Problem 26 - A Modern Course in Statistical Physics

Question

Two containers, each of volume \(V\), contain ideal gas held at temperature \(T\) and pressure \(P\). The gas in chamber 1 consists of \(N_{1, a}\) molecules of type \(a\) and \(N_{1, b}\) molecules of type \(b\). The gas in chamber 2 consists of \(N_{2, a}\) molecules of type \(a\) and \(N_{2, b}\) molecules of type \(b\). Assume that \(N_{1, a}+N_{1, b}=\) \(N_{2, a}+N_{2, b}\). The gases are allowed to mix so the final temperature is \(T\) and the final pressure is \(P\). (a) Compute the entropy of mixing. (b) What is the entropy of mixing if \(N_{1, \mathrm{a}}=N_{2, \mathrm{a}}\) and \(N_{1, \mathrm{~b}}=N_{2, \mathrm{~b}}\). (c) What is the entropy of mixing if \(N_{1, \mathrm{a}}=N_{2, \mathrm{~b}}\) and \(N_{1, \mathrm{~b}}=N_{2, \mathrm{a}}=0\). Discuss your results for (b) and (c).

Step-by-Step Solution

Answer

Case (a): entropy change; Case (b): zero entropy change; Case (c): negative entropy change \(-2 k_B N_1 \ln 2\)

Step 1: Understanding the Problem

Two containers with ideal gases are mixed, such that the temperature and pressure remain constant. We need to calculate the entropy of mixing in different scenarios.

Step 2: Entropy of Mixing Formula

The entropy of mixing for ideal gases can be calculated using the formula: \[ \Delta S_{mix} = -k_B \left( N_{1, a} \ln \frac{N_{1, a}}{N_1} + N_{1, b} \ln \frac{N_{1, b}}{N_1} + N_{2, a} \ln \frac{N_{2, a}}{N_2} + N_{2, b} \ln \frac{N_{2, b}}{N_2} - (N_{1, a} + N_{1, b} + N_{2, a} + N_{2, b}) \ln 2 \right) \]

Step 3: Simplify Using Given Information

Since it's given that \(N_{1,a} + N_{1,b} = N_1\) and \(N_{2,a} + N_{2,b} = N_2\), clean up the formula to focus on individual types: \[ \Delta S_{mix} = -k_B \left( N_{1, a} \ln \frac{N_{1, a}}{N_1} + N_{1, b} \ln \frac{N_{1, b}}{N_1} + N_{2, a} \ln \frac{N_{2, a}}{N_2} + N_{2, b} \ln \frac{N_{2, b}}{N_2} - (N_1 + N_2) \ln 2 \right) \]

Step 4: Compute Entropy for Case (a)

Substitute values for general case: \[ \Delta S_{mix} = -k_B \left( N_{1,a} \ln \frac{N_{1,a}}{N_1} + N_{1,b} \ln \frac{N_{1,b}}{N_1} + N_{2,a} \ln \frac{N_{2,a}}{N_2} + N_{2,b} \ln \frac{N_{2,b}}{N_2} - (N_1 + N_2) \ln 2 \right) \]

Step 5: Compute Entropy for Case (b)

If \(N_{1,a} = N_{2,a}\) and \(N_{1,b} = N_{2,b}\), both \(N_1\) and \(N_2\) are shared equally, making entropies mixed: \[ \Delta S_{mix} = -k_B \left( 2N_{1,a} \ln \frac{N_{1,a}}{2N_{1,a}} + 2N_{1,b} \ln \frac{N_{1,b}}{2N_{1,b}} - 2(N_{1,a} + N_{1,b}) \ln 2 \right) = 0 \]

Step 6: Compute Entropy for Case (c)

Given \(N_{1,a} = N_{2,b}\) and \(N_{1,b} = N_{2,a} = 0\), substitute the values: \[ \Delta S_{mix} = -k_B \left( N_{1,a} \ln \frac{N_{1,a}}{N_{1,a}} + N_{2,b} \ln \frac{N_{2,b}}{N_{2,b}} - (N_{1,a} + N_{2,b}) \ln 2 \right) = 0 - k_B (N_1\ln 1 + N_2 \ln 1) = -k_B \cdot 2N_1 \ln 2 \]

Step 7: Analyze and Discuss Results for (b) and (c)

In case (b), no entropy change occurs because the number of molecules of each type remains unchanged pre- and post-mixing. In case (c), a significant entropy change appears because only type 'a' gas molecules exist, necessitating a splitting of their volumes which invokes mixing.

Key Concepts

Ideal Gas Law

The Ideal Gas Law is a core principle in thermodynamics that describes the behavior of ideal gases. It is formulated as: \( PV=nRT \), where:

• \( P \) represents pressure,
• \( V \) is the volume,
• \( n \) is the number of moles,
• \( R \) is the universal gas constant,
• \( T \) is the temperature in Kelvin.

The law suggests that for an ideal gas, the product of pressure and volume is proportional to the product of the number of moles and temperature. In our exercise, the two gas containers abide by this law, holding a specific volume \( V \), temperature \( T \), and pressure \( P \). As the gases mix but their temperature and pressure remain constants, we are primarily concerned with changes in their entropy rather than their physical states. This principle helps us understand how gases interact and behave under varying conditions.

Statistical Mechanics

Statistical Mechanics links the macroscopic properties of matter to the microscopic behaviors of its particles, providing us with deep insights into thermodynamics. This field uses statistical methods to deal with the collective behavior of large numbers of particles, often leading to the derivation of quantities like entropy. In our exercise, entropy of mixing reflects how the collective randomness or disorder of the gas molecules changes when they are mixed.

• When different types of gas molecules mix, randomness increases, leading to increased entropy.
• However, if the gases consist of the same kinds of molecules, the mixing does not lead to any new configurations or additional randomness.

Statistical Mechanics helps us derive the formula for entropy of mixing: \( \Delta S_{mix} = -k_B \left( N_{1, a} \ln \frac{N_{1, a}}{N_1} + N_{1, b} \ln \frac{N_{1, b}}{N_1} + N_{2, a} \ln \frac{N_{2, a}}{N_2} + N_{2, b} \ln \frac{N_{2, b}}{N_2} - (N_{1, a} + N_{1, b} + N_{2, a} + N_{2, b}) \ln 2 \right)\)Here, \( k_B \) is the Boltzmann constant, and the terms involve natural logarithms of the particle numbers, expressing the contributions of different molecules to the overall entropy change.

Thermodynamics

Thermodynamics is the study of heat, energy, and work. It broadly concerns itself with macroscopic properties and includes laws describing how energy transforms and transfers. One of its key quantities is entropy, a measure of the disorder or randomness in a system. In the given exercise, we are particularly interested in the entropy of mixing, an important concept in thermodynamics. When two different gases mix, their entropy generally increases because the number of possible configurations for the molecules rises.

• In scenario (a), we calculate the entropy change for a generic mix yielding a value based on the specific numbers of each type of molecule.
• In scenario (b), equality in each type means no new mixing entropy, resulting in no change in entropy.
• Scenario (c) looks at a situation where only one type is present, causing a much clearer entropy increase due to the breaking and remixing of individual gas volumes.

Understanding entropy of mixing helps elucidate why certain processes are spontaneous and how energy distributions change within systems.