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.

  • $\begingroup$ why does it strictly have to be isothermal or isobaric? $\endgroup$ Oct 10, 2019 at 14:05
  • $\begingroup$ 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? $\endgroup$
    – Vedvart1
    Oct 10, 2019 at 14:13
  • $\begingroup$ In the isothermal process, you remove weights from the piston. $\endgroup$ Oct 10, 2019 at 14:13
  • $\begingroup$ 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 $\endgroup$ Oct 10, 2019 at 14:33
  • 1
    $\begingroup$ 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. $\endgroup$
    – hyportnex
    Oct 10, 2019 at 16:43

1 Answer 1


in an isothermal process- for example, the isothermal compression of a gas by a piston in a cylinder- the process is performed in such a way that at every incremental step of the process, there are no thermal gradients anywhere in the system. This means that at each step, the whole system is at thermal equilibrium, or nearly so. In the compression example, this means that if you compress the gas by pushing the piston through some tiny distance (delta x) where (delta x) is vanishingly small, you do it so slowly that the rise of temperature (delta T) of the gas caused by the compression work (delta W) has plenty of time to diffuse out the walls of the container.

  • $\begingroup$ If the whole system is in thermal equilibrium at each step, why does it do anything at all? For instance, in isothermal expansion, what's causing pressure to convert to volume (to keep ideal gas law satisfied), and why doesn't the atmospheric pressure then restore the PV distribution to something where the pressures are equal? $\endgroup$
    – Vedvart1
    Oct 10, 2019 at 18:05
  • $\begingroup$ in an isothermal expansion, heat is moving from the environment into the expanding gas. This is the basis for mechanical refrigeration and heat pumps. $\endgroup$ Oct 10, 2019 at 20:15
  • $\begingroup$ Is isothermal expansion spontaneous? If so:: heat flows into the gas from a heat bath. This addition of temperature raises both pressure and volume. Then the added heat dissipates back into the walls of the container. Would this not lower pressure and volume back to where they were? What is the catalyst for gaining volume and lowering pressure here, as opposed to just staying constant? Both occurances would satisfy the ideal gas law. $\endgroup$
    – Vedvart1
    Oct 11, 2019 at 2:47
  • $\begingroup$ 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. $\endgroup$
    – hyportnex
    Oct 11, 2019 at 13:03

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.