1 / 20

The Meaning of Absolute Zero

The Meaning of Absolute Zero. The Boltzmann distribution shows us what is special about absolute zero When T = 0 , All molecules in the system are in their lowest energy (ground) state.

heleneb
Download Presentation

The Meaning of Absolute Zero

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  1. The Meaning of Absolute Zero The Boltzmann distribution shows us what is special about absolute zero When T = 0, All molecules in the system are in their lowest energy (ground) state. That is, n0* = N at T = 0. There can be no temperature lower than absolute zero, because molecules cannot go into a lower energy state. for i ≠ 0

  2. The Molecular Partition Function The equilibrium probability distribution (Boltzmann distribution) tells us the probability of finding a molecule in state i. This may be written as The sum in the denominator of this expression plays an important role in the equilibrium properties of large collections of molecules. It is called the molecular partition function, q(T) The partition function is a function of temperature and the energy levels of the molecule.

  3. Properties of the Molecular Partition Function What kind of physical interpretation can we make of q(T)? It is a weighted sum of the occupation numbers relative to the ground state occupation number, and it increases as the occupation number of excited states goes up. q(T) increases with increasing temperature. If T 0, then q(T) 1 – all molecules are the same ground state. If T ¥, then q(T) the number of molecular states. That is, more (higher-energy) states are occupied as T increases. The molecular partition function, q(T) is, roughly, the number of thermally accessible states per molecule.

  4. Sample Problems: Equilibrium Configuration 1) Given a set of occupation numbers ni and molecular state energies i, how would you determine whether or not the configuration corresponds to an equilibrium configuration? To answer this question you need to make use of a property of the equilibrium distribution, i.e. This expression can be rearranged to give (j - i)-1 ln(ni*/nj*) = 1/kBT = b, a constant for all possible i and j. If you pick any two molecular states and insert their values of the occupation numbers and energies, then this configuration is an equilibrium configuration only if the left hand side of the equation equals a constant that does not change when the same calculation is performed on a different pair of states. Alternatively, you can make a graph of ln(ni*) against i. If the configuration is an equilibrium one, then this will be a straight line with slope b.

  5. Sample Problems: Equilibrium Configuration 2) Having established that a given configuration is an equilibrium configuration, how can you determine the temperature? This follows pretty easily from the previous question. Having established (or been told) that the configuration is an equilibrium one, the temperature is obtained from inverting the right hand side of the equation above and dividing by kB. Make sure your units of energy are consistent with the units of the Boltzmann constant. The alternative approach gives:- ln(ni*) = ln(n0*) + 0/kBT - i /kBT So, a graph of ln(ni*)versus ei is a straight line for the equilibrium distribution, and the slope of that line is -b, or -1/kBT.

  6. Temperature …is a property of an equilibrium configuration. What happens if the configuration isn’t an equilibrium configuration? Can you still define a temperature for such a configuration? The answer is no. Temperature is a property of equilibrium systems only (you can regard it as being defined by the equilibrium occupation numbers). Away from equilibrium, temperature, strictly, ceases to exist and even if there is a nonequilibrium most probable configuration, such as in a pumped laser cavity, the configuration can never be described by a single quantity like T.

  7. Flash Quiz • If energy levels were closer together, would the partition function go up, down, or stay the same?

  8. Answer • If energy levels were closer together, would the partition function go up, down, or stay the same? The occupancy of each level, ni*, would go up, because ei is now closer in energy to e0. So the overall partition function would also go up.

  9. Degeneracy of Molecular States So far we have always referred to molecular states. It if often more natural to talk about molecular energy levels. Spectroscopy, for example, specifically resolves differences in energy levels. The difference between states and energy levels arises because a number of different states may have the same energy and hence share the same energy level. Such states are referred to as degenerate states. The number of states that have the same energy is called the degeneracy of that level. The degeneracy of the energy level  is denoted by g(). A molecule with a doubly-degenerate first excited state would be represented as shown in this energy-level diagram. The occupation probability of each of these states, considering only the most probable configuration, depends only on the energy of the state: Pr1 = n1*/N = n1’*/N. On the other hand, the probability that energy level 1 is occupied is given by Pr(1) = n1*/N + n1*/N = g(1) Pr1.

  10. There are two equivalent ways of writing the partition function q(T), depending on whether you sum over the individual molecular states i or molecular energy levels i, summing over states summing over energy levels Similarly, you may write the Boltzmann distribution by summing over the molecular energy levels i, Homework Problem Consider two energy levels at i= 2.05 x 10-21J and j= 2.64 x 10-21J. The occupation numbers of these in a particular system at T = 200K are ni = 4,761 and nj = 11,535. Energy i is singly-degenerate. (That is, there is only one state of energy i.) kB = 1.381 x 10-23 J K-1 How many states are there with energy j?

  11. Work and Heat The First Law of Thermodynamics decomposes the change in energy of a system into two parts, work and heat. We can write this as dE = dEwork + dEheat Using our definition of the energy of a system at equilibrium as , we can ask “How can a small change in the energy of a system come about?” We write a change in E as DE as We will focus on a very small change in the total energy. To represent this very small change we shall replace  by d (the differential) and we have (16) where we have neglected the third term on the right hand side as it will be considerably smaller than the first two terms since it involves multiplying two very small quantities together.

  12. Work and Heat Putting this expression into words - a small change in E can be accomplished by small changes in the molecular energy levels themselves and by small changes in the equilibrium configuration. Work: Small changes in the molecular energy levels. Heat: Small changes in the occupation numbers of each of the molecular energy levels. With respect to work we can now ask, “How is it possible to change the energies of individual molecular states?”.

  13. The Ideal Gas We will build our ideal gas using the particle in a box model. The energy levels of a 1-D particle in a box are m = mass of particle, L = length of 1-D box, and Planck’s constant h = 6.626 × 10-34 J s. Clearly we can change the energy levels by changing the size of the box, L. As L is decreased and the particle more confined, the energies increase. A particle in a 3-D box in the form of a cube of side-length L has similar energy levels, but now we require three quantum numbers to specify the state. As translation is the only possible degree of freedom for an atomic (or monatomic) gas, the allowed energies are all specified. We can now calculate the molecular partition function, energy, and other properties of the system.

  14. The Ideal Gas Using the known energies, we can calculate the total energy of the ideal gas system as …(18) The molecular partition function is similarly given by a sum over all energy states (with =0) (19) …and the equilibrium occupation number if a state is given by These summations over huge numbers of states are not especially convenient, so we’d like to replace them with an analytical expression. To do this we need the sums to converge conveniently (they don’t) or to approximate the discrete values with a continuous function. That is, we assume that very many states are occupied. In order to decide this we need the concept of a characteristic temperature.

  15. The Characteristic Temperature A characteristic temperature  = (1 - 0)/kB is, roughly, the lowest temperature at which the first excited state is significantly populated. At T>> , many states are occupied. At T << , only the ground state is significantly occupied. How big is trans? For He in a 1m3 box, trans ~ 10-17K. For heavier particles, the characteristic temperature is even lower. Using the partition function we can estimate the number of occupied states of such an ideal gas a T = 1K to be q(1K) ~ 108. Under the condition of T>> , …(20) And the energy of the system is (from equation (15)) (21)

  16. …back to Work Remember that we want to find an expression for the work, the energy change associated with changing allowed (translational) energy levels in our ideal atomic gas. Just a little bit of calculus does this for us. Recall that We will just write this as Which we now substitute into dEwork This derives the ideal gas equation, and the macroscopic relationship between pressure applied to compress a system and increase its energy.

  17. Work and The Ideal Gas - Summary The energy of particle confined in a box of length L goes like i.e. decreasing L means increasing εi Since L = V1/3 we could change the volume to change the energy levels. It turns out (see online notes) that the change in εi is related to the change in volume by and dEwork = -PdV where P = nRT/V, the ideal gas equation

  18. Summary • You should now • Be able to explain the absolute (Kelvin) temperature scale, and what is so special about absolute zero. • Define degeneracy, molecular partition function, characteristic temperature, heat and work • Describe the assumptions and properties of an ideal gas as a theoretical model. • Do the homework problem. Next Lecture • Heat and heat capacity • Separation of degrees of freedom and partition functions

  19. Some Additional Notes – not examinable

  20. The Total Energy of a System We will see that the molecular partition function is a convenient route from allowed energies to many macroscopic system properties. The total energy of a system of molecules is defined by (5) For a system at equilibrium, we may write this as However q(T) = i exp[-(i - 0)], so we can write Rearranging this gives (15) =

More Related