Work in isobaric processes

while researching, I found a lot of conflicting information about this topic. Let me explain my thought process so far and talk about an example I came up with.

An isobaric process is defined as one where the pressure inside a gas remains the same throughout a state change from state 1 to state 2. That would imply that dp=0, and the pressure inside the gas is constant. That would mean that in theory, isobaric processes should be possible, where the external pressure is not equal to the pressure of the gas itself.

For the definition of pressure-volume work, following integral is used:

Where it is specifically stated, that P is the pressure inside of the system, which I interpret as the pressure of the gas itself, but that could also be wrong.

Now that is where the problems begin. During free expansion of an ideal gas, there is no external pressure, against which the gas needs to expand. Assuming the the piston which held the gas in place before free expansion is massless, no work should be done by the gas.

But now we used P(ext) to calculate the total work done by the gas(which is zero); which pressure is it now, pressure of gas or the external pressure (on the other side of the piston) to calculate the gas work?

I read somewhere that we can visualize the pressure-volume work the gas does as some kind of fighting against an external resisting pressure, so W=-p(ext)*dV (negative sign convention will be used here.).

However, on Wikipedia, it states that p is the pressure of the gas and work can be represented by:

W=-pdV=-nR*dT (isobaric case)

Using the pressure of the system itself to calculate work doesn't make much sense to me to be honest. And it doesn't fit the description of work in a free expansion. However, to calculate work from a pV diagram for an isobaric process, we usually use p of the gas:

So here's an example. A molar ideal gas at 273.15K and at 1bar expands isobarically (conncected to heat source) against a constant pressure of 0.5bar to 2 times it's volume.

In this example, to calculate the work done by the gas, should we use the pressure of the gas (1bar), which means W=-10^5 Pa * (V2-V1), or should we use the pressure on the other side of the piston, which means W=-0.5*10^5 Pa * (V2-V1)? I really can't tell.

I reckon this question only applies to isobaric processes, as in isothermal reversible processes, pressure of gas and external pressure is always the same, and in isochoric processes, no work is done at all.

• I can help you figure this out on you own by asking you some leading questions. But stack-exchange frowns on the use of leading question. If you would like me to help you figure this out on you own, please resubmit this question at PhysicsForuns.com. I won't disappoint you. Commented Mar 23, 2022 at 20:17

An isobaric process is defined as one where the pressure inside a gas remains the same throughout a state change from state 1 to state 2.

That is only true for a reversible isobaric process, one that is carried out so slowly that the gas pressure is always in equilibrium with the external pressure. For an irreversible isobaric process the external pressure is constant but the gas has pressure and temperature gradients within. For an irreversible process the gas pressure is undefined, except at the boundary where it equals the external pressure (per Newton's 3rd law).

For the definition of pressure-volume work, following integral is used: Where it is specifically stated, that P is the pressure inside of the system, which I interpret as the pressure of the gas itself, but that could also be wrong.

The statement only applies to a reversible isobaric process where the gas is always in equilibrium with the surroundings. Otherwise, the pressure is only the external pressure. The more general integral is

$$W=\int_{V1}^{V2}P_{ext}dV$$

Where $$P_{ext}$$ is a constant external pressure. For a reversible process $$P_{ext}=P$$ where $$P$$ is the equilibrium pressure of the gas.

But now we used P(ext) to calculate the total work done by the gas(which is zero); which pressure is it now, pressure of gas or the external pressure (on the other side of the piston) to calculate the gas work?

By free expansion of an ideal I assume you mean expansion against a vacuum. Such a process is irreversible and therefore the pressure of the gas is undefined. No work is done against a vacuum.

However, on Wikipedia, it states that p is the pressure of the gas and work can be represented by:

W=-pdV=-nR*dT (isobaric case)

Wikipedia is referring to a reversible isobaric process for an ideal gas. The ideal gas equation only applies to an ideal gas in equilibrium. It does not apply during the free expansion of an ideal gas because the gas is not in internal equilibrium. It only applies to the initial and final equilibrium states.

Your PV diagram applies to both a reversible and irreversible isobaric process because $$P$$ is always the external pressure. Only for a reversible process is the pressure also the gas pressure.

Hope this helps.

• Wow, that actually cleared up all the confusion I had regarding this matter. One small question though; in my last physical chemistry exam (3rd semester) we were given a pV diagram of a gas and one step was an isobaric expansion of the gas; I just calculated the work done by gas as -pdV, where p was the pressure of the gas. Following your expanation, that would only apply to reversible processes. How could we decide whether an isobaric process is reversible or irreversible just looking at the pV diagram?
– 冰淇淋
Commented Mar 24, 2022 at 16:58
• @冰淇淋 If you are given a PV diagram you can always calculate the work done using the P of the diagram without worrying about whether the process is reversible. My only point was that you understand P may not be the gas pressure. Regarding determining whether the process is reversible or irreversible by looking at the pV diagram, it depends on the type of irreversible process Commented Mar 24, 2022 at 19:39
• If it involves a sudden drop in the external pressure, that would be evident on the diagram due to a vertical line. But if it involves irreversible heat transfer it wouldn’t be evident. If you can’t tell from the diagram, and you are not told the process is irreversible and how it is, then by default you can assume it is reversible Commented Mar 24, 2022 at 19:39
• To sum it up, whether or not the isobaric process is reversible is not so important for calculating work as it is for determining the entropy change of the surroundings. Commented Mar 24, 2022 at 19:40
• Thank you for the thorough answer.
– 冰淇淋
Commented Mar 24, 2022 at 21:48