What is ##Tds## in this equation : ##dU = Tds - PdV##

[MatinSAR]

Homework Statement

What is $Tds$ in this equation : $dU = Tds - PdV$. Is it the transferred heat in a reversible change or it is equal to transferred heat in any change even irreversible changes?

Relevant Equations

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More explanation :
First law is given by $dU=dQ+dW$.
For a reversible change we have:
$dQ = Tds$
$dW = - PdV$
So I rewrite first law as :
$dU=Tds - PdV$
As mentioned before this $Tds$ is the heat transferred in a reversible change. And the $-PdV$ is the work done by system in a reversible change. Then why we use this formula for any change even irreversible changes?

This question was asked by my professor in class and I should answer it in one page. He said we can use in our proof :
$dQ_r/T >= dQ/T$

My idea :
Because dU in the equation is independent of path it is true for any change or process. But that Tds is not equal to the heat transferred in a irreversible change. Sadly what I've said is not logical because if dU is the same for both reversible and irreversible changes so how their dQ can have different values?

 

[Chestermiller]

The equation dU=TdS-PdV describes the relationship between dU, dS, and dV between two closely neighboring thermodynamic equilibrium states , one at (U,S,V) and the other at (U+dU, S+dS, V+dV). For a long and convoluted irreversible path between two arbitrary end states, the equation does not apply at every point along the path. But, if the initial and final thermodynamic equilibrium states are closely neighboring states, the equation does apply to these two end states, although it doesn't apply to all intermediate states (that are not thermodynamic equilibrium states).

 

[MatinSAR]

The equation dU=TdS-PdV describes the relationship between dU, dS, and dV between two closely neighboring thermodynamic equilibrium states , one at (U,S,V) and the other at (U+dU, S+dS, V+dV). For a long and convoluted irreversible path between two arbitrary end states, the equation does not apply at every point along the path. But, if the initial and final thermodynamic equilibrium states are closely neighboring states, the equation does apply to these two end states, although it doesn't apply to all intermediate states (that are not thermodynamic equilibrium states).

Good point. Thanks @Chestermiller . But what about my answer? Was my reasoning wrong? I guess the answer should be something similar to what I've said in post 1.
Becuase I've found something similar to it in a book which my professor uses.

The last paragraph ...

 

[Chestermiller]

Good point. Thanks @Chestermiller . But what about my answer? Was my reasoning wrong? I guess the answer should be something similar to what I've said in post 1.
Becuase I've found something similar to it in a book which my professor uses.
View attachment 338073
The last paragraph ...

All I can say is that the equation does not apply to irreversible paths, except for the initial and final end points which, if they are differentially separated and thermodynamic equilibrium states, the equation can be applied to the difference between these states. Usually, S is considered not definable for non-equiibrium states (i.e., states along an irreversible path).

 

[MatinSAR]

All I can say is that the equation does not apply to irreversible paths, except for the initial and final end points which, if they are differentially separated and thermodynamic equilibrium states, the equation can be applied to the difference between these states. Usually, S is considered not definable for non-equiibrium states (i.e., states along an irreversible path).

I was thinking about your answer. I think it is similar to what I have in my mind especially this part of your answer :

the equation can be applied to the difference between these states

@Chestermiller I appreciate your time.

 

[kuruman]

Consider an ideal gas which expands to, say, twice its volume at constant temperature in two different ways

1. Isothermal expansion in a cylinder in contact with a reservoir at temperature $T$ by moving a piston..
2. Free expansion in an insulated container partitioned into two halves with the gas in one half and vacuum in the other. The partition is removed and the gas expands to fill the entire volume. No heat flows in or out the gas and the gas does no work. The volume doubles, the pressure is halved and the temperature does not change.

In the two expansions above the final and initial states of the gas are the same. What's the difference?

The first process is reversible and can be analyzed using the first law in the form $\Delta U=Q- W$ where $W$ is the work done by the gas. For the isothermal process $\Delta U=0$, therefore $$Q=W=\int_V^{2V}p~dV=nRT\int_V^{2V}\frac{dV}{V}=nRT\ln(2).$$The entropy change of the gas is $$\Delta S_{\text{gas}}=\frac{Q}{T}=nR\ln(2).$$Meanwhile, the entropy change of the reservoir that loses heat $Q$ at constant $T$ is $\Delta S_{\text{res.}}=-\dfrac{Q}{T}=-nR\ln(2).$ Thus, the total entropy change is $$\Delta S_{\text{total}}=\Delta S_{\text{gas}}+\Delta S_{\text{res.}}=0$$ in this reversible process.

In the case of the free expansion in the insulated container, the change in internal energy is zero, the heat entering the system is zero and the work done by the gas is also zero. The first law, as written above, is useless. Now look at what $dU=TdS-pdV$, sometimes referred to as the combined first and second law, can do. The internal energy change is again zero, so we write $$dS_{\text{gas}}=\frac{p}{T}dV=\frac{nR}{V}dV\implies \Delta S_{\text{gas}}=nR\int_V^{2V}\frac{dV}{V} =nR\ln(2)$$ same as in the isothermal expansion because the entropy is a state function. However, the entropy change in the environment outside the container is zero because there is no transfer of heat and/or work between the two. Thus, the total entropy change is $$\Delta S_{\text{total}}=\Delta S_{\text{gas}}+0=nR\ln(2)>0$$in this irreversible process.

By comparing the two processes, you can see that $T\Delta S_{\text{gas}}$ in the irreversible free expansion can be viewed as energy "lost" in the sense that it could have been extracted from the gas if the process between the same two states were a reversible isothermal expansion instead of an irreversible free expansion.

So now you see that the answer to your thread title question, is that $TdS$ is always $TdS$ but may become $dQ$ in reversible processes.

 

[Chestermiller]

Consider an ideal gas which expands to, say, twice its volume at constant temperature in two different ways

1. Isothermal expansion in a cylinder in contact with a reservoir at temperature $T$ by moving a piston..
2. Free expansion in an insulated container partitioned into two halves with the gas in one half and vacuum in the other. The partition is removed and the gas expands to fill the entire volume. No heat flows in or out the gas and the gas does no work. The volume doubles, the pressure is halved and the temperature does not change.

In the two expansions above the final and initial states of the gas are the same. What's the difference?

The first process is reversible and can be analyzed using the first law in the form $\Delta U=Q- W$ where $W$ is the work done by the gas. For the isothermal process $\Delta U=0$, therefore $$Q=W=\int_V^{2V}p~dV=nRT\int_V^{2V}\frac{dV}{V}=nRT\ln(2).$$The entropy change of the gas is $$\Delta S_{\text{gas}}=\frac{Q}{T}=nR\ln(2).$$Meanwhile, the entropy change of the reservoir that loses heat $Q$ at constant $T$ is $\Delta S_{\text{res.}}=-\dfrac{Q}{T}=-nR\ln(2).$ Thus, the total entropy change is $$\Delta S_{\text{total}}=\Delta S_{\text{gas}}+\Delta S_{\text{res.}}=0$$ in this reversible process.

In the case of the free expansion in the insulated container, the change in internal energy is zero, the heat entering the system is zero and the work done by the gas is also zero. The first law, as written above, is useless.

The first law tells us that the change in internal energy of the system is zero, and, from that, the temperature change is zero.

Now look at what $dU=TdS-pdV$, sometimes referred to as the combined first and second law, can do. The internal energy change is again zero, so we write $$dS_{\text{gas}}=\frac{p}{T}dV=\frac{nR}{V}dV\implies \Delta S_{\text{gas}}=nR\int_V^{2V}\frac{dV}{V} =nR\ln(2)$$ same as in the isothermal expansion because the entropy is a state function.

It doesn't follow from the equation dU=TdS-PdV. It follows from evaluating the entropy change for the alternative reversible path in case 1.

 

[kuruman]

It doesn't follow from the equation dU=TdS-PdV. It follows from evaluating the entropy change for the alternative reversible path in case 1.

Are you saying that $dU=TdS-pdV$ is not applicable to the free expansion case and cannot be used to find the entropy change?

 

[Chestermiller]

Are you saying that $dU=TdS-pdV$ is not applicable to the free expansion case and cannot be used to find the entropy change?

Yes. It can not be applied to every step during an irreversible process. It can only be used for the change between two closely neighboring thermodynamic equilibrium states.

 

[kuruman]

I thought that it could be used in this case because we know that $dU=0$ and in the equation that ensues, $dS=nR\dfrac{dV}{V}$, both sides are exact differentials. In that case, the result depends only on the end points and one can integrate without making any assertions or assumptions about what's going on between the end ponts.

 

[MatinSAR]

So now you see that the answer to your thread title question, is that $TdS$ is always $TdS$ but may become $dQ$ in reversible processes.

Thanks for taking time to write this. Good example! So $Tds$ is not equal to $dQ$ in irreversible processes.

Yes. It can not be applied to every step during an irreversible process. It can only be used for the change between two closely neighboring thermodynamic equilibrium states.

Why closely neighboring thermodynamic equilibrium states? This is the only difference between my answer and your guidance. U(internal energy) is a function of state and it is path independent. So why I cannot use a reversible path to calculate it using $TdS-PdV$ for far thermodynamic equilibrium states? Then I don't need to care if it is experiencing an irreversible process or not because U is path independent and depend only on initial and final points.

I thought that it could be used in this case because we know that $dU=0$ and in the equation that ensues, $dS=nR\dfrac{dV}{V}$, both sides are exact differentials. In that case, the result depends only on the end points and one can integrate without making any assertions or assumptions about what's going on between the end ponts.

I was thinking this way ...

 

[vela]

I thought that it could be used in this case because we know that $dU=0$ and in the equation that ensues, $dS=nR\dfrac{dV}{V}$, both sides are exact differentials. In that case, the result depends only on the end points and one can integrate without making any assertions or assumptions about what's going on between the end ponts.

The problem is that variables like $T$, $V$, and $p$ describe thermodynamic equilibrium states of the system. During a free expansion, the system isn't in an equilibrium state, so $T$, $p$, and $V$ for the system aren't well defined. For example, right after the barrier is removed, what's the pressure of the system? At the empty end of the box, there's no force exerted on the wall because the particles haven't made it over to that end, but at the other end, there is a force exerted because some particles are colliding with the wall. There's no one value of $p$ you can assign to the system. If the variables aren't well defined, you can't really use $dU=T\,dS - p\,dV$.

To find the entropy change, however, all you need to do is look at the initial and final equilibrium states. Because $S$ is a state variable, you can calculate $\Delta S$ using any reversible path connecting the same two states. That's essentially what you did. You found $\Delta S$ for a quasistatic isothermal expansion.

 

[MatinSAR]

Thank you to everyone who contributed to this topic. Your comments improved my answer.

 

[kuruman]

The problem is that variables like T, V, and p describe thermodynamic equilibrium states of the system.

I agree and I understand that the intensive variables $T$ and $p$ are not well-defined unless the gas is in equilibrium. I thought that the volume is well-defined. It's $V$ before the partition is removed and $2V$ after. However, to get from $dU=TdS-pdV$ when $dU=0$ to $TdS=nR\dfrac{dV}{V}$ one has to invoke the ideal gas law which contains the poorly-defined pressure and volume. Is that the argument?

Finally, one last question. Is it fair to say then that $dU=TdS-pdV$ is applicable to irreversible processes but not very useful for calculations unless one finds a reversible path from the initial to the final state? Sort of like Gauss's law without the symmetry. I saw all this stuff decades ago and never revisited it.

 

[vela]

I agree and I understand that the intensive variables $T$ and $p$ are not well-defined unless the gas is in equilibrium. I thought that the volume is well-defined. It's $V$ before the partition is removed and $2V$ after.

I don't think the volume is well defined while the gas is expanding either. Is it the actual volume the particles occupy at some instant in time or the total volume they could potentially occupy?

However, to get from $dU=TdS-pdV$ when $dU=0$ to $TdS=nR\dfrac{dV}{V}$ one has to invoke the ideal gas law which contains the poorly-defined pressure and volume. Is that the argument?

Finally, one last question. Is it fair to say then that $dU=TdS-pdV$ is applicable to irreversible processes but not very useful for calculations unless one finds a reversible path from the initial to the final state? Sort of like Gauss's law without the symmetry. I saw all this stuff decades ago and never revisited it.

No and no. The argument is that the relation only applies to equilibrium states. Since the gas is not in equilibrium during the irreversible process of free expansion, the relation doesn't apply.

You might recall when you first encountered thermo in intro physics there was typically a mention of that processes were assumed to be quasi-static—that is, they happen slowly that the gas was essentially in equilibrium during the entire process. This assumption is needed so that we could use relations like $dU = T\,dS-p\,dV$ in analyzing a process.