Pressure exerted by an ideal gas

[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:

 

[PainterGuy]

Hi again,

Everything is kept the same other the length of pipe in this setup. The water enters from the left and exit on the right end. In Fig. 1 the length of pipe is 1 km and in Fig. 2 it is 2 km. Do you think that the length of pipe would affect the height of water column in Tower 1, Tower 2 and Tower 3 in Fig. 2?

Moreover, in my book the Bernoulli's equation is given as P1 + ½ ρv₁^2 + ρgh₁ = P2 + ½ ρv₂^2 + ρgh₂ where P1 and P2 are pressure. If h₁ and h₂ are the same then the factors ρgh can be neglected. In the view of this equation, I don't see why the height of p1 and p3 in this picture should differ. Could you please guide me? Thanks.

 

[jbriggs444]

Moreover, in my book the Bernoulli's equation is given as P1 + ½ ρv₁^2 + ρgh₁ = P2 + ½ ρv₂^2 + ρgh₂ where P1 and P2 are pressure. If h₁ and h₂ are the same then the factors ρgh can be neglected. In the view of this equation, I don't see why the height of p1 and p2 in this picture should differ. Could you please guide me? Thanks.

https://en.wikipedia.org/wiki/Bernoulli's_principle:

"The following assumptions must be met for this Bernoulli equation to apply:[9]

• the flow must be steady, i.e. the fluid velocity at a point cannot change with time,
• the flow must be incompressible – even though pressure varies, the density must remain constant along a streamline;
• friction by viscous forces has to be negligible."

Can you think of an assumption above which is not always realistic?

 

[Herman Trivilino]

What really troubles me is that if a molecule makes collisions with other molecules on its journey back and forth, it would affect the force exerted on wall drastically.

When two identical molecules have a perfectly elastic collision they simply trade momenta. So the molecule that you called molecule_1 in your later post switches places, in effect, with the one you called molecule_2 and they go on their merry way, as if there were no collision at all and they had traded identities. It has no effect on the average force exerted on the wall.

 

[PainterGuy]

Thank you.

All of those three assumptions are ideal.

But in this setup do you think that the length of pipe would affect the height of water column in Tower 1, Tower 2 and Tower 3 in Fig. 2? Everything is kept the same other the length of pipe in this setup. The water enters from the left and exit on the right end. In Fig. 1 the length of pipe is 1 km and in Fig. 2 it is 2 km.

I do understand that if the location of Tower 3 is moved to point A in Fig 2, there will be be further decrease in the height. Thanks.

 

[jbriggs444]

All of those three assumptions are ideal.

Right. In real life, one or all of them may be violated. The effects of the violation may or may not be negligible. It depends on how much error you are willing to neglect.

But in this setup do you think that the length of pipe would affect the height of water column in Tower 1, Tower 2 and Tower 3 in Fig. 2?

It depends on how much error you are willing to neglect. And pipe diameter. And pipe length. And flow rate.

 

[PainterGuy]

Thank you.

In both cases, Fig 1 and Fig 2, error tolerance is the same. Also the pipe diameter and flow rate is the same in both cases. The only difference is length - in Fig 1 it's 1 km and in Fig 2 it's 2 km. In comparative terms, would it affect the height of water column in Tower 1, Tower 2 and Tower 3 in Fig. 2 compared to Fig 1? Thank you.

 

[jbriggs444]

In both cases, Fig 1 and Fig 2, error tolerance is the same. Also the pipe diameter and flow rate is the same in both cases. The only difference is length - in Fig 1 it's 1 km and in Fig 2 it's 2 km. In comparative terms, would it affect the height of water column in Tower 1, Tower 2 and Tower 3 in Fig. 2 compared to Fig 1? Thank you.

There is no way to know from the limited information that you have provided.

Given the flow rate that you have not specified, the pipe diameter that you have not specified and the viscosity of water which you can look up, how much pressure drop would the Poiseuille equation call for?

https://en.wikipedia.org/wiki/Hagen–Poiseuille_equation

 

[PainterGuy]

Thank you.

The Hagen–Poiseuille equation is ΔP=(8μLQ)/(πR⁴) and ΔP is is the pressure difference between the two ends. I think that "the two ends" don't refer to the two ends of pipe. Could ΔP be defined as the pressure difference between two points along the pipe? Thanks.

 

[jbriggs444]

The Hagen–Poiseuille equation is ΔP=(8μLQ)/(πR⁴) and ΔP is is the pressure difference between the two ends. I think that "the two ends" don't refer to the two ends of pipe. Could ΔP be defined as the pressure difference between two points along the pipe?

Yes.

You seem strangely resistant to providing numbers or doing calculations. Say you drop three diameters off of each end of your one kilometer pipe. Can you apply Poiseuille's equation to that length?

 

[PainterGuy]

Thank you.

Is it okay? Please let me know. Thanks.

 

[PainterGuy]

Hi @jbriggs444,

Could you please confirm if the calculation is correct? Thank you.