Can the isothermal expansion of an ideal gas be irreversible?

[Philip Koeck]

For the reversible expansion of an ideal gas the heat flowing out of the surroundings and into the system is equal to the work done by the system. Since both system and surroundings have the same constant temperature the entropy increase of the system is equal to the entropy decrease of the surroundings giving a total entropy change of zero.
What would have to be different in an irreversible expansion and could this process still be isothermal?
An example I can come up with is: If the temperature of the surroundings is higher than that of the system then the entropy decrease of the surroundings is smaller than the increase in the system so the process is irreversible. However, I don't see how it still can be isothermal.

 

[Chestermiller]

Please describe in more detail the irreversible process you are envisioning.

 

[Philip Koeck]

Please describe in more detail the irreversible process you are envisioning.

A cylinder with a piston contains an ideal gas at temperature Ts.
If I now place this cylinder in thermal contact with an environment at temperature Te which is slightly higher than Ts it will expand and do work. At the same time heat will flow into the system.
I would say this is a spontaneous and therefore irreversible process.
We could also assume that this process proceeds very slowly so we can describe it as quasi-static.
However, this expansion can't be isothermal, since in the end the temperatures of system and environment have to become equal.
This is just one example of trying to imagine a process that is isothermal and quasi-static, but irreversible.
Could it be that all isothermal processes that are sufficiently slow have to be reversible?

 

[Chestermiller]

A cylinder with a piston contains an ideal gas at temperature Ts.
If I now place this cylinder in thermal contact with an environment at temperature Te which is slightly higher than Ts it will expand and do work. At the same time heat will flow into the system.
I would say this is a spontaneous and therefore irreversible process.
We could also assume that this process proceeds very slowly so we can describe it as quasi-static.
However, this expansion can't be isothermal, since in the end the temperatures of system and environment have to become equal.
This is just one example of trying to imagine a process that is isothermal and quasi-static, but irreversible.
Could it be that all isothermal processes that are sufficiently slow have to be reversible?

You are asking us to conceive of processes that are isothermal, quasi-static, but irreversible. I can't think of any except possibly for one that has friction between the piston and the cylinder (in a specially defined way, such as where the coefficients of static- and kinetic friction between the piston and cylinder are equal, so there is a smooth transition). Another case would be one in which chemical reaction is occurring within the gas.

 

[Philip Koeck]

You are asking us to conceive of processes that are isothermal, quasi-static, but irreversible. I can't think of any except possibly for one that has friction between the piston and the cylinder (in a specially defined way, such as where the coefficients of static- and kinetic friction between the piston and cylinder are equal, so there is a smooth transition). Another case would be one in which chemical reaction is occurring within the gas.

I want to stay away from reactions since I'm only thinking about an ideal gas, but I'll give the friction-case a go.
If I start with a frictionless, reversible, quasi-static, isothermal expansion the work done by the system (gas in the cylinder) is equal to the heat transferred from the environment to the system.
The inner energy of the system and thus its temperature are constant.
Now I'll add a little friction between the cylinder wall and the piston, but I still assume a very slow (quasi-static) expansion.
The system now has to do a bit of extra work to overcome the friction. Where does this work come from?
One possibility would be that the inner energy of the system decreases to provide the work, but then the expansion is not isothermal.
The other possibility is that the environment provides the energy for the extra work in form of extra heat going into the system. That would lead to an additional decrease of the environment's entropy compared to the reversible case. Not sure how that works. There would have to be some extra increase in entropy somewhere else to give an increase of the total entropy required for an irreversible process.

 

[Chestermiller]

I want to stay away from reactions since I'm only thinking about an ideal gas, but I'll give the friction-case a go.
If I start with a frictionless, reversible, quasi-static, isothermal expansion the work done by the system (gas in the cylinder) is equal to the heat transferred from the environment to the system.
The inner energy of the system and thus its temperature are constant.
Now I'll add a little friction between the cylinder wall and the piston, but I still assume a very slow (quasi-static) expansion.
The system now has to do a bit of extra work to overcome the friction. Where does this work come from?
One possibility would be that the inner energy of the system decreases to provide the work, but then the expansion is not isothermal.
The other possibility is that the environment provides the energy for the extra work in form of extra heat going into the system. That would lead to an additional decrease of the environment's entropy compared to the reversible case. Not sure how that works. There would have to be some extra increase in entropy somewhere else to give an increase of the total entropy required for an irreversible process.

My buddy @Robert Davidson and I have been working on ideal gas adiabatic expansion with friction included in the following thread: https://www.physicsforums.com/threads/thermodynamics-gas-expansion-with-piston-friction.963282/. Lately, Bob appears to have lost enthusiasm for continuing. But, you are welcome to join us. Extending this to contact with a constant temperature reservoir wouldn't be too complicated.