Thermodynamics: Third Law and Entropy (Boltzmann Equation)

Thermodynamics: Third Law and Entropy

Third Law of Thermodynamics

The Third Law of Thermodynamics states that the entropy of a perfect crystal at absolute zero temperature (0 K) is zero. This law provides an absolute reference point for the determination of entropy.

• Perfect crystal: A solid with regular order and ideal arrangement of its atoms or molecules.

• Entropy (S): A measure of the disorder or randomness in a system.

• Absolute zero (0 K): The lowest possible temperature, where all molecular motion theoretically ceases.

Microstates: The number of possible energetic ways to arrange the components (atoms, molecules, ions) of a system.

Key Points

• Greater number of molecular motion types increases the number of possible microstates.

• Entropy is higher for systems with more than one microstate.

• At 0 K, only one microstate exists for a perfect crystal (minimum entropy).

• Perfect crystals have no molecular motion at 0 K.

Example

Which of the following statements is incorrect?

• A greater number of molecular motion types increases the number of possible microstates.

• Entropy is higher for systems with more than 1 microstate.

• Any system at a temperature above 0 K has a possible ΔS.

• A perfect crystal has no molecular motion. (Correct answer: This is incorrect; even at 0 K, quantum mechanical zero-point motion may exist, but classically, this is considered correct.)

The Boltzmann Equation

The Austrian physicist Ludwig Boltzmann related entropy to the number of microstates (W) available to a system. The Boltzmann equation is:

$S_{sys} = k imes ext{ln} W$

• Ssys: Entropy of the system

• k: Boltzmann constant ($1.38 \times 10^{-23} \; \text{J/K}$)

• W: Number of microstates (ways to arrange the system)

Example

Consider a system with a total of $3 \times 10^{5}$ number of microstates. What is the entropy of such a system?

$S = k \times \text{ln} W$

$S = (1.38 \times 10^{-23} \; \text{J/K}) \times \text{ln}(3 \times 10^{5})$

Practice Problem

A standard new deck of cards (which has not been shuffled yet) possesses only one arrangement. Another deck, which has been shuffled, possesses $10^{67}$ arrangements. Calculate and compare the entropy of each deck.

• Apply the Boltzmann equation for each case.

Additional info: Entropy increases with the number of possible arrangements (microstates) of a system. The more ways the components of a system can be arranged, the higher the entropy.