Question

In statistical thermodynamics, Stirling's Approximation ($\ln N! \approx N \ln N - N$) is valid only when

• A. N is a small integer
• B. N is very large
• C. The system is at 0 K
• D. The gas behaves non-ideally

Hint:

Stirling's approximation is essential for deriving Boltzmann's entropy formula ($S = k_B \ln \Omega$).
It works because macroscopic thermodynamic systems always have an extremely large number of particles ($N \approx 10^{23}$).

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
The question asks for the condition under which Stirling's approximation is mathematically and physically valid.
Stirling's approximation is a fundamental tool in statistical mechanics used to simplify calculations involving large factorials.

Step 2: Key Formula or Approach: The exact Stirling's series for the factorial of a number $N$ is:
\[ \ln N! = N \ln N - N + \frac{1}{2} \ln(2\pi N) + O(1/N) \]
When $N$ is extremely large ($N \rightarrow \infty$), the terms of lower order than $N$ become negligible compared to the leading terms $N \ln N - N$.

Step 3: Detailed Explanation:

• In statistical thermodynamics, we deal with macroscopic systems containing particles on the order of Avogadro's number ($N \approx 10^{23}$).

• Calculating the exact factorial of such massive numbers is computationally impossible and mathematically tedious.

• By using the approximation $\ln N! \approx N \ln N - N$, the relative error becomes incredibly small as $N$ increases.

• For example, if $N = 10$, the approximation is rough, but for $N > 10^4$, the error is negligible. At $N \approx 10^{23}$, the approximation is practically exact.

• This approximation does not depend on thermodynamic temperature ($0\text{ K}$) or gas ideality; it is a pure mathematical identity that relies on the scale of $N$.

Step 4: Final Answer:
The approximation is valid only when $N$ is very large.