Stat-Mech problem: pressure from a partition function

In summary: I think the key idea is that the gas molecules can occupy surface states on the wall in addition to occupying states in the gas phase. When the gas comes to equilibrium with the wall, the pressure will be slightly reduced because some of the gas molecules are no longer moving around and colliding with the wall. Instead, they are stuck to the wall in surface states.In summary, the problem involves a vessel containing dilute helium gas that is in thermal equilibrium with the surroundings at a temperature T. After some time, a number of helium atoms adhere to the walls of the vessel, occupying surface states with binding energy Δ. The partition function for the system is given by Z = q^(N-M) * (N0e^(
  • #1
TroyElliott
59
3

Homework Statement


A vessel having a volume ##V## initially contains ##N## atoms of dilute (ideal) helium gas in thermal equilibrium with the surroundings at a temperature ##T##, with initial pressure ##P_{i} (T ,V ) = \frac{NRT}{V}## . After some time, a number of helium atoms adhere to the walls of the vessel, each occupying one of ##N_{0}## available surface states having binding energy ##\Delta##, where ##N_{0}>>N.## When ##M## atoms are adsorbed on the surface, the partition function for the system is given by

$$Z = \frac{q^{N-M}(N_{0}e^{\beta \Delta})^{M}}{M!(N-M)!}$$

where ##q = V\sqrt{\frac{mkT}{2\pi\hbar^{2}}}.##

Show that the final pressure is ##P_{f} = P_{i}(1+\frac{N_{0}}{q}e^{\beta \Delta})^{-1},##

after equilibrium is reached between the gas and the surface.

Homework Equations


##P_{final} = \frac{1}{\beta}\frac{\partial \ln{Z}}{\partial V}##

The Attempt at a Solution


After using ##P = \frac{1}{\beta}\frac{\partial \ln{Z}}{\partial V},## I get ##P = \frac{(N-M)}{\beta V}.## I am not seeing how I can write this such that the answer appears like ##P_{f} = P_{i}(1+\frac{N_{0}}{q}e^{\beta \Delta})^{-1}.## Any ideas on where to go from here?

Edit: From here I thought that I could relate ##N-M## to the following $$<M> = \frac{Ne^{\beta \Delta}}{Z},$$

or

$$<N-M> = \frac{N}{Z}$$
 
Last edited:
Physics news on Phys.org
  • #2
TroyElliott said:
$$Z = \frac{q^{N-M}(N_{0}e^{\beta \Delta})^{M}}{M!(N-M)!}$$
where ##q = V\sqrt{\frac{mkT}{2\pi\hbar^{2}}}.##

I hope I'm not leading you astray. But, I believe the actual partition function for the system would be obtained by summing the above expression for ##Z## over all allowable values of M. The sum can be carried out without much difficulty and the resultant ##Z## appears to lead to the given expression for the final pressure.

The reason for summing over M is that M will vary in time as molecules in the gas phase randomly attach to the wall and then get kicked off the wall by collisions from other molecules in the gas phase. So, the possible micro-states of the system will include states with various values of M.
 
  • Like
Likes TroyElliott
  • #3
TSny said:
I hope I'm not leading you astray. But, I believe the actual partition function for the system would be obtained by summing the above expression for ##Z## over all allowable values of M. The sum can be carried out without much difficulty and the resultant ##Z## appears to lead to the given expression for the final pressure.

The reason for summing over M is that M will vary in time as molecules in the gas phase randomly attach to the wall and then get kicked off the wall by collisions from other molecules in the gas phase. So, the possible micro-states of the system will include states with various values of M.
TSny said:
I hope I'm not leading you astray. But, I believe the actual partition function for the system would be obtained by summing the above expression for ##Z## over all allowable values of M. The sum can be carried out without much difficulty and the resultant ##Z## appears to lead to the given expression for the final pressure.

The reason for summing over M is that M will vary in time as molecules in the gas phase randomly attach to the wall and then get kicked off the wall by collisions from other molecules in the gas phase. So, the possible micro-states of the system will include states with various values of M.

Thank you for the suggestion. I think it would be strange for the writer of the problem to purposely call ##Z## the partition function, but then not include a summation over the states. I will take a look at your suggestion and see if it leads to the correct answer. Thanks!
 
  • #5

1. What is the partition function in statistical mechanics?

The partition function in statistical mechanics is a mathematical function that describes the distribution of particles in a system at a given temperature. It takes into account the energy levels and degeneracy of the particles in the system.

2. How is pressure calculated from the partition function?

Pressure can be calculated from the partition function by taking the derivative of the logarithm of the partition function with respect to the volume of the system. This is known as the virial theorem and is derived from the statistical mechanics equations.

3. What factors affect the pressure of a system according to the partition function?

The pressure of a system according to the partition function is affected by the temperature, volume, and number of particles in the system. It is also influenced by the interactions between particles and any external forces acting on the system.

4. How does the partition function relate to thermodynamic properties?

The partition function is a fundamental concept in statistical mechanics and is used to calculate thermodynamic properties such as internal energy, entropy, and free energy. These properties can then be used to predict the behavior of a system at different temperatures and volumes.

5. Can the partition function be used for any type of system?

Yes, the partition function can be used for any type of system, including gases, liquids, and solids. It is a versatile tool in statistical mechanics and can be applied to a wide range of physical systems to predict their thermodynamic properties.

Similar threads

Replies
1
Views
777
  • Advanced Physics Homework Help
Replies
1
Views
974
  • Advanced Physics Homework Help
Replies
3
Views
857
  • Advanced Physics Homework Help
Replies
2
Views
1K
  • Advanced Physics Homework Help
Replies
4
Views
977
  • Advanced Physics Homework Help
Replies
3
Views
1K
  • Advanced Physics Homework Help
Replies
1
Views
913
  • Advanced Physics Homework Help
Replies
1
Views
1K
  • Advanced Physics Homework Help
Replies
8
Views
1K
  • Advanced Physics Homework Help
Replies
9
Views
3K
Back
Top