The Gibbs paradox is a thought experiment in statistical mechanics that highlights a seeming contradiction when considering the mixing of identical particles. It was first discussed by physicist Josiah Willard Gibbs in the late 19th century.
The paradox arises from the following scenario: Consider two identical containers, Container A and Container B, each containing the same ideal gas at the same temperature and pressure. Now, suppose we remove the partition separating the two containers, allowing the gases to mix and reach equilibrium.
According to classical statistical mechanics, the total number of microstates for the combined system should increase when the gases mix. This increase in the number of microstates would suggest that the entropy of the system should also increase. However, the classical definition of entropy implies that identical particles are indistinguishable, meaning that if we exchange the particles between the two containers, there should be no change in the entropy of the system. In other words, the mixing of identical particles should not affect the entropy.
This apparent contradiction between the increase in the number of microstates and the indistinguishability of particles is known as the Gibbs paradox.
The resolution of the Gibbs paradox lies in the quantum mechanical nature of particles. Quantum mechanics treats identical particles as indistinguishable, and the indistinguishability leads to a restriction on the counting of microstates. When considering the quantum mechanical nature of particles, the total number of microstates for the mixed system does not increase as significantly as in the classical case. Therefore, the paradox is resolved, and the increase in entropy upon mixing is consistent with the principles of statistical mechanics.
The Gibbs paradox serves as an important reminder that classical statistical mechanics, while useful in many cases, has limitations and must be reconciled with quantum mechanics when dealing with systems of identical particles.
The mathematical derivation of the Gibbs paradox involves considering the entropy change when mixing identical gases and accounting for the indistinguishability of particles. Here's a brief outline of the derivation:
Start with two containers, Container A and Container B, each containing N identical particles of the same gas.
Assume that the particles are non-interacting and indistinguishable.
The total number of microstates for the combined system is given by the product of the individual microstates for each container:
Ω_total = Ω_A * Ω_B
Clarification
[For each container, let's assume there are two available states for each particle, labeled 1 and 2.
In Container A, we have three particles, and each particle can be in one of the two available states. So, the total number of microstates for Container A, denoted as Ω_A, is given by:
Ω_A = 2 * 2 * 2 = 2^3 = 8
Similarly, in Container B, we also have three particles, and each particle can be in one of the two available states. Therefore, the total number of microstates for Container B, denoted as Ω_B, is also 8.
Now, to find the total number of microstates for the combined system, we multiply the individual numbers of microstates:
Ω_total = Ω_A * Ω_B = 8 * 8 = 64
In this example, the total number of microstates for the combined system is 64.]
If we consider that the particles are distinguishable, the number of microstates for each container is given by the classical expression:
Ω_A = (Ω_single)^N
Ω_B = (Ω_single)^N
where Ω_single represents the number of microstates for a single particle in a container.
The total number of microstates becomes:
Ω_total = (Ω_single)^N * (Ω_single)^N = (Ω_single)^(2N)
The entropy for the combined system is then:
S_total = k_B * ln(Ω_total)
Now, consider the case where we exchange particles between Container A and Container B. In the classical case, this exchange does not change the distinguishability of particles, so the number of microstates remains the same:
Ω_total_classical = Ω_total = (Ω_single)^(2N)
S_total_classical = k_B * ln(Ω_total_classical)
However, in reality, identical particles are indistinguishable, and we must account for this indistinguishability. The correct expression for the number of microstates, considering indistinguishability, is given by the Bose-Einstein statistics (for bosons) or Fermi-Dirac statistics (for fermions).
For bosons (particles with integer spin), the correct expression for the number of microstates is:
Ω_total_indistinguishable = (Ω_single/N)!^2
This accounts for the fact that we cannot distinguish individual particles, so we need to divide by the factorial of N to avoid overcounting.
The entropy for the indistinguishable case becomes:
S_total_indistinguishable = k_B * ln(Ω_total_indistinguishable)Comparing the classical and indistinguishable cases, we have:
S_total_classical = k_B * ln(Ω_total) = 2N * k_B * ln(Ω_single)
S_total_indistinguishable = k_B * ln(Ω_total_indistinguishable) = 2N * k_B * ln((Ω_single/N)!)Simplifying the expressions, we find:
S_total_classical = 2N * k_B * ln(Ω_single)
S_total_indistinguishable = 2N * k_B * [ln(Ω_single) - ln(N!)]The difference between the classical and indistinguishable entropies is:
ΔS = S_total_indistinguishable - S_total_classical = -2N * k_B * ln(N!)The term ln(N!) represents the Stirling approximation, which for large N can be approximated as:
ln(N!) ≈ N * ln(N) - NSubstituting this approximation into the expression for ΔS, we get:
ΔS ≈ -2N * k_B * (N * ln(N) - N) = -2N * k_B * N * ln(N) + 2N^2 * k_BIn the thermodynamic limit (N → ∞), the second term dominates, and we have:
ΔS ≈ 2N^2 * k_BThe paradox arises from the fact that this entropy change is positive, indicating an increase in entropy upon mixing. However, intuitively, one would expect no change in entropy when mixing identical particles.
The resolution of the Gibbs paradox lies in recognizing that the assumption of distinguishability in the classical case leads to an overcounting of microstates. When correctly accounting for the indistinguishability of particles, the entropy change becomes consistent with the principles of statistical mechanics.
The paradox arises from the following scenario: Consider two identical containers, Container A and Container B, each containing the same ideal gas at the same temperature and pressure. Now, suppose we remove the partition separating the two containers, allowing the gases to mix and reach equilibrium.
According to classical statistical mechanics, the total number of microstates for the combined system should increase when the gases mix. This increase in the number of microstates would suggest that the entropy of the system should also increase. However, the classical definition of entropy implies that identical particles are indistinguishable, meaning that if we exchange the particles between the two containers, there should be no change in the entropy of the system. In other words, the mixing of identical particles should not affect the entropy.
This apparent contradiction between the increase in the number of microstates and the indistinguishability of particles is known as the Gibbs paradox.
The resolution of the Gibbs paradox lies in the quantum mechanical nature of particles. Quantum mechanics treats identical particles as indistinguishable, and the indistinguishability leads to a restriction on the counting of microstates. When considering the quantum mechanical nature of particles, the total number of microstates for the mixed system does not increase as significantly as in the classical case. Therefore, the paradox is resolved, and the increase in entropy upon mixing is consistent with the principles of statistical mechanics.
The Gibbs paradox serves as an important reminder that classical statistical mechanics, while useful in many cases, has limitations and must be reconciled with quantum mechanics when dealing with systems of identical particles.
The mathematical derivation of the Gibbs paradox involves considering the entropy change when mixing identical gases and accounting for the indistinguishability of particles. Here's a brief outline of the derivation:
Start with two containers, Container A and Container B, each containing N identical particles of the same gas.
Assume that the particles are non-interacting and indistinguishable.
The total number of microstates for the combined system is given by the product of the individual microstates for each container:
Ω_total = Ω_A * Ω_B
Clarification
[For each container, let's assume there are two available states for each particle, labeled 1 and 2.
In Container A, we have three particles, and each particle can be in one of the two available states. So, the total number of microstates for Container A, denoted as Ω_A, is given by:
Ω_A = 2 * 2 * 2 = 2^3 = 8
Similarly, in Container B, we also have three particles, and each particle can be in one of the two available states. Therefore, the total number of microstates for Container B, denoted as Ω_B, is also 8.
Now, to find the total number of microstates for the combined system, we multiply the individual numbers of microstates:
Ω_total = Ω_A * Ω_B = 8 * 8 = 64
In this example, the total number of microstates for the combined system is 64.]
If we consider that the particles are distinguishable, the number of microstates for each container is given by the classical expression:
Ω_A = (Ω_single)^N
Ω_B = (Ω_single)^N
where Ω_single represents the number of microstates for a single particle in a container.
The total number of microstates becomes:
Ω_total = (Ω_single)^N * (Ω_single)^N = (Ω_single)^(2N)
The entropy for the combined system is then:
S_total = k_B * ln(Ω_total)
Now, consider the case where we exchange particles between Container A and Container B. In the classical case, this exchange does not change the distinguishability of particles, so the number of microstates remains the same:
Ω_total_classical = Ω_total = (Ω_single)^(2N)
S_total_classical = k_B * ln(Ω_total_classical)
However, in reality, identical particles are indistinguishable, and we must account for this indistinguishability. The correct expression for the number of microstates, considering indistinguishability, is given by the Bose-Einstein statistics (for bosons) or Fermi-Dirac statistics (for fermions).
For bosons (particles with integer spin), the correct expression for the number of microstates is:
Ω_total_indistinguishable = (Ω_single/N)!^2
This accounts for the fact that we cannot distinguish individual particles, so we need to divide by the factorial of N to avoid overcounting.
The entropy for the indistinguishable case becomes:
S_total_indistinguishable = k_B * ln(Ω_total_indistinguishable)Comparing the classical and indistinguishable cases, we have:
S_total_classical = k_B * ln(Ω_total) = 2N * k_B * ln(Ω_single)
S_total_indistinguishable = k_B * ln(Ω_total_indistinguishable) = 2N * k_B * ln((Ω_single/N)!)Simplifying the expressions, we find:
S_total_classical = 2N * k_B * ln(Ω_single)
S_total_indistinguishable = 2N * k_B * [ln(Ω_single) - ln(N!)]The difference between the classical and indistinguishable entropies is:
ΔS = S_total_indistinguishable - S_total_classical = -2N * k_B * ln(N!)The term ln(N!) represents the Stirling approximation, which for large N can be approximated as:
ln(N!) ≈ N * ln(N) - NSubstituting this approximation into the expression for ΔS, we get:
ΔS ≈ -2N * k_B * (N * ln(N) - N) = -2N * k_B * N * ln(N) + 2N^2 * k_BIn the thermodynamic limit (N → ∞), the second term dominates, and we have:
ΔS ≈ 2N^2 * k_BThe paradox arises from the fact that this entropy change is positive, indicating an increase in entropy upon mixing. However, intuitively, one would expect no change in entropy when mixing identical particles.
The resolution of the Gibbs paradox lies in recognizing that the assumption of distinguishability in the classical case leads to an overcounting of microstates. When correctly accounting for the indistinguishability of particles, the entropy change becomes consistent with the principles of statistical mechanics.
Comments
Post a Comment