Reversible / Irreversible Fundamental Equation

  • #1
aimLow
2
0
As far as I learned, the following statements should be correct (for closed systems, no chemical reactions), irrespective whether the process is reversible or irreversible (since S and V are state variables):

dU=TdS−pdV
dU=dQ+dW

Does this imply, that the statement:

dU≤TdS−pdV

is wrong?
This is e.g. written in [Reichl, L. 2016, Modern Course in Statistical Physics]). Reichl says that "The equality holds for reversible changes, and the inequality holds for changes which are spontaneous."
 

Answers and Replies

  • #2
22,427
5,265
The equality applies to two closely neighboring (differentially separated) thermodynamic equilibrium states. If you follow a tortuous irreversible path between the same two closely neighboring thermodynamic equilibrium states, then you cannot integrate the equation along this irreversible path. Only if the tortuous path is reversible can you do this. In either case, the differential expression gives the correct result for the change from the initial to the final thermodynamic thermodynamic equilibrium states if they are differentially separated.
 
  • #3
aimLow
2
0
Thanks for you answer. I agree with you but I'm still a little bit confused by Reichls statement. However, I would also agree with that if the following is true (closed system, no chemical reactions, Volume work being the only work interaction coordinate).
\begin{align*}
dU &= dQ + dW ~~~(1) \\
dU &= dQ_{rev} + dW_{rev} ~~~ (2) \\
dU &= TdS - pdV ~~~ (3) \\
\end{align*}
Then I can insert Clausius defintion of Entropy ## dS = dQ_{rev}/T ## into eq. (2):
$$ dU = TdS + dW_{rev} $$
Comparing to eq. (3) yields the relation for reversible work:
$$ dW_{rev} = - pdV $$
For the irreversible case, ##dS \ge dQ/T##, in fact, one could write:
$$ dS = dQ/T + dS_{gen} ~~~ (4) $$
with ##dS_{gen}## being the generated entropy during the process. Inserting eq. (4) into eq. (1) yields:
$$dU = TdS - TdS_{gen} + dW ~~~ (5)$$
or, equivalently, due to the fact that ##dS_{gen}## is always positve:
$$dU \le TdS + dW$$
This last equation is equal to that of Reichl (I wrongly cited it, Reichl did not use -pdV, but rather a general work term).
From eq. (5) and eq. (3) it follows, that:
$$dW = -pdV + TdS_{gen}$$
This would imo make sense, when e.g. compressing a gas irreversibly, one does not only need work to overcome the internal pressure ##-pdV##, but also some work to overcome friction (assuming that ##TdS_{gen}## is actually friction).

Can someone confirm, that this is true?
 
  • #4
22,427
5,265
I never work with differentials for irreversible processes. Plus, for irreversible processes, the Clausius inequality should read $$dQ=T_BdS+T_BS_{gen}$$not TdS, where ##T_B## is the boundary temperature of the system with its surroundings rather than the average temperature.
 
Last edited:

Suggested for: Reversible / Irreversible Fundamental Equation

Replies
12
Views
235
  • Last Post
Replies
1
Views
263
Replies
4
Views
366
Replies
2
Views
217
Replies
6
Views
444
Replies
3
Views
604
Replies
1
Views
941
  • Last Post
Replies
19
Views
2K
Top