P = P0 - mg/A, as the internal pressure cannot exceed atmospheric pressure without causing the plunger to move.

PREREQUISITES
  • Understanding of gas laws, specifically the ideal gas law (pV = constant).
  • Knowledge of pressure calculations involving force and area (P = F/A).
  • Familiarity with basic physics concepts such as gravitational force and equilibrium.
  • Ability to interpret and plot logarithmic data (Log(L) vs. Log(F)).
NEXT STEPS
  • Review the ideal gas law and its applications in experimental physics.
  • Study the principles of pressure and force equilibrium in fluid mechanics.
  • Learn about the effects of atmospheric pressure on gas behavior in closed systems.
  • Explore advanced topics in thermodynamics related to gas laws and pressure variations.
USEFUL FOR

Students and educators in physics, particularly those focusing on fluid mechanics and gas laws, as well as anyone conducting experiments involving pressure and force measurements.

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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