Derivation of the change of air pressure with height

[Dale]

if we take a slab of air then molecules from above would cause collisional pressure on the slab and the weight of the air above would cause additional pressure

This is incorrect. All of the pressure is “collisional”, including whatever pressure might be related to the weight of the gas. There is no additional pressure.

 

[Adesh]

This is incorrect. All of the pressure is “collisional”, including whatever pressure might be related to the weight of the gas. There is no additional pressure.

But here you have said that it’s not same

No, it is not the same. The weight is a gravitational force. The pressure is a “collisional” force as you put it. Even in situations where the magnitude of the forces are the same they are still distinct forces originating from different physical interactions

 

[Dale]

Yes. The pressure is not the weight. All of the pressure is collisional, all of the weight is gravitational. What is at all unclear here?

I cannot believe that this is confusing. Draw a free body diagram for the slab of air.

 

[Adesh]

Yes. The pressure is not the weight. All of the pressure is collisional, all of the weight is gravitational. What is at all unclear here?

I cannot believe that this is confusing. Draw a free body diagram for the slab of air.

Wouldn’t gravity going to attract the air above the slab and since the slab lies below therefore it has to bear that force by which gravity is pulling the air above. So slab must have be having some pressure from below which should balance both collisional pressure and force of gravity of air above.

 

[Dale]

Please draw the free body diagram, as requested. Also, label each force and identify:
1) the type of force
2) the object providing the force (I.e. the other object in the third law pair for each force)

 

[Chestermiller]

Plain old pressure? Sir I haven’t understood your allusion.

Collisional pressure is the pressure caused by fluids, such as on piston or walls of container containing it. So, if we take a slab of air then molecules from above would cause collisional pressure on the slab and the weight of the air above would cause additional pressure. This is the problem, why we didn’t take that pressure from weight into our account.

If you do a force balance in the fluid above, the pressure from below times the area is equal to the weight of the fluid above.

 

[Adesh]

Please draw the free body diagram, as requested. Also, label each force and identify:
1) the type of force
2) the object providing the force

That Weight_{air} is weight of the air which lies above the slab .

 

[Dale]

$Weight_{air}$ does not act on the slab. It does not belong on the free body diagram for the slab.

 

[Adesh]

If you do a force balance in the fluid above, the pressure from below times the area is equal to the weight of the fluid above.

Yes sir.

 

[Adesh]

$Weight_{air}$ does not act on the slab. It does not belong on the free body diagram for the slab.

It acts on air above but slab is bearing that.

 

[Dale]

It acts on air above but slab is bearing that.

So what? That is completely irrelevant for the free body diagram of the slab. Only forces acting on the body itself are included in a free body diagram. Were you not taught this?

 

[Adesh]

So what? That is completely irrelevant for the free body diagram of the slab.

Why? Weight of air is pushing the slab downwards, collisional pressur eis pushing the slab downwards and the grravity too is attracting it downwards. Why my words are not making sense to you?

 

[Dale]

Why?

Because in a free body diagram you only include forces directly acting on the free body.

Why my words are not making sense to you?

This is not a communication problem. You have a conceptual physics misunderstanding and instead of correcting it you are trying to explain why you are right. You are wrong. You are analyzing the situation incorrectly. The mistake you are making is demonstrated by the incorrect inclusion of the $Weight_{air}$ force in the free body diagram of the slab. It doesn’t belong there and including it is wrong.

 

[Chestermiller]

Yes sir.

OK. So, if z is the elevation at the top of the slab, this means that $$p(z)=\int_z^{\infty}{\rho(\xi) g d\xi}$$where $\xi$ is a dummy variable of integration (representing elevations above z). This says that the weight per unit horizontal area of the fluid above z (right hand side) is equal to the pressure at z. Do you agree with this equation?

 

[DrStupid]

There may be two kinds of pressure:-

1. Defined by you as the contact force.

2. Pressure due to collision of molecules with the walls of container that we use in thermodynamics . So the air above the slab must be colliding with our slab and hence may be causing some collisional pressure.

They are identical.

 

[Dale]

@Adesh let me explain how to analyze forces and draw a free body diagram.

Each force represents an interaction between two bodies. A contact force is a surface interaction between two bodies that are touching. Tension is a similar interaction between a rope or cable and another body. Pressure is a “collisional” interaction between a fluid and another body. Weight is a gravitational interaction between the Earth and a body, etc.

For each interaction we have in a scenario we can describe the type of interaction and the two bodies that are interacting.

In this case we have four bodies: the earth, the slab, the air above, and the air below. There is a gravitational interaction between the Earth and the air above. There is a gravitational interaction between the Earth and the slab. There is a gravitational interaction between the Earth and the air below. There is pressure between the air above and the slab. There is pressure between the slab and the air below. There is pressure between the air below and the earth.

This is a complete list of the important interactions in this scenario. Note that there is no pressure interaction between bodies that are not touching, and there is no gravitational interaction between the various air bodies.

Each interaction involves two forces, one acting on each of the two different bodies, and Newton’s third law says that these forces are equal and opposite. When you draw a free body diagram you include all of the forces that act on that body and none of the forces acting on other bodies.

Note, in the list of interactions I highlighted each occurrence of the slab. The slab occurred in three interactions, so there should be exactly three forces on its free body diagram. One is the pressure from the top, one is the pressure from the bottom, and the other is the weight of the slab.

The weight of the top air is not included because it does not act on the slab. It is an interaction between the top air and the earth, so it would be included in a free body diagram for the top air or for the earth. It is wrong to include it on a free body diagram of the slab.

I hope this systematic approach helps you understand how to analyze a scenario and correctly draw the free body diagrams. Once you have correctly drawn a free body diagram then you can apply Newton’s second law to analyze the dynamics of that body.

 

[Adesh]

OK. So, if z is the elevation at the top of the slab, this means that $$p(z)=\int_z^{\infty}{\rho(\xi) g d\xi}$$where $\xi$ is a dummy variable of integration (representing elevations above z). This says that the weight per unit horizontal area of the fluid above z (right hand side) is equal to the pressure at z. Do you agree with this equation?

Wow. What a great explanation! I’ll always be grateful to you.

 

[vanhees71]

It's not as simple since to close the hydrodynamic equations you need an equation of state, i.e., a relation between pressure and density. See my posting #25 for the full calculation, using they polytropic ansatz (valid, e.g., for ideal gases for an adiabatic or isothermal atmosphere).

 

[Herman Trivilino]

What about the pressure caused due to collision of molecules?

The force caused by the collision of molecules is the contact force I referred to in Post #17. That is the force you feel. You can't possibly feel the weight $mg$ of the sack. You can prove this to yourself by setting the sack down on the floor. The weight $mg$ of the sack doesn't go away when you do that, but the force exerted on your head by the sack does go away.

Forces are interactions between objects. They are not properties of objects.

 

[Chestermiller]

Wow. What a great explanation! I’ll always be grateful to you.

You should really be thanking @Dale and the others. This equation that I wrote is fully consistent with what they have been saying. You can see that now, right?

 

[Adesh]

Thanks everyone for giving their precious time and intellect to me. I’m really grateful to everyone. I mean it.