In whatever resources I have consulted to study reversible thermodynamic processes, a common statement seems to be :

The system and surroundings (if the boundary allows) are in mechanical ($P_{sys}=P_{surr})$ and thermal $(T_{sys}=T_{surr})$ equilibrium at all instants.

However, in various answers on the Chemistry stack exchange, and in physical chemistry by peter atkins, it is claimed that the "surrounding" is an infinite reservoir: so the temperature can be assumed to be constant.

1.Is this not contradictory to the "thermal equilibrium" I mentioned in the beginning?

2.Thermal equilibrium of a system with its surroundings means that the entropy change of the surroundings , should be given by $\int\dfrac{-dq_{sys}}{T_{surr}}$. However, everywhere, this equation is simply given as $\dfrac{-Q_{sys}}{T_{surr}}$, which seems to be explained by the claim that the Temperature of the surroundings doesnt change. This equation, according to me should be applicabe only to Irreversible processes happening quickly. (or to isothermal processes).

so what exactly should we consider: $T(sys)=T(surr)$ at each instant, or $T(surr)$=constant?

  • 1
    $\begingroup$ You might want to elaborate on point #1. What exactly is the contradiction? $\endgroup$
    – Buck Thorn
    Aug 3, 2020 at 20:26
  • $\begingroup$ If the system changes its temperature from T1 to T2,(T2 $\neq T1$) then the surroundings also change from T1 to T2, in a reversible process. Doesnt this clearly contradict the statement :"surrounding temperature remains the same"? $\endgroup$
    – satan 29
    Aug 4, 2020 at 7:39

2 Answers 2


In a cases where where the temperature is changing during the reversible process, instead of just using a single reservoir at a constant temperature, what you do is use a continuous sequence of reservoirs, all at slightly different constant temperatures (over the range from the beginning to the end of the process), with a tiny (differential) amount of heat exchanged with each. So, for each reservoir, you have $$dS_{surr}=\frac{dq_{syst}}{T_{surr}}$$ and, if you add all these differential changes in entropy up, you end up with $$\Delta S_{surr}=-\int{\frac{dq_{syst}}{T_{surr}}}$$


With regard to question 2, the correct equation to describe the entropy change during a finite reversible process is

$$\Delta S_{\text{surr}} = \int\dfrac{-dq_{sys}}{T_{surr}} \tag{1}$$

which follows from the statement $$dS_{\text{surr}} = \dfrac{-dq_{sys}}{T_{surr}} \tag{2}$$

Only for an isothermal process does Eq. (1) become

$$\Delta S_{\text{surr}}=\dfrac{-Q_{sys}}{T_{surr}} \tag{3}$$

It is possible to alter the temperature of the surroundings during a process (in practice and conceptually), so Eq. (3) is not a general statement.

The series of pseudo-stationary states traversed during a reversible process are equilibrium states, attainable under the limiting condition that the process is carried out infinitely slowly. At each step during a reversible process we can write $T_{sys}=T_{surr}$, but this does not require the temperature to be equal for each state$^\dagger$. The use of the word infinity is meant to convey an ideal process that can only be approached approximately in reality. An infinitely large heat reservoir means that it has an infinitely large heat capacity, so that changes in the system do not alter the temperature of the surroundings:

$$dT = \frac{dq}{C_{\text{surr}}}=\frac{dq}{\infty} =0 \tag{4}$$

A textbook example of a process carried out reversibly while altering the temperature is the cycle of heating and cooling a pure substance before and after bringing it to a phase transition temperature, in order to compute the entropy change for the spontaneous process under supercooled conditions.

$^\dagger$ If this appears to be like one of Zeno's paradoxes, well, it is.


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.