Lecture 15 first law of thermodynamics

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Lecture 15 first law of thermodynamics

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Lecture 15 first law of thermodynamics

  1. 1. Lecture 15 First law of thermodynamics.
  2. 2. Internal energy State variables: p, T, V They only depend on the “state” of the system They do NOT depend on the “history” Total energy of the molecules is also a state variable: Internal energy: U = KE all + PE all KE : kinetic energy of atoms, random motion PE : interactions, e.g. attraction between molecuses For ideal gases, no interaction: U = KE all
  3. 3. Internal energy for some systems Nk = nNAk = nR Monoatomic ideal gas: 3  U = KE all = N  kT ÷ 2  3 U = nRT 2 Diatomic ideal gas: 5  U = KE all = N  kT ÷ 2  U = Monoatomic solid crystal: U = KE all + PE all 3  = N  kT ÷× 2 2  5 nRT 2 U = 3nRT In general, internal energy can also depend on p, V. But it never depends on the history of the system.
  4. 4. Changes in internal energy Temperature change is associated to heat transfer. Change in internal energy transfer of energy We already know that: When heat is absorbed/released by a system, its internal energy increases/decreases. Other ways of transferring energy? Work!! (we learned it in 221)
  5. 5. Work done by gas volume change A gas with pressure p expands by pushing a piston by a distance dx p Force by gas on piston F = pA (A = area of the piston) dx Work by gas: dW = pAdx = pdV As volume goes from Vi to Vf Vf W = ∫ pdV Vi
  6. 6. Work is the area under pV curve But be careful with the sign! Expansion (Work done by gas) > 0 Compression (Work done by gas) < 0
  7. 7. ACT: Three processes - Work A gas can go from state 1 to state 2 through three different processes. In which process does gas do the least amount of work ? A C B D: It’s the same for all three
  8. 8. Even if the initial and final states are the same, work depends on the path taken by the process A B W is NOT a state variable. It is only defined when the state changes. C
  9. 9. ACT: Three processes – Internal energy In which case does the internal energy of the gas change the most ? A D: It’s the same for all three C B U is a state function and does not depend on the process. ΔU = U1 − U2 for all processes.
  10. 10. In-class example: Work in closed cycle Which of these processes represents the most work done by the system per cycle? P P P V V A C V E P P Work = area inside cycle. Note that WE < 0 V V B D CW ⇔ W > 0 CCW ⇔ W < 0
  11. 11. First law of thermodynamics If heat is absorbed by a system, Q>0 ΔU > 0 If heat is released by a system, Q<0 ΔU < 0 W>0 ΔU < 0 If work is done on the gas (compression), W < 0 ΔU > 0 If work is done by the gas (expansion), ∆U = Q −W Remember the sign convention! W > 0 work done by system Q > 0 heat absorbed by system
  12. 12. Example: Cyclic process A gas (not necessarily ideal) goes through the cycle shown in the pV diagram below. A Process 2 3 Process 1 B Pr oc es s P Data: C VA = 2.0 m3 VC = 4.0 m3 PA = 1.0 × 105 Pa PC = 2.0 × 105 Pa a) Determine the work done by the gas in each of the three parts of the cycle. V 1 (A to B): W1 = 0 (constant volume) 5 2 (B to C): W2 = p ∆V = pB (VC −VB ) = 2 × 10 J 5 3 (C to A): W2 = − ( yellow area ) = −3 × 10 J
  13. 13. P A C 3 Process 2 Pr oc es s Process 1 B Data: V VA = 2.0 m3 VC = 4.0 m3 PA = 1.0 × 105 Pa PC = 2.0 × 105 Pa b) For the entire cycle, what are the work done by the gas, the change in internal energy of the gas and the heat exchanged with the surroundings? Is this heat absorbed or released by the gas? Wcycle = W1 +W2 +W3 = −1 × 105 J ∆Ucycle = Uf − Ui = UA − UA = 0 ΔUclosed cycle = 0 Qcycle = ∆Ucycle +Wcycle = −1 × 105 J Q < 0: heat is released by the gas
  14. 14. Overall, in this cycle: Wcycle < 0 Qcycle < 0 work Work done on the system. System releases heat. system P B C heat A We do work on the system, we obtain heat. (This could be used to warm up a room…) Reverse cycle (A→C →B): Wcycle > 0 Work done by the system. Qcycle > 0 System absorbs heat. work system heat System absorbs heat and produces work. (Like some kind of steam motor…) V P B C A V

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