Is entropy a state property in thermodynamics?

[kelvin490]

The usual "proof" entropy is a state property is like that:

"Consider a system which undergoes a reversible process from state 1 to state 2 along path A, and let cycle be completed along path B, which is also reversible. Since the cycle is reversible we can write:

∫_1-2 δQ / T + ∫_2-1 δQ / T = 0

Now let cycle be completed along path C, but paths B and C represent arbitrary reversible processes. So ∫_2-1 δQ / T is the same for all reversible paths between states 2 and 1."

My question is, isn't the equation above already assume entropy is a state property? Only if it is a state property it can go around a cycle without changes. How can it be valid to prove entropy is state property if it is already assumed from the beginning?

 

[kelvin490]

I think it's just because the equation ∫1-2 δQ / T + ∫2-1 δQ / T = 0 itself is a statement of 2nd law and change of entropy for reversible process is defined as ∫ δQ / T. A direct result from these two points is that entropy doesn't depend on path and is a state property.

 

[td21]

the equation above already does not assume entropy is a state property. It is because the change of entropy from state 1 to 2 is the same as the negative of that from state 2 to 1 for any different paths. Therefore, the change is path independent and depends on the state only.

 

[kelvin490]

the equation above already does not assume entropy is a state property. It is because the change of entropy from state 1 to 2 is the same as the negative of that from state 2 to 1 for any different paths. Therefore, the change is path independent and depends on the state only.

I think we still need the statement of the 2nd law in order to complete the proof. It's because "change of entropy from state 1 to 2 is the same as the negative of that from state 2 to 1 for any different paths" is just the definition of state property, we cannot simply use the definition to "prove" entropy is a state property.

 

[Chestermiller]

Hi Kelvin 490.

I tend to agree with what you are saying. This seems to be a chicken/egg kind of thing. In the final analysis, does it really matter, and who really cares? For a different perspective on entropy as a state function (more aligned with your own conceptual thinking), see my PF Blog.

Chet

 

[Jano L.]

I think it's just because the equation ∫1-2 δQ / T + ∫2-1 δQ / T = 0 itself is a statement of 2nd law and change of entropy for reversible process is defined as ∫ δQ / T. A direct result from these two points is that entropy doesn't depend on path and is a state property.

No, the equation

$$
\int_1^2 \frac{dQ}{T}+\int_2^1 \frac{dQ}{T} = 0
$$

is not a statement of 2nd law. It is a statement that can be derived for reversible cycles from similar equation valid for Carnot cycle (which too is reversible). Based on this equation, entropy may be immediately defined as a state function.

2nd law is a different thing. It says that for any cycle that begins and ends at the same equilibrium state $A$, reversible or not,
$$
\int_1^2 \frac{dQ}{T}+\int_2^1 \frac{dQ}{T} \leq 0
$$
This inequality cannot be derived just from reversible cycles. Usually it is derived from experience that heat does not flow from colder to warmer body spontaneously, or impossibility of complete transformation of accepted heat into work in a cyclic process.

 

[Andrew Mason]

My question is, isn't the equation above already assume entropy is a state property? Only if it is a state property it can go around a cycle without changes. How can it be valid to prove entropy is state property if it is already assumed from the beginning?

For a reversible process, ∂W = PdV. So the first law becomes ∂Q_rev = dU + PdV

Since U, P and V are state variables, ∂Q_rev depends only on the state of the system. And, since T is also a state function, it follows that ∫∂Q_rev/T = ∫dS is a state function.

AM

 

[Andrew Mason]

For a reversible process, ∂W = PdV. So the first law becomes ∂Q_rev = dU + PdV

Since U, P and V are state variables, ∂Q_rev depends only on the state of the system. And, since T is also a state function, it follows that ∫∂Q_rev/T = ∫dS is a state function.

AM

If you thought my "proof" was too simple, you were right. Thanks to Chestermiller for gently pointing this out.

This works only for infinitesimals, where we assume that P does not change. The actual general proof that entropy is a state function is quite difficult. It is fairly easy to show for an ideal gas because:

ΔS = ∫d_Qrev/T = ∫dU/T + ∫PdV/T = nC_v∫dT/T + nRT/T∫dV/V = nC_vln(T_f/T_i) + nRln(V_f/V_i)

i.e. ΔS depends only on the final state not the path.

But to show this to be true in general requires several pages of difficult analysis and was not done, apparently, until Caratheodory did it early in the 20th C. The analysis is presented by Chandrasekar in Chapter 1 of his 1939 book An Introduction to the Study of Stellar Structure


So I apologize for making it out to be a simple matter. Unless one wants to tackle the analysis, it is something one just has to accept for the time being.

AM

 

[Jano L.]

The actual general proof that entropy is a state function is quite difficult...
But to show this to be true in general requires several pages of difficult analysis and was not done, apparently, until Caratheodory did it early in the 20th C...

The fact that entropy is a state function is not that difficult to derive from experience. I believe this was done by Clausius based on the 1st and 2nd law of thermodynamics and the idea of Carnot cycles.

Caratheodory proved mathematical theorem that is related to this, but it is largely irrelevant in thermodynamics because it assumes abstract mathematical property of the internal energy function :


In every neighborhood of any state there exists another state that cannot be reached from the first state by adiabatic process.


From this he proved that for system of any number of degrees of freedom, there exists integrating denominator $T$- function of state variables - that will make it possible to express heat $dQ$ as $TdS$ where also $S$ is function of state variables.

His assumption has little motivation in experience and it was argued that it is not equivalent to the second law of thermodynamics.
In thermodynamics, it is much easier and makes much more sense to derive existence of entropy from the 1st and 2nd law.