# Conceptual question about gas expanding against piston

## Main Question or Discussion Point

I'm trying to consolidate my understanding of gas thermodynamics. If anywhere my reasoning is wrong, please correct me.

Consider a cylindrical piston in which an ideal gas is sealed. The piston is well insulated, and it is assumed that all points in the gas are at temperature T and this temperature is fixed throughout the experiment.

Assume that the piston is weightless and frictionless. Also assume that there is no mass on the piston and that atmospheric pressure is 0.

Since the gas pressure is P, and the external pressure is 0, the gas will expand against the piston, and its volume will increase. Simultaneously, its pressure will decrease.

This process will technically occur for ever since the pressure P will never reach 0.

But what will be the work done by the gas from volume V1 (initial volume) to V2 assuming that the gas undergoes expansion ? Will it be positive or negative or even zero? What about the work done by the surroundings on the gas in this same process? Will it be positive or negative?

One argument tells me that the work done by the gas will be positive, because the gas pressure is directed against the piston, and the displacement of the piston is parallel to the gas pressure. Another argument tells me that the work done is zero, since the piston is weightless to begin with. Yet another argument tells me that the work done is positive, since the gas molecules themselves are being propelled in the direction of the gas pressure. Which of these is correct?

What about the work done by the gas as it expands from volum V1 to a volume of infinity? If this number can be expressed as a definite integral, does the integral diverge in this case?

These are all questions that are hindering but at the same time exciting my knowledge of physics. I appreciate all the help offered. Thanks!

BiP

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Andrew Mason
Homework Helper
I'm trying to consolidate my understanding of gas thermodynamics. If anywhere my reasoning is wrong, please correct me.

Consider a cylindrical piston in which an ideal gas is sealed. The piston is well insulated, and it is assumed that all points in the gas are at temperature T and this temperature is fixed throughout the experiment.

Assume that the piston is weightless and frictionless. Also assume that there is no mass on the piston and that atmospheric pressure is 0.

Since the gas pressure is P, and the external pressure is 0, the gas will expand against the piston, and its volume will increase. Simultaneously, its pressure will decrease.

This process will technically occur for ever since the pressure P will never reach 0.

But what will be the work done by the gas from volume V1 (initial volume) to V2 assuming that the gas undergoes expansion ? Will it be positive or negative or even zero? What about the work done by the surroundings on the gas in this same process? Will it be positive or negative?
Does the gas do work on the surroundings? How? If it does no work on the surroundings, what can you say about the internal energy of the (ideal) gas?

One argument tells me that the work done by the gas will be positive, because the gas pressure is directed against the piston, and the displacement of the piston is parallel to the gas pressure. Another argument tells me that the work done is zero, since the piston is weightless to begin with. Yet another argument tells me that the work done is positive, since the gas molecules themselves are being propelled in the direction of the gas pressure. Which of these is correct?
Neither. How can the gas do work on a massless, frictionless piston?

What about the work done by the gas as it expands from volum V1 to a volume of infinity? If this number can be expressed as a definite integral, does the integral diverge in this case?

These are all questions that are hindering but at the same time exciting my knowledge of physics. I appreciate all the help offered. Thanks!
Have a look at the thermodynamics of "free expansion".

AM

Does the gas do work on the surroundings? How? If it does no work on the surroundings, what can you say about the internal energy of the (ideal) gas?

Neither. How can the gas do work on a massless, frictionless piston?

Have a look at the thermodynamics of "free expansion".

AM
Thank you! So the gas does work only when it lifts a mass !

BiP

Andrew Mason
Homework Helper
Thank you! So the gas does work only when it lifts a mass !

BiP
A gas does work when it expands against an external force. It does not have to lift a mass to do that. Any external force will suffice.

AM

Some remarks:
"it is assumed that all points in the gas are at temperature T and this temperature is fixed throughout the experiment"
There is no such thing as "temperature at all points in the gas". Temperature is a global physical quantity, characterizing the whole ensemble of molecules and implying a certain velocity distribution of them. In microscopic physics temperature is associated with the intensity of molecular motion. But even in a thermally homogeneous medium there are fluctuations and some molecules move or vibrate faster or slower than others.

If the piston has no mass then from the conservation of momentum it follows that the piston will fly away instantaneously and at infinite speed as soon as the first molecule hits it. From that time the gas will expand freely in space without any counter force. As if the air in a room had all been confined in one of the corners and the separation wall had been suddenly disappeared. The air would spread out to fill the whole room. If you suppose that the temperature is set constant throughout the experiment then the ever increasing volume will cause an ever decreasing pressure. The work done by the gas due to expansion is covered by the energy you pump into it by heating the gas to maintain the constant temperature.

But the problem with this setup is that you cannot really follow it by using quasi-equilibrium thermodynamics. Yes, this is what you learn in high school or at universities until undergrad level. Abrupt and instantaneous changes (like the disappearing of the piston or the wall) do not fit into the framework of quasi-equilibrium thermodynamics. The expansion of the gas from the piston in vacuum is a very rapid process. In reality, supposing that you are capable of maintaining a constant temperature during this experiment is unphysical or irrational. It is rather an adiabatic process than any other, i.e. it happens without any heat transfer between the gas and its surroundings. The gas will expand and cool down very rapidly. If you ever saw a CO2 cartridge opened up, you know what I am talking about.

Some remarks:
"it is assumed that all points in the gas are at temperature T and this temperature is fixed throughout the experiment"
There is no such thing as "temperature at all points in the gas". Temperature is a global physical quantity, characterizing the whole ensemble of molecules and implying a certain velocity distribution of them. In microscopic physics temperature is associated with the intensity of molecular motion. But even in a thermally homogeneous medium there are fluctuations and some molecules move or vibrate faster or slower than others.

If the piston has no mass then from the conservation of momentum it follows that the piston will fly away instantaneously and at infinite speed as soon as the first molecule hits it. From that time the gas will expand freely in space without any counter force. As if the air in a room had all been confined in one of the corners and the separation wall had been suddenly disappeared. The air would spread out to fill the whole room. If you suppose that the temperature is set constant throughout the experiment then the ever increasing volume will cause an ever decreasing pressure. The work done by the gas due to expansion is covered by the energy you pump into it by heating the gas to maintain the constant temperature.

But the problem with this setup is that you cannot really follow it by using quasi-equilibrium thermodynamics. Yes, this is what you learn in high school or at universities until undergrad level. Abrupt and instantaneous changes (like the disappearing of the piston or the wall) do not fit into the framework of quasi-equilibrium thermodynamics. The expansion of the gas from the piston in vacuum is a very rapid process. In reality, supposing that you are capable of maintaining a constant temperature during this experiment is unphysical or irrational. It is rather an adiabatic process than any other, i.e. it happens without any heat transfer between the gas and its surroundings. The gas will expand and cool down very rapidly. If you ever saw a CO2 cartridge opened up, you know what I am talking about.
So how are you supposed to integrate the pressure with respect to the velocity?

BiP

Andrew Mason
Homework Helper
Some remarks:
If the piston has no mass then from the conservation of momentum it follows that the piston will fly away instantaneously and at infinite speed as soon as the first molecule hits it.
The term "massless" is understood to mean simply that the mass is negligible. Collisions between gas molecules and the piston do not result in the gas molecules losing energy.

From that time the gas will expand freely in space without any counter force. As if the air in a room had all been confined in one of the corners and the separation wall had been suddenly disappeared. The air would spread out to fill the whole room. If you suppose that the temperature is set constant throughout the experiment then the ever increasing volume will cause an ever decreasing pressure. The work done by the gas due to expansion is covered by the energy you pump into it by heating the gas to maintain the constant temperature.
Are you supposing that the room was initially empty or was it filled with air?

. In reality, supposing that you are capable of maintaining a constant temperature during this experiment is unphysical or irrational. It is rather an adiabatic process than any other, i.e. it happens without any heat transfer between the gas and its surroundings. The gas will expand and cool down very rapidly. If you ever saw a CO2 cartridge opened up, you know what I am talking about.
What if you had compressed helium in the cartridge and it was opened in a vacuum. Would it cool?

AM