Gibbs Paradox

Derive the statistical expression of entropy for a thermodynamic system ( $S=k \ln W$)

Derivation : We have

\begin{aligned} \Delta S & = \dfrac{dQ}{T} \\ &= \dfrac{\int P\, dV}{T}\\ &= \dfrac{\int \frac{nRT}{V}\, dV}{T}\\ &= nR \int \dfrac{1}{V}\, dV\\ &= nR \ln \left( \dfrac{V_2}{V_1} \right)\\ &= n k N_A \ln \left( \dfrac{V_2}{V_1} \right) \\ &= N k \ln \left( \dfrac{V_2}{V_1} \right)\\ &= k \ln \left( \dfrac{V_2}{V_1} \right)^N\\ &= k \ln \left( \dfrac{V_2^N}{V_1^N} \right)\\ &= k \ln \left( \dfrac{W_2}{W_1} \right)\\ \end{aligned}

(Since thermodynamic probability is proportional to volume available) \[ = k \ln W_2 - k \ln W_1 \] \[ \Rightarrow S_2 - S_1 = k \ln W_2 - k \ln W_1 \] \[ \Rightarrow S = k \ln W \]

Prove that $s_{A}+s_{B}=s_{A B}$ or prove that entropy is additive in nature.

Proof.
Consider a vessel divided into two equal halves, A and B, in which there are two different gases at the same temperature $ T $ and pressure $ P $. Let $ W_A $ and $ W_B $ be the thermodynamic probabilities of the two halves. Entropy of the first half A:

\begin{equation} S_A = k \ln W_A \end{equation}

Entropy of the Second half B:

\begin{equation} S_B = k \ln W_B \end{equation}

Total entropy of (A + B):

\begin{aligned} S_{AB} &= k \ln W_{AB} \\ &= k \ln (W_A W_B)\\ &= k \ln W_A + k \ln W_B \\ &= S_A + S_B \end{aligned}

Thus, entropy of a system is the sum of the entropies of its subsystems. That’s why it is an extensive parameter like volume, area, length, and internal energy.

What is Gibbs' Paradox? How is it resolved by Gibbs?

Gibbs' Paradox arises from classical statistical mechanics, where the entropy appears to increase upon mixing two identica gases — violating the expected additive nature of entropy. This paradox occurs due to the \textbf{incorrect assumption of distinguishability} of identical particles in classical statistics.

Resolution

The paradox is resolved using \textit{quantum statistics}, which treats identical particles as \textbf{indistinguishable}. The meaningless overcounting in classical Boltzmann statistics is corrected by dividing the number of microstates by $ \mathbf{n!} $, where $ n $ is the number of molecules.

Explanation Consider two systems A and B, each at the same temperature $ T $. Let the systems be characterized by:

\[ (N, V, T, m) \quad \text{and} \quad (N, V, T, m) \] where:
$N$ : Number of molecules,
$V$ : Volume,
$m$ : Mass of each molecule,
$T$ : Temperature
Assume the partition separating systems A and B is both insulating and impermeable Then the entropy of the combined system (A + B) is:

\begin{aligned} S_{A+B} &= S_A + S_B \\ &= \left[nk \ln \left( \frac{V}{h^3}(2\pi m k T)^{3/2} \right) + \frac{3}{2} nk \right] \\ &\quad + \left[nk \ln \left( \frac{V}{h^3}(2\pi m k T)^{3/2} \right) + \frac{3}{2} nk \right] \\ &= 2 \left[nk \ln \left( \frac{V}{h^3}(2\pi m k T)^{3/2} \right) + \frac{3}{2} nk \right] \end{aligned}

Now, if we remove the partition and allow the gases to mix (even though they are identical), then: Total number of molecules becomes $ 2n $, Total volume becomes $ 2V $. Then, the entropy becomes:

\[ \begin{aligned} S_{AB} &= (2n)k \ln \left( \dfrac{2V}{h^3}(2\pi m k T)^{3/2} \right) + \dfrac{3}{2}(2n)k \\ &= 2 \left[ nk \ln \left( \dfrac{V}{h^3}(2\pi m k T)^{3/2} \right) + \dfrac{3}{2} nk \right] + 2nk \ln 2 \end{aligned} \]

Conclusion: The additional term $ 2nk \ln 2 $ implies an artificial increase in entropy upon mixing, which is paradoxical since the gases are identical. Quantum mechanics resolves this by acknowledging the indistinguishability of particles and dividing by $ n! $, thereby canceling the extra $ \ln 2 $ term. Hence it is obserbed that before and after removal of Partition the entropy differes by a value $2 n k \ln 2$ violating the additive nature of entropy.

Removal of Gibbs' Paradox

The partition function for one molecule is given by: \[ Z = \dfrac{V}{h^{3}} (2 \pi m k T)^{3/2} \] The partition function for $ n $ molecules (assuming distinguishable particles) is: \[ Z_n = \left( \dfrac{V}{h^{3}} (2 \pi m k T)^{3/2} \right)^n \] \[= \dfrac{V^n}{h^{3n}} (2 \pi m k T)^{3n/2} \] However, since the molecules are indistinguishable, the corrected partition function should be: \[ Z_n = \dfrac{1}{n!} \cdot \dfrac{V^n}{h^{3n}} (2 \pi m k T)^{3n/2} \] Now, the correct entropy is:

\[ \begin{aligned} S &= k \ln Z_n + \dfrac{3}{2} n k \\ &= k \ln \left( \dfrac{V^n}{n! h^{3n}} (2 \pi m k T)^{3n/2} \right) + \dfrac{3}{2} n k \\ &= k \ln \left( \dfrac{V^n}{h^{3n}} (2 \pi m k T)^{3n/2} \right) - k \ln n! + \dfrac{3}{2} n k \\ &\approx k \ln \left( \dfrac{V^n}{h^{3n}} (2 \pi m k T)^{3n/2} \right) - n k \ln n + n k + \dfrac{3}{2} n k \\ &= k \ln \left( \dfrac{V^n}{n^n h^{3n}} (2 \pi m k T)^{3n/2} \right) + \dfrac{5}{2} n k \\ &= k \ln \left( \left( \dfrac{V}{n h^3} (2 \pi m k T)^{3/2} \right)^n \right) + \dfrac{5}{2} n k \\ &= n k \ln \left( \dfrac{V}{n h^3} (2 \pi m k T)^{3/2} \right) + \dfrac{5}{2} n k \end{aligned} \]

Thus, the final expression for entropy is:

\[ S = n k \ln \left( \dfrac{V}{n h^3} (2 \pi m k T)^{3/2} \right) + \frac{5}{2} n k \] or \[ S = n k \ln \left( \dfrac{V}{n h^3} (2 \pi m k T)^{3/2} e^{5/2} \right) \]

This is the famous Sackur–Tetrode equation Now, using this equation, the entropy of the system (A + B) before removal of the partition is:

\begin{aligned} S_{A+B} &= S_A + S_B \\ &= 2 \left[ n k \ln \left( \frac{V}{n h^3} (2 \pi m k T)^{3/2} \right) + \frac{5}{2} n k \right] \end{aligned}

And after removal of the partition, the entropy becomes:

\begin{aligned} S_{AB} &= (2n) k \ln \left( \dfrac{2V}{2n h^3} (2 \pi m k T)^{3/2} \right) + \frac{5}{2} (2n) k \\ &= 2 \left[ n k \ln \left( \dfrac{V}{n h^3} (2 \pi m k T)^{3/2} \right) + \frac{5}{2} n k \right] \end{aligned}

Conclusion: There is no change in entropy before and after mixing identical gases — which resolves the Gibbs' Paradox. The paradox vanishes when quantum indistinguishability is properly accounted for.Which proves both are equal and hence the additive nature of entropy.

What is Nernst's Heat Theorem ?

Ans: This theorem states that at absolute zero temperature, entropy of a system vanishes i.e at $\mathrm{T}=0, \mathrm{~S}=0$. This is also called the third law of thermodynamics. As temperature decreases , the system comes to the lowest quantum state for which the statistical weight W becomes 1 which makes $\ln \mathrm{W}=0$ thus $\mathrm{S}=0$. This fact verifies the therem. But from Sackur Tetrode formula for entropy we find $S$ decreases with decrease in $T$ but at $T=0$ , it is not Zero. Hence Sackur-Tetrode Formula for a Perfect gas though free feom Gibb’s Paradox does not satisy third law of thermodynamics.