Statistical Mechanics: Ideal Gas

In summary, to find the entropy for the mixing of two ideal gases using the microcanonical ensemble, we first need to determine the number of microstates (Ω) for the given energy of the system by integrating over all possible momentum states for both gases. Then, we can use the equation S = kB ln(Ω) to calculate the entropy, making sure to use the correct value of kB for our units.
  • #1
splitringtail
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Homework Statement



Using the microcanonical ensemble, find the entropy for the mixing of two ideal gases, but we need to compute all at once instead of separately for each gas and adding the two.

Homework Equations



[tex]\Omega(E)=\int_{H<E}d\overline{p}d\overline{q}[/tex]

[tex]S(E,V)= k_{B} (Ln[\Omega(E)])[/tex]

The Attempt at a Solution



Now we did an N particle ideal gas in class. I feel this is very similar, but I have the
Hamiltonian

[tex] H = \stackrel{N_{a}}{\sum}} \frac{p^{2}_{i}}{2 m_{a}} + \stackrel{N_{b}}{\sum}} \frac{p^{2}_{j}}{2 m_{b}} [/tex]

Well, since it is independent of the position, then before mixing

[tex]\Omega(E)=V^{N_{a}}_{a}V^{N_{b}}_{b} \int_{H < E} \frac{\stackrel{N_{a}}{\prod}d\overline{p}_{i}}{h^{3N_{a}}}\frac{\stackrel{N_{b}}{\prod}d\overline{p}_{j}}{h^{3N_{b}}}[/tex]

The remaining integral should be a [tex]3(N_{a}+N_{b})[/tex]-dimensional hypersphere w/ a radius of [tex]\sqrt{2 m_{a} m_{b} E}[/tex]

The rest should fall in place, but I wondering if it is that straightforward.
 
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  • #2


Hello, thank you for your post. It seems like you are on the right track with your solution. The microcanonical ensemble is used to calculate the entropy of a system at a fixed energy, volume, and number of particles. In this case, we are considering the mixing of two ideal gases, which means that the total energy, volume, and number of particles of the system are fixed.

To calculate the entropy, we need to first determine the number of microstates (Ω) that correspond to the given energy of the system. This can be done by integrating over all possible momentum states for both gases, as you have done in your attempt at a solution. However, I would suggest using the product rule for integrals to make the calculation a little easier. This would give you an integral over the momentum states for gas A and an integral over the momentum states for gas B, which can then be multiplied together to get the total number of microstates for the system.

Once you have determined the number of microstates, you can use the equation S = kB ln(Ω) to calculate the entropy. Just be sure to use the correct value of kB for your units.

I hope this helps. Let me know if you have any further questions. Good luck with your calculations!
 

Related to Statistical Mechanics: Ideal Gas

1. What is statistical mechanics?

Statistical mechanics is a branch of physics that uses statistical methods to understand the behavior of a large number of particles, such as molecules in a gas. It is based on the idea that the macroscopic properties of a system can be determined by studying the statistical behavior of its microscopic components.

2. What is an ideal gas?

An ideal gas is a theoretical gas consisting of a large number of identical particles that do not interact with each other except through collisions. This means that the particles have no volume and do not exert any forces on each other. Ideal gases are used as a simplified model for real gases in statistical mechanics.

3. How is the behavior of an ideal gas described by statistical mechanics?

Statistical mechanics describes the behavior of an ideal gas by using the Boltzmann distribution and the ideal gas law. The Boltzmann distribution gives the probability of finding a particle in a particular energy state, while the ideal gas law relates the pressure, volume, temperature, and number of particles in the gas.

4. What is the significance of the Boltzmann distribution in statistical mechanics?

The Boltzmann distribution is significant because it allows us to calculate the probability of particles being in different energy states in an ideal gas. This distribution is derived from the principles of statistical mechanics and is used to explain many macroscopic properties of gases, such as temperature and pressure.

5. How does statistical mechanics explain the relationship between temperature and pressure in an ideal gas?

According to statistical mechanics, the temperature of an ideal gas is related to the average kinetic energy of its particles. As the temperature increases, the particles move faster and exert more force on the walls of the container, resulting in an increase in pressure. This relationship is described by the ideal gas law, which states that the pressure of an ideal gas is directly proportional to its temperature and number of particles, and inversely proportional to its volume.

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