# Confusion on entropy change calculations for irreversible process

1. Feb 10, 2015

### Shan K

I was studying 2nd law thermodynamics. In that context found the clausius's inequality saying closed integral of dQ/T <0 for irreversible process. And from the reversible process entropy was defined. And from that view they said that for irreversible process dS>dQ/T. Now when I saw some examples of calcutating entropy for irreversible process, they connected two states by a reversible path and calculated entropy change because entropy is a state function . My question is at one side clausius inequality is saying that for irreversible process dS>dQ/T and at other hand they are calculating entropy change in irreversible process by dS=dQ/T.
Sorry for my bad english and as I am posting from mobile I can't able to use latex for mathematics so apolozise for that .

2. Feb 10, 2015

### Quantum Defect

Entropy is a state function. In calculating the change in entropy for a system, you can use this fact to your advantage.

Delta S (of the universe) is always greater than zero for irreversible processes.

Two different things -- the system (you can define this) and the universe (system plus surroundings)

When you calculate the change in entropy of the system, you do not need to calculate this along the path that the system actually takes -- just like any other state function. Since Entropy is a state function for the system, this difference does not matter.

3. Feb 10, 2015

### Shan K

Ok . But the book was saying that integral of dQ/T between two state points for reversible process is greater than the integral of dQ/T between the same two points in irreversible process . Then how can one define a proper entropy change for irreversible process . because it was defined as dS= dQ/T for reversible process.

4. Feb 10, 2015

### Useful nucleus

I think you are confusing "quasi-static" with "reversible". Quasi-static means sufficiently slow and in that case dS=dQ/T whether the process is reversible or irreversible. On the other hand reversible means change in S is zero whereas irreversible means change in S is > 0.

5. Feb 10, 2015

### Staff: Mentor

I sympathize with your confusion. The integral of dQ/T along an irreversible path is not the change in entropy for a system between the initial and final equilibrium states. The change in entropy for a system between the two equilibrium states is the maximum value of the integral of dQ/T over all possible paths, both reversible and irreversible. Fortunately, this maximum value is also the same as the integral over each and every reversible path between the two equilibrium states. So, to determine the change in entropy, you can't use the actual irreversible path. You need to dream up a reversible path between the same two equilibrium states (any reversible path will do), and calculate the value of the integral of dQ/T over that path.

I have written up some notes on this that I think may be helpful to you. Here are the notes, which are really pretty short:

FIRST LAW OF THERMODYNAMICS

Suppose that we have a closed system that at initial time ti is in an initial equilibrium state, with internal energy Ui, and at a later time tf, it is in a new equilibrium state with internal energy Uf. The transition from the initial equilibrium state to the final equilibrium state is brought about by imposing a time-dependent heat flow across the interface between the system and the surroundings, and a time-dependent rate of doing work at the interface between the system and the surroundings. Let $\dot{q}(t)$ represent the rate of heat addition across the interface between the system and the surroundings at time t, and let $\dot{w}(t)$ represent the rate at which the system does work on the surroundings at the interface at time t. According to the first law (basically conservation of energy),
$$\Delta U=U_f-U_i=\int_{t_i}^{t_f}{(\dot{q}(t)-\dot{w}(t))dt}=Q-W$$
where Q is the total amount of heat added and W is the total amount of work done by the system on the surroundings at the interface.

The time variation of $\dot{q}(t)$ and $\dot{w}(t)$ between the initial and final states uniquely characterizes the so-called process path. There are an infinite number of possible process paths that can take the system from the initial to the final equilibrium state. The only constraint is that Q-W must be the same for all of them.

If a process path is irreversible, then the temperature and pressure within the system are inhomogeneous (i.e., non-uniform, varying with spatial position), and one cannot define a unique pressure or temperature for the system (except at the initial and the final equilibrium state). However, the pressure and temperature at the interface can be measured and controlled using the surroundings to impose the temperature and pressure boundary conditions that we desire. Thus, TI(t) and PI(t) can be used to impose the process path that we desire. Alternately, and even more fundamentally, we can directly control, by well established methods, the rate of heat flow and the rate of doing work at the interface $\dot{q}(t)$ and $\dot{w}(t)$).

Both for reversible and irreversible process paths, the rate at which the system does work on the surroundings is given by:
$$\dot{w}(t)=P_I(t)\dot{V}(t)$$
where PI(t) is the pressure at the interface and $\dot{V}(t)$ is the rate of change of system volume at time t. However, if the process path is reversible, the pressure P within the system is uniform, and

$P_I(t)=P(t)$ (reversible process path)

Therefore, $\dot{w}(t)=P(t)\dot{V}(t)$ (reversible process path)

Another feature of reversible process paths is that they are carried out very slowly, so that $\dot{q}(t)$ and $\dot{w}(t)$ are both very close to zero over then entire process path. However, the amount of time between the initial equilibrium state and the final equilibrium state (tf-ti) becomes exceedingly large. In this way, Q-W remains constant and finite.

SECOND LAW OF THERMODYNAMICS

In the previous section, we focused on the infinite number of process paths that are capable of taking a closed thermodynamic system from an initial equilibrium state to a final equilibrium state. Each of these process paths is uniquely determined by specifying the heat transfer rate $\dot{q}(t)$ and the rate of doing work $\dot{w}(t)$ as functions of time at the interface between the system and the surroundings. We noted that the cumulative amount of heat transfer and the cumulative amount of work done over an entire process path are given by the two integrals:
$$Q=\int_{t_i}^{t_f}{\dot{q}(t)dt}$$
$$W=\int_{t_i}^{t_f}{\dot{w}(t)dt}$$
In the present section, we will be introducing a third integral of this type (involving the heat transfer rate $\dot{q}(t)$) to provide a basis for establishing a precise mathematical statement of the Second Law of Thermodynamics.

The discovery of the Second Law came about in the 19th century, and involved contributions by many brilliant scientists. There have been many statements of the Second Law over the years, couched in complicated language and multi-word sentences, typically involving heat reservoirs, Carnot engines, and the like. These statements have been a source of unending confusion for students of thermodynamics for over a hundred years. What has been sorely needed is a precise mathematical definition of the Second Law that avoids all the complicated rhetoric. The sad part about all this is that such a precise definition has existed all along. The definition was formulated by Clausius back in the 1800's.

Clausius wondered what would happen if he evaluated the following integral over each of the possible process paths between the initial and final equilibrium states of a closed system:
$$I=\int_{t_i}^{t_f}{\frac{\dot{q}(t)}{T_I(t)}dt}$$
where TI(t) is the temperature at the interface with the surroundings at time t. He carried out extensive calculations on many systems undergoing a variety of both reversible and irreversible paths and discovered something astonishing. He found that, for any closed system, the values calculated for the integral over all the possible reversible and irreversible paths (between the initial and final equilibrium states) was not arbitrary; instead, there was a unique upper bound (maximum) to the value of the integral. Clausius also found that this result was consistent with all the "word definitions" of the Second Law.

Clearly, if there was an upper bound for this integral, this upper bound had to depend only on the two equilibrium states, and not on the path between them. It must therefore be regarded as a point function of state. Clausius named this point function Entropy.

But how could the value of this point function be determined without evaluating the integral over every possible process path between the initial and final equilibrium states to find the maximum? Clausius made another discovery. He determined that, out of the infinite number of possible process paths, there existed a well-defined subset, each member of which gave the same maximum value for the integral. This subset consisted of what we call today the reversible process paths. So, to determine the change in entropy between two equilibrium states, one must first dream up a reversible path between the states and then evaluate the integral. Any other process path will give a value for the integral lower than the entropy change.

So, mathematically, we can now state the Second Law as follows:

$$I=\int_{t_i}^{t_f}{\frac{\dot{q}(t)}{T_I(t)}dt}\leq\Delta S=\int_{t_i}^{t_f} {\frac{\dot{q}_{rev}(t)}{T(t)}dt}$$
where $\dot{q}_{rev}(t)$ is the heat transfer rate for any of the reversible paths between the initial and final equilibrium states, and T(t) is the system temperature at time t (which, for a reversible path, is equal to the temperature at the interface with the surroundings). This constitutes a precise mathematical statement of the Second Law of Thermodynamics.

6. Feb 10, 2015

### Staff: Mentor

This is not correct. dS=dQ/T only for a reversible path between the initial and final equilibrium states of a system.
This is true only for an isolated system.

7. Feb 10, 2015

### Useful nucleus

Chestermiller, the heat transfer from a hot body to a cold body is an example of an irreversible process for which dS=dQ/T.

8. Feb 10, 2015

### Staff: Mentor

Please tell me you don't really think that.

Chet

9. Feb 11, 2015

### Shan K

Thanks for the beautiful description .
So you are saying that for every process the entropy change is fixed whether we take the reversible or irrevesible one. But for reversible paths thhe integral give us just the change in entropy for two states but for the irreversible paths we get less than that of the change in entropy . So it is the mathematical notion which is different for those two paths but the changing of entropy is fixed if the states are fixed. Right ?

10. Feb 11, 2015

### Shan K

Ya. But it is not the whole description . We use dS=dQ/T for calculating the entropy change . But in this process the system goes from states (say) a to b but the surrounding is not it is going to some other states thats why when we take the system from b to a ,it able to recover its initial state but as the surrounding didnt transit between states a to b as by the system it will not able to come to its initial value thats why the total eenntropy of the universe increases.

11. Feb 11, 2015

### Staff: Mentor

Yes. Exactly.

Chet

12. Feb 11, 2015

### Staff: Mentor

Hi Useful nucleus. I think what you are saying here is that, for a process path involving the irreverisble transfer of heat from a hot body to a cold body, the integral of dQ/T along the irreversible process path gives you the correct change of entropy? If this is what you are saying, I can show you why this statement is incorrect. If you are willing to work with me, I can precisely define such a process and then, with your help, determine both the integral of dQ/T along the reversible path as well as the change in entropy. We can then compare the results. Are you willing to work with me and try this?

Chet

13. Feb 11, 2015

### Useful nucleus

Hi Chestermiller, sure I'm happy to work on that with you but please allow sometime for me to get back to you in this thread. Just also to clarify, in this example I mentioned , the two bodies are perfectly isolated from the surrounding. All what is allowed is heat transfer from the hot to the cold body "quasi-statically". My claim is that the process is irreversible, and my second claim is that TdS=dQ. The second claim holds in its differential form and no need to expect that integrating two equal differentials will give different quantities. Thank you for your well to clarify!

14. Feb 11, 2015

### Staff: Mentor

OK. Let's see where this takes us. We have two identical flat slabs of the same material, one at temperature TH and the other at temperature TC. At time zero, we bring the bodies into direct contact with one another and allow them to thermally equilibrate. Is this what you had in mind, or do we need to add something else to obtain your quasi-static constraint? Or, is this OK to start with?

Chet

15. Feb 11, 2015

### Useful nucleus

That sounds good, Chet. Although for such thought experiments I tend to use (ideal) gases and adiabatic or diathermal walls for simplicity , but the example you brought is perfectly fine. I may add the isolation from the surrounding to focus on the two bodies.

16. Feb 12, 2015

### DrDu

Given two objects with temperature $T_1$ and $T_2$ exchanging heat, the entropy produced is $dS=\delta Q/T_1-\delta Q/T_2$, assuming that $T_2>T_1$, $\delta Q>0$ and $dS>0$, too.

17. Feb 12, 2015

### Staff: Mentor

Yes. Isolated as a combination.
Now, what I intend for us to be showing is that, for this irreversible process as posed, the Clausius inequality applies to the combination of the two bodies as well as to each of the bodies individually (between the initial and final equilibrium states of the system): $ΔS>\int{\frac{dQ}{T_I}}$, where TI is the temperature at the interface between the system and the surroundings.

So let's begin. I hope that this is not too elementary for you.

What is the final equilibrium temperature of the two identical slabs?
If m is the mass of each slab and C is the heat capacity of each slab, what is the amount of heat that gets transferred from the hot slab to the cold slab?
At all times after the instant that the two identical flat slabs are brought into contact with one another, what is the temperature at the interface between the two slabs?
What is the change in entropy of the initially hot slab?
What is the change in entropy of the initially cold slab?

This is where I'll stop for now.

Chet

18. Feb 18, 2015

### Staff: Mentor

Useful Nucleus, are you still out there? Is there anyone else out there who would like to offer answers to the questions I posed in my previous post. Is there anyone out there who would like to see me post the answers to the questions I posed my my previous post?

Chet

19. Feb 18, 2015

### Useful nucleus

Chet, I'm definitely still interested in responding and also to see your answers, but as I mentioned above it might take me sometime to respond. Sorry if took longer than expected. But if there is interest to post your replies soon , please feel free to do so.

20. Feb 18, 2015

### Useful nucleus

I agree with this, DrDu. But I'll try to find an intuitive argument to reach to this conclusion.