Pressure Experiment - P=P0±mg/A

In summary: This equation is based on the fact that the net force on the plunger is zero and that the downward force due to the mass (mg) is the only other force, and that force is p0 * A - mg. This force must be balanced by the upward pressure force (p0) * (area of plunger) and that force is p0 * A, so the net force is p0 * A - mg. But the pressure force (pressure is force / area) is p0, so p0 = force / area, and thus force = p0 * A. Now just substitute that into p = p0 - mg/A and you get p = p0 + mg/A.
  • #1
charlottexo
6
0
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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  • #2
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?
 
  • #3
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
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?
 
  • #5
Hello Charlotte,

First of all, great job on conducting your pressure experiment! It sounds like you have a good understanding of the gas law and how to calculate force using the formula F=P0*A-mg.

To answer your question, the correct expression for pressure in terms of P0, A, M, and g is indeed P=P0-mg/A. This is because pressure is defined as force per unit area, so in this case, the pressure of the air would be equal to the force (P0-mg) divided by the cross-sectional area (A).

I'm not sure where the confusion may have come from, but it's possible that there was a typo or mistake in the original question or formula given to you. It's always a good idea to double check your calculations and make sure your units are consistent.

I hope this helps clarify things for you. Keep up the good work in your experiments!

Best,
 

1. What is the purpose of the "Pressure Experiment - P=P0±mg/A"?

The purpose of this experiment is to measure the pressure exerted on an object due to the force of gravity. This can be used to understand the effects of weight and surface area on pressure.

2. How is pressure calculated in this experiment?

The pressure, P, is calculated by dividing the weight of the object, mg, by the surface area, A, on which it is acting. This can be expressed as P = mg/A.

3. What does "P0±mg/A" stand for in the experiment?

"P0" represents the initial pressure before any additional weight is added to the object, while "mg/A" represents the change in pressure due to the added weight. The ± symbol indicates that the change in pressure can be either positive or negative, depending on the direction of the force.

4. How does the surface area affect pressure in this experiment?

The surface area, A, directly affects the pressure exerted on an object. As the surface area decreases, the pressure increases, as there is less area for the force to be distributed over. Similarly, as the surface area increases, the pressure decreases.

5. What are some real-world applications of the "Pressure Experiment - P=P0±mg/A"?

This experiment can be applied to various real-world situations, such as understanding the pressure exerted on an object at different depths in a body of water, or calculating the tire pressure needed for a certain weight of vehicle. It can also be used in engineering and construction to determine the weight-bearing capacity of structures.

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