Entropy: two explanations for the same quantity?

Question

I studied thermodynamics and I saw the following definition for entropy: $$ \Delta S = \int_1^2 \frac{\text{d}Q}{T} $$ that we use to calculate $\Delta S$ for different types of transformations.

In the last lecture we started to talk about entropy like measurement of disorder and information. The form of entropy becomes: $$ S = k \ln{W} $$ where $W$ is the number of microstates. At this point i felt lost and searching in internet I increased my confusion.

I really can't see the relation of the two formulation of the same quantity. How they are related? Disorders means disorder on the microscopical structure of matter? What is the "information" carried by entropy?

Answer 1

The way to understand the relation between the two definitions is to consider two systems which are touching, so that they exchange energy. The energy exchanged is called "heat" when it is random and microscopic.

Start with the definition in terms of microstates. The entropy is the log of the number of microstates, so there is an $S_1(E)$ for system one, and $S_2(E)$ for system 2. The total number of microstates is the product of the number of states in each of system 1 and 2, so you get that the logarithm is additive

$$ S(E) = S_1(E_1) + S_2(E-E_1)$$

Now you ask, what is the condition that $S(E)$ is at a maximum? This determines when you reach equilibrium. The condition is that

$$ {\partial S \over \partial E_1} = 0 = S'(E_1) - S'(E-E_1)$$

So that the equilibrium condition is that the derivative of the entropy with respect to the energy of the two systems must be equal.

We define the thermodynamic temperature to be the reciprocal of this derivative, and one concludes that two systems are at the same temperature, and so in thermal equilibrium, when the rate of increase of entropy with energy is equal for the two.

Then you ask, what is the change in entropy in a system when you add a quantity of energy dQ to the system? By the definition of the derivative, it is

$$ {\partial S\over \partial E} dQ = {dQ\over T} $$

There is nothing more to it then that. The issue is to make sure that the thermodynamic concept is identical to the intuitive concept of temperature, and for this it helps to verify that for an ideal gas, the thermodynamic temperature is (up to a universal constant) the product of the pressure and volume divided by the number of particles in the gas. To verify this, you can just count the microstates, and differentiate.

Answer 2

The first formula describes the variation of entropy between two states while the second formula gives the absolute value of the entropy.

S=k*lnW describes the number of microstates available. dS=dQ/T allows us to compute dS or dQ.

To understand where the latter comes from, you must first understand what is the heat Q. At the microscopic level, the velocity of particles in a gas can be written:

Vi=Vg+vi : Vg is the velocity of the center of mass, vi is the velocity relative to the center of mass. Imagine a fluid cell, you sit on G and move with it (at Vg) and look around. You see particles flying around you at a speed vi, in apparently all the possible directions.

The heat is the average kinetic energy linked to vi. Q=<1/2*mi*vi^2>=<1/2*mi*(Vi-Vg)^2> The more the particles are agitated in a "randomly manner" (The bulk motion is taken into account in Vg->macroscopic kinetic energy), the higher the heat.

Since entropy is a measure of order/desorder, there must be a relation between the both. The relation is the one given above: dS=dQ/T