# Entropy of an irreversible process

supermesh
Can entropy change be zero for a closed system in an irreversible process? If yes under what conditions? Thanks!

Is it ture that it would be zero in an adiabatic, process carried in consant temperature?

Staff Emeritus
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Can entropy change be zero for a closed system in an irreversible process? If yes under what conditions? Thanks!
Entropy is a state function and as such the integral;

$$\int^{f}_{i}\frac{dQ_R}{T}$$

is path independent. What does that tell you?

perebal
Is the irreversible part of entropy state function? Or is the total entropy (included the rev. and irrev. part) state function?

Fernsanz
Hello.

As Hotenanny has pointed out, entropy is a function of state. That means that its value depends only on the (equilibrium) state of the system under consideration and no on the process carried out to get the system to that state. So, think of any two isoentropic states. Then, no matter by mean of which process you connect those states, the entropy of the system under consideration at those states will be the same.

I have highlighted the system under consideration because even though the entropy change in a irreversible process can be zero for the thermodynamical system you are considering, the total entropy change will be strictly positive. Total means the entropy change of the system plus the entropy change of the surrounding universe. And this is in fact always positive for an irreversible process. So, the answer is yes for the parcel of the universe that you agree to call "your system", and no for the total entropy which accounts for all changes made in the universe.

In an adiabatic process you have dQ=0. If, additionaly, the process is isotermal then it is reversible (because talking of temperatures involves talking of equilibrium). So, in an adiabatic isothermal process you can apply the integral with dQ=0 and hence the entropy change is zero.

Rainbow Child
... In an adiabatic process you have dQ=0. If, additionaly, the process is isotermal then it is reversible (because talking of temperatures involves talking of equilibrium). So, in an adiabatic isothermal process you can apply the integral with dQ=0 and hence the entropy change is zero. ...

You can't have an adiabatic isothermal process which is simultaneously reversible.

$$p\,V^\gamma=const.\Rightarrow T\,V^{\gamma-1}=const.$$

which tells you that the temperature is not constant.

An adiabatic isothermal process can be only irreversible, so the entropy change is positive. Consider for example the following situation

The box is made of adiabatic walls, and in the left hand side there is an ideal gas at temperature T. If we remove the black inside wall then the gas will fill all the space. Since the walls are adiabatics $$Q=0$$, and no work is done $$W=0$$, from the 1st law of thermodynamics we have

$$Q=\Delta U+W\Rightarrow \Delta U=0 \Rightarrow n\,C_V\,\Delta T=0 \Rightarrow T=const.$$

Thus the process is adiabatic and isothermal but irrevesible.
In order to calculate the entropy's change, since we cannot apply $$\int^{f}_{i}\frac{dQ}{T}$$ we use the fact that entropy is a state funtion as Hootenanny posted.
That means we can imagine an revesible processure which has the same initial and final states. That could be an isothermal from the volume $V_1$ to the volume $V_1+V_2$. The heat for this process is $$Q=n\,R\,T\,\ln\frac{V_1+V_2}{V_1}$$,so

$$\Delta S=n\,R\,\ln\left(1+\frac{V_2}{V_1}\right)>0$$

Fernsanz
You can't have an adiabatic isothermal process which is simultaneously reversible.

$$p\,V^\gamma=const.\Rightarrow T\,V^{\gamma-1}=const.$$

which tells you that the temperature is not constant.

An adiabatic isothermal process can be only irreversible, so the entropy change is positive. Consider for example the following situation

In general it is possible to have adiabatic isothermal reversible processes. You have stuck to a system consisting only of ideal gases that is when Poisson's law is valid.

It depends on which boundaries you specify for your system. Depending of the boundaries one same form of energy can be considered heat or work. For example, if you put an electrical resistance inside the thermodynamical system, is it that electrical energy heat or work? If the boundary of the system does not contain the resistance, then it is heat what you are giving to the system. However, if the boundary contains the resistance then it is work that you are performing to the system by mean of electric charges in movement.

So, put a resistance inside the system, consider it as part of the system and you can have an isothermal adiabatic reversible process because the only interchange of energy will be in form of work.

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Rainbow Child
In general it is possible to have adiabatic isothermal reversible processes. You have stuck to a system consisting only of ideal gases that is when Poisson's law is valid.

It depends on which boundaries you specify for your system. Depending of the boundaries one same form of energy can be considered heat or work. For example, if you put an electrical resistance inside the thermodynamical system, is it that electrical energy heat or work? If the boundary of the system does not contain the resistance, then it is heat what you are giving to the system. However, if the boundary contains the resistance then it is work that you are performing to the system by mean of electric charges in movement.

In the first case is heat as you say. In the second case is electical energy provided to the ends of the resistance.

So, put a resistance inside the system, consider it as part of the system and you can have an isothermal adiabatic reversible process because the only interchange of energy will be in form of work.

What makes you think that the above process is reversible?? How can you bring the system back to it's initial state?
And you also think that this process is isothermal??
Could you write down for me the the equation that describes the state of the system $$f(p,V,T)=0[/itex] in order to calculate the change of entropy by [tex] \Delta S=\int^{f}_{i}\frac{dQ}{T}$$

Rainbow Child
Here is the Clausius-Duhem inequality: http://en.wikiversity.org/wiki/Continuum_mechanics/Clausius-Duhem_inequality_for_thermoelasticity

Which entropy ($$\eta$$) is in it? The reversible part, the irreversible part or the sum of them?

I can't understand that wiki page!

But as far as I know the Clausius-Duhem inequality of classical thermodynamics is a statement of the 2nd. It applies to thermodynamic processes during which a system evolves from one equilibrium state (i) to another (f) and it says that the integrated heat absorbed by the system, divided by the temperature at which that heat is absorbed, is bounded from above by the net change in the entropy of the system:

$$\int^{f}_{i}\frac{dQ}{T}\leq\Delta S=S_f-S_i$$

The equality stands for a reversible process.

Fernsanz
In the second case is electical energy provided to the ends of the resistance.

Exactly, and since we agree in that is not heat, it must be work. The first law says that energy can be only heat or work.

What makes you think that the above process is reversible?? How can you bring the system back to it's initial state?
And you also think that this process is isothermal??
Could you write down for me the the equation that describes the state of the system $$f(p,V,T)=0[/itex] in order to calculate the change of entropy by [tex] \Delta S=\int^{f}_{i}\frac{dQ}{T}$$

I was not referring to any particular process. But if you want one consider the reversible separation of gases by mean of semipermeable membranes. There you will find an isothermal adiabatic reversible process.

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Rainbow Child
Exactly, and since we agree in that is not heat, it must be work. The first law says that energy can be only heat or work.

But it isn't work. It's electrical energy provided to the resistance. If you want to include to your system the resistance as you say

However, if the boundary contains the resistance then it is work that you are performing to the system by mean of electric charges in movement.

then this is not a closed system because the charges are moving through the boundaries of your system.

I was not referring to any particular process.

So, put a resistance inside the system, consider it as part of the system and you can have an isothermal adiabatic reversible process because the only interchange of energy will be in form of work.

...You have stuck to a system consisting only of ideal gases ...
...But if you want one consider the experiment described in the Gibbs' paradox. There you will find an isothermal adiabatic reversible process.

The Gibbs paradox in order to decsribe adiabatic reversible process, has to deal with ideal gases. See for instance

"Explanation of the Gibbs paradox within the framework of quantum thermodynamics" at
http://arxiv.org/abs/quant-ph/0507145

Fernsanz
But it isn't work. It's electrical energy provided to the resistance. If you want to include to your system the resistance as you say

Well, it is energy given to the system and it is not heat, so what could it be? First law reads

$$dU= dQ + dW_{mechanical} + dW_{electrical} + \ldots$$​

In fact first law says defines the heat as "if it is not work it is heat". So, if it is not heat it is work. So in the case of the resistance it is electrical work. It is exceedingly cut-clear.

then this is not a closed system because the charges are moving through the boundaries of your system

I don't remember anyone impossing it has to be closed. I was just giving an example of an adiabatic isothermal reversible process.

?

The Gibbs paradox in order to decsribe adiabatic reversible process, has to deal with ideal gases

First of all, sorry because I didn't mean to say Gibbs's paradox. I mean reversible separation of gases. which, in any case, deals with ideal gases. Here you will have an adiabatic isothermal reversible process and even more: with no work! $$dQ=dW=dU=0$$; and is a closed system. What else do you need? As I suspect you are more interested in discussing than in understanding my arguments, I have decided to attach a picture of the page of the Atkin's book which explain this. The figure showing the arrangement is also attached.

I think with this, discussion is over and it is clear and laid-down that there is adiabatic isothermal reversible processes.

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