B4-2. Statistical and Macroscopic EntropyHL Only
1. Macroscopic Definition of Entropy
Macroscopically, entropy ($S$) measures the degradation of thermal energy—meaning it quantifies how much energy in a system is scattered and unavailable to perform useful mechanical work. For a reversible process occurring at a constant absolute temperature, the change in entropy ($\Delta S$) is defined as the ratio of heat transferred to that absolute temperature.
Macroscopic Entropy Equation:
- $\Delta S$: Change in the system's entropy in Joules per Kelvin (J K⁻¹)
- $\Delta Q$: Thermal energy reversibly added or removed from the system in Joules (J)
- $T$: Absolute constant temperature of the system during the process in Kelvin (K)
Directional Behavior: When heat enters a system ($\Delta Q > 0$), $\Delta S$ is positive, and system entropy increases. Conversely, when heat is removed ($\Delta Q < 0$), $\Delta S$ is negative, and system entropy decreases.
2. The Second Law of Thermodynamics
The Second Law of Thermodynamics dictates that natural processes possess a fundamental directionality. It states that:
The total entropy of an isolated system (or the entire universe) must always increase or remain constant over time for any spontaneous process. It can never naturally decrease.
While a local system can experience a decrease in entropy (e.g., water freezing inside a freezer), the heat rejected to the surroundings will always increase the entropy of the environment by a value that exceeds the internal system's drop.
3. Microscopic (Statistical) Definition of Entropy
Statistically, Ludwig Boltzmann established that entropy is a direct logarithmic measure of the microscopic disorder of a system. A macroscopic state (defined by $p, V, T$) corresponds to a specific number of unique, invisible arrangements of molecular positions and velocities, known as microstates ($\Omega$).
Boltzmann Entropy Formula:
When a system transitions from an initial state with $\Omega_i$ microstates to a final state with $\Omega_f$ microstates, the change in microscopic entropy is given by:
- $S$: Microscopic entropy of the configuration (J K⁻¹)
- $k_B$: Boltzmann constant ($1.38 \times 10^{-23}$ J K⁻¹)
- $\Omega$: Total number of microstates available to the system (dimensionless)
4. Entropy Change via Isothermal Free Expansion
When a gas expands isothermally from an initial volume $V_i$ to a final volume $V_f$, its particles have access to a larger physical space, increasing the number of available spatial microstates. Because the probability of finding a single molecule in a specific spatial sub-volume scales with total volume, the ratio of final to initial microstates for $N$ independent molecules becomes $(\frac{V_f}{V_i})^N$. Substituting this into Boltzmann's formula yields:
Isothermal Volume Expansion Entropy Formula:
Example 1: Rigorous Microscopic Entropy Analysis
Problem: A vessel is separated into two equal compartments, each of volume $V$. The left compartment holds $3.01 \times 10^{23}$ molecules of an ideal gas, while the right compartment is entirely evacuated. The central partition is punctured, allowing the gas to spontaneously undergo free expansion to fill both chambers (total volume $2V$) at a constant temperature. Calculate the ratio of final to initial microstates ($\frac{\Omega_f}{\Omega_i}$) and determine the exact change in entropy ($\Delta S$).
Solution:
- Step 1: Determine the microstate ratio ($\Omega_f / \Omega_i$).
The initial volume $V_i = V$ and the final volume $V_f = 2V$. The total number of molecules $N = 3.01 \times 10^{23}$.
$$\dfrac{\Omega_f}{\Omega_i} = \left(\dfrac{V_f}{V_i}\right)^N = \left(\dfrac{2V}{V}\right)^N = \mathbf{2^{(3.01 \times 10^{23})}}$$ - Step 2: Apply Boltzmann's equation to calculate $\Delta S$.
$$ \begin{aligned} \Delta S &= k_B \ln\left(\dfrac{\Omega_f}{\Omega_i}\right) = k_B \ln\left(2^N\right) = N k_B \ln(2) \\ &= (3.01 \times 10^{23}) \times (1.38 \times 10^{-23} \text{ J K⁻¹}) \times \ln(2) = 4.1538 \times 0.69315 \\ &\approx \mathbf{2.88 \text{ J K⁻¹}} \end{aligned} $$