# Reversibiliy and equilibrium

My textbook states "A system is in thermodynamic equilibrium if it is in Thermal, Mechanical, and chemical equilibrium simultaneously".

The Wikipedia article on "reversible processes" states that the system is thermodynamic equilibrium with the surroundings throughout the process. This means, as per my initial statement, $$\ce{Tsys=Tsurr}$$ at every instant of the process.

Atkins-Physical Chemistry, page 77 states, while explaining the entropy change of the surroundings: "furthermore, because the temperature of the surroundings is constant whatever the change, for a measurable change,$$\ce{\Delta}Ssurr=\dfrac{q_{surr}}{T_{surr}}$$".

But then this treatment seems to contradict the fact that the system and surrounding are in thermal equilibrium! Since $$\ce{Tsys=Tsurr}$$ implies that $$\ce{Tsurr}$$ is not constant as long as the process isn't isothermal.

I therefore believe that my first statement is incorrect.If it isn't incorrect,then what am I misinterpreting?

• I don't see the problem. The temperature is the same and is constant in the system and in the surroundings in all your statements. So they are all correct. There are no contradictions. – Maurice Jun 30 '20 at 18:57
• I really dont get what you mean – satan 29 Jun 30 '20 at 18:59
• One paragraph (referencing atkins) clearly states that the surrounding temperature is constant. – satan 29 Jun 30 '20 at 19:00
• While Tsys=Tsurr means that Tsurr is not constant. If the temperature of the system varies, This equation ensures that the temparature of the surroundings will also vary. – satan 29 Jun 30 '20 at 19:01
• @Maurice would you please elaborate upon your comment? – satan 29 Jun 30 '20 at 19:02

When a system is in thermodynamic equilibrium, there is no tendency for spontaneous change. In other words everything stays the same, the composition, temperature, pressure do not change over time, etc.

Therefore:
Tsurr is constant and
Tsys is constant.

If you make an infinitesimal change, the system will tend to revert to the equilibrium.

So there is no issue.

When you are talking about a reversible process in thermodynamic terms, you're talking about a system which can be taken from its initial state to another state, and then back to the initial state without any net change to either the system or the surroundings.

The idea being that you make an infinitesimal change to some aspect of the surroundings, the system changes to a new state which is again in equilibrium with the surroundings, lather, rinse, repeat. Of course it would take infinite time for the process to be completed.

The process described in Atkins must refer to an isothermal process, since otherwise the appropriate expression to use in computing the change in entropy of the surroundings is

$$\ce{\Delta}S_{surr}=\int\dfrac{dq_{surr}}{T_{surr}}=-\int\dfrac{dq}{T_{surr}}$$

where $$dq$$ represents an inexact differential.

This becomes equal to the statement you quote only if the temperature of the surroundings remains constant during the process.

The entropy change during a reversible process, during which there is thermal balance between system and surroundings at each step ($$T_{surr}=T_{sys}=T$$), is given by

$$\ce{\Delta}S_{sys}=-\ce{\Delta}S_{surr,rev}=\int\dfrac{dq}{T}$$

My textbook states "A system is in thermodynamic equilibrium if it is in Thermal, Mechanical, and chemical equilibrium simultaneously".

Since you are talking about thermodynamic equilibrium with the surroundings, I would replace that statement (which is fine by itself, though it is not comprehensive) with something like: "A system is in thermodynamic equlilibrium with its surroundings when it is in equilibrium with respect to all forms of work or heat flow that can move across its boundary".

Thus if the boundary is rigid, even a system is in thermodynamic equilibrium, it would not be in mechanical equilibrium with its surroundings, because it is not communicating with its surroundings mechanically. If the boundary is adiabatic, a system at thermodynamic equilibrium would not be in thermal equilibrium with its surroundings, because it is not communicating thermally with its surroundings (e.g., if the walls are adabiatic, the system could be in thermodynamic equilibrium at, say T = 300 K, while the temperature of the surroundings could have any value (yes, above absolute zero, of course).

The Wikipedia article on "reversible processes" states that the system is thermodynamic equilibrium with the surroundings throughout the process. This means, as per my initial statement, $$\ce{Tsys=Tsurr}$$ at every instant of the process.

Correspondingly, the above is not generally true; it's only true for isothermal reversible processes (you may be misrepresenting the content of the Wikipedia article here by not including the isothermal restriction in your statement—I don't know, since you didn't provide the reference). In an adiabatic reversible process, by contrast, the system is not in thermal equilibrium with the surroundings and, typically, $$\ce{Tsys \ne Tsurr}$$ (except perhaps at one point, or in the case that you have an adiabatic process with no temperature change, such as the free expansion of an ideal gas).

Since $$\ce{Tsys=Tsurr}$$ implies that $$\ce{Tsurr}$$ is not constant as long as the process isn't isothermal.

You have it backwards. For $$\ce{\Delta}S_{\text{surr}}=\dfrac{q_{\text{surr}}}{T_{\text{surr}}}$$ to apply, the surroundings must be an infinite heat bath. I.e., the surroundings are so vast that their intensive properties (e.g., temperature) are unaffected by exchange of heat, work, or matter with the system. Thus, for isothermal processes in a system that is in thermal equilibrium with its surroundings, it is the temperature of the surroundings that determines the temperature of the system, not the other way around.