1. Lecture # 7 Cassandra Paul Physics 7A Summer Session II 2008

2. Today Equation Confusion Heat capacities and their role in Particle model Start Intro to Thermodynamics Quiz

3. Equation confusion… The reason I want you to memorize a few equations is not (contrary to popular belief) so you can do physics somewhere you don’t have access to your notes…

4. Equations contain Ideas What is gravitational potential energy? – We can increase potential energy by increasing the height or mass of an object. – We can also increase potential energy if we have greater gravity. Hence, PE=mgh The equation reminds you of the concepts.

5. What about Kinetic Energy? KE= ½ mv 2 Faster objects have more kinetic energy More massive objects have more kinetic energy Kinetic energy can’t be negative. Ok… so you have to remember the ½. But that’s not the important part

6. Bond Energy Why do we have three different equations for bond energy? Because each are useful at different times. If we want to talk about the Macroscopic properties of solids liquids and gasses, we want to use ΔE bond = ± l ΔmΔH l. Why? Because we aren’t talking about individual particles here, only the energy associated with changing the phase of the system.

7. Bond Energy When do we use the equation: E bond = Σ all pairs (PE pair-wise ) When we know something about the number of particles in the system and we have a small number of bonds where we can count. Why? That’s the exact definition of bond energy at the microscopic level.

8. Bond Energy When do we use this equation? E bond = (tot # n-n bonds) (-ε) When there are so many bonds in the system that we can’t count so we estimate.

9. Cassandra, the blue pages in my book say: E bond = (tot # n-n bonds) (-ε) But in DL we were taught: E bond = (nn)/2 (-ε) (tot # of atoms) Which is correct!? Cassandra, the blue pages in my book say: E bond = (tot # n-n bonds) (-ε) But in DL we were taught: E bond = (nn)/2 (-ε) (tot # of atoms) Which is correct!? Let’s ask your classmates…..

10. Which is the correct approximation? A.E bond = (tot # n-n bonds) (-ε) B.E bond = (# of n-n)/2 (-ε) (tot # of atoms) C.E bond = (# of n-n)/2 (-ε) (Avogadro's #) D.A B and C are all correct E.A and B are correct

11. Which is the correct approximation? A.E bond = (tot # n-n bonds) (-ε) B.E bond = (# of n-n)/2 (-ε) (tot # of atoms) C.E bond = (# of n-n)/2 (-ε) (Avogadro's #) D.A B and C are all correct E.A and B are correct A and B are both correct, and say EXACTLY the same thing. When is C correct? When the total number of atoms is equal to 1 mole.

12. Why do you have us memorize any equations? If you know the ideas you can work out the (derive) the equations. If you know the equations, you can work out the concepts. This way you aren’t stuck. Even if you forget something, you can check it against what you know. Notice… many times I will give you the equations even if I tell you that you must memorize them. This is not because I’m a ‘push over’, it’s to reduce test anxiety. If you don’t know what I’m going to give you on the exam, you’ll work a little harder to organize your thoughts.

13. Let’s talk E thermal now. From the beginning of the course, we know: ΔE thermal = mcΔT This tells us that the amount of energy it takes to change the temperature of a substance depends on how massive it is, how much of a temperature change we are talking about and finally what the heat capacity of that substance is.

14. In the Particle Model of Thermal Energy… E thermal = (amount of energy per mode) (Total number of Ethermal modes) E thermal =( ½ k b T)(tot # particles)(tot # E th modes per particle) What if we want to know the change in E th ? ΔE thermal =( ½ kb ΔT)(tot # particles)(tot # E th modes per particle) We place a delta in front of the Temperature because this is our indicator for when Eth is changing!

15. Cassandra, in the beginning of the course we talked about heat capacity being a part of Eth. In the particle model of thermal energy, it doesn’t seem like heat capacity is represented, why is this? Ahhh but heat capacity is represented… it is just hidden in the concept of MODES. Let’s see how…

16. KE mode PE mode Total Solids336 Liquids336 Monatomic gasses303 Diatomic gasses3+2+117 Equipartition tells us that the energy per mode is 1/2 k B T. Also it tells us that each mode gets the same amount of energy.

17. KE mode PE mode Total Solids336 Liquids336 Monatomic gasses 303 Diatomic gasses3+2+117 Equipartition tells us that the energy per mode is 1/2 k B T. Question: Does it take More or Less energy to raise the temperature of diatomic gas compared to monatomic gas? (a) More (b) Less (c) Equal (d) Can’t Determine

18. Let’s think of modes as ‘buckets’ and Energy and Temperature as Water. If I have two systems, one system has two buckets and another has three buckets, which system is it harder to raise the water level of? (Remember, equipartition says that all modes get the same amount of energy!)

19. Energy addedEnergy added System 1 System 2

20. System 1 System 2

21. If the same amount of energy is added to each system… Less modes, lower heat capacity, easier to change T More modes, higher heat capacity, harder to change T ΔT 1 ΔT 2

22. ΔEth = ½ k b ΔT (tot # of modes) In other words… Lots of Modes = smaller temp change… Less Modes = larger temp change… It’s like the red and blue car!

23. Let’s Get a more quantitative Idea about how heat capacity and modes are related….

24. In DL on Friday… Think back to the beginning of the course… Heat Capacity Definition: C=Q/ΔT We want to make a definition of heat capacity at the microscopic level, where should we start? We know that Eth is present in both the microscopic and macroscopic models And we know that Q sometimes = ΔE th What assumptions must we make to assume that Q=ΔE th ?

25. ΔE th + ΔE bond = Q + W No Change in E bond No work Done ONLY then is ΔE th = Q So operating under those assumptions: C = Q/ΔT  C v = ΔE th /ΔT = ( ½ k b ΔT)(tot # particles) (tot # E th modes per particle)*/ΔT C v = (½ k b )(tot # of particles)(# E th modes per particle) This is Heat Capacity, what if we want specific heat in MOLAR units? (per mole) C vm = (½ k b )(Avagadro’s #)(# E th modes per particle) *From slide 13

26. Deriving Molar Specific Heat Definition C vm = (½ k b )( Avagadro’s #)(# E th modes per particle We could be done here but Na (Avagadro’s #) multiplied by k b = R (gas constant) (1.38x10-23)*(6.02x1023) = 8.3076 (or 8.31 J/K!) So… C vm = (½ R)(# E th modes per particle)

27. And now we have a definition of molar specific heat for the particle model of matter. C vm = (½ R)(# E th modes per particle)

28. DL sections Swapno: 11:00AM EversonSection 1 Amandeep: 11:00AM Roesller Section 2 Yi: 1:40PM Everson Section 3 Chun-Yen: 1:40PM Roesller Section 4