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Physics
Classical Physics
Thermodynamics
Reversible / Irreversible Fundamental Equation
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[QUOTE="aimLow, post: 6834616, member: 731840"] 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? [/QUOTE]
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Classical Physics
Thermodynamics
Reversible / Irreversible Fundamental Equation
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