Homework Solution Draw out all 15 possible microstates for a system of three molecules with equally-spaced (1 quantum) energy levels and four quanta of energy (i.e. N=3, E=4). Group the microstates by their occupation numbers into their 4 possible configurations. Demonstrate using Equations (4) and (5) that each configuration satisfies the total N and E constraints on the system. n4 = 1 n3 = 0 n2 = 0 n1 = 0 n0 = 2 1 2 4 3 1 2 4 3 1 2 4 3 1 2 4 3 1 2 4 3 1 2 4 3 n3 = 0 n2 = 2 n1 = 0 n0 = 1 1 2 4 3 1 2 4 3 1 2 4 3 1 2 4 3 1 2 4 3 1 2 4 3 n3 = 1 n2 = 0 n1 = 1 n0 = 1 1 2 4 3 1 2 4 3 1 2 4 3 n0 = 0 n2 = 1 n1 = 2

Homework Solution Draw out all 15 possible microstates for a system of three molecules with equally-spaced (1 quantum) energy levels and four quanta of energy (i.e. N=3, E=4). Group the microstates by their occupation numbers into their 4 possible configurations. Demonstrate using Equations (4) and (5) that each configuration satisfies the total N and E constraints on the system. n4 = 1 n3 = 0 n2 = 0 n1 = 0 n0 = 2 1 2 4 3 1 2 4 3 1 2 4 3 1 2 4 3 1 2 4 3 1 2 4 3 n3 = 0 n2 = 2 n1 = 0 n0 = 1 1 2 4 3 1 2 4 3 1 2 4 3 1 2 4 3 1 2 4 3 1 2 4 3 n3 = 1 n2 = 0 n1 = 1 n0 = 1 1 2 4 3 1 2 4 3 1 2 4 3 I N = 2 + 0 + 0 + 0 + 1 = 3 E = 1x4 = 4 II N = 1 + 0 + 2 + 0 + 0 = 3 E = 2x2 = 4 III N = 1 + 1 + 0 + 1 + 0 = 3 E = 1x1 + 3x1= 4 IV N = 0 + 2 + 1 + 0 + 0 = 3 E = 2x1 + 1x2 = 4 n0 = 0 n2 = 1 n1 = 2

Can we predict the probability of a configuration? To do this, we make one of the fundamental assumptions of statistical thermodynamics, The Equal Probability Assumption. Every microstate that satisfies the given N and E occurs with equal probability. That is, whatever processes occur to redistribute energy amongst all the possible microstates of the system, there is no bias that favours some microstates over others. How does this help? Each microstate is one possible outcome when a system distributes available energy among available molecules. All the microstates together describe all the possible outcomes. So a microstate can be viewed as simply one outcome amongst all the possible outcomes, like pulling one card from a deck…

Probability as the fraction of successful outcomes When drawing a single card from a fair deck of 52 cards, each card represents one outcome out of the 52 possible outcomes. The probability of drawing, say, the two of clubs is thus 1/52. There is only one way of choosing the two of clubs (one outcome) out of the 52 possible outcomes. The probability of drawing any club (any one of 13 equally-probable outcomes) is 13/52, or 1/4. There are thirteen ways of choosing a club (13 outcomes) out of the 52 possible outcomes. The probability of drawing a two of any suit (one of 4 equally-probable outcomes) is 4/52, or 1/13. There are four ways of choosing a two (4 outcomes) out of the 52 possible outcomes.

Can we predict the probability of a configuration? The probability of any particular microstate is 1/Wtot. There is only one way of distributing the energy among the available molecules (one outcome) out of the Wtot possible outcomes. The probability of any configuration is Pr(n0, n1, n2,…) = W(n0, n1, n2,…)/Wtot. There are W(n0, n1, n2,…) ways of distributing the available energy to give the same set of occupation numbers W(n0, n1, n2,…) outcomes) out of the Wtot possible arrangements. The sum of all the probabilities for all configurations must be one, as this accounts for all possible outcomes. i.e.

Number of Microstates W Increase the size and available energy (N and E) and things get simpler! Let’s look at a slightly larger system with N=10 and E=5. Configuration Label Occupation Numbers Number of Microstates W Probability W/Wtotal I n0=9, n5=1 10 0.005 II n0=8, n1=1, n4=1 90 0.045 III n0=8, n2=1, n3=1 IV n0=7, n1=2, n3=1 360 0.180 V n0=7, n1=1, n2=2 VI n0=6, n1=3, n2=1 840 0.420 VII n0=5, n1=5 252 0.126 Wtotal 2002 1.001 one configuration (VI) occurs with a much higher probability (42%) unlike ‘simple’ configurations like I and VII, the most probable configuration has molecules distributed among the greatest number of states

Equilibrium as the Most Probable Configuration As N and E become large, one configuration – the configuration with the maximum number of microstates – occurs with a probability approaching one. (Law of Large Numbers) The equilibrium configuration is the one which maximises W for a given N and E. Justification: No matter what microstate the system starts out in, the exchange of energy between molecules leads it towards the most probable configuration because every other configuration occurs with near-zero probability. Once a system reaches its most probable configuration, it is extremely improbable that it will undergo any further change in configuration. This stability to change is what we call equilibrium. The system will continue to exchange energy and adopt different microstates, but the probability is overwhelming that these will all have the same configuration.

Finding the Most Probable Configuration For large systems at equilibrium we can neglect all but the most probable configuration. We will denote the occupation numbers for this configuration by n0*, n1*, n2*,… The problem of finding the equilibrium configuration for any chemical system can now be posed as a purely mathematical optimization problem: What are the values of ni* that maximise W for a given N and E? Answer: where ei is the energy of state i. This formula says that the occupation number of the most probable configuration decreases exponentially with increasing energy above the ground state (e0)

Most Probable Configuration The equation for the most probable configuration contains two unknowns, b and n0*. We solve or at least interpret these by applying the two constraints of fixed N and E, which apply to all configurations. b completely determines the distribution of molecules among the various energy levels, and depends on the total available energy, E. If b is large, then only the lowest energy levels will be occupied (the exponential function decays quickly). If b is small, then higher energy levels (excited states) are occupied. As E increases, b decreases.

Flash Quiz! If the energy in a system is increased: there will come a point at which there are more particles in energy level i=1 than in i=0. TRUE or FALSE, why? there will come a point at which there are more excited particles than those in the ground state. TRUE or FALSE, why?

Answer If the energy in a system is increased: there will come a point at which there are more particles in energy level i=1 than in i=0. FALSE: there will come a point at which there are more excited particles than those in the ground state. TRUE, if the energy is high, adding up all the occupation of all the excited states will give you a higher total than the occupation number of the ground state.

What is Temperature? We define a quantity called temperature as follows: If we then choose to make the constant = 0, then this gives Where kB is the Boltzmann constant, and T is the absolute or Kelvin scale of temperature. Notice that b and therefore T is a property of an equilibrium configuration. It is a collective property of the system, and further is only defined for a system at equilibrium.

The Boltzmann Distribution The equilibrium or most probable configuration can be described as Or, equivalently in terms of occupation probability, ni*/N This formula says that the ratio of occupation numbers in excited state i and the ground state 0 increase as temperature increases. That is, more excited states are occupied at higher temperatures.

Ludwig Boltzmann

Boltzmann is considered the father of statistical thermodynamics. Ludwig Boltzmann Boltzmann is considered the father of statistical thermodynamics. He obtained his PhD at age 22. He then worked in this field for over 40 years. Then one summer holiday he hung himself. Enjoy Statistical Thermodynamics!

The Boltzmann Distribution How does the equilibrium distribution look for a system containing a large number (2 million) molecules with equally spaced energy levels (5 x 10-22 J separation) at 50 and 200K? At higher temperature, say 200K, the number of molecules in higher energy states is greater, and of course n0* is smaller.

Summary Next Lecture You should now Be able to calculate the number of microstates corresponding to a particular configuration. Explain how the most probable configuration defines the equilibrium state of a chemical system, and the Law of Large Numbers. Define temperature as a property of the most probable configuration. Do the homework problem. Next Lecture The molecular partition function Degenerate states Heat and Work Ideal Gases