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Goal: to understand Thermodynamics. Objectives: To learn the first law of Thermodynamics To learn about the PV diagram To learn about work done on a gas To learn about work at constant pressure To learn about work at a constant volume To learn about work at a constant Temperature
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Goal: to understand Thermodynamics Objectives: To learn the first law of Thermodynamics To learn about the PV diagram To learn about work done on a gas To learn about work at constant pressure To learn about work at a constant volume To learn about work at a constant Temperature To learn about Adiabatic processes To examine heat engines and heat pumps
First Law • ΔU = Q + W • U = internal energy • Q = heat • W = work done ON the gas
PV diagram • Will show the starting Pressure and Volume of both the initial state and the final state. • W = - PaveΔV
Constant Pressure • P = Pave • So, W = -P ΔV
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 * 105 Pa then find the work done ON the balloon
Constant Volume • W = -P ΔV • If ΔV = 0 then W = 0J • So, ΔU = Q
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
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)?
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?
Adiabatic Process • In this case there will be no heat flow. • That is to say Q = 0 • So, W = ΔU = 1.5 nR ΔT
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?
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
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
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.
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.
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.
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.
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.
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
Conclusion • We have learned about heat engines • We have learned about heat pumps/refrigerators • We have learned about efficiency