About entropy in Clausius inequality

Question

From what I know, entropy can be derived from Clausius theorem but this theorem put a constraint in the temperature of the surroundings so, how is $\Delta S\ge0$ in an isolated system valid even with variable temperature of the surrounding?

Answer 1

If the system is isolated, it is not interacting with the surroundings by definition, and therefore the variable temperature of the surroundings is not a concern.

Answer 2

In its simplest form, given an isolated system $\Delta S \ge 0$ means that starting from equilibrium and ending in equilibrium you remove internal constraints within the system. These constraints initially keep the system in equilibrium then by removing the constraint you initiate an internal equilibration process and when it stops you compare the total final to the total initial entropy. An example is an adiabatic rigid impregnable/impenetrable wall separating two parts of a gaseous fluid isolated from the outside. Both parts are in equilibrium separately maintained by the isolating wall between them, but upon removing the wall you let the parts settle, molecules diffuse, achieve common temperature and pressure, etc., and there will be a new total entropy that is larger (or equal) than the sum of the entropies in the separate subsystems at start.

Answer 3

Entropy is not derived from the Clausius Inequality. The Clausius Inequality is just a tool we have for establishing an upper bound to the entropy change of a closed system experiencing a thermodynamic process between two thermodynamic equilibrium states.

In my judgment, a more useful version of the Clausius Inequality, applicable to a closed system, (no mass entering or exiting the system) is $$\Delta S\geq\int{\frac{dQ_{Interface}}{T_{Interface}}}$$where $dQ_{Interface}$ is a differential amount of heat flow passing across the interface (boundary) between the system and its surroundings during the process and $T_{Interface}$ is the absolute temperature at the interface through which $dQ_{Interface}$ flows at the time that $dQ_{Interface}$ enters the system. The equal sign applies if the process takes place reversibly, and the > sign applies if the process takes place irreversibly. For an adiabatic process, for which no heat flow is allowed ($dQ_{Interface}=0$) and for which $\int{\frac{dQ_{Interface}}{T_{Interface}}}=0$ (irrespective of the temperature at the interface, the inequality reduces to $\Delta S\geq0$. For an isolated system which, of course, is adiabatic, we have no control over any internal processes that are occurring (such that they all must be irreversible), we must have $\Delta S>0$.

Answer 4

Your "I know" is not correct. The Clausius theorem does not define entropy or concern entropy directly; it states that this integral over one cycle of any cyclic process, including those that may not be representable using equilibrium states at all, is lower or equal to zero:

$$ \int_{cycle} \frac{dQ}{T_r} \leq 0, $$

where $dQ$ is heat accepted by the system from the surroundings during infinitesimal displacement in the cycle and $T_r$ is temperature of the surroundings. In words, sum of reduced heats in a cyclic process can't be positive. Reduced heat transferred over boundary with temperature $T_r$ is defined by the ratio $Q/T_r$.

If surroundings do not have single temperature, or do not have temperature at all, and heat is transferred, the theorem is not applicable.

The law of non-decrease of entropy of an adiabatically isolated system (work can be exchanged), often written as

$$ \Delta S \geq 0 $$ is about a different situation. We have a system that does not exchange heat, but can exchange work. And we do not consider cyclic process, but any process satisfying the condition that no heat is transferred. Then change of entropy of the system cannot be negative.

These two things are connected, but they do not directly follow from each other. One first has to define entropy of any equilibrium state $\mathbf X$ (shorthand for array $(X_1,...X_k)$ of state variables) as the integral in the space of equilibrium states

$$ S(\mathbf X) = \int_{\gamma(\mathbf X_0,\mathbf X)} \frac{dQ}{T} $$

where:

• $\gamma(\mathbf X_0,\mathbf X)$ is any curve joining the two states $\mathbf X_0 , \mathbf X$;

• $dQ = dU-dW$, where work $dW$ is expression of the form $\sum_k -F_k(\mathbf X)dX_k$, where $U$ and all $X_k$ are equilibrium state parameters and $F_k$ are those functions of $\mathbf X$ that lead to correct expression for infinitesimal work $dW$ for that system;

• $T$ is absolute temperature of the system, a function of state $\mathbf X$.

Since in the definition, we integrate over equilibrium states, the curve can be made a part of reversible cycle, for which the Clausius integral is zero. This proves the definition is unique and does not depend on the choice of the curve $\gamma$, only on the endpoints.

Then one can show that for this definition of $S$, if the adiabatic process starts and ends in an equilibrium state, then $\Delta S \geq 0$.