Revision on thermodynamics


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Revision on thermodynamics

  1. 1. Revision on Thermodynamics
  2. 2. First Law of Thermodynamics <ul><li>Statement: </li></ul><ul><li>Energy can not be created or destroyed and the total energy of a system is always constant. </li></ul><ul><li>Mathematical formula for a closed system: </li></ul><ul><li>∆ U = Q + W </li></ul><ul><li>Where; </li></ul><ul><li>∆ U: change in internal energy </li></ul><ul><li>Q: heat transferred to the system. </li></ul><ul><li>W: work done by the system. </li></ul>
  3. 3. First Law of Thermodynamics <ul><li>Sign convention: </li></ul><ul><li>If energy transferred to the system as q or w: </li></ul><ul><li>Q = +ve, W = +ve </li></ul><ul><li>If energy transferred from the system as q or w: </li></ul><ul><li>Q = -ve, W = -ve </li></ul>
  4. 4. First Law of Thermodynamics <ul><li>Recall: </li></ul><ul><li>- System </li></ul><ul><li>The part of the universe that we will study. It is separated from its surroundings by boundaries. </li></ul><ul><li>System may be: </li></ul><ul><li>a. Open ; mass & heat can transfer </li></ul><ul><li>b. Closed ; no mass transfer </li></ul><ul><li>c. Isolated ; no mass or heat transfer </li></ul>
  5. 6. First Law of Thermodynamics <ul><li>Internal Energy </li></ul>
  6. 7. First Law of Thermodynamics <ul><li>Work: </li></ul><ul><li>work arising from a change in volume. </li></ul><ul><li>work done by a gas as it expands and drive back the atmosphere. </li></ul><ul><li>dw = -P ext .dV </li></ul><ul><li>For reversible process: dw = - PdV </li></ul><ul><li>P = system pressure </li></ul>
  7. 8. First Law of Thermodynamics <ul><li>Heat: </li></ul><ul><li>Q = m C ∆T or Q = n C ∆T </li></ul><ul><li>C = specific heat ( J/mol. K) , (J/gm. K) </li></ul><ul><li>C v = specific heat at constant volume </li></ul><ul><li>C p = specific heat at constant pressure </li></ul>
  8. 9. First Law of Thermodynamics <ul><li>Molar heat capacity: </li></ul><ul><li>energy of one mole of a substance. (J/mol. K). </li></ul>
  9. 10. First Law of Thermodynamics <ul><li>Enthalpy: H = U + PV </li></ul><ul><li>Heat content at constant pressure. </li></ul><ul><li>∆H = </li></ul><ul><li>∆H = n Cp ∆T; if Cp is independent on T </li></ul>
  10. 11. Ideal gas processes ( rev process) <ul><li>For different processes: </li></ul>n C p ∆T n C v ∆T n Cv ∆t Zero Adiabatic n C p ∆T n C v ∆T - P.ΔV -nR ∆T n Cp ∆T= ∆H Isobaric n C p ∆T n C v ∆T Zero n Cv ∆T= ∆U Isochoric Zero Zero -Q nRT ln(V f /V i )= nRT ln (P f /P i ) Isothermal ∆ H ∆ U W Q
  11. 12. First Law of Thermodynamics <ul><li>PVT Relation: (for ideal gas) </li></ul><ul><li>Isothermal : PV = const. </li></ul><ul><li>Isobaric: V/T = Const </li></ul><ul><li>Isochoric: P/T = Const </li></ul><ul><li>Adiabatic: T 2 /T 1 = (V 1 /V 2 ) γ-1 </li></ul><ul><li>P 1 V 1 γ = P 2 V 2 γ ; C P /C V = γ </li></ul><ul><li>C P – C V = R </li></ul>
  12. 13. Sheet (1) <ul><li>A sample containing of 1.0 mole of argon expands isothermally at 0º C from 22.4 dm 3 to 44.8 dm 3 . Calculate Q, W, ΔU, and ΔH if the gas expands: </li></ul><ul><li>reversibly </li></ul><ul><li>against a constant external pressure equal to the final pressure of the gas </li></ul><ul><li>freely </li></ul>
  13. 14. Sheet (1) <ul><li>2. The constant pressure heat capacity of a sample of a prefect gas was found to vary with temperature according to the expression: </li></ul><ul><li>C P / (J/K) = 20.17 + 0.3665T </li></ul><ul><li>Calculate Q, W, ΔU, and ΔH when the temperature is raised from 25ºC to 200ºC : </li></ul><ul><li>(a) at constant pressure, </li></ul><ul><li>(b) at constant volume. </li></ul>
  14. 15. Sheet (1) <ul><li>3. An ideal gas undergoes the following sequences of mechanically reversible processes in a closed system; </li></ul><ul><li>a) From an initial state 70°C and 1 bar, it is compressed adiabatically to 150° C. </li></ul><ul><li>b) It is then cooled from 150° to 70°C at constant pressure. </li></ul><ul><li>c) Finally, it is expanded isothermally to its original state. </li></ul><ul><li>Calculate W, Q, ΔU, and ΔH of each of the three processes and for the entire cycle. Take Cp = (5/2)R and Cv = (3/2) R </li></ul>
  15. 16. Second Law of Thermodynamics <ul><li>Direction of energy : </li></ul><ul><li>A spontaneous process is a process in which the final state is more probable than the initial state. </li></ul><ul><li>Every system which is left to itself will, on average, change toward a system of maximum entropy. </li></ul>
  16. 17. Second Law of Thermodynamics <ul><li>Entropy definition: </li></ul><ul><li>A measure of disorder of the system. </li></ul><ul><li>ΔS is equal to the heat Q it absorbs, divided by T. </li></ul>
  17. 18. Second Law of Thermodynamics Zero Reversible adiabatic Cp ln (T 2 /T 1 ) - R ln( P 2 /P 1 ) Any Process Cv ln (T 2 /T 1 ) Isochoric Cp ln (T 2 /T 1 ) Isobaric nR ln (V 2 /V 1 ) Isothermal ∆ S Process
  18. 19. Sheet 1 <ul><li>4. A 40 kg casting (Cp= 0.5 kJ/kg k ) at of 450° C is quenched in a 150 kg of oil (Cp= 2.5 kJ/kg k) at 25° C. if there are no heat losses, what is the change in entropy of the casting, the oil, and both considered together. </li></ul>
  19. 20. Third Law <ul><li>Law of zero entropy </li></ul><ul><li>S (at any T) = S° (at 298 K) + </li></ul>
  20. 21. Sheet 1 <ul><li>5. Calculate the absolute entropy for water at 400 K, knowing that Cp(liq)=75.29 J/mol K, and Cp(vap) =28.85+0.012 T+ (0.1×10 6 ) /T 2 . Knowing that the standard entropy for water equal 69.91 J/mol K and heat of vaporization ΔHv =44.016 kJ/mol. </li></ul>
  21. 22. Thermodynamic equilibrium <ul><li>S is a measure of equilibrium during reversible adiabatic process or isentropic process. </li></ul><ul><li>To define equilibrium in process other than isentropic process we use free energies. </li></ul>
  22. 23. <ul><li>Gibbs free energy </li></ul><ul><li>G = H – TS </li></ul><ul><li>H = total energy </li></ul><ul><li>TS = unavailable energy </li></ul><ul><li>dG = VdP – SdT = fn ( P, T) </li></ul><ul><li>∆ G = 0 at constant T and P </li></ul><ul><li>∆ G = -ve (The process will be spontaneous) </li></ul>
  23. 24. <ul><li>Isothermal Process: </li></ul><ul><li>dG = VdP ∆G = RT ln(P 2 /P 1 ) </li></ul><ul><li>Isobaric process: </li></ul><ul><li>dG= -SdT ∆G= ʃ- SdT </li></ul><ul><li>Isochoric Process: </li></ul><ul><li>∆ G = VdP - ʃ- SdT </li></ul>
  24. 25. <ul><li>Helmholtz free energy </li></ul><ul><li>A = U – TS </li></ul><ul><li>U = Total energy </li></ul><ul><li>TS = unavailable energy </li></ul><ul><li>dA = -PdV – SdT </li></ul><ul><li>∆ A = 0 at constant V and T </li></ul><ul><li>G = H – TS = U + PV- TS = A + PV </li></ul><ul><li>∆ G = ∆A + ∆ (PV) </li></ul>
  25. 26. Sheet 1 <ul><li>6. Calculate ΔG, ΔA, and ΔS for each of the following processes: </li></ul><ul><li>a) Reversible melting of 36 gm of ice at 1 atm and 0°C. </li></ul><ul><li>b) Reversible vaporization 39 gm of C 6 H 6 at its normal boiling point of 80 </li></ul>
  26. 27. Sheet 1 <ul><li>7. A sample consisting of 1 mole of argon is taken through the following cyclic process: </li></ul><ul><li>a) Isobaric expansion at 1 atm from 25 to 50 liters. </li></ul><ul><li>b) Isochoric cooling at 50 liters from 1 to 0.5 atm. </li></ul><ul><li>c) Isothermal compression to the initial state. </li></ul><ul><li>Calculate ΔU, ΔH,∆S ΔG, ΔA for each step and for the cycle. Argon may be considered ideal gas with C v =3/2R, C p =5/2R </li></ul>
  27. 28. Thank you