Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

I understand the meaning of pressure on a surface to be force acting per unit area.

But when it comes to understanding pressure in the context of fluids in motion or at rest I think I am having some trouble grasping the concept.

When we say the pressure of a gas in a container (which may be of any arbitrary blobby shape) is say 2 Pascals, what does it really mean?

1) Does it mean the pressure of the gas on ANY small area considered on the surface of the container == 2 pascals?

2) Or Does it mean the pressure on a thin small wafer (possible imaginary?) placed anywhere inside the gas == 2 pascals?

I am really confused!

To add to my problem of understand pressure, is a sentence from Chorin and Marsden's a Mathemtical introduction to fluid dynamics when they define an ideal fluid.

*"Lets us define an ideal fluid with the following property: For any motion of the fluid there is a function p(x,t) called the pressure such that if S is a surface in the fluid with a chosen unit normal n, the force of stress exerted across the surface S per unit area at x in S at time t is p(x,t) n i.e.

force across S per unit area=p(x,T)n"*

Here is my problem: Suppose we have an ideal fluid and a time and position varying pressure function as above. For a thin small wafer placed at x at time t, there are two equal and oppsite forces acting across its surface i.e. p(x,t)n and -p(x,t)n since the surface has 2 forces in the direction of the two normals to it.

This reasoning means the force on the thin small wafer is zero. Which means that the force at any point the the fluid at any time is zero. Which is totally counterintuitive. Why would they define an ideal fluid like that?

share|improve this question
    
Horrible! How can one read a "mathematical introduction ...." without any basics in experimental physics? What kind of mis(e)ductional system does cause that? –  Georg Nov 18 '11 at 9:55
1  
My physics fundamentals are pretty weak. But I feel this is a perfectly valid question, which does not seem to be answered in most physics books which I have read. When I asked a prof for a quick introduction to fluid mechanics he recommended me Chorin MArsden. IF you can recommend me some good books/sources on experimental physics that would be great. Thanks! –  smilingbuddha Nov 19 '11 at 19:46
add comment

4 Answers 4

1) Roughly speaking pressure of a gas means that x gas molecules hit your wall unit with y speed per time unit. This intuitively corresponds to the ideal gas law: $$P\ V = N\ k_B\ T$$ You can increase pressure by either adding more gas molecules or increasing the temperature (kinetic energy) of the gas molecules. Pressure is defined as force per area, so if you put a thin wafer in a container with gas it will experience the same pressure at the front and the back side. There is no difference here between the pressure on the wafer or on the inside walls of the container.

2) For the ideal fluid things are a little different. Here without motion the pressure is zero. According to your definition $\vec{F} = p\ \vec{n}$ the force would always be parallel to the normal surface. This cannot be correct, as you clearly experience when you put the wafer at an angle to your fluid motion $\vec{x}$. (Or alternatively the direction of the force is independent of the fluid motion direction, which is also not true.)

share|improve this answer
    
Thank you for the answer. I have quoted Chorin Marsden exactly and there the pressure function is given to be independent of the orientation of the wafer at position x. (i.e. p(x,t) and not p(x,t,n) ). Even I feel this definition is incorrect, but thats the way it is printed :( –  smilingbuddha Nov 18 '11 at 2:18
    
I found a preview at google books and on page 7 there is the answer. With the pressure $p$ they mean hydrostatic pressure, so with the fluid at rest. This pressure acts on the surface like a gas, i.e. parallel to $\vec{n}$. –  Alexander Nov 18 '11 at 2:42
add comment

I think the wafer thought experiment is a little self-defeating, the reason being that the force exists equally on both sides of the wafer. Maybe you'll say, well that's not a problem, because there's still a measurable force between the two surfaces. But that force also exists on the edges, so we would change the problem of defining/measuring pressure in a gas to that of measuring it in a solid. It doesn't make headway to answer the question.

My definition:

Consider a rigid container that contains a perfect vacuum in the inside. This container must be placed within the gas in question, and it must not be moving relative to the gas. The pressure of the gas is then defined as the force per unit area on any flat surface of this container. In the case of an infinitely flat right cylinder, this can be directly measured by compressive force between the two flat circular sides.

Impractical? Yes. However, I think this answers the question. There is an objective physical concept of a perfect vacuum. Without introducing that, there can never be any definition of absolute pressure. Otherwise, who's to say what zero pressure is?

share|improve this answer
add comment

The easiest way to get un-confused regarding this is to consider the notion of momentum flows.

When you have a current of 5 A, you know 5 Coulombs are flowing through the wire every second. This means that 5 Coulombs are flowing through any surface which intersects the wire, no matter how crumpled or deformed, every second. You don't get confused in this case.

Charge is a scalar conserved quantity, so its flows are a vector. The momentum is a vector conserved quantity, so its flow, its current is a vector-of-vectors. This is called the stress tensor. The T_xy stress is the flow of the x-component of momentum in the y direction. Just like any current, it is best thought of as composed of a current density, which you integrate over a surface to get the total current.

The pressure is simply a diagonal stress tensor--- T_xx=T_yy=T_zz = p. The flow of x momentum is in the x direction, the flow of y momentum is in the y direction, and the flow of z momentum is in the z direction. This is a rotationally invariant current of momentum, and it is the only rotationally invariant momentum flow possible (because only the on-diagonal 1,1,1 tensor is rotationally invariant).

This point of view makes it very intuitive why pressure doesn't depend on the crumply surface, or on anything else. It is just like current, except for a vector momentum.

share|improve this answer
add comment

Pressure is external force exerted on a body per unit area and stress is internal resistance exerted by a body per unit area.

Pressure exerted by gas on cylinder walls means the sum of force the gas molecules exert, per unit area, on the walls; 2 Pascals means on every $1 m^2$ area of cylinder, the gas exerts a force of 2 Newtons. Now if your area varies so will the force exerted, but it will be 2 Pascal always.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.