Is it possible that a system undergoes an irreversible change while the surroundings undergo a reversible change?

Question

Context

I was solving questions from the 5th edition of "A Textbook of PHYSICAL CHEMISTRY: Thermodynamics and Chemical Equilibrium" by Kapoor, K. L., which is the second of several excellent volumes the author has published on physical chemistry, where I came across this:

At 263.15, water may exist as a super-cooled liquid, as stated in the book. This super-cooled liquid water will spontaneously freeze to ice at the same temperature:

$$ \text{H}_2\text{O (l)} \xrightarrow[\text{spontaneous}]{263.15} \text{H}_2\text{O (s)} $$

Naturally, for such a process, $(\Delta S)_\text{universe} > 0$. In order to do this calculation, the process was modelled into three distinct steps:

$$ \text{H}_2\text{O (l)} \xrightarrow[\Delta]{263.15 \text{ to } 273.15} \text{H}_2\text{O (l)} \xrightarrow{273.15} \text{H}_2\text{O (s)} \xrightarrow[-\Delta]{273.15 \text{ to } 263.15} \text{H}_2\text{O (s)} $$

where $\Delta$ means to heat the system. $(\Delta S)_\text{sys}$ for the three processes can be calculated easily. However, to calculate $(\Delta S)_\text{universe}$, $(\Delta S)_\text{surr}$ must be calculated. Here, the author has made the following assumption:

When evaluating $(\Delta S)_\text{surr}$, we assume that the surroundings receive heat equal to the heat of crystallization at $263.15$ K in a reversible manner at temperature equal to $263.15$ K.

Further, this calculation can be made using the Kirchoff's law. And upon doing these calculations, we do obtain $(\Delta S)_\text{universe} > 0$, as expected.

Question

This procedure seems to be followed for various such questions where, even when a system is undergoing an irreversible change, the surroundings are assumed to undergo a reversible change. This seems rather odd. According to me, if a system is undergoing an irreversible change, so should the surroundings. The author has stated that this is an assumption, but I wanted to know its validity and, accordingly, my question: is it possible that a system undergoes irreversible change while the surroundings undergo a reversible change?

Answer 1

Yes, it is possible if the surroundings involve ideal constant temperature reservoirs, for which $\Delta S$ is always Q/T. Imagine such a reservoir to consist of a constant temperature bath containing pure liquid and corresponding melting/freezing pure solid at equilibrium.

Answer 2

In thermo-statics or thermo-dynamics, you should think of the "surroundings" as being the boundary conditions for the equations. Their role is very similar to a hypothetical ideal battery or voltage/current source in an electric circuit with resistors, capacitors and inductors described by Kirchhoff's laws. In real life, there is no such a thing as a source whose voltage is independent of its load and is also independent of temperature, pressure, humidity, etc., in which it is imbedded, as anybody trying to start an engine in the winter learns it. Of course, if we did not make such idealization we would not be able to calculate the current through even the simplest linear resistor...

An ideal battery has zero internal resistance, has infinite amount of freely moving electric charge, and can absorb from or release in the circuit to which it is connected any amount of said charge all at the same voltage.

An ideal thermal reservoir has infinite amount of internal charge called entropy, its temperature is independent of anything the reservoir is attached to and with it can exchange any amount of thermal charge, i.e., entropy, always at the same temperature. Notice the analogy with an ideal battery. From the point of view of the thermal reservoir entropy is like electric charge and temperature is like voltage. In the case of a battery the simplest model to the load-dependent voltage drop is a non-zero internal resistance that may depend on its environment (temperature, pressure, humidity, age, etc.). For an ideal thermal reservoir connected to a thermodynamic system one may employ a similar "internal resistance" by specifying the interface between it and the system to be a thermal conductor whose one end is at a fixed temperature as defined by the ideal reservoir but its other end is connected to the system and its temperature is variable.

The analogy is not superficial, it only fails in the sense that electric charge is unconditionally conserved always and everywhere while entropy is only conditionally conserved. Total entropy along with entropy change in detail is conserved in a reversible process but is only "semi-conserved", that is lower bounded in an irreversible process.

As you can imagine such idealizations can lead to apparent paradoxes, contradictions, etc., unless one is careful how to use them. For example, one may not ever directly connect two ideal batteries of different voltages in parallel. The battery itself has infinite energy and infinite charge, etc., so one cannot apply energy conservation to the complete circuit including the battery and thereby calculate the total energy, only energy changes are meaningful when an ideal battery is included. Exactly the same way one may never connect two reservoirs of different temperatures directly, only energy changes are meaningful not total energy, etc.

And finally, one may go beyond the concept of thermal reservoirs and introduce work reservoirs. These are similar boundary conditions to temperature but for the mechanical, electrical, magnetic, etc., intensive parameters, such as pressure, tension, stress, gravitational potential, electric potential, magnetic potential, etc. It is assumed that the reservoir holds an infinite amount of the corresponding extensive quantity, that is charge: volume, density, concentration, mass, etc., all with the underlying assumption that the finite amount of charge that is exchanged with the system does not effect the potential itself at which the exchange takes place.

Having said all this it should be clear now that the process in an ideal reservoir must be reversible irrespective of what happens to the system it is connected because this is how we set it up. In the mechanical reservoirs there is nothing "thermal", it has no temperature nor entropy, everything is reversible by definition. In an ideal thermal reservoir we cannot define the increase of the total entropy for that is infinite but we define the reservoir in such a way that whatever entropy is absorbed at its boundary is what goes in it, and whatever leaves its boundary is what goes in the system to which it is connected, all this by definition, so there again the entropy generation if there is one happens outside the reservoir, again by definition, so the reservoir's processes are reversible.

Answer 3

It is an idealization, but as @Chet Miller says, it is indeed possible. To explore the conditions for this idealization consider the following scenario.

A box is divided into two parts, each containing a gas at different pressure and temperature. We place the box in a bath at temperature $T$, break the wall that separates the gas and remove the insulation that maintained the temperatures initially. Clearly what happens inside this box is irreversible.

How are the surroundings affected? Apparently some heat is exchanged with the bath, which means there is a temperature gradient between the box and the bulk of the bath that moves this heat in the proper direction. This is essentially the only irreversible feature in the interaction with the surroundings. To the extent that we my ignore it we declare the transfer of heat to be reversible with respect to the bath and calculate the entropy of the bath as $$ \Delta S_\text{bath} = \pm\frac{|Q|}{T_\text{bath}} $$ with the appropriate sign for trasnfer to or from the bath. The entropy generation for the process is $$ S_\text{gen} = \Delta S_\text{sys} \pm\frac{|Q|}{T_\text{bath}} . $$ where both $\Delta S_\text{sys}$ and $Q$ can be calculated from information about the changes in the system.

In this example the system interacts with the surroundings only via heat exchange. If $PV$ work is involved we require that this work is quasistatic, i.e., it does not involved viscous losses, shock waves or other mechanisms that pass some of that work into generation of entropy.

Answer 4

I do not have the complete statement of the exercise but as you present it, it seems incomplete to me. The second principle can be presented in the following form: $$d S=\delta S_e+\delta S_c$$ with $$\delta S_e= \iint_{\partial S} \frac{\delta Q}{T}$$ $\partial S$ is the boundary of the system and $T$ the temperature on this boundary. The second principle imposes $$\delta S_c \geq 0$$

This formulation seems to me preferable to that involving "the entropy of the universe" because it emphasizes the local character of the creation of entropy. In the thermodynamics of irreversible phenomena, not too far from equilibrium, we have relationships that allow us to calculate the rate of creation of entropy per unit volume associated with thermal conduction, viscosity, etc.....

But often, these phenomena are too complicated and one is satisfied to calculate the entropy created by making the difference : $$S_c=\Delta S -S_e$$ The first term $\Delta S$ is the variation of a state function and does not depend on the path followed by the system. But the second $S_e$ depends on the detail of the transformation. In the case of your exercise, you find $S_e$ assuming that throughout the transformation, the system is at a uniform temperature which is varied in a quasistatic way. By doing this you calculate the entropy created inside the system for this particular transformation. And it has to be positive regardless of what's going on outside.

You could also have carried out the transformation by fixing the surface temperature at $T_0=$ 263.15 K. In this case, the exchanged entropy would have been different. As we are at constant pressure, the heat exchanged would have been $Q=\Delta H$ , variation of a state function, and the entropy exchanged $$S_e=\frac{\Delta H}{T_0}$$ In the end, the entropy created within the system would have been different. But always positive if the transformation is possible.

To return to the initial question: I insist on the fact that the entropy creation rate is a local quantity which must be positive independently of what is happening outside.

When we forget this point, we do not understand the following joke: I want to build an engine that extracts heat from the atmosphere to turn it into work. Unfortunately, this is contrary to Kelvin Planck's statement of the second principle:

"it is impossible to currency a heat engine that takes heat from the hot reservoir ( Q H ) and converts all the energy into useful external work without losing heat to the cold reservoir"

This engine would decrease the entropy of the universe.

But I have a brilliant idea : I'm creating a very unethical company that simultaneously burns oil in a very distant country to compensate for the universe's diminishing entropy. Globally, the second principle would be satisfied! But locally this would not be the case. (In addition, when you know a little about relativity, the notion of simultaneity raises questions.....).

Hope it can help and sorry for my poor english.