Pressure in giant ball of water floating in space

Question

In this question:

Swimming in a ball of water in space

They seem to have reached the conclusion that the pressure inside a giant ball of water in space is zero.

Quoting the second answer:

As a conclusion, you would be swimming in a bubbling sphere of water, feeling no pressure at all, having a bit less difficulty in moving your arms and legs since the water would be full of bubbles, however I am assuming it would be harder to move around for the very same reason.

Quoting the first answer:

So, the bottom line is swimming in a big ball of water pretty much feels like swimming very slowly in space - until the ball of water gets big enough (2.68 km). Then it just feels like swimming in a giant pool on a distant planet. For practicality, the ball of water doesn't work, but the lunar swimming pool is awesome.

However, I'm still confused about one thing. So far as I know, although the cause of pressure here at the surface of the earth may be the gravitational weight of all the air above us (and the cause of more pressure deeper in the ocean is the weight of all the water above it) at the local level, pressure is really a cause of the random motion of the particles, smashing into the particles around them, and delivering their momentum.

After all, that's why rising temperatures cause the pressure to rise.

Assuming this giant ball of water can hold itself together due to cohesion, wouldn't you still feel the pressure from...well, simply the water molecules themselves, moving randomly in all directions?

---

Edit:

To help readers, and as a response to the criticism from the person who answered (criticism taken kindly, you're right, I should clarify. Although you could've said it in a nicer way...), this is why I'm asking this question:

I'm trying to understand pressure.

I'm trying to understand whether it originates fundamentally from having to "sustain" a fluid in place due to gravitational forces, or from something intrinsic to the fluid itself (its random motion.)

Although I know that at a local level, the random motion of particles is the cause of pressure, what I'm trying to understand is whether the random motion is caused by something else, or is intrinsic to the fluid itself.

In short, I want to know if there would be pressure without gravity, and this is the best thought experiment I could come up with to explain what I mean by that.

Thanks.

Answer 1

Assuming this giant ball of water can hold itself together due to cohesion, wouldn't you still feel the pressure from...well, simply the water molecules themselves, moving randomly in all directions?

This is a pretty unrealistic assumption, and showing what would happen should help explain how.

The cohesive forces allow for a surface tension, which can maintain a pressure difference between the sphere of water and the outside. The pressure difference due to surface tension between an inside and outside fluid and gas surface is known as the Laplace pressure. The Laplace pressure for a sphere is given by the equation $$\Delta P = \gamma \frac 2R$$

where $\Delta P$ is the pressure difference between the curved surfaces, $\gamma$ is the surface tension of the liquid, and $R$ is the radius of the sphere. We can assume in the vacuum of space that the external pressure is 0, so the value of $\Delta P$ will represent the total pressure inside the sphere, if we assume only the cohesive forces are acting.

Now if we look at the surface tension of water, $\gamma_{\text{water}}=71.97 \ \frac{\text{mN}}{\text{m}}$ (I'm assuming standard conditions to illustrate the point; but realistically due to reasons below, I don't think you can calculate the actual surface tension of liquid water in the vacuum of space) and the Laplace pressure equation, we can see part of the problem. Let's assume the sphere is 2 m in radius, since that is likely the smallest radius you could even consider it swimming. $$\Delta P = \frac {2}{2 \ \text{m}} \cdot71.97 \ \frac{\text{mN}}{\text{m}} = 71.97 \frac{\text{mN}}{\text{m}^2}$$

which is only $0.07197 \ \text{Pa}$. Atmospheric pressure is 1.4 million times greater (and it only gets lower with increasing radius unless you consider gravity). So to explain that aspect, if a giant ball of water could keep itself together through cohesion alone, it wouldn't really feel like any pressure at all to swim inside it.

But that probably doesn't solve all of your confusion, which relates to what I mentioned at the beginning. The unrealistic assumption is more that water would remain a liquid in these conditions at all. It cannot hold itself together due to cohesion, as liquid water at these pressures. It will want to change phases, as mentioned in the other answer. This will all depend on the thermodynamic effects of the fluid, not as much the cohesive effects. It should be pretty easy to see that at low pressure, (such as the vacuum of space with minimal cohesive force) you cannot even have a liquid phase of water. see here for an image

Answer 2

Let me first address the general issue that you raise: on how to understand pressure.

As we know, the macroscopic view and the microscopic view must corroborate each other. As I read your question that is what you are doing; you are trying to match the macroscopic and the microscopic view.

Matter consists of atoms, which means (as you point out) that transfer of pressure thoughout a medium happens in the form of atoms (or molecules) colliding with each other.

Case 1:
Water in pressurized environment, in weightlessness

The kind of footage is familiar: an astronaut in a space station allows some water to flow out a drinking bag, and a ball of water just floats there. That ball of water remains liquid because it is subject to air pressure from the surrounding air. The pressurized environment sustains the liquid state (The weightlessness is not a factor, it is just visually striking.)

Case 2:
Water in zero pressure environment.
To simplify, consider a very, very small droplet of water, just a couple of thousand water melecules. What happens when a droplet like that is released in a zero pressure environment? The droplet would instantanously expand into water vapor. Would a large droplet expand instantaneously too? Well, with a large droplet the inertia of the mass as a whole would come in as a factor.

Large amount of water case:
A ball of water, released in zero pressure environment. The surface area would instantaneously expand to water vapor. That first water vapor creates a shell around the ball of non-zero pressure that will temporarily sustain a liquid state of the remaining ball. Also, the liquid water and the water vapor will both become colder. The water vapor cools down because it is expanding, and the liquid water becomes colder because in the process of evaporation the liquid water is all the time losing its fastest molecules.

A comparison:
Have you seen demonstrations of the behavior of $CO_2$ at the pressure where it readily liquifies? You have a glass tube, inner diameter a couple of milimeters, length 10 centimeters or so, pure $CO_2$ inside, at high density, the tube is sealed. Above 31 degrees Celcius (about 90 Fahrenheit) all of the $CO_2$ is in gaseaous form. But below 32 degrees C. the $CO_2$ molecules are slow enough to form a liquid. The usual setup is that the tube has been filled to a density of $CO_2$ so that when conditions for liquid $CO_2$ are met then about half the length of the tube is showng liquid $CO_2$

I recommend you look up demostrations of that, and that you make sure you understand it. For instance this demo of supercritical CO2 by Ben Krasnow, who runs the Youtube channel Applied science

Now back to the water:
When liquid water is released to a zero pressure environment you really shouldn't think of it as a liquid anymore. For a brief period the inside of the ball of water would still have the density of liquid water, but that should be attributed to inertia. The evaporation front will travel from the outside of the ball to the center of mass at a rapid rate. (The process will slow down somewhat due to the ball surface and water vapor becoming ever colder.)

Answer 3

The misconception that is probably causing your confusion is that

at a local level, the random motion of particles is the cause of pressure,

Random motion of particles is measured by temperature; the higher the temperature, the more intense the random motion.

If we are to talk about causes, the cause of pressure on some wall is first and foremost mutual interaction of the particles and the wall. The fact that the particles move randomly is secondary. True, in gases increase of pressure often goes with increase in this random motion, because the increase of gas pressure can be done only by putting in substantial energy. But in liquids, it is possible to increase the pressure substantially with negligible amount of work and so with negligible change in intensity of this random motion.

Pressure of such liquid is due to force interaction of the particles with walls and each other, not necessarily due to their random motion. It suffices that particles push or pull each other. They do not have to move rapidly. You can have high pressure in very cold water or in ice cold at 1 K.

When pressure of a liquid water is increased, say, by moving a piston in a blocked syringe filled with water, water temperature increase is very small and is usually neglected.

Now to your question - gravity isn't necessary for pressure either. What is necessary to increase pressure is some other body that will squeeze the gas or liquid into smaller volume. On Earth, this body is the Earth with its gravity, but the same pressure is achieved in a closed vessel, such as the International Space Station, simply by making it robust enough to withstand the pressure and pushing in enough amount of gas. There is no effective gravity there, but there is pressure close to 100kPa, due to walls not allowing the gas to escape.

Answer 4

We know that ideal vacuum can not be created. The pressure in outer space is of the order 10^-17 Pa. Some how we put the ball of water there. We know that boiling point if water decreases with decrease in pressure, As the pressure decreased a lot boiling point decreases a lot. The boiling point is approximately -68 degrees celsius. But the temperature there is -273.15 decrees celsius. The difference is around 205 degrees celsius. This difference is greater than the difference at room temperature. That means it will become more denser. That mean the movement of water particles becomes less. However there will occur some collisions. Hence pressure is not zero in the water, but nearly negligible. Also cohesive forces exist...