Question

According to the Boltzmann Equation $S = k_B \ln \Omega$, if a system is in a state of "Perfect Order" (a perfect crystal at $0\text{ K}$), the number of microstates $\Omega$ and the entropy $S$ are

• A. $\Omega = 0, S = 0$
• B. $\Omega = 1, S = 0$
• C. $\Omega = 1, S = 1$
• D. $\Omega \to \infty, S \to \infty$

Hint:

The Third Law of Thermodynamics states that the entropy of a perfect crystal at absolute zero is exactly zero.
This corresponds to a single microstate ($\Omega = 1$), representing zero randomness.

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
This question relates statistical thermodynamics to macroscopic thermodynamic behavior.
It asks for the statistical definition of entropy under the specific condition of perfect order at absolute zero, which forms the physical basis of the Third Law of Thermodynamics.

Step 2: Key Formula or Approach:
The Boltzmann entropy equation relates entropy to statistical microstates:
\[ S = k_B \ln \Omega \] Where:
$S$ is the macroscopic entropy.
$k_B$ is the Boltzmann constant.
$\Omega$ is the thermodynamic probability, representing the number of microstates corresponding to the macrostate.

Step 3: Detailed Explanation:

• At absolute zero temperature ($0\text{ K}$), a system is in its lowest energy configuration (the ground state).

• For a perfect crystal in a state of "Perfect Order", there is only one unique microscopic arrangement of atoms that satisfies this macroscopic state.

• Therefore, the number of microstates is $\Omega = 1$.

• Substituting $\Omega = 1$ into the Boltzmann equation:
\[ S = k_B \ln(1) \]
• Since the natural logarithm of 1 is zero ($\ln 1 = 0$), the entropy becomes:
\[ S = 0 \]
• This confirms that the entropy of a perfect crystal at absolute zero is zero.

Step 4: Final Answer:
Under a state of perfect order, $\Omega = 1$ and $S = 0$, which corresponds to option (B).