Pressure exerted by an ideal gas

[sophiecentaur]

To understand how particle collisions take place and what the kinetic theory of ideal gases is about, I recommend to read the textbook "PHYSICS" by David Halliday and Robert Resnick. There is a chapter about collisions between particles in two or three dimensions and there is a chapter about the kinetic theory of gases and the microscopic definition of an ideal gas.

I don't have that book but I do have several A Level Physics textbooks and I have a Degree Level Thermodynamics book (Zemanski) somewhere in a cupboard. Collision between two spheres was something we did when I was at School and I would be surprised if you could find anywhere where it says that the final speeds become nearer equal than the initial speeds. Can you photograph a page where that is written?
The derivation of the simple gas laws is not hard and it does not result in all molecules acquiring the same speed - which is what I challenged you one initially. (Perhaps you didn't mean what you wrote in that sentence). The distribution of mean square velocities will remain appropriate for the temperature of the gas and container.

 

[Lord Jestocost]

...which is what I challenged you one initially. (Perhaps you didn't mean what you wrote in that sentence)...

See my comment#31

 

[sophiecentaur]

See my comment#31

Jeez, I had to read that several times but, I can go along with it in a practical case. The problem with the ideal gas case is that there can be no collisions between molecules because, by definition, they have zero volume so they will not collide - the walls are the only thing that they collide with so they will each remain with the same kinetic energy.

 

[Lord Jestocost]

Jeez, I had to read that several times but, I can go along with it in a practical case. The problem with the ideal gas case is that there can be no collisions between molecules because, by definition, they have zero volume so they will not collide - the walls are the only thing that they collide with so they will each remain with the same kinetic energy.

From https://en.wikipedia.org/wiki/Ideal_gas

The ideal gas model depends on the following assumptions:

• The molecules of the gas are indistinguishable, small, hard spheres
• All collisions are elastic and all motion is frictionless (no energy loss in motion or collision)
• Newton's laws apply
• The average distance between molecules is much larger than the size of the molecules
• The molecules are constantly moving in random directions with a distribution of speeds
• There are no attractive or repulsive forces between the molecules apart from those that determine their point-like collisions
• The only forces between the gas molecules and the surroundings are those that determine the point-like collisions of the molecules with the walls
• In the simplest case, there are no long-range forces between the molecules of the gas and the surroundings

From [PDF]Here - University of Southampton

Think about a single molecule inside a gas which is in thermal equilibrium. Our chosen molecule continually suffers collisions; sometimes it gains energy from a collision and rebounds at high speed; sometimes it is briefly brought almost to a standstill. Our molecule (the system) is continually exchanging energy with the gas (the reservoir) - we say that its microstate is changing, even though in the macroscopic world all the measurable physical properties of the gas as a whole stay the same in time. Although we cannot predict the velocity of our molecule at any moment, we can say precisely that the relative probability of a state of velocity v is given by the Boltzmann distribution...

 

[PainterGuy]

Thank you.

As I learn more I believe that my ability at using algebra and abstraction would improve. I understand your point but I didn't even attempt to use any numbers or symbols because I didn't think there was any need for it. My question was general, and at the same time some other members were already able to help me in previous posts. Nonetheless, your point is well taken and I agree that it does help to formulate and track the information if proper symbols and numbers are used.

@Lord Jestocost: The balance would measure the weight of gas plus the mass atop the piston. The piston imparts the velocity along a single axis but with time all this 'additional' velocity gets distributed along all the axes assuming the piston is thermally insulated. Do I make any sense? Thank you.

 

[jbriggs444]

The balance would measure the weight of gas plus the mass atop the piston. The piston imparts the velocity along a single axis but with time all this 'additional' velocity gets distributed along all the axes assuming the piston is thermally insulated. Do I make any sense? Thank you.

Momentum is strictly conserved in every direction independently. It does not redistribute in any other direction. The piston imparts momentum into the system. If there is no countering force, the system gains momentum which cannot be lost in any other direction.

 

[PainterGuy]

Thank you.

I do understand that the momentum should be conserved in every direction. Doesn't the piston impart velocity to the molecules along a single axis?

If there is no countering force, the system gains momentum which cannot be lost in any other direction.

Could you please elaborate on it a little? Thank you.

 

[jartsa]

Let us consider a box-shaped tank containing gas. And momentum.

So, every wall is absorbing momentum from the gas, an alternative way to say that is that every wall is giving momentum to the gas. Yet another way to say that is to say that the gas pushes the wall, or the wall pushes the gas.

Each wall can push the gas only in one direction, because there is no friction between the wall and the gas.

So, if the 'ceiling' wall gives the gas some momentum, that momentum can be absorbed by the 'floor' wall, but not the side walls.

 

[PainterGuy]

Thank you.

So, if the 'ceiling' wall gives the gas some momentum, that momentum can be absorbed by the 'floor' wall, but not the side walls.

So, doesn't this imply that the molecules will have greater velocity in that particular direction? Greater velocity along a certain direction also implies a larger pressure along that direction. Thank you for your help.

 

[jartsa]

So, doesn't this imply that the molecules will have greater velocity in that particular direction? Greater velocity along a certain direction also implies a larger pressure along that direction. Thank you for your help.

Let's consider gas molecules that don't collide with each other, in a box-shaped container that has perfectly smooth walls. In this special case it is true that extra vertical pressure stays extra vertical pressure forever. Are we interested about this case?

The above situation is very abnormal. Normally pressure becomes isotropic very quickly. Like when walls are not smooth. A rough side wall exerts vertical forces on gas molecules, not on gas, but on individual gas molecules.

 

[Dadface]

Thank you.

I do understand that the momentum should be conserved in every direction. Doesn't the piston impart velocity to the molecules along a single axis?.

Yes it does. The molecules bounce off of the advancing piston with increased velocity and as a result the temperature of the gas increases. The amount of change depends on how the compression is carried out, for example the quicker the compression the greater the velocity change.

 

[PainterGuy]

Thank you.

Let's consider gas molecules that don't collide with each other, in a box-shaped container that has perfectly smooth walls. In this special case it is true that extra vertical pressure stays extra vertical pressure forever. Are we interested about this case?

The above situation is very abnormal. Normally pressure becomes isotropic very quickly. Like when walls are not smooth. A rough side wall exerts vertical forces on gas molecules, not on gas, but on individual gas molecules.

You have rightly understood my question. The first case with ideally smooth walls is not practical because as you say that extra pressure would stay like that so I'm interested in the second case.

Now I could rephrase my original question like this that what's the reason that pressure would become isotropic? I understand that the even the majority of molecules which run into the moving piston at 90° won't reflect back at 90° due to the bumpiness of piston surface. For example, if you throw a ball at 90° toward a smooth wall, it would reflect back along the incident path at 90°. But if the wall has a slope as shown in the attachment, the ball won't bounce back along the incident path. Would you like to elaborate on it a little more like why pressure becomes isotropic? Thank you.

 

[jartsa]

Now I could rephrase my original question like this that what's the reason that pressure would become isotropic? I understand that the even the majority of molecules which run into the moving piston at 90° won't reflect back at 90° due to the bumpiness of piston surface. For example, if you throw a ball at 90° toward a smooth wall, it would reflect back along the incident path at 90°. But if the wall has a slope as shown in the attachment, the ball won't bounce back along the incident path. Would you like to elaborate on it a little more like why pressure becomes isotropic? Thank you.

Well I don't know, but let's consider a piston and a cylinder that look like this as seen from the side: |V|

And then let's ask some questions:

1: what kind of momentum are we giving to the gas when we push the piston down?

2: what kind of speed are we giving to the molecules when we push the piston down?

3: Some question about pressure when we push the piston down

 

[jartsa]

Well I don't know, but let's consider a piston and a cylinder that look like this as seen from the side: |V|

And then let's ask some questions:

1: what kind of momentum are we giving to the gas when we push the piston down?

2: what kind of speed are we giving to the molecules when we push the piston down?

3: Some question about pressure when we push the piston down

1: Downwards momentum. Sideways momentums cancel out each other.

2: Molecules get extra speed into the direction of the normal of the surface, if the surface is perfectly smooth, if rough then all directions.

3: Molecules get more speed so pressure increases. And volume decreases so pressure increases for that reason too. Now I should say something about why pressure increases in all directions ... well if the molecules don't collide with each other then the molecules are like light. So if we shine a light beam into container with perfectly smooth walls and this shape: |V| what happens? ... Well the beam can't be perfectly collimated, so the initial randomness is amplified as the beam bounces around, and the light becomes diffuse very quickly.

 

[PainterGuy]

1: Downwards momentum. Sideways momentums cancel out each other.

2: Molecules get extra speed into the direction of the normal of the surface, if the surface is perfectly smooth, if rough then all directions.

3: Molecules get more speed so pressure increases. And volume decreases so pressure increases for that reason too. Now I should say something about why pressure increases in all directions ... well if the molecules don't collide with each other then the molecules are like light. So if we shine a light beam into container with perfectly smooth walls and this shape: |V| what happens? ... Well the beam can't be perfectly collimated, so the initial randomness is amplified as the beam bounces around, and the light becomes diffuse very quickly.

Thank you. I was trying to understand Bernoulli's equation from conceptual point of view and was stuck at that point where I couldn't understand how pressure became isotropic. Thanks for helping to come over that obstacle.

 

[PainterGuy]

Hi

The equation PV=nRT for ideal gases is also applicable in an atmosphere free from gravity. How do we find the pressure exerted by a liquid in an atmosphere free from gravity? The formula ρgh cannot be used because it uses gravitational constant. Thank you.

 

[jbriggs444]

The equation PV=nRT for ideal gases is also applicable in an atmosphere free from gravity. How do we find the pressure exerted by a liquid in an atmosphere free from gravity? The formula ρgh cannot be used because it uses gravitational constant. Thank you.

This liquid is just sitting there, stationary?

How much pressure are you exerting on it? That's how much pressure it is exerting on you.

 

[PainterGuy]

Thank you.

Yes, the liquid is just sitting there stationary.

Let's start like this. There is a 1-liter container. The container is filled up with water to the maximum. The water molecules are hitting the faces of container just like gas molecules and this motion of the molecules is not dependent on the gravity. When this container is put in an environment free from gravity, what pressure will the water molecules exert on the container faces? I'm mentally visualizing water molecules moving around like gas molecules except that the water molecules need to stick together into a single lump. For example, mean free path for water molecules is 0.2 nm while for the air molecules at 25 C and 760 mm Hg pressure is almost 70 nm. Where am I going wrong? Thanks.

 

[sophiecentaur]

I couldn't understand how pressure became isotropic.

That idea is totally correct for a gas not under gravity; in a container in deep space.
It only works for 'small volumes' of a gas under gravity. Imagine a 1m³ cube at sea level. Let it be in a cuboidal bubble of massless material, to make it easier. That cube of air will have a mass of about 1.2kg (weight of 12N) and the weight will be acting downwards, spread over the 1m². So the pressure on the lower face will be 12Pa greater than the pressure acting on the upper face. The pressure inside the cube will vary by 12Pa! so it's not actually isotropic but what is the fractional difference if the overall pressure is 100kPa?? (=12/10⁵! Could you easily measure that (the answer is 'only just')? Now consider a 1l cube; the pressure difference would be 1.2Pa. The limit, as the cube gets smaller and smaller is Zero so, at a given point, the pressure can be said to be the same in all directions (isotropic) because the 'bubble' has no volume..

 

[sophiecentaur]

Hi
The equation PV=nRT for ideal gases is also applicable in an atmosphere free from gravity. How do we find the pressure exerted by a liquid in an atmosphere free from gravity? The formula ρgh cannot be used because it uses gravitational constant. Thank you.

Liquids are the hardest to consider; the pressure exerted by an ideal solid, floating in weightless conditions, is Zero because the forces of repulsion and attraction are in equilibrium (if it is at constant temperature). I think the best description of the 'pressure' at the surface of a liquid a liquid would be the Vapour Pressure; the surface molecules are constantly leaving due to their kinetic energy so, if you contain it in a bubble, the bubble will continue to expand and lose energy (lowering its temperature) so it wouldn't be an equilibrium situation. With a warm liquid bubble in an atmosphere, the pressure will exceed the atmospheric pressure by its vapour pressure until it has reached equilibrium through loss of its internal energy.
I think this is a topic that requires one to think around all the possible situations oneself and to identify the details of the substance and its environment. A sort of Private, Internal PF Thread. The answers to many of those 'what if' questions can best be found within your own mind. It can make the brain ache, of course.

 

[Dadface]

I think in the absence of gravity and other external forces the forces between the molecules themselves will be dominant and the water will tend to take a spherical shape. Surface tension theory can be used to calculate the pressure difference across the water surface.

 

[Nidum]

Look up the videos showing water drops on ISS .

 

[jbriggs444]

here is a 1-liter container. The container is filled up with water to the maximum.

Statically indeterminate.

If an ideal rigid container is filled completely with an ideal incompressible liquid, the pressure on the container walls is not determined.

Edit: If the container were to shrink by any finite amount, the pressure would become infinite. If the liquid were to expand by any finite amount, the pressure would become infinite. If the container were to grow by any finite amount, a void would be created. If the liquid were to shrink by any finite amount, a void would be created.

 

[sophiecentaur]

I think in the absence of gravity and other external forces the forces between the molecules themselves will be dominant and the water will tend to take a spherical shape. Surface tension theory can be used to calculate the pressure difference across the water surface.

Evaporation will occur for as long as the vapour pressure in the surrounding region of air is lower than the vapour pressure of the droplet.
Like I said, liquid is a difficult state to understand. I guess that the vapour pressure of some solids would need to be considered under some conditions.

 

[PainterGuy]

Thank you, everyone.

Now I can imagine that why it's difficult to determine the pressure exerted by a liquid on the walls of a container in an environment free from gravity assuming that the liquid 'just' touches the walls. I wouldn't say that no pressure would be exerted by the liquid when it's 'barely' touching the walls of container, it's just that the pressure is indeterminate.

I really appreciate your help.

 

[PainterGuy]

Please have a look on the attachment. In Fig 1 the water is stationary and has an equal level all along. As they say that water seeks its own level and the reason for this is the fluid pressure due to gravity.

In Fig 2 the water is flowing from left to right. The height of column water in the tubes vary which represents pressure differential because of varying speeds.

Is the height of water in the tubes affected by whether the tubes are exposed to atmospheric pressure or have vacuum?

I believe that it does matter. The atmospheric pressure finds its way everywhere even in underground water reservoirs. The flowing water in the pipe is already affected by atmospheric pressure. Therefore, the pressure at point C is only due to the weight of column of water and the atmospheric pressure gets canceled with each other. Please see Fig 2.

The water in a column is raised to almost 10 m height by atmospheric pressure therefore when the tubes have a vacuum, their height should be enough to compensate for 10 m height plus additional increase in height due to the pressure of flowing water.

Thank you.

 

[jbriggs444]

Is the height of water in the tubes affected by whether the tubes are exposed to atmospheric pressure or have vacuum?

If you increase the pressure of both tubes by 1 atmosphere and the pressure in the water at all points by 1 atmosphere, why would you expect anything to change?

 

[PainterGuy]

Thank you.

I don't understand your statement but I see that my own wording was confusing in previous post so let me elaborate.

The atmospheric pressure finds its way everywhere around us even in deep underground water reservoirs. I'm saying this because one could think that water in the pipe is not affected by atmospheric pressure as if the atmospheric pressure cannot 'penetrate' inside the pipe.

I'm considering a case when the tubes are sealed at top and have vacuum. In such a case when the water is not flowing the height of column of water in those tubes would be almost 10 m. Because as I said that atmospheric pressure finds its way everywhere therefore the water inside a pipe is pressed by atmospheric pressure and it gets raised to the height of 10 m in both tubes.

A flowing fluid has its own pressure and is a function of fluid's speed along its path of flow as stated by Bernoulli's principle so when the water starts flowing, it exerts more pressure in region with larger cross section compared to smaller cross-sectional region. Let's assume that in larger cross-sectional region this 'flowing water' pressure is equivalent to 0.1 m height of column of water therefore total height of column of water in that region would be 10m+0.1m=10.1 m. Likewise, the smaller cross-sectional region could be assumed to have 'flowing water' pressure equivalent to 0.05 m height of column of water therefore total height of column of water in that region is 10m+0.05m=10.05 m. Do I have it right? Thanks a lot.

 

[jbriggs444]

"Flowing water pressure" is negative.

We could speculate that the water is flowing from a large reservoir on the left which is maintained at atmospheric pressure and flowing into a large reservoir on the right which is also maintained at atmospheric pressure. We could neglect the viscosity of water so that the flow can be maintained by a negligible pressure difference between the two reservoir.

Given that setup, the pressure in the wide pipe section under the left vacuum tower will be slightly less than atmospheric pressure. Bernoulli's principle tells us that this must be so. We might choose a flow rate and pipe diameter so that this pressure is "50 cm H20" less than atmospheric pressure. Similarly, we might choose a pipe diameter so that this same flow rate gives a pressure under the right hand vacuum tower that is "100 cm H20" less than atmospheric pressure.

In this case, the water height in the left column would be 50cm less than the nominal 10 meters and the water height in the right column would be 100 cm less.

Does that answer the question you were trying to ask?

Edit: Note that water pressure in pipes is not always equal to atmospheric pressure. What is depicted here are pipes.

 

[PainterGuy]

Thank you.

Yes, it did help and I believe that we are headed in the right direction.

Could you please elaborate a little on ""Flowing water pressure" is negative"? I had thought that the flowing water (or, fluid) has its own pressure. For example, if we have those pipes in an environment free from atmosphere even then as the water start flowing, the pressure can be detected.

Thanks.

 

[sophiecentaur]

""Flowing water pressure" is negative"?

That is negative relative with the ambient pressure. This is another issue and you should Google Bernouli Effect when you have put this lot to bed. We are dealing with static pressure at the moment.

 

[PainterGuy]

Thank you.

I do have conceptual understanding of Bernoulli effect but as you start looking at things from a different angle, you could gain more understanding about them.

Let's have a look at it again considering the setup from post #4 but assuming that there is no atmosphere and hence no atmospheric pressure. We can assume that a pump is pushing water from one reservoir to another reservoir. The water is flowing with a certain flow rate and both reservoirs are the same level. In the far left and right towers water would be at the same height and in the middle tower it would be at half the height. Difference in height signifies pressure differential and I believe that this is 'dynamic' pressure. Am I right? Thanks a lot for your help.

 

[Lord Jestocost]

Let's have a look at it again considering the setup from post #4

Have a look at this picture:

From: RWTH Aachen, Germany (https://web.physik.rwth-aachen.de/~.../Exscript/9Kapitel/IX2Kapitel.html#VersuchIX2)

Do you see the difference with regard to your sketch (vacuum doesn't matter in principle).

 

[PainterGuy]

Thank you.

This is the first time I have seen that the height varies along the pipe in areas even with the same cross section. This is the mostly encountered depiction. But it looks like that there is contradictory statement and wanted to verify with you. I was using Google translate. Doesn't the highlighted statement contradict the height shown for pipe 1 and pipe 3 in the figure on left because the height varies for both which means different hydrostatic pressure? Thanks for your help.

 

[Lord Jestocost]

Doesn't the highlighted statement contradict the height shown for pipe 1 and pipe 3 in the figure

The sentence means that the hydrostatic pressure in pipe 1 and pipe 3 is the same as in the reference case: