# What determines whether a process is isothermal or isobaric?

This is motivated by a confusion surrounding the specifics of how isothermal processes actually occur.

In an isobaric process, if the gas is free to expand and heat is added, the pressure will stay the same (because the piston will move to the point where the pressure on either side is equal, so the net force is equal).

In an isothermal process, if the gas is free to expand and heat is added (but temperature is constant?), the gas will stay the same but the pressure will decrease.

Suppose you have a gas in a container with a frictionless piston on one end, so it can expand freely. On the other end is a heat reservoir. Is this isothermal or isobaric? On the one hand, the pressure MUST stay the same, or the piston would have a net force on it and the pressure would equalize again. On the other hand, the TEMPERATURE must stay the same, because the heat reservoir is keeping it heated to $$T_h$$.

All clarification is appreciated.

EDIT: If this process is neither strictly isobaric nor strictly isothermal, how does one achieve an isothermal or isobaric process? I came across this scenario as a example of both a fundamental isobaric process AND a fundamental isothermal process.

• why does it strictly have to be isothermal or isobaric? Oct 10, 2019 at 14:05
• I suppose it doesn't, necessarily, but it seems like this ideal scenario is given as the "base case" of sorts for both isothermal and isobaric processes. So if this is neither strictly, how does one achieve an isothermal or isobaric process? Oct 10, 2019 at 14:13
• In the isothermal process, you remove weights from the piston. Oct 10, 2019 at 14:13
• remember that true isobaric and isothermal processes don't exist in the real world - but you can get pretty close by allowing the interaction a really long or really short time to occur respectively Oct 10, 2019 at 14:33
• Oh I see; I made a stupid comment, ignore it if you can... Even the ideal gas law seems to be irrelevant to your question for one can say that in general one has $f(p,V,T)=0$ for some $f(.)$ function of the internal variables. Then what seems to be left is that keeping both boundary conditions on, $p_{ext}=const$ and $T_{ext}=const$, you must violate at least one of the conditions of either $T=const$ or $p=const$ where these are the internal variables. Oct 10, 2019 at 16:43

• I do not think you are answering the question that at first I also misunderstood. To me the question of @Vedvart1 means that given a thermodynamic system characterized by the equation of state $f(p,V,T)=0$ what reversible processes are possible if it is externally constrained by $p_{ext}=const$ and $T_{ext}=const$. I believe that the correct answer is that there is no such process. Oct 11, 2019 at 13:03