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The first law of thermodynamics for open systems states that a change in internal energy is given by

$$ d U = d Q + d W + \sum_{i} \mu_i d N_i. $$

When the system undergoes a reversible process, we can set $d Q = T d S$, while when the system performs only pressure-volume work, we set $ dW = - p d V$, to obtain

$$ d U = T d S - p d V + \sum_{i} \mu_i d N_i. $$

Now, on wikipedia, it says

Since U, S and V are thermodynamic functions of state, the above relation holds also for arbitrary non-reversible changes.

Is this statement correct?

One would think that, since for irreversible processes it holds that $T d S > d Q$, we obtain

$$ d U < T d S + p d V + \sum_{i} \mu_i d N_i, $$

which is similar to R.E. Reichl, A modern course in statistical physics, Equation (3.25)

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  • $\begingroup$ Denbigh is much more precise about the exact nature of the irreversible process that is being implied, particularly with regard to the temperature and pressure. Please check it out, because this is the source of your confusion. $\endgroup$ – Chet Miller Sep 13 '18 at 11:28
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Yes it is absolutely correct. A function of state is a property that depends only on the current state of the system, not how it got there. The volume of a system does not "know" whether the system was expanded reversibly or irreversibly. This means we can compute changes in state variables during irreversible processes by finding a reversible process with the same initial and final states and calculating the change in during that process. Since state variables depend only on the start and end points, this must give the same result.

In terms of the implications of $dQ \le TdS$, the first law and the fundamental relation together tell us that $$ dQ + dW = TdS - pdV + \sum_i \mu_i dN_i $$ Clausius inequality then implies that the work on the system satisfies $$ dW \ge -pdV + \sum_i \mu_i dN_i $$ in other words the maximum work done by the system occurs in the case when the equality holds, i.e. in a reversible process.

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  • $\begingroup$ Thank you for your clarifications. I think I understand, however, my understanding may be flawed since I can't figure out the problem with the following argumentation: $\endgroup$ – deadcruiser Sep 13 '18 at 10:58
  • $\begingroup$ Thank you for your clarifications. I think I understand, however, I can't figure out the problem with the following argumentation: Let us consider a closed system. The entropy change during an irreversible process is given by $$ d S = \dfrac{d Q_{rev}}{T} = \dfrac{d Q}{T} + d_i S, $$ where $d_i S >0$ is the internal entropy production due to irreversible processes. Rearranging this equation for $d Q$ and combining with the fundamental equation, we get $$ dU = T(d S - d_i S) -p dV+\sum_i \mu_i d N_i, $$ and since $d_i S > 0$, $$ dU < T d S- p d V + \sum_i \mu_i d N_i, $$ $\endgroup$ – deadcruiser Sep 13 '18 at 11:05
  • $\begingroup$ $dQ$ does not appear in the fundermental relation, it appears in the first law, and for irreversibly processes you cannot mix and match terms from the 2. Explicitly, for an irreversible process, $dQ \le TdS$, so $dU \ge dQ - pdV +\sum\mu_idN_i$. This means when you substitute in your expression for $dQ$ there is no problem. $\endgroup$ – By Symmetry Sep 13 '18 at 11:40
  • $\begingroup$ I think I get it. I made a mistake in my comment; what i wanted to write was: [...] Rearranging this equation for $dQ$, combining with the first law, and writing $W=-p dV$, we get $$ d U = T (d S - d_i S)-p dV + \sum_i \mu_i d N_i $$ [...] I guess my mistake here is to set $W=-p dV$, since when the process is irreversible, this will not be true? $\endgroup$ – deadcruiser Sep 13 '18 at 12:03
  • $\begingroup$ Exactly, for an irreversible process $dW >- pdV$ to compensate for $dQ<TdS$ $\endgroup$ – By Symmetry Sep 13 '18 at 12:58

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