# Why can we say that $\bar{d}Q=TdS$?

When we introduce entropy we do this by saying that: $$\bar{d}Q=TdS.$$ Now I was wondering why this should be true? I know that by looking at a Carnot cycle, we do get this relation for reversible processes. But what about a general process?

This relation is not true for general processes. For a closed system, the general relation is $\delta Q \leq TdS$, as is illustrated by the Clausius Theorem (http://en.wikipedia.org/wiki/Clausius_theorem).

Another way of writing it is $dS=\delta Q/T + dS_{irr}$ where $dS_{irr}$ is the entropy change due to irreversibilitiy in the closed system transformation.

• In statistical physics we thus always study reversible processes? Since the above form is always considered? – Nick Oct 21 '14 at 10:58
• Statistical physics usually deals with static systems. Non-equilibrium thermodynamics is the branch that explores those issues further. – Whelp Oct 21 '14 at 11:16
• but does non-equilibrium not start from the equilibrium-version and build upon it? Everywhere I look I see $dU=-pdV+TdS$ which they treat as general. I always tought that every proces was then approximated by a reversible proces. – Nick Oct 21 '14 at 12:07
• This equation is one form of the fundemental thermodynamic equation. It is valid between equilibrium states. That is because quantities such as $U$ or $T$ are not well defined for non-equilibrium states, since there is no statistical ensemble associated with those states. However, it is often possible to have quasi-equilibrium states that are closely approximated by equilibrium states for which thoese quantities are well defined. This is why those thermodynamic relations can be broadly applied. – Whelp Oct 21 '14 at 12:36
• In non-equilibrium thermodynamics, one common assumption is that of local thermodynamic equilibrium, where each infinitesimal cell in the medium is assumed to be approximated by an equilibrium state, even though there is no equilibrium at the larger scale. The addition of Onsager relations then allows for a description of how those local states interact and evolve. – Whelp Oct 21 '14 at 12:36

$$\bar{d}Q=T\,dS$$
We move from point $P_1$ to $P_2$ in our macrostate space: even if we do so irreversibly, we could, in principle do this reversibly. And, for the whole set of reversible paths linking $P_1$ to $P_2$, there is only one value of the integral $\int_{P_1}^{P_2} \frac{{\rm d}\,Q}{T}$ independent of path, by dint of equality in the Clausius theorem in this case.
What this means is that, once we have defined an $S_0$ at some point $P_0$, then there must be a function of macrostate alone defined by $S(P) = S_0 + \int_{\Gamma(P_0,\,P)}\frac{{\rm d}\,Q}{T}$ where $\Gamma(P_0,\,P)$ is any reversible path between $P_0$ and $P$.
If we move from $P_0$ to $P$ irreversibly, then we "create" entropy $S_{irr}$ in Whelp's notation and thus add $S_{irr} = \int_{P_0}^{P}\frac{{\rm d}\,Q}{T} - S(P)$ (where $S(P)$ is the function of state we've just defined) to the reservoirs in doing so.