Equivalence of Entropy maximization principle and Clausius' Inequality

Question

From Clausius' inequality,
$\oint\frac{dQ}{T}\leq 0$
From this, we can show that $\frac{dQ}{T}\leq dS$

For an isolated system having adiabatic walls, $dQ=0$
So, $dS\geq 0\tag{1}$
So, a isolated system when move towards equilibrium state, its entropy increases (spontaneous process maximizes the entropy).

In Callen's Themodynamics and an Introduction to Thermostatics, the maximum entropy principle is given as

The equilibrium value of any unconstrained internal parameter is such as to maximize the entropy for the given value of the total internal energy.

Mathematically, for an isolated system
if $S(U,x)$, where $x$ is an extensive independent coordinate $\frac{\partial S}{\partial x}\Bigg\rvert_U=0$ and $\frac{\partial^2 S}{\partial x^2}\Bigg\rvert_U<0$

I have the following doubt-

We know that the Clausius' inequality and Entropy maximization principle both are the statements of Second Law of Thermodynamics. I am not able to prove Entropy maximization principle from the Clausius' inequality.
Like (1) is the consequence of Clausius' inequality, but it suggests that entropy in spontaneous process of isolated system increases (maximizes). But this shows that
$\frac{\partial S}{\partial x}\Bigg\rvert_U=0$ and $\frac{\partial^2 S}{\partial x^2}\Bigg\rvert_U<0$ or $\frac{\partial S}{\partial U}\Bigg\rvert_x=0$ and $\frac{\partial^2 S}{\partial U^2}\Bigg\rvert_x<0$ or both.
But entropy maximization principle tells that $\frac{\partial S}{\partial x}\Bigg\rvert_U=0$ and $\frac{\partial^2 S}{\partial x^2}\Bigg\rvert_U<0$ (there is a coordinate x for which system attains maximum entropy at a particular internal energy) holds for sure. Like why instead of maximum entropy at a praticular internal energy, it is not the case that system attains maximum entropy at a particular coordinate for sure?

Answer 1

The entropy is an extensive quantity $$ S = S(E, V, N) $$ and as well all its variables are extensive. That means they will grows linearly as the system grows big: $S \uparrow$ as $E \uparrow$ or $V \uparrow$ or $N \uparrow$. Therefore \begin{align} \frac{\partial S}{\partial E} &\ne 0;\\ \frac{\partial S}{\partial V} &\ne 0;\\ \frac{\partial S}{\partial N} &\ne 0;\\ \end{align}

But, $S$ entropy has to be maximized under the given constrains $E=U$. You then maximum the entropy using Lagrangian un-determine multiplier:

$$ \frac{\partial}{\partial E} \left\{ S - \lambda_E \left(E-U\right) \right\}=0 $$ which then gives $$ \frac{\partial S}{\partial E}\Big\vert_{E=U} = \lambda_E. $$ Compare with Maxwell relation, we conclude that $\lambda_E = \frac{1}{T}$.

Similarly, we maximize $S$ entropy under the given constrains $V=V_0$: $$ \frac{\partial}{\partial V} \left\{ S - \lambda_v \left(V-V_0\right) \right\}=0 $$ which then gives $$ \frac{\partial S}{\partial V}\Big\vert_{V=V_0} = \lambda_v \to \frac{P}{T}. $$

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Let me address more about the 3 principles.

1. What is the meaning of $\frac{\partial S}{\partial E}\Big\vert_{E=U} = \frac{1}{T}$. It is $\Delta F = 0$.

Lets rewrite this equation as : \begin{align} T \Delta S =& \Delta U;\\ \Delta U - T \Delta S =& 0;\\ \Delta F =& 0. \end{align}

For a constant temperature, the equilibrium of an isolated system is determined by the minimum of Helmholtz free energy $F = U - TS$. It quantify the well known two counter balance factors: minimum energy and maximum randomness.

2. Maximum entropy (maximum configurations)

In thermodynamics, the equilibrim of a state is not determined by the maximum of entropy. Then when to apply the maximum entropy principle? The maximum entropy is used in statistical mechanics to determine the distribution function. For microcanonical ensemble. The maximum entropy (maximum configurational number) is the equal probability, every micorstates has equal accessing probability. And for canonical ensemble, the maximum entropy leads to the Boltzmann distribution $p(E) \propto e^{-\beta E}$, and thus the minimum of free energy $F = -KT \ln Z$.

3. About the second law $\Delta S \ge 0$.

This relation is referring to the entropy change of the system or/and the surrounding during a thermal process. A thermal process always involve something exchanged with reservoirs. This law cannot apply to an isolated state. This is mentioned by Bod D. The idea that the thermal processes intent to grow larger the universal total entropy. The "maximization" of the universal entropy is nothing to do with the equilibrium rule of a thermal state, and not related to the statistical maximum entropy rule.

Answer 2

We know that the Clausius' inequality and Entropy maximization principle both are the statements of Second Law of Thermodynamics. I am not able to prove Entropy maximization principle from the Clausius' inequality.

The Clausius equality

$$\oint\frac{dQ}{T}\leq 0$$

applies to any real heat engine cycle, where $Q$ is the heat entering the system at any point during the cycle and $T$ is the temperature at the point of heat entry. Since heat enters the system in the Clausius's inequality, it does not apply to an isolated or adiabatic system. As such I'm not sure you can use the Clausius inequality to imply or prove the entropy maximization principle, which pertains to an isolated system.

On the other hand, it can be shown that the Clausius inequality leads to the increase in entropy principle of the second law, or

$$\Delta S_{tot}=\Delta S_{sys}+\Delta S_{sur}>0$$

The Clausius inequality means that for a real (irreversible) heat engine the entropy transferred to the surroundings by the system in the form of heat is larger than the entropy transferred to the engine from the hot reservoir in the form of heat, the difference being entropy generated in the system.

And since, for any cycle (reversible or not), we always have

$$\Delta S_{sys}=0$$

Then, for an irreversible cycle,

$$\Delta S_{sur}>0$$

Hope this helps.