First Law of Thermodynamics for Irreversible Processes

[omberlo]

Hello everyone.

I have found quite a lot of conflicting information about the first law applied to irreversible processes and in particular whether the dS term in the equation dU = TdS - pdV only accounts for the entropy transferred or for the total entropy, i.e. dS_transferred + dS_created.

Can someone clarify this for me?

 

[omberlo]

I'll answer my own question and raise more doubts:
http://www.av8n.com/physics/thermo/boundary.html explains that the TdS term is really an interior property and not a boundary-flow term, thus dS must account for the total entropy change, i.e. even the entropy created from sctratch due to irreversibility.

My next doubt is the following:
for an isochoric process dV = 0 and thus dU = TdS.
Many sources state that for this kind of processes dU = heat transferred = T dS_transferred, but is this also true for irreversible processes where some entropy is created and not transferred?
The same goes for enthalpy in an isobaric process (dp = 0).

Thank you.

 

[Studiot]

If, by S, you are referring to entropy the property does not enter into the first law of Thermodynamics.

The equation you offer is sometimes referred to as the 'combined first and second laws' and carries restrictions applicable to both.

go well

 

[omberlo]

Yes, S is entropy.
Yes, that equation does carry restrictions that apply to the first and second law, but (ir)reversibility should not be one of them, since the whole equation involves only state functions which behave equally well for both reversible and irreversible processes.

So, can someone explain to me mathematically why heat transferred equals change in internal energy (dq = dU) for constant volume processes even when going through an irreversible process?

To repeat where exactly my doubt lies:
for an irreversible isobaric process
dU = T(dS_total) = T(dS_transferred + dS_created) = TdS_transferred + TdS_created
where as dQ = T(dS_transferred)
So dU ≠ dQ ?

Thanks.

 

[Studiot]

It is often forgotten that the complete statement of the second law is

\Delta S \ge \frac{{\Delta q}}{T}

Where the equality only holds for reversible processes.

Look up 'the inequality of Clausius'

\oint {ds} < 0

It is true that the entropy change of a body (system in question) is fixed by the use of the state variables the inequality above must be used in the above expression for the second law. This means that your derived equation contains an inequality.
In other words, for a given change of state of a body at uniform temperature the entropy change is fixed, but the heat change depends upon the path.

You should also be aware that the variation of entropy with work is not complete either.

dS = {\left( {\frac{{dS}}{{dp}}} \right)_v} + {\left( {\frac{{dS}}{{dv}}} \right)_p}

Again this is path dependent.

 

[Ken G]

So, can someone explain to me mathematically why heat transferred equals change in internal energy (dq = dU) for constant volume processes even when going through an irreversible process?

It might help to consider an example where you know that dQ = dU for an irreversible process at constant volume: when heat crosses a T difference. Imagine two gases at different T that are brought briefly into thermal contant-- heat dQ will cross from the higher T to the lower T, and this is irreversible. There's no change in V here (both gases are in boxes), so the only way to get heat dQ to come across in the first place is to have a T difference, i.e., for the process to be reversible. So what's actually hard is to come up with a process that has dV=0 and a finite dQ = dU but the process is reversible! That handles the "TdS transferred" part, but your question seemed to be around the "TdS created", which deals with entropy changes without heat transfer. I can't think of any examples of that where dV=0, what do you have in mind there?

 

[netheril96]

If the irreversibility is a result of not being quasi-static, then none of T, S, P, V is well defined, and this equation fails.
If the irreversibility is due to friction, then obviously
{\rm{d}}U = T{\rm{d}}S - p{\rm{d}}U + \boldsymbol{f}\cdot\boldsymbol{x}