# What is gas pressure?

1. Aug 2, 2015

### rogerk8

Hi!

I wonder what gas pressure really is.

This I have learned from Wikipedia:

1) Pressure is a scalar (which means it has no direction).
2) Since a system under pressure has potential to perform work on its surroundings, pressure is a measure of potential energy stored per unit volume.
3) Gauge Pressure is relative to normal air pressure (atmospheric pressure).
4) In a static gas, the gas as a whole does not appear to move. The individual molecules of the gas, however, are in constant random motion.

The most fascinating quality I think is 2).

A more seldom use for pressure unit is J/m^3 (even though the SI-unit is N/m^2).

I think this fact is really interesting.

A gas has an energy density.

Within one cubic meter it has Joule as unit.

So more specific, how can a unit volume of gas contain pure energy?

And what is this energy, really?

Yes, we could say that the gas has the "potential of perfoming work" (like in the Otto engine f.i)

Yes, we know that.

But what is the mechanism?

We know that the internal Ek is around

$$Ek=\frac{mv^2}{2}\propto kT [J]$$

So speed of the molecules are closely related to temperature.

This is one property.

But how about the other called pressure

$$p=nkT [J/m^3]$$

where n is the particle density.

This just says that pressure is related to density and temperature.

But what is pressure, really?

If we can say that speed of the particles is closely related to temperature according to the first formula, what is then pressure?

If it's not speed it has to be something other.

And I simply don't understand what.

A mental experiment I did not so long ago made me come to this conclusion:

$$P=\rho_s\frac{dv}{dt} [N/m^2]$$

where rho simply is the surface density of the gas.

This is nothing other than the use of Newton's first law applied on a gas.

So a gas has the ability to generate a force on f.i a window by "rushing" to it with a certain surface density (and acceleration).

This along with 2) above gives me some kind of clue of what gas pressure is.

The only enigma here is the (de)acceleration of the surface density.

Because how can we know that?

How far from the truth am I?

Best regards, Roger

2. Aug 2, 2015

### nasu

There is no meaningful difference between "What something is" and "what something really is".
This sort of questions appears quite often and usually gets nowhere.
If you know the definition of pressure, this is what it is. Really, actually and in every "deeper" sense.

Pressure is also a result of the motion of the gas molecules. When they hit the walls of the container they transfer some of their (kinetic) energy to that wall. In equilibrium, the wall also transfers back some energy so the gas does not loose or gain energy. If one of the walls is mobile, the energy balance is broken. The gas transfers more than it gets back, some of its kinetic energy is lost. The gas cools down.

Not different, in principle, from the work done by a moving object when it collides with another object. Do you wonder about the energy of a moving projectile? You can divide that by its volume and find an energy density. Each cm cube will "have" some energy in Joules. :)

3. Aug 2, 2015

### rogerk8

I like this explanation!

Especially the last paragraph.

I could never have come up with that thought :)

Yet, I think I miss something important in understanding the gas pressure concept.

And I think I have found the reason for "myself".

The energy within a gas may be written

$$U=Ek+Ep$$

As "always".

But the potential energy as mentioned in 2) above is really hard to understand.

I think I get the kinetic part, but what is the potential part?

What is it really?

Best regards, Roger

4. Aug 2, 2015

### nasu

For ideal gas the potential part is zero so there is nothing to understand.
The entire internal energy is kinetic.

5. Aug 2, 2015

### Staff: Mentor

The potential energy is the effect of the gas molecules interacting with one another as they get closer together. At low pressures, there is only kinetic energy with negligible potential energy (so you have ideal gas type behavior). As the gas is compressed and the molecules get closer together, they exert more significant forces on one another, as sometimes characterized by a so-called 6-12 potential, referring to exponents on the radial spacings between the molecules. So the potential energy of the system of gas molecules changes.

Chet

6. Aug 3, 2015

### rogerk8

This explanation is really interesting, thanks!

Repeating for myself:

A) Low pressure, only kinetic energy (ideal gas behaviour)
B) High pressure/compressed gas, intermolecular forces (varies as well as Ep)

A stupid question here is what makes low and high pressure respectively.

Take air for instance, the ambient pressure is 1 atm.

Is air an ideal gas (I imagine 1 atm to be of "low pressure")?

Consider now that you have a can with a piston in it and you pull the piston fully out while opening a valve to let "1atm" in.

The air in the can is now of same pressure as ambient (which I consider low).

While the pressure is low, A applies.

Now, pushing the piston in gives compression and thus "high pressure".

While the pressure now is high, B applies.

Hope this is true.

Today I had an appifany (actually I suddenly remember it from my former PF blog).

Consider Hydrogen for simplicity (two equal charges of opposit sign).

$$W=\int Fdr$$

What this means is that work has to be done to move the electron a distance from the nuclei.

While doing this, the electron gets that potential energy.

It is actually as simple as mgh because in both cases there are attractive forces which has to be overcome.

So we have the relationship that charged particles behave in this way.

And close to the nuclei, potential energy is low.

It is hard to se the analogy for ordinary (say ideal) gas where the molecules/atoms are neutral(?)

What are the intermolecular forces then?

"Gravity", or?

If my reasoning has some truth, compressing the gas according to 2) makes the atoms come even closer together which points at even less potential energy.

I must confess that part of this reasoning is from PF.

Best regards, Roger
PS
Your colleque Sophiecentaur told me to take a course in thermo dynamics (which I'm almost finished with). There I learned a very interesting thing. If you take ordinary ice and heat it up slowly while measureing the temperature vs time. The ice is just melting and temperature climbs linearly with supplied heat. But at certain point the temperature stops climbing while still supplying heat. The temperature just stands still for a while. This plateu is, according to course, due to the change of state.

On the plateu we have a change of potential energy (only), I think.

This change of Ep may be due to different binding energies of the different states (my guess).

It is interesting to note that the state that is easiest to change (ice to water) comes first and is thus least dependent of amount of heat supplied.

In other words, Ep should be at its lowest for ice (where the molecules are closest to eachother)

7. Aug 3, 2015

### rogerk8

If we look at mgh for air we have that Ep increses with hight but as it increases pressure, according to

$$P=\rho gh$$

actually decreases.

I saw now in my notes from Andrew Mason that pressure of a gas actually has this property.

High pressure: low Ep

Low pressure: High Ep

So what you Chestermiller told me above, is not fully true.

I think I've almost proven the above facts i.e that Ep is low close to the molecule and high further away, and from this it is simple to see that if the molecules are far apart, pressure is low and vice versa.

Best regards, Roger

8. Aug 3, 2015

### nasu

You are mixing different things in hard to understand posts.
It may be better to stay on one question until is clarified.

The PE in gravitational field is not the same as the internal potential energy due to inter-molecular forces. Chestermiller was talking about this last one.
If you have problems with the basics of the pressure concept, stay away from potential energy. Try first to learn and understand the ideal gas model and the pressure described by this model.

Then maybe you can look at the corrections to pressure introduced by inter-molecular forces, by looking at the van der Waals model, for example.

9. Aug 3, 2015

### rogerk8

Do I dear ask what these inter-molucular forces are?

To me a gas is often neutral in nature, so how can there be inter-molecular forces?

Inter-molecular "hits" sure, but forces?

I have actually read some about van der Waal but no explanation to the correction koefficient was given.

Best regards, Roger

10. Aug 3, 2015

### nasu

This is why is recommend to keep with ideal gas model and forget about PE for now. :)

Yes, the forces are small because the atoms are neutral in normal conditions. One such force is the so called van der Waals force (or London dispersion force).
The Wiki article looks to explain pretty clearly the meaning of the coefficients.

But again, you don't need this to understand pressure.
You are the one that keep bringing the PE in the discussion that started as being about pressure of a gas.

11. Aug 3, 2015

### rogerk8

Just want to say that I have now read the Wiki article about the van der Waals force.

Did not understand so much, though :D

But I found the part of polarization interesting.

Obviously molecules (like HCL) can function as a dipole (two interesting questions are why and how :) ).

But if we consider that charge is uneven spread (somehow) over the molecule the molecule will then constitute a positively charged side and a negatively charged side and if there are two equal molecules they can bind because of this uneven spread of charge.

So one aspect of the van der Waals force is that it still uses the electromagnetic force to bind.

And finally, I am confident that in such cases Ep is at it's lowest when the molecules are close together (i.e high pressure).

Best regards, Roger

12. Aug 3, 2015

### sophiecentaur

The forces only come into play when the molecules are 'near' and in what you'd call the collision zone. The Van der Waal's forces (mentioned above) have the effect of modifying the basic model of an ideal gas, in which the particles are infinitely small and the collisions are perfectly elastic. The forces make the molecules behave a bit like rubber balls, which compress a bit on impact and slow up the molecules during the impact so some of the Kinetic Energy of the ideal molecules becomes stored as Potential Energy (as in a compressed spring). The higher the pressure (closer together the molecules) the greater proportion of the total internal energy is Potential. In the limit, of course, the molecules can stick together and you will get condensation. (High enough pressure and low enough temperature / KE)
But, as stated above, you should first sort out the basics of an ideal gas first and ignore these forces and the 'behaviour of 'real' gases.
Incidentally, the basic Kinetic Theory of gases discusses the pressure in terms of Momentum changes as the molecules hit an ideal, infinitely massive wall. There was mention of KE exchange at a wall, further up in the thread but momentum (mass times velocity) is reversed on impact, producing a force (on average because of the high number of collisions). If the temperature is held constant, the KE is neither gained nor lost at the wall.
PS You might like to browse through this link. http://hyperphysics.phy-astr.gsu.edu/hbase/kinetic/kinthe.html
It gives the standard 'Mechanical' approach to how pressure is set up in gases.

13. Aug 3, 2015

### Staff: Mentor

With all due respect to Wikipedia, this is not quite correct. For any arbitrarily oriented surface (either a real solid surface, or the interface between two adjacent parcels of material in a gas or liquid), the direction that the pressure force acts is always perpendicular to the surface. So pressure actually does have directionality in a well defined sense. At a more advanced level, pressure is equal to the magnitude of the isotropic part of the stress tensor (as second order tensor). And second order tensors definitely do have directionality. For a gas or liquid in static equilibrium, the pressure force per unit area acting on an arbitrarily oriented element of surface area is equal to the stress tensor dotted with a unit normal to the surface, and turns out to be equal to the magnitude of the pressure times the unit normal.

Chet

14. Aug 4, 2015

### rogerk8

Ok.
Ok.
Interesting.
Ok.
Interesting.
I keep hearing this ;)
Your nice link bellow told me exactly what you say and it has the Momentum View of it, which I like.
Regarding the Momentum View, how can

$$F\Delta t=\Delta p$$

I can't for my life remember when

$$[Ns]=[kgm/s]$$

I also had a hard time understanding the average Force formula

$$F=\frac{mv_{1x}^2}{L}$$

But that was explained to me with the help of the Momentum View.

Best regards, Roger
PS
Now I see it :)

$$F=ma=[kgm/s^2]$$

15. Aug 4, 2015

### rogerk8

Very intersting even though, in my layman terms, I see pressure as a force acting perpendicular to a surface.

But maybe you can see it the other way around also (like I think you do)?

I don't know if this is even close to the truth :)

Ok, even though I have no clew about what a tensor is, isotropic as a concept I understand.
Ok, I'll take your word for it.
But how do I relate to "pressure force"?

Is it the force that pressure generates per unit area?

Dotted, gives the component that is the aligned multiplied magnitude of both the vectors, right?

Like

$$C=A\cdot B=ABcos\alpha$$

Where alpha is the angle between them both.

You are amazing, Chet!

You actually take the time writing to me about tensors :D

I feel honored!

I have actually come in contact with tensors once but I never understood them (and it was no requirement) so I just saw them write them down on the blackboard some 20 years ago.

Best regards, Roger
PS
Here is where Wikipedia might be wrong, then: https://en.wikipedia.org/wiki/Pressure#Definition

Last edited: Aug 4, 2015
16. Aug 4, 2015

### Staff: Mentor

I would call it the force that the gas exerts per unit area.
Roger,

It sounds like you are really ready to start learning about 2nd order tensors (You already know about first order tensors, namely vectors). If you are, then here is something to get you started.

Suppose you write down two vectors $\vec{A}$ and $\vec{B}$ right next to one another (in juxtaposition), with no operation like the cross product or the dot product implied to be acting between them. That is $\vec{C}=\vec{A} \vec{B}$. The way this relationship is written, the quantity $\vec{C}$ has no physical or geometric meaning. However, consider what would happen if we dotted the quantity $\vec{C}$ with an arbitrary vector $\vec{D}$ in either of the two following possible ways:
$$\vec{C}\centerdot \vec{D}=\vec{A} \vec{B}\centerdot \vec{D}=\vec{A}(\vec{B}\centerdot \vec{D})\tag{1}$$
$$\vec{D}\centerdot \vec{C}=\vec{D}\centerdot \vec{A} \vec{B}=(\vec{D}\centerdot \vec{A})\vec{B} \tag{2}$$

According to Eqn. 1, dotting the entity $\vec{C}$ on the right hand side by the vector $\vec{D}$ has the effect of mapping the vector $\vec{D}$ into a new vector $\vec{A}(\vec{B}\centerdot \vec{D})$ that has the same direction as the vector $\vec{A}$, but with a magnitude equal to the magnitude of $\vec{A}$ times the scalar $(\vec{B}\centerdot \vec{D})$. Similarly, according to Eqn. 2, dotting the entity $\vec{C}$ on the left hand side by the vector $\vec{D}$ has the effect of mapping the vector $\vec{D}$ into a new vector $(\vec{D}\centerdot \vec{A})\vec{B}$ that has the same direction as the vector $\vec{B}$, but with a magnitude equal to the magnitude of $\vec{B}$ times the scalar $(\vec{A}\centerdot \vec{D})$. So, by performing such operations, we can use the quantity $\vec{C}=\vec{A} \vec{B}$ to map an arbitrary vector $\vec{D}$ into a new vector featuring a different magnitude and direction.

The quantity $\vec{C}=\vec{A} \vec{B}$ is an example of a second order tensor. When a second order tensor is written exclusively in terms of just two vectors placed in juxtaposition, we call it a dyad. Sightly more advanced forms of second order tensors can be written as a linear combination of dyads (i.e., the linear sum of scalar constants times dyads). This is analogous to writing a vector (first order tensor) as a linear summation of scalar components times basis vectors. When we write a tensor as a linear combination of dyads, we call such a quantity a dyadic.

You now know all the mathematics necessary to apply second order tensors in mechanics.

I'm going to stop here and let you digest what I have written so far.

Chet

17. Aug 5, 2015

### rogerk8

Chet,

This is the first time here on physics forums people haven't told me to go study basics, thanks!

One more thing, you have also never supplied a link fo me to read (which is easy to produce and always contains more than bargained for), thanks!

This just means that I preferre to chat rather than be directed to (basic) links (but in the end I will probably have to study links anyway :) )

So once again, I'm honored that you now even take the time to try to explain second order tensors (vectors=first order tensors, my note) for me.

Amazingly, I actually think I understand a bit of what you are saying :)

What these dyads (two vectors, my note) or linear combination of dyads actually are (because there are no operators between them) is however an enigma to me but I understand that they are called tensors which could consist of more than linear combinations of just pairs of vectors, right?

$$\vec{C}\centerdot \vec{D}=\vec{A} \vec{B}\centerdot \vec{D}=\vec{A}(\vec{B}\centerdot \vec{D})\tag{1}$$

$$\vec{D}\centerdot \vec{C}=\vec{D}\centerdot \vec{A} \vec{B}=(\vec{D}\centerdot \vec{A})\vec{B} \tag{2}$$

In 1) A has the direction (and the magnitude is the B dot D (scalar, my note) times magnitude of A).

In 2) B has the direction (and the magnitude is the D dot A times magnitude of B).

While C was AB, we have in 1) dotted from the right releasing B dot D while vector A remains.

In a similar manner we have in 2) dotted from the left releasing D dot A while vector B remains.

A stupid question I have here is if for instance B dot D = D dot B or not?

But while it gives a scalar (and do not change sign like crossproduct) it should not matter, I think.

So maybe your example is just "pedagogic" :)

Here I quote you last remark:

I think I see this now.

I also see that D will never be more than a part of a scalar.

But it will change both direction and magnitude while either A or B has the direction of it all.

It strikes me now that the only directions actually are A or B, so its more of choosing direction while manipulating magnitude.

Am I right?

Best regards, Roger
PS
What makes the "release" I so unprofessionally call it?

18. Aug 5, 2015

### Staff: Mentor

As I said, there is no clearcut physical meaning to second order tensors. They only fulfill the their purpose in life when they are used to linearly map vectors into new vectors.
No. They are just linear combinations of vector pairs, and nothing more.

Yes, the dot product of two vectors is still independent of their order (I think the math guys call this commutative).
Don't read too much into these examples. My only purpose here was to (a) demonstrate how the mathematical manipulations are carried out and (b) to show how a tensor can be used to map a vector into another vector.

I don't understand this question.

If you are comfortable with what we have done so far, I'm ready to continue on to practical applications.

Chet

19. Aug 6, 2015

### rogerk8

Yes Chet, I think I am, so please continue :)

First, you may ignore that last question you did not understand, I think I understand now (no clearcut physical meaning to second order tensors...).

Second, where you say No... above, I think I may have been ahead of you (in a naive way) because I was thinking of even higher order tensors and I saw linear combinations of 3 or more vectors as part of a tensor. This is either totally wrong or my english is bad :)

I can't wait for your practical applications!

This will be fun!

Best regards, Roger
PS
When I saw your "Like" yesterday I sadly thought this nice chat was over :)

20. Aug 6, 2015

### Staff: Mentor

Wire In Tension
One of the things many students struggle with is the idea that the tension in a wire seems to have two directions, one to the left and one to the right. But that is only because the tension in a wire has a bidirectional (tensorial) nature to it (which we will now analyze).

Consider a wire laid out along the x-axis, featuring a tension of magnitude T. The tension in the wire (which is in equilibrium) is established by applying a force of $T\vec{i}_x$ to the right hand end of the wire and a force of $-T\vec{i}_x$ to the left hand end of the wire, where $\vec{i}_x$ is the unit vector in the +x direction. So the wire is in equilibrium. Now suppose we consider a cross section of the wire at location x0 along the wire, and ask the question "what is the force exerted by the part of the wire at x = x0+ on the part of the wire x = x0- at the cross section." We also ask the question "what is the force exerted by the part of the wire at x = x0- on the part of the wire x = x0+ at the cross section." We can do this with confidence using tensor analysis by writing the "tension tensor" $\vec{T}$ as
$$\vec{T}=T\vec{i}_x\vec{i}_x$$

What do you get if you dot this tensor with the unit vector in the + x direction, $\vec{i}_x$? What do you get if you dot this tensor with the unit vector in the - x direction, $-\vec{i}_x$?

Chet

21. Aug 6, 2015

### rogerk8

I'm amazed you think I can solve this but let me try (I'm bad at this PF syntax though, using TEX instead):

$$\vec{T}=T\vec{i}_x\vec{i}_x$$

If you dot this with

$$\vec{i}_x$$

from the right you get

$$\vec{T}=T\vec{i}_x$$

In other words, you get the tension in the +x-direction because

$$\vec{i}_x$$

is a unit vector which comes down to 1 while dotted with yet another unit vector in the same direction

Dotting with a negative unit vector will of course give a negative tension in the same manner.

This is either too simple or I haven't understood a thing :D

Best regards, Roger

22. Aug 6, 2015

### Staff: Mentor

Roger,

You've done exactly what I was hoping you would do, and you got the right answer. Now I'd like to generalize what you've found.

If you have an area cross section within a wire under tension, and you want to find the tensile force exerted by the material on one side of the cross section on the material on the other side of the cross section, you draw a unit normal vector from the "on" side of the cross section to the "by" side of the cross section and take the dot product of this unit normal with the "tension tensor." This "algorithm" will give you the results that obtained above.

This approach is very important. Does it make sense?

Chet

23. Aug 7, 2015

### rogerk8

Chet,

I actually think it does.

Best regards, Roger
PS
I was very pleased hearing you say that I got the right answer :)

24. Aug 7, 2015

### Staff: Mentor

For the case of a wire in tension, we can also express the "tension tensor" as follows:
$$\vec{T}=T\vec{i}_x\vec{i}_x=A\left(\frac{T}{A}\right)\vec{i}_x\vec{i}_x$$
where A is the cross sectional area of the wire. The quantity T/A in this equation is called the "tensile stress" in the wire. So another way of writing this is:
$$\vec{T}=T\vec{i}_x\vec{i}_x=A(σ_{xx}\vec{i}_x\vec{i}_x)=A\vec{σ}$$
where $\vec{σ}$ is called the stress tensor (for the special case of tension in a wire) and $σ_{xx}=T/A$ is called the x-x component of the stress tensor.

Please note that the stress tensor $\vec{σ}$ is a more general quantity than the tension tensor $\vec{T}$ (which applies only to the case of a wire, or a bar under pure uniaxial loading). Therefore, we will be working exclusively with the stress tensor from this point on in our development. This will be particularly useful when we consider the pressure p within a fluid or a gas.

The way that we apply the stress tensor in more general situations is very much analogous to the way we outlined for the tension tensor in a rod:

If you have a differential element of surface area dA of arbitrary spatial orientation within a liquid or a gas (or at the solid boundary of a liquid or gas) and you want to find the tensile force exerted by the material on one side of the area element on the material on the other side of the area element, you draw a unit normal $\vec{n}$ from the "on" side to the "by" side of the area element, take the dot product of the stress tensor $\vec{σ}$ with the unit normal, and then multiply it by the area dA. Thus, the algorithm for getting the force acting on the area element is:
$$\vec{dF}=(\vec{σ}\centerdot \vec{n}) dA$$
This equation is called the Cauchy stress relationship. It is extremely powerful and important. It tells us that the vector force per unit area on a surface is equal to $\vec{σ}\centerdot \vec{n}$. So all we need to do to get the force per unit area is figure out how to represent the unit normal vector to a given surface in component form.

Before I stop for now, let me give you one more thing to think about. For a liquid or gas in hydrostatic equilibrium, the stress tensor can be expressed in component form by the dyadic sum
$$\vec{σ}=-p\vec{i}_x\vec{i}_x-p\vec{i}_y\vec{i}_y-p\vec{i}_z\vec{i}_z$$
where p is the local pressure.

Chet

25. Aug 7, 2015

### rogerk8

Chet,

You write very interesting and well and I love the way you are trying to teach me!

I have now read your nice post three times and I thought for a wile that I should not answer until tomorrow.

But life is short :)

I think that the notation "x" for all the components in $\vec{T}$ is just an example how the stress tensor may be designed, right?

Looking at

$$\vec{dF}=(\vec{σ}\centerdot \vec{n}) dA$$

It is obvious that there has to be a normal unit vector for dA (called $\vec{n}$).

Pressure is dF/dA so we have

$$p=\vec{σ}\centerdot \vec{n}$$

left.

Your last example kind of defines the stress tensor (at least its complexity).

Here I quote you:
This is hard to understand.

Because how can we do that in a gas?

If the walls were stiff we will easily have normals to them.

But in my dream, we have no walls.

I don't know if these reflections are that good but at least I tried :)

MVH/Roger