AP Physics Chp 15

1. AP Physics Chp 15

2. Thermodynamics – study of the relationship of heat and work • System vs Surroundings • Diathermal walls – allow heat to flow through • Adiabatic walls – do not allow heat to flow

3. Zeroth Law of Thermodynamics • Two systems in thermal equilibrium with a third are also in equilibrium with each other

4. First Law of Thermodynamics • Internal energy changes based on the amount of heat and/or work done by/on the system. • ∆U = Q – W W= ∆PV • Q is positive when it goes in (endothermic) • W is positive when the system does work

5. What is the change in the internal energy if you supply 15 kJ to a 35 m3 sample of helium at 101150 Pa and it is allowed to expand to 52 m3?

6. ∆U = Q – W • ∆U = 15000 J – (101150 Pa)(52 m3 – 35 m3) • ∆U =

7. If a process is slow enough then the P and T are uniform. • When P is constant it’s called an isobaric process. • W = P∆V Why is W negative when work is done on a system?

8. Isochoric processes occur at constant volume • This is the bomb calorimeter idea.

9. At constant T its an isothermal process • Adiabatic processes occur without the transfer of any heat

10. One way to relate work for a system is to plot the P vs V graph and compare the area under the “curve”.

11. How much work is done in compressing the gas from 4 m3 to 3 m3? Why is it more than 9 m3 to 8 m3?

12. What would a graph for an isochoric process look like? Why does it show no work being done?

14. What about isobaric, how’s it’s graph look and is there any work?

16. Isothermal process – Expansion or Compression • Since T is constant the internal energy is constant so • Q = W

17. Any work done by the gas results in heat flowing out to the surroundings and vice versa.

18. Adiabatic Processes – Expansion/Compression • Since no heat is transferred the internal energy is related only to the work • ∆U = -W

19. When the gas does work the T decreases and the internal energy of the gas has decreased

20. If 2 moles of an ideal gas expands from 0.020 to 0.050 m3 at a pressure of 101300 Pa, how much work is done? • W = P∆V • W = 101300Pa(0.050 m3 -0.020 m3) • W = 3039Pa m3 = J

21. If the temperature is allowed/forced to remain constant how has the internal energy changed? • 0 • U = 3/2 nRT so with no change in T there is no change in internal energy

22. How much heat was transferred? • The same as the work. Q = W • Q = 3039 J

23. What is the temperature of the gas? • 3039J = (2n)(8.31J/nK)T ln(0.050/0.020) • T = 199.6 K

24. Specific Heat Capacities • Gases use a molar heat capacity at constant pressure and another for constant volume • Cp and Cv

25. Ideal Gases • At constant pressure the heat is related to both the change in internal energy and work thus Cp = 5/2R • At constant volume its only the internal energy and Cv = 3/2R • So Cp – Cv = R

26. Isobaric (P const) W = P∆V • Isochoric (V const) W = 0 • Isothermal (T const) W = nRTln(Vf/Vo) • Adiabatic (no Q) W = 3/2nR(To – Tf)

27. 2nd Law of Thermodynamics • Heat flows spontaneously from a higher temperature to a lower temperature

28. Heat engines use heat to perform work. • Heat comes from a hot reservoir • Part of the heat is used to perform work • The remainder is rejected to the cold reservoir • Efficiencey e = W/QH

30. Efficiency can be multiplied by 100 to make it a percentage. • Since QH = W + QC W = QH – QC • e = 1 – QC/QH

31. Carnot created a principle that says that a irreversible engine can not have a greater efficiency than a reversible one operating at the same temperatures. • For a Carnot engine QC/QH = TC/TH • ecarnot = 1 – TC/TH

32. If absolute zero could be maintained while depositing heat in then a 100% efficiency would be possible but it’s not.

33. If my truck operates at a running temperature of 94 oC and the outside air is only -5 oC, what is the maximum efficiency for the engine?

34. TH = 273 +94 = 367 K • TC = 273 + -5 = 268 K • e = 1 – TC/TH • e = 1 – 268K / 367 K = 0.27 or 27%

35. Refrigerators, Air Conditioners, Heat Pumps • All of these take heat from the cold reservoir and put it into the hot reservoir by doing a certain amount of work. • It’s the reverse of the heat engine.

37. Why can’t you cool your house by running an air conditioner without having it exhaust outside? • Coefficient of performance = QC/W • Heat pumps warm up a space by moving heat from the cold outside to the warm inside.

38. Seems kind of weird that the cold outside has heat. • If you use a Carnot heat pump to deliver 2500 J of heat to your house to achieve a temperature of 20 oC while it is -5oC outside, how much work is required?

39. W = QH – QC and QC/QH = TC/TH • So QC = QHTC/TH and • W = QH – QHTC/TH • W = QH(1-TC/TH) • W = 2500J (1- 268 K/293K) = 210 J

40. Entropy • Randomness or disorder gas>>>liquids>solids • The entropy of the universe increases for irreversible process but stays constant for reversible

42. Since carnot engines are reversible • QC/TC = QH/TH Thus

43. If we set the hot coffee pot at 372K on the table at 297K and they exchange 4700 J of heat, how much has the entropy of the universe changed?

45. What happens to the energy in irreversible processes? • Since the ∆Suniv increases the increase is due to the energy being removed from being able to do any work

46. Wunavailable = Tc∆Suniv • So how much energy was “lost” to do work in the earlier example? • Wunav = (295K)(3.3J/K) = 970 J