What does the differential of U mean in an irreversible process?

[zenterix]

Homework Statement

For a closed system the 1st law in differential form is

$$dU=\delta Q+\delta W\tag{1}$$

Relevant Equations

For a reversible process we have

$$\delta Q_{rev}=TdS\tag{2}$$

$$\delta W_{rev}=-PdV\tag{3}$$

and so

$$(dU)_{rev}=TdS-PdV\tag{4}$$

For an irreversible, closed system we have

$$(dU)_{irrev}=\delta Q_{irrev}+\delta W_{irrev}\tag{5}$$

My question is about the following statement

Because the internal energy is a state function we can write

$$(dU)_{irrev}=(dU)_{rev}=TdS-PdV\tag{6}$$

or

$$dU=TdS-PdV\tag{7}$$

Sure, internal energy is a state function. I still don't understand, however, what it means to have a differential change in an irreversible process.

Here is my understanding right now.

$U$ of a closed system is a state function of two of the variables $P,V,$ and $T$.

However, from what I understand, implicit in this function existing is the notion that the system is in equilibrium so that the variables $P,V,$ and $T$ are actually defined for the system.

Suppose we have a Joule expansion occurring in a closed system of volume $V$ which is split into two volumes $V_1$ and $V_2$. The gas is initially in volume $V_1$ and then suddenly fills volume $V_2$.

What does equation (6) mean in this context?

 

[Chestermiller]

Homework Statement: For a closed system the 1st law in differential form is

$$dU=\delta Q+\delta W\tag{1}$$
Relevant Equations: For a reversible process we have

$$\delta Q_{rev}=TdS\tag{2}$$

$$\delta W_{rev}=-PdV\tag{3}$$

and so

$$(dU)_{rev}=TdS-PdV\tag{4}$$

For an irreversible, closed system we have

$$(dU)_{irrev}=\delta Q_{irrev}+\delta W_{irrev}\tag{5}$$

My question is about the following statement



Sure, internal energy is a state function. I still don't understand, however, what it means to have a differential change in an irreversible process.

Here is my understanding right now.

$U$ of a closed system is a state function of two of the variables $P,V,$ and $T$.

However, from what I understand, implicit in this function existing is the notion that the system is in equilibrium so that the variables $P,V,$ and $T$ are actually defined for the system.

Suppose we have a Joule expansion occurring in a closed system of volume $V$ which is split into two volumes $V_1$ and $V_2$. The gas is initially in volume $V_1$ and then suddenly fills volume $V_2$.

What does equation (6) mean in this context?

For an irreversible process, we cannot write dU=TdS-PdV unless this applies to two equilibrium end states that are infinitesimally separated from one another. Otherwise, for an irreversible process, we can only write $$\Delta U=Q-W$$

 

[vela]

What does equation (6) mean in this context?

The point is that if you want to find $\Delta U$ for an irreversible process, you can instead calculate $\Delta U$ for a reversible process that connects the same initial and final states.

 

[zenterix]

The point is that if you want to find $\Delta U$ for an irreversible process, you can instead calculate $\Delta U$ for a reversible process that connects the same initial and final states.

I understand that this is true because if we go from an initial state to a final state the change in $U$ is path-independent.

This concept seems to be expressed by

$$\Delta U=Q+W$$

which is an empirical law and does not seem to specify anything about the intermediate states.

However, the differential form (6) seems to imply intermediate equilibrium states. In addition, it seems that we would be able to integrate (6).

 

[Chestermiller]

I understand that this is true because if we go from an initial state to a final state the change in $U$ is path-independent.

This concept seems to be expressed by

$$\Delta U=Q+W$$

which is an empirical law and does not seem to specify anything about the intermediate states.

However, the differential form (6) seems to imply intermediate equilibrium states. In addition, it seems that we would be able to integrate (6).

Oh really. Do you think the ideal gas law is valid for irreversible processes? If not, how would you integrate to get the work?

 

[zenterix]

The ideal gas law is a state equation. It represents equilibria of an ideal gas. How we get from one equilibrium to another is through some process.

If the process is reversible, then every intermediate state of the process is an equilibrium and so the ideal gas law is true for all such states.

If the process is irreversible, then there are no intermediate equilibrium states and the ideal gas law is not applicable to the intermediate states.

Which brings me to my original question. Here is the portion of the book that generated my question

My question is about equation (5.21) above.

 

[zenterix]

It seems that the idea is that if we go from an initial state with internal energy $U_1$ to a second state with internal energy $U_1+dU$ then since $dU_{rev}=TdS-pdV$ and $dU=dU_{rev}=dU_{irrev}$ then it must be that $dU_{irrev}=TdS-pdV$.

If we keep going with such infinitesimal changes in $U$ then we would be able to say that $dU_{irrev}=TdS-pdV$ for each such step.

But somehow this doesn't make sense to me since the irreversible process doesn't seem to ever proceed in such an infinitesimal manner.

 

[Chestermiller]

It seems that the idea is that if we go from an initial state with internal energy $U_1$ to a second state with internal energy $U_1+dU$ then since $dU_{rev}=TdS-pdV$ and $dU=dU_{rev}=dU_{irrev}$ then it must be that $dU_{irrev}=TdS-pdV$.

If we keep going with such infinitesimal changes in $U$ then we would be able to say that $dU_{irrev}=TdS-pdV$ for each such step.

But somehow this doesn't make sense to me since the irreversible process doesn't seem to ever proceed in such an infinitesimal manner.

it is valid only if the initial and final states are closely neighboring thermodynamic equilibrium states. In terms of TdS and PdV, it is not valid at all points along an irreversible path.

 

[zenterix]

So is the book vague, misleading, or incorrect?

 

[Chestermiller]

So is the book vague, misleading, or incorrect?

Judgment call. I regard it as both incorrect and misleading.

 

[zenterix]

I've been trying to study from multiple Thermodynamics books. They are all misleading to some extent multiple sections. It is probably the most frustrating topic to study I've ever had. :(

The best book is Zemansky and Dittman but it is a bit advanced as a first pass through the subject.

Anyways...

 

[Chestermiller]

Try Moran et al, Fundamentals of Engineering Thermodynamics