1. Goal: to understand Thermodynamics Objectives: 1)To learn the first law of Thermodynamics 2)To learn about the PV diagram 3)To learn about work done on a gas 4)To learn about work at constant pressure 5)To learn about work at a constant volume 6)To learn about work at a constant Temperature 7)To learn about Adiabatic processes 8)To examine heat engines and heat pumps

2. First Law ΔU = Q + W U = internal energy Q = heat W = work done ON the gas

3. PV diagram Will show the starting Pressure and Volume of both the initial state and the final state. W = - P ave ΔV

4. Constant Pressure P = Pave So, W = -P ΔV

5. Sample A air filled balloon is placed inside a freezer by mistake. The balloon shrinks from a volume of 0.004 cubic meters to 0.003 cubic meters. If the air pressure remains a constant 1.0 * 10 5 Pa then find the work done ON the balloon

6. Constant Volume W = -P ΔV If ΔV = 0 then W = 0J So, ΔU = Q

7. Constant Temperature In this case there will be no change in internal energy (U = 1.5 kT) So, Q + W = 0 or, W = -Q Using fancy math I won’t replicate it turns out that you will get that: W = nRT ln(Vi/Vf) R = gas constant = 8.314 J/(mol K) n = # of moles

8. Example A balloon is attached to a rock and tossed into an ocean which has the same temperature as the air. The balloon sinks to a depth of 10 m at which point the outside pressure has doubled. As a result – before we get into the problem – what will happen to the balloon (hint thing net force)?

9. Example continued A balloon is attached to a rock and tossed into an ocean. The balloon sinks to a depth of 10 m at which point the outside pressure has doubled. You now know what will happen to the volume (that is to say the value of Vi/Vf) In this particular balloon there were 200 moles of an ideal gas. What will the work done on the balloon be?

10. Adiabatic Process In this case there will be no heat flow. That is to say Q = 0 So, W = ΔU = 1.5 nR ΔT

11. Example The balloon from the previous example is cut from the rock tied to it and can now shoot upward very quickly. What is the work done on the balloon if it does not have time to exchange any heat?

12. Heat Engines Takes energy in some form and coverts it to heat so that you can transform the energy to what you want/need. They process in a cycle such that you get a net work out of it. That is W = -P ΔV for each step You add up the steps to get a net work

13. Combustion Engine Is one form of engine You have a piston in a chamber that changes the size of the chamber Step 1 You start with a small volume and up the temperature to create a large pressure. W = -P ΔV = 0J as V has not changed

14. Step 2 You push in the piston out increasing the volume. W = -P ΔV since V drops this will be a negative work done ON THE GAS Which means positive work done on the piston. Step 3: push out the gas, which has a lower pressure than before so the change in volume will produce little work.

15. Step 4 Piston pulled back out at a constant pressure. This negates step 3. Final step: piston goes back in to recompress the gas. However it is done at a lower pressure so the work done in this step is far lower than the work done in step 2 so the net is that work is done on the piston, and therefore the car.

16. Efficiency E = Wnet / Qused This just tells you what fraction of the energy is used for what you want. The rest is wasted as exhaust, ect.

17. Heat Pumps / Refrigerators Work in the reverse They try to exhaust MORE heat. You compress a fluid. This heats it. That heat is then radiated or pumped via a fan outside the system.

18. Efficiency Is usually greater than 1 (many are 9 to 10) The reason, you are using a little bit of energy to toss out a LOT of energy. In other words you are just moving heat around.

19. Temperature difference – reversible engine How cold you get the fridge is found by: e = 1 – (Tc / Th) Tc and Th must be done in Kelvin Only works for e < 1

20. Conclusion We have learned about heat engines We have learned about heat pumps/refrigerators We have learned about efficiency