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Plan for Today (AP Physics 2)

Plan for Today (AP Physics 2). Wrap Up Notes on 1 st Law First Law Worksheet Homework: First Law Worksheet, Review Worksheet due ___________. Note.

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Plan for Today (AP Physics 2)

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  1. Plan for Today (AP Physics 2) • Wrap Up Notes on 1st Law • First Law Worksheet • Homework: First Law Worksheet, Review Worksheet due ___________

  2. Note • For a cyclic process the net work done per cycle by a gas on its surroundings is equal to the area enclosed by the path representing the process on a PV diagram

  3. AB: Heated at constant V to 400 K. Example Problem: A 2-L sample of Oxygen gas has an initial temp-erature and pressure of 200 K and 1 atm. The gas undergoes four processes: • BC: Heated at constant P to 800 K. • CD: Cooled at constant V back to 1 atm. • DA: Cooled at constant P back to 200 K.

  4. B 400 K 800 K PB A 200 K 1 atm 2 L PV-DIAGRAM FOR PROBLEM How many moles of O2 are present? Consider point A: PV = nRT

  5. B 400 K 800 K PB A 200 K 1 atm PA P B = 2 L TA T B 1 atmP B P B = 2 atm = 200 K400 K or 203 kPa PROCESS AB: ISOCHORIC What is the pressure at point B?

  6. B 400 K 800 K PB A 200 K 1 atm 2 L Q = +514 J U = +514 J W = 0 PROCESS AB: Q = U + W Analyze first law for ISOCHORIC process AB. W = 0 Q = U = nCv T U = (0.122 mol)(21.1 J/mol K)(400 K - 200 K)

  7. B 400 K 800 K PB C 200 K VB V C 1 atm D = TB T C 2 L 4 L 2 LV C V C = V D = 4 L = 400 K800 K PROCESS BC: ISOBARIC What is the volume at point C (& D)?

  8. B 400 K 800 K 2 atm C 200 K 1 atm 2 L 4 L U = +1028 J FINDING U FOR PROCESS BC. Process BC is ISOBARIC. P = 0 U = nCv T U= (0.122 mol)(21.1 J/mol K)(800 K - 400 K)

  9. B 400 K 800 K 2 atm C 200 K 1 atm 2 L 4 L W = +405 J FINDING W FOR PROCESS BC. Work depends on change in V. P = 0 Work = PV W = (2 atm)(4 L - 2 L) = 4 atm L = 405 J

  10. B 400 K 800 K 2 atm C 200 K 1 atm 2 L 4 L Q = 1433 J U = 1028 J W = +405 J FINDING Q FOR PROCESS BC. Analyze first law for BC. Q = U + W Q = +1028 J + 405 J Q = +1433 J

  11. B 400 K 800 K PB C A 200 K D 1 atm PC P D = 2 L TC T D 2 atm1 atm T D = 400 K = 800 KTD PROCESS CD: ISOCHORIC What is temperature at point D?

  12. 800 K 400 K C PB 200 K 400 K 1 atm D 2 L Q = -1028 J U = -1028 J W = 0 PROCESS CD: Q = U + W Analyze first law for ISOCHORIC process CD. W = 0 Q = U = nCv T U = (0.122 mol)(21.1 J/mol K)(400 K - 800 K)

  13. 400 K 800 K 2 atm 200 K 400 K A 1 atm D 2 L 4 L U = -514 J FINDING U FOR PROCESS DA. Process DA is ISOBARIC. P = 0 U = nCv T U = (0.122 mol)(21.1 J/mol K)(400 K - 200 K)

  14. 400 K 800 K 2 atm A 200 K 400 K 1 atm D 2 L 4 L W = -203 J FINDING W FOR PROCESS DA. Work depends on change inV. P = 0 Work = PV W= (1 atm)(2 L - 4 L) = -2 atm L = -203 J

  15. 400 K 800 K 2 atm A 200 K 400 K 1 atm D 2 L 4 L Q = -717 J U = -514 J W = -203 J FINDING Q FOR PROCESS DA. Analyze first law for DA. Q = U + W Q = -514 J - 203 J Q = -717 J

  16. For all processes: DQ = DU + DW PROBLEM SUMMARY

  17. +404 J B C B C -202 J 2 atm 2 atm Neg 1 atm 1 atm 2 L 4 L 2 L 4 L B C Area = (1 atm)(2 L) Net Work = 2 atm L = 202 J 2 atm 1 atm 2 L 4 L NET WORK FOR COMPLETE CYCLE IS ENCLOSED AREA

  18. Example 2: A diatomic gas at 300 K and 1 atm is compressed adiabatically, decreasing its volume by 1/12. (VA = 12VB). What is the new pressure and temperature? ( = 1.4)   B PAVA = PBVB PB A PA PAVA PBVB = TA T B VB VA ADIABATIC EXAMPLE: Q = 0

  19.  B PAVA = PBVB PB Solve for PB: 300 K 1 atm A Q = 0 VB 12VB PB = 32.4 atm or 3284 kPa ADIABATIC (Cont.): FIND PB

  20. TB=? B 32.4 atm 300 K 1 atm A Solve for TB Q = 0 VB 12VB (1 atm)(12VB)(32.4 atm)(1 VB) = (300 K)T B TB = 810 K ADIABATIC (Cont.): FIND TB

  21. B 810 K Since Q = 0, W = - U 32.4 atm 300 K 1 atm A Q = 0 8 cm3 96 cm3 PV RT n = ADIABATIC (Cont.): If VA= 96 cm3and VA= 8 cm3, FIND W W = - U = - nCVT & CV= 21.1 j/mol K Find n from point A PV = nRT

  22. PV RT (101,300 Pa)(8 x10-6 m3) (8.314 J/mol K)(300 K) n = = B 810 K 32.4 atm 300 K 1 atm A W = - 3.50 J 8 cm3 96 cm3 ADIABATIC (Cont.): If VA= 96 cm3and VA= 8 cm3, FIND W n = 0.000325 mol & CV= 21.1 j/mol K T = 810 - 300 = 510 K W = - U = - nCVT

  23. TheFirst Law of Thermodynamics:The net heat taken in by a system is equal to the sum of the change in internal energy and the work done by the system. Q = U + W final - initial) Summary • Isochoric Process: V = 0, W = 0 • Isobaric Process: P = 0 • Isothermal Process: T = 0, U = 0 • Adiabatic Process: Q = 0

  24. The Molar Specific Heat capacity, C: Q n T c = Q = U + W U = nCv T PV = nRT Summary (Cont.) Units are:Joules per mole per Kelvin degree The following are true for ANY process:

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