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Homework Solution 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.

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Presentation on theme: "Homework Solution 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."— Presentation transcript:

1 Homework Solution Consider two energy levels at i = x 10-21J and j = 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 = x J K-1 How many states are there with energy j? If the occupation number of of the singly-degenerate state i is 4761, then the occupation number of one state of energy j is That is, there are11,535/3845 or ~3 states of energy j.

2 Ideal Gas “Road Map” Our model of an ideal atomic gas begins with the quantum mechanical model of a wave/particle in a cubic box of side length L. Solving the Schrödinger Equation yields the translational energy states, ei:- From here we form the molecular partition function:- For mathematical convenience (to make the equation solvable), we would like to replace the sums over discrete states with a continuous function. However this is only physically reasonable if the number of occupied states is very large. This is equivalent to saying that the temperature must be much greater than the characteristic temperature for translation, T>> trans. This gives an analytical expression for the molecular partition function:- We obtain the energy of an ideal gas as a function of temperature:-

3 qtrans Notice that qtrans is a high-temperature approximation.
As T  0, q should  1, but at T=0, As T  0, E should  Ne0, but at T=0, Both of these failings are expected from the assumptions and approximations made in developing this model.

4 Heat We described heat flow by the following microscopic change
If we maintain our constraint that N is fixed, then changing the values of ni* must mean a change in the most probable configuration, and hence a change in temperature. Heat flow is related to temperature change. We define this relationship macroscopically using the heat capacity at constant volume, CV, written as the amount of energy required to raise the temperature of a body by 1K.

5 CV = CV(s) x mass of sample = CV(M) x No. of moles in sample
The heat capacity may be reported in a couple of ways such as The specific heat, CV(s), is the heat capacity of 1g of material. The molar heat capacity, CV(M), is the heat capacity of 1 mole of material The heat capacity of a particular sample is then given by CV = CV(s) x mass of sample = CV(M) x No. of moles in sample

6 Statistics of Heat Flow and Microstates
To consider the statistics of spontaneous change we will consider two equilibrium systems of particles, A and B, initially isolated from one another. Then we shall place in them in some kind of contact and allow them to come to their new equilibrium state and see what changes take place. Consider two systems, A and B, each with 10 molecules with the single degree of freedom. System A has 6 quanta of energy and System B has 4 quanta of energy. The most probable configuration for A is n0=5, n1=4,n2=1 with 1260 microstates The most probable configuration for B is n0=7, n1=2, n2=1 with 360 microstates The total number of microstates for the two systems is WA×WB = We multiply the number of microstates in the two systems because for every microstate in A we could pick any microstate in B to make up a new microstate in the combined A+B system.

7 Statistics of Heat Flow and Microstates
What happens to the total number of microstates as 1 quanta moves from A to B? Both systems now have 5 quanta and 840 microstates in their most probable configuration. (We showed this in Lecture 2.) So WA ×WB = 8402 = i.e an increase over the before the energy transfer. Therefore – this energy flow will occur spontaneously because it moves A+B to a more probable configuration. This energy transfer only involved changes in occupation numbers and so we refer to it as heat. The difference between the original configurations in A and B could be characterised by their different temperatures with TA > TB. If allowed to, heat will flow spontaneously from a high temperature system to a low temperature system in order to increase the total number of microstates available to the two systems.

8 Heat Capacity and Temperature after Heat Flow
Consider two lumps of material, A and B, with heat capacities CVA and CVB, respectively. A is at 100°C and B is at 50°C. What happens when A and B are placed in contact with each other? In a heat flow problem, we use the fact that energy must be conserved. That is, dEA = -dEB, to determine the final temperature of the combined system. We write this in terms of (constant) heat capacities as DEA = -DEB CVA(Tf – TiA) = -CVB(Tf – TiB) For the special case of a system in which CVA = CVB, this reduces to Tf – TiA = TiB –Tf Tf – 100 = 50 –Tf, or 2 Tf = 150 => Tf = 75°C.

9 Flash Quiz! How might the concept of heat capacity explain how a human can walk on coals at temperatures around 800ºC? Photograph thanks to Aidan Jones of Oxford, U.K.

10 Answer The heat capacity of the surface layer of the foot is roughly equivalent to the heat capacity of many grams of water. Water has a high heat capacity (and thermal conductivity). The ash on top of the coals has a low heat capacity (and is a poor heat conductor). So the equilibrium temperature is closer to that of the original foot than that of the original coal. Photograph thanks to Aidan Jones of Oxford, U.K.

11 Homework Problem Initially A is at 100ºC and B is at 50ºC. Calculate the equilibrium temperature of the combined A+B system when (a) CVA = 2CVB and (b) 2CVA = CVB.

12 What Determines Heat Capacity?
What are the microscopic parameters that determine the heat capacity? Consider first the ideal atomic gas. We determined the energy of that gas to be Using the definition of heat capacity as We obtain That is, the heat capacity of a monatomic gas is a constant at all temperatures. This agrees with observations for atomic gases, but is not the case for many other materials including molecular gases, liquids and solids. Typically the heat capacity is constant at high temperatures but falls away at low T. Why does this occur?

13 CV and Characteristic Temperature
The heat capacity of an atomic gas is derived from the high temperature approximation for the energy. The correct behaviour for the energy shows a decrease in slope towards zero near the characteristic temperature We therefore expect that the heat capacity will decrease near a characteristic temperature for any degree of freedom. We now need to consider other kinds of motions.

14 Typical characteristic vibrational temperatures
Vibrations Vibrational motion will be considered in this course using a harmonic oscillator approximation. That is, a model that describes small motions around the equilibrium bond length of a molecule. It will not describe energies approaching the dissociation energy of the molecule. The allowed energies of the harmonic oscillator are εl = (l + ½)hn where n is the frequency of the bond, proportional to the square root of the bond force constant. The characteristic temperature for vibration is thus Qvib = hn/kB Typical characteristic vibrational temperatures Ar(s) 40K Pb(s) 60K SO2(g) stretch 2000K H2O (g) stretch 5000K i.e. the ‘stiffer’ the bond, the higher is it’s characteristic temperature

15 Partition Function for Vibrations
Because the energies of a harmonic oscillator are evenly spaced, the partition function is a geometrical series. This can be solved exactly to give the analytical expression From this expression we can readily calculate the energy and heat capacity of a vibrating system. Of course many systems will exhibit multiple degrees of freedom. Our next question is how to deal with such systems, and we will do this using the example of a diatomic gas.

16 Summary Next Lecture You should now be able to
Define or explain characteristic temperature Explain the high-T approximation used for translational motions Understand the statistical/quantum origin of work Define heat, heat capacity, and related terms Calculate the equilibrium temperature for two bodies in contact given heat capacities and initial temperatures Explain the approximations used for vibrational motion Next Lecture Separation of independent degrees of freedom Heat capacity of a diatomic gas and more complex systems


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