- #1

supermesh

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

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- Thread starter supermesh
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- #1

supermesh

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

- #2

Hootenanny

Staff Emeritus

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

[tex]\int^{f}_{i}\frac{dQ_R}{T}[/tex]

is path

- #3

perebal

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- #4

Fernsanz

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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

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

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.

Do I answer your question?

- #5

Rainbow Child

- 365

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... 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

In a reversible adiabatic process Poisson's law reads

[tex] p\,V^\gamma=const.\Rightarrow T\,V^{\gamma-1}=const. [/tex]

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 [tex]Q=0[/tex], and no work is done [tex]W=0[/tex], from the 1st law of thermodynamics we have

[tex]Q=\Delta U+W\Rightarrow \Delta U=0 \Rightarrow n\,C_V\,\Delta T=0 \Rightarrow T=const.[/tex]

Thus the process is adiabatic

In order to calculate the entropy's change, since we cannot apply [tex]\int^{f}_{i}\frac{dQ}{T}[/tex] we use the fact that entropy is a state funtion as Hootenanny posted.

That means we can

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

- #6

Fernsanz

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You can't have anadiabatic isothermalprocess which is simultaneouslyreversible.

In a reversible adiabatic process Poisson's law reads

[tex] p\,V^\gamma=const.\Rightarrow T\,V^{\gamma-1}=const. [/tex]

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

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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- #7

perebal

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

- #8

Rainbow Child

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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, 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.inside the thermodynamical system

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 anbecause the only interchange of energy will be in form of work.isothermal adiabatic reversible process

What makes you think that the above process is

And you also think that this process is

Could you write down for me the the equation that describes the

[tex] \Delta S=\int^{f}_{i}\frac{dQ}{T}[/tex]

- #9

Rainbow Child

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Which entropy ([tex]$\eta$[/tex]) 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

[tex] \int^{f}_{i}\frac{dQ}{T}\leq\Delta S=S_f-S_i[/tex]

The equality stands for a

- #10

Fernsanz

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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?? How can you bring the system back to it's initial state?reversible

And you also think that this process is??isothermal

Could you write down for me the the equation that describes theof the system [tex]f(p,V,T)=0[/itex] in order to calculate the change of entropy bystate

[tex] \Delta S=\int^{f}_{i}\frac{dQ}{T}[/tex]

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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- #11

Rainbow Child

- 365

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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

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

I was not referring to any particular process.

And that about that?

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

http://arxiv.org/abs/quant-ph/0507145

- #12

Fernsanz

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Butwork. It's electrical energy provided to the resistance. If you want to include to your system the resistance as you sayit isn't

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

[tex]dU= dQ + dW_{mechanical} + dW_{electrical} + \ldots[/tex]

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 isanotbecause the charges are moving through the boundaries of your systemclosed system

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

And that about that?

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

First of all, sorry because I didn't mean to say Gibbs's paradox. I mean

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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