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By the second law of Thermodynamics we can define the state function Entropy up to an additive constant via

$ \Delta S := \int{\frac{\delta Q_{rev}}{T}} $

At the same time the second law of TD gives us the inequality:

$dS := \frac{\delta Q_{rev}}{T} > \frac{\delta Q_{irrev}}{T}$

Hence when taking a transition from an equilibrium A to a thermodynamic equilibrium B that is irreversible, we produce Entropy - So entropy is a state function that is not conserved.

When I look at an isolated System that is described by the state variables $(U,V,N) = const.$ in which entropy is produced via internal irreversible processes (i.e. the whole universe) - where does it go? Isn't entropy supposed to be consant as it is a function of state and $(U,V,N)$ don't change?

$S=S(U,V,N)$

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  • $\begingroup$ I suppose part of the confusion arises from the fact that, while $\text dQ$ is an exact form, $\frac1T\text dQ$ isn't, and therefore the notation $\text dS$ is improper. It is better, imho, to write $\delta S = \frac1T\text dQ$. $\endgroup$ – Phoenix87 Aug 10 '16 at 11:08
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While for an isolated system $U,V,N$ are all well defined and remain constant with time, entropy function $S=f(U,V,N)$ is well defined for an equilibrium state only, which means that properties must be uniform throughout the system. An isolated system in equilibrium (under constraints) may yet contain two subsystems each of which is separately in equilibrium, but one in a different thermodynamic state than the other. Say, you have a container partitioned into two halves by an adiabatic wall, containing gas of different temperature and pressure in each half. In such a case you cannot directly calculate entropy for the entire system, knowing its total $U,V,N$, because properties are not uniform over the entire system. Here we must calculate entropy of the entire system by summing the entropy of all subsystems, $S=S_1(U_1,V_1,N_1)+S_2(U_2,V_2,N_2)$.

If now some process occurs in this isolated system, it means that you have relaxed some constraint within the system. Say you made the wall conducting, so heat flows, and temperature of both subsystems change until they become equal. See that in this process $U_1,U_2$ of the two subsystems change even though $U=U_1+U_2$ for the entire system remains constant. Therefore $S_1,S_2$ also change and so does $S=S_1+S_2$, even though $U,V,N$ for the entire system remains unchanged. In fact $S$ attains the maximum value consistent with the current constraints on the system.

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In the irreversible process you describe for an isolated system, entropy is generated within the system as a result of viscous dissipation of mechanical energy, heat transfer under the action of finite temperature gradients, and mass diffusion under the action of finite concentration gradients. These mechanisms for entropy generation are not present (i.e., not significant) in reversible processes. If you want to determine how much entropy was generated in your irreversible change, you need to devise a reversible path that takes your system between the same two thermodynamic equilibrium states. However, the reversible path does not have to be adiabatic or to involve no work. For this reversible path, the change in entropy between the two states will be the integral of dQ/T. This game plan enables you to determine the amount of entropy that was generated during the irreversible process.

Entropy is not something that needs to be conserved (even though it is a state function). It is something that can be generated, but not destroyed.

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