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I am having trouble reconciling the following statements I have recently learned about thermodynamics:

  1. The temperature of the surroundings is taken to be constant (with only infinitesimal changes)

  2. In a reversible process, $T_{sys}=T_{surr}$

Combining we get $T_{sys}=T_{surr}=\text{constant}$

Therefore, $T_{sys}$ will need to be constant making isothermal as the only process possible. Also, the isothermal process will need to be at same temperature as the surroundings. Where have I gone wrong?

Also, do the above 2 points hold for an irreversible process?

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  • $\begingroup$ The surroundings are so large that even if you add heat, their temperature stays pretty much constant. Such object is also called a heat bath. $\endgroup$
    – FusRoDah
    Jun 30, 2022 at 10:40
  • $\begingroup$ @FusRoDah I know that and that it is my first point, but how does it answer the question? $\endgroup$
    – Boson
    Jun 30, 2022 at 10:41
  • $\begingroup$ And if the system and the surroundings are at a different temperature, then heat exchange will take place, which is not a reversible process. Unless the system is isolated, but then the surroundings don't matter. $\endgroup$
    – FusRoDah
    Jun 30, 2022 at 10:41
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    $\begingroup$ I do not understand what is your question. Is the process characterized by an isothermal surrounding? Or do you think that a reversible thermodynamic process may only be at a constant surrounding temperature? $\endgroup$ Jun 30, 2022 at 10:42
  • $\begingroup$ @GiorgioP from what I've learnt, we assume the surrounding has infinite mass*heat capacity so its temperature remains constant $\endgroup$
    – Boson
    Jun 30, 2022 at 10:43

2 Answers 2

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This is an example of 2+2=5.

First of all, not all reversible processes are isothermal. You have reversible isobaric, isochoric and adiabatic processes.

For a reversible adiabatic process, the temperature of the surroundings is irrelevant.

For a reversible isobaric and isochoric process, the temperature of the system and surroundings are always in equilibrium (actually, differ infinitesimally) but they are not constant.

As far as irreversible heat transfer processes are concerned, the temperature of the system and surroundings is not the same, except at the boundary. Within the system temperature gradients exist.

Finally, with regard to the (essentially) constant temperature surroundings, the mass or heat capacity need not be "infinite". It need only be large enough so that, given the amount of heat transfer involved, its change in temperature would be infinitesimal.

You say "For a reversible isobaric and isochoric process, the temperature of the system and surroundings are always in equilibrium (actually, differ infinitesimally) but they are not constant." But isn't that contradicting your last statement that the temperature change of the surroundings is infinitesimal i.e the temperature change of the surroundings is not constant?

No. Because for the reversible isobaric and isochoric processes the surroundings is an infinite series of reservoirs ranging in temperature between the initial and final temperature, not a single reservoir as in an isothermal process.

Consider the reversible isochoric heat addition. The temperature of each reservoir is infinitesimally greater than the previous reservoir and infinitesimally greater than the gas temperature. So essentially we have $T_{sys} = T_{sur}$ = constant for each reservoir, meeting your criteria 2, but the system and surroundings temperature is not constant over the entire process because there are multiple reservoirs. So it is not an isothermal process.

Hope this helps.

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  • $\begingroup$ You say "For a reversible isobaric and isochoric process, the temperature of the system and surroundings are always in equilibrium (actually, differ infinitesimally) but they are not constant." But isn't that contradicting your last statement that the temperature change of the surroundings is infinitesimal i.e the temperature change of the surroundings is not constant? $\endgroup$
    – Boson
    Jun 30, 2022 at 14:19
  • $\begingroup$ @Boson No. Because for those reversible processes the surroundings is an infinite series of reservoirs ranging in temperature between the initial and final temperature, not a single reservoir as in an isothermal process $\endgroup$
    – Bob D
    Jun 30, 2022 at 14:30
  • $\begingroup$ This seems new, could you edit your answer to elaborate on it? $\endgroup$
    – Boson
    Jun 30, 2022 at 14:31
  • $\begingroup$ Sure but will have to be later when I’m back at my computer $\endgroup$
    – Bob D
    Jun 30, 2022 at 14:33
  • $\begingroup$ Sure, no problem $\endgroup$
    – Boson
    Jun 30, 2022 at 14:33
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Maybe you are confusing the general processes involving a thermodynamic system and the particular case of a system in contact with a thermal reservoir at fixed temperature (for example, the typical system described in Statistical Mechanics via the canonical ensemble). You do not have to confuse the two things.

In general, it is possible to have any process (reversible and irreversible, isothermal, isobaric, isochoric, or even with all the thermodynamic variables changing simultaneously).

The situation of a thermodynamic system at equilibrium with a thermostat at a fixed temperature is a possible situation but not necessary. In no case a reversible process requires a constant temperature. For example, when studying phase transitions, one increases or decreases the thermostat temperature so slowly to ensure that the temperature of the sample and the thermostat are always as closest as possible to each other.

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  • $\begingroup$ Please see my comment on Bob's answer $\endgroup$
    – Boson
    Jun 30, 2022 at 14:26

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