2
$\begingroup$

I need to have a fully functioning understanding of entropy so I've got a few questions to ask

  • What are the conditions for an increase in entropy to be reversible?
    I (think) I know that any process where the temperature is constant is assumed to be reversible, but what other constraints can I put on other thermodynamic variables to make a process reversible?

  • I've read that $dS=dS_{i}+dS_{e}$
    Where $dS_{e}$ is the entropy inflow or outflow to the environment sorrounding the system and $dS_{i}$ is the rate of entropy production in the system. Why is is that it is assumed that $dS_{e}$ is always given by a reversible process? (https://courses.physics.ucsd.edu/2020/Fall/physics210b/Non-Eqbrm%20Thermo_Demirel%20and%20Gerbaud.pdf)

  • I've read the problem of heat exchange between two bodies at fixed volumes
    It is assumed that for the first and second body in contact at fixed volumes $dS_{1}=\frac{dQ}{T_{sys}}$ and $dS_{2}=\frac{-dQ}{T_{env}}$ so that $dS_{tot}=dS_{1}+dS_{2}= \frac{dQ}{T_{sys}} + \frac{-dQ}{T_{env}}> 0$ but we also have $TdS_{k}=Cv_{k}dT$ which implies that $\Delta S_{tot}=C_{v,env}ln(T_{f}/T_{env})+C_{v,sys}ln(T_{f}/T_{sys})$, and $T_{f}=\frac{Cv_{1}T_{1}+Cv_{2}T_{2}}{Cv_{1}+Cv_{2}}$ and I don't understand how these expressions are related. (Thermodynamics of Nonequilibrium Processes. Authors: S. Wisniewski, B. Staniszewski, R. Szymanik, (p.12))

    I would very much appreciate clarification of all these.

$\endgroup$
2

2 Answers 2

4
$\begingroup$
  • What are the conditions for an increase in entropy to be reversible? I (think) I know that any process where the temperature is constant is assumed to be reversible, but what other constraints can

The other constraints are that the process has to be carried out very slowly so that the system is always in equilibrium with its surroundings, and that there can be no form of friction that would generate heat.

  • I've read that $dS=dS_{i}+dS_{e}$ Where $dS_{e}$ is the entropy inflow or outflow to the environment sorrounding the system and $dS_{i}$ is the rate of entropy production in the system. Why is is that it is assumed that $dS_{e}$ is always given by a reversible process?

It is not assumed that $dS_e$ is for a reversible process. It is the entropy transferred to the system and environment at the boundary between them. $dS$ is the total system entropy change for either a reversible or irreversible process since $S$ is a system property. Only when there is no entropy generated (the process is reversible) is when $dS=dS_{e}$. In terms of the total entropy change of a body we can write $dS=dS_{e}+dS_{i}$ as

$$\Delta S=\int_{Ti}^{Tf}\frac{\delta Q}{T_{B}}+S_{G}\tag{1}$$

where

$\Delta S$ = total entropy change of the body

$\frac{\delta Q}{T_B}$= entropy transfer at the boundary, or interface, between the bodies due to heat $\delta Q$ crossing the boundary.

$T_B$= the temperature at the boundary, or interface, between the system and environment.

$S_G$= the entropy generated within the body due to any irreversible heat and/or work.

Note that for two semi infinite bodies having perfect thermal contact and thermal properties that are uniform and independent of temperature, it can be shown that the temperature at the interface when brought into contact will be constant and equal to the final equilibrium temperature, $T_f$ given by the equation you presented. Therefore we have

$T_{B}=T_{f}$= constant

For further background see: Heat Transfer between 2 bodies

Equation (1) can now be written for the two bodies under consideration where body 1 has been designated the "system" and body 2 the "environment":

$$\Delta S_{sys}=\int_{T1}^{Tf}\frac{\delta Q}{T_{f}}+S_{G1}\tag{2}$$

$$\Delta S_{env}=\int_{T2}^{Tf}\frac{\delta Q}{T_{f}}+S_{G2}\tag{3}$$

  • I've read the problem of heat exchange between two bodies at fixed volumes I don't understand how these expressions are related.

Essentially the expressions obtained from the reference involve application of equations (2) and (3) to the two bodies.

$\Delta S$ is a state property so its change between two equilibrium states does not depend on the process. That means even though the actual process is irreversible (due to the finite temperature difference between the two equal volumes) we can devise any convenient reversible path between the the initial and final temperatures of each body and apply the following definition of the entropy change for a reversible transfer of heat:

$$\Delta S=\int_{Ti}^{Tf}\frac{\delta Q_{rev}}{T}\tag{4}$$

Substituting equation (4) into equations (2) and (3) we obtain

$$\Delta S_{sys}=\int_{Ti}^{Tf}\frac{\delta Q_{rev}}{T_{sys}}=\int_{T1}^{Tf}\frac{\delta Q}{T_{f}}+S_{G1}\tag{5}$$

$$\Delta S_{env}=\int_{Ti}^{Tf}\frac{\delta Q_{rev}}{T_{env}}=\int_{T2}^{Tf}\frac{\delta Q}{T_{f}}+S_{G2}\tag{6}$$

In order to calculate the left side of each equation we imagine individually exposing each body to an infinite series of external thermal reservoirs incrementally ranging from the initial temperature of each volume to its final temperature, so that each body is always in thermal equilibrium with the reservoir. The concept is shown in the figure below. This gives us

$$\Delta S_{sys}=\int_{Ti}^{Tf}\frac{\delta Q_{rev}}{T_{sys}}=C_{v1}\ln\frac{T_{f}}{T_1}\tag{7}$$

$$\Delta S_{env}=\int_{Ti}^{Tf}\frac{\delta Q_{rev}}{T_{env}}=C_{v2}\ln\frac{T_{f}}{T_2}\tag{8}$$

Given $T_f$ =constant, the entropy transfers between the bodies becomes,

For body 1 (system),

$$\int_{T1}^{Tf}\frac{\delta Q}{T_{f}}=\frac{C_{v1}(T_{f}-T_{1})}{T_f}\tag {9}$$

For body 2 (environment)

$$\int_{T2}^{Tf}\frac{\delta Q}{T_{f}}=\frac{C_{v1}(T_{f}-T_{2})}{T_f}\tag {10}$$

Finally, substituting equations (7) through (10) into equations (5) and (6) we obtain the final equations for the change in entropy of each body as follows:

$$\Delta S_{sys}=C_{v1}\ln\frac{T_{f}}{T_1}=\frac{C_{v1}(T_{f}-T_{1})}{T_f}+S_{G1}\tag{11}$$

$$\Delta S_{env}=C_{v2}\ln\frac{T_{f}}{T_2}=\frac{C_{v1}(T_{f}-T_{2})}{T_f}+S_{G2}\tag{12}$$

EXAMPLE:

To simplify the math let the two bodies have the same constant volume heat capacity, $C_{v1}=C_{v2}=C_v$

Let

$T_1$=300 K, $T_2$=500 K, then $T_f$=400 K

Substituting the data into equation (11) we have

$$\Delta S_{sys}=C_{v}\ln\frac{400}{300}=C_{v}\frac{400-300}{400}+S_{G1}\tag{13}$$ $$0.287C_{v} = 0.25 C_{v} + S_{G1}$$ $$S_{G1}=0.0376 C_{v}$$

Substituting the data into equation (12) we have

$$\Delta S_{env}=C_{v}\ln\frac{400}{500}=C_{v}\frac{400-500}{400}+S_{G2}\tag{14}$$ $$-0.223 C_{v}=-0.25 C_{v}+S_{G2}$$ $$S_{G2}=0.0269 C_{v}$$

Adding the results of equations (13) and (14) we get

$$\Delta S_{tot}=\Delta S_{sys}+\Delta S_{env}=S_{G1}+S_{G2}=+0.0645 C_{v}\gt 0\tag{15}$$

For a comprehensive understanding of how to determine the change in entropy for an irreversible process, I highly recommend the tutorial in the link that Chet Miller provided to you.

Hope this helps.

enter image description here

$\endgroup$
16
  • $\begingroup$ Thank you for your reply! However I'm still having trouble understanding why $dS_{e}=dS$ implies the process is reversible, as in the example of heat conduction between bodies I would think that the entropy increase in the system would be due to $dS_{e}$ rather than the system producing entropy in the form $dS_{i}$, irreversibly. $\endgroup$
    – manoroli
    Mar 30 at 19:03
  • 1
    $\begingroup$ @manoroli I think you are confusing the actual heat transfer process between the two volumes where the actual border temperature is not constant, with the made up reversible heat transfers between each volume with the temporary series of constant temperature thermal reservoirs not part of either volume with each reservoir constituting a constant temperature border with the volume immersed in it. $\endgroup$
    – Bob D
    Mar 30 at 20:14
  • 1
    $\begingroup$ @manoroli Did you read the tutorial recommended by Chet Miller? It explains that that the reversible process(es) used to calculate the entropy change need not bear any resemblance to the actual irreversible process as long as it is between the same two equilibrium states. $\endgroup$
    – Bob D
    Mar 30 at 20:17
  • 1
    $\begingroup$ @manoroli See my updated answer in which I have tried to tie the concepts together. Hope it helps. $\endgroup$
    – Bob D
    Apr 1 at 13:40
  • 1
    $\begingroup$ @manoroli has my update answered your questions? $\endgroup$
    – Bob D
    Apr 2 at 0:17
3
$\begingroup$

What are the conditions for an increase in entropy to be reversible ?

Any physical change in a system is reversible if the physical laws that govern the behaviour of the system are time reversible. The key question is how likely is the change to be reversed ? A broken egg can spontaneously reassemble itself using random energy that it harvests from the surrounding environment - this process breaks no physical laws, but it is incredibly unlikely.

Why is is that it is assumed that $dS_e$ is always given by a reversible process ?

See above. The standard underlying assumption is that physical laws themselves are reversible.

I don't understand how these expressions are related

The first expression assumes that the system and the environment are kept at constant temperatures $T_{sys}$ and $T_{env}$. The second expression allows temperatures to change. $T_1$ and $T_2$ are the initial temperatures of the two systems and $T_f$ is their (common) equilibrium temperature.

$\endgroup$
1
  • $\begingroup$ But then why would $T_{sys}$ and $T_{env}$ be lfixed if we're modelling the diferential increase in entropy of the system not just at the start, but generally? I've seen authors claim that in irreversible heat exchange in the problem I mentioned would be implied by the differential form to be $\Delta S_{tot} = \frac{\Delta Q}{T_{env}}+ \frac{\Delta -Q}{T_{env}}$. Is this only true when $T_{env} \approx T_{sys}$ then? $\endgroup$
    – manoroli
    Mar 30 at 19:20

Not the answer you're looking for? Browse other questions tagged or ask your own question.