4
$\begingroup$

Containers holding gas at a high pressure don't slowly lose the internal energy of the gas. It seems like the high speed particles would collide with the metal walls and slowly transfer their energy to the slower particles outside the container.

Even if the pressure is from more particles in the container, they can do work when released so they have energy. Shouldn't that energy dissipate over time?

$\endgroup$
2
  • 4
    $\begingroup$ But outside the container it is rather hot! $\endgroup$ Jul 9, 2016 at 15:41
  • 1
    $\begingroup$ When they fill scuba tanks with compressed air, they do it under water to prevent the tank overheating. So the metal walls do heat up. $\endgroup$
    – user108787
    Jul 9, 2016 at 15:47

5 Answers 5

9
$\begingroup$

Gases in containers at high pressures have those pressures because there are more molecules in them than in the same container at atmospheric pressure, not because there is a difference between the molecular energies. At the same temperature, two containers with different numbers of molecules in them have the same probability distribution of energies.

The pressure difference is owing to the difference between the collision frequencies with the walls in each case, the collision frequency being proportional to the number of molecules.


Question from OP

Even so shouldn't the stored energy dissipate over time? You can let the high pressure gas out to do work so it is stored energy.

Firstly, it seems that you may be confusing temperature with concentration of energy (energy per volume). Temperature is wholly about the probability distribution of a system's particles, not about how much total energy there is. I'll try the following argument to try to show why it is temperature difference between the gas and its surroundings, and not the concentration of energy, that determines whether energy escapes through heat flow, which is the only way it can escape if the bottle is sealed.

Think of things from one particle's standpoint. From time to time it bumps into other gas particles, and also into the thermalized particles that make up the bottle walls. Sometimes these particles will have more energy than our lone particle, sometimes less. But, over the long term, the expected rate of transfer of energy from the particle is nought - that's what we mean, by definition, when we say that the system is at thermodynamic equilibrium. This zero expected rate depends wholly on the probability distributions of the system particle energies, it does not depend on how often the particles collide. If there were only one gas particle in the bottle (so you had a very hard vacuum), its mean kinetic energy would be set by the kinetic energies of the particles making the bottle wall up: it would reach a point where a collision with the wall were equally likely to lose or gain energy. And that expected energy would be the mean energy of the particles in the wall. Energy cannot simply come rushing in because it is more concentrated in the walls, the transfer is governed by stochastic, passive processes.

$\endgroup$
7
  • $\begingroup$ Addition: Pressure could be high because of 2 reasons either more no. of molecules or higher temperature/kinetic energy. The first case has been explained. Case 2: If they had different molecular energies the temperature would have been different and then the kinetic energy of the gas would be transferred to the container to equalize the temperature. In that case the gas would loose its internal energy and cool down. $\endgroup$
    – oshhh
    Jul 9, 2016 at 16:03
  • $\begingroup$ This is a great answer and it definitely needs the above comment to be complete. Thanks, @OsheenSachdev. $\endgroup$
    – BoddTaxter
    Jul 9, 2016 at 20:09
  • $\begingroup$ Even so shouldn't the stored energy dissipate over time? You can let the high pressure gas out to do work so it is stored energy. $\endgroup$
    – BoddTaxter
    Jul 9, 2016 at 20:38
  • $\begingroup$ @AdamG See my update. $\endgroup$ Jul 10, 2016 at 2:28
  • 1
    $\begingroup$ @AdamG If there is no nozzle, and no outlet valve, then yes the energy concentration would linger on forever. This is a system in a local energy minimum, so its like a spring held sprung by a clamp, or like an atomic nucleus other than $^{56} {\rm Fe}$. Supply enough energy to those systems by cutting the clamp or fusing / fissioning as appropriate and the system can get to a lower energy state. But they need outside energy to raise them out of their local minimum states before they can do that. $\endgroup$ Jul 10, 2016 at 3:45
2
$\begingroup$

There are two ways you can change the internal energy of a gas, one is macroscopic, that is, performing work on or by the gas, if the gas either expands or contracts. The other is microscopically through heat. If the compressed gas is at the same temperature than the outside gas, these microscopic collisions will not result in an exchange of energy, because the speed distribution of the particles is a function of the temperature, not of the pressure, so it is not correct to assume that the particles inside the container are faster than those outside, and thus there will be no net transfer of energy.

$\endgroup$
1
$\begingroup$

It seems like the high speed particles would collide with the metal walls and slowly transfer their energy to the slower particles outside the container.

The mechanism you describe is correct, but you have to keep in mind that average kinetic energy, $\langle K \rangle$, is only proportional to temperature:

$$\langle K \rangle = \frac 3 2 N k T$$

So (since the volume is fixed and therefore macroscopic work is ruled out) there will be a net energy transfer only if the temperature inside the vessel is different from the temperature of the outside environment and if heat transfer is allowed (i.e. if the walls of the containers are not adiabatic).

$\endgroup$
1
$\begingroup$

re: "Why don't high pressure gases stored in containers lose energy?"

They can gain & lose energy:

Energy (heat) is lost from a gas as the gas is compressed (whether thru mechanical compression or thru cooling compression (e.g. passing a gas thru a tube that is immersed in a very cold liquid -- like liquid nitrogen).

Energy (heat) is gained by a gas when a very cold gas is transferred to a warmer tank (e.g. especially from a high pressure tank to a lower pressure tank that isn't super insulated).

Energy (heat) can also be lost or gained while a gas is stored in a tank as ambient external temperatures fluctuate.

$\endgroup$
0
$\begingroup$

This question has not been answered and I understand why the original questioneer still asks the question. I don't know why people are bringing temperature into the picture. The question is, based on the principle of the conservation of energy, energy imposed on and object must go somewhere. So for example if a person tries to push a refrigerator with insufficient force, say it weighs 100 pds and he only pushes with 80 pds of pressure we ask where does all that energy go if the fridge is not budging? The answer is it works against the person himself in the form of muscle tension, chemical reactions, and heat. The problem becomes where does all the energy the the gas is exerting against the walls of the container go? And we know no energy is lost since a metal canister under 50,000 psi from a gas will retain the same amount of energy after years of sitting there. Shouldn't the energy have to go somewhere since it is pushing against the walls like a person pushing against a fridge which does not move??? I had the same question and don't have the answer. Everyone here is trying to answer and they have no idea what they are talking about!

$\endgroup$
1
  • $\begingroup$ This does not really answer the question. If you have a different question, you can ask it by clicking Ask Question. To get notified when this question gets new answers, you can follow this question. Once you have enough reputation, you can also add a bounty to draw more attention to this question. - From Review $\endgroup$
    – Miyase
    Oct 20, 2023 at 8:37

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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