Revision on thermodynamics
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Transcript of Revision on thermodynamics
- 1.Revision on Thermodynamics
2. First Law of Thermodynamics
- Statement:
- Energy can not be created or destroyed and the total energy of a system is always constant.
- Mathematical formula for a closed system:
- U = Q + W
- Where;
- U: change in internal energy
- Q: heat transferred to the system.
- W: work done by the system.
3. First Law of Thermodynamics
- Sign convention:
- If energy transferred to the system as q or w:
- Q = +ve, W = +ve
- If energy transferred from the system as q or w:
- Q = -ve, W = -ve
4. First Law of Thermodynamics
- Recall:
- -System
- The part of the universe that we will study. It is separated from its surroundings by boundaries.
- System may be:
- a. Open ; mass & heat can transfer
- b. Closed ; no mass transfer
- c. Isolated ; no mass or heat transfer
5. 6. First Law of Thermodynamics
- Internal Energy
7. First Law of Thermodynamics
- Work:
- work arising from a change in volume.
- work done by a gas as it expands and drive back the atmosphere.
- dw = -P ext .dV
- For reversible process: dw = - PdV
- P = system pressure
8. First Law of Thermodynamics
- Heat:
- Q = m C T or Q = n C T
- C = specific heat ( J/mol. K) , (J/gm. K)
- C v= specific heat at constant volume
- C p= specific heat at constant pressure
9. First Law of Thermodynamics
- Molar heat capacity:
- energy of one mole of a substance. (J/mol. K).
10. First Law of Thermodynamics
- Enthalpy: H = U + PV
- Heat content at constant pressure.
- H =
- H = n Cp T; if Cp is independent on T
11. Ideal gas processes ( rev process)
- For different processes:
n C p T n C vT n Cv t Zero Adiabatic n C p T n C vT - P.V -nR T n Cp T= H Isobaric n C p T n C vT 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 12. First Law of Thermodynamics
- PVT Relation: (for ideal gas)
- Isothermal :PV = const.
- Isobaric: V/T = Const
- Isochoric: P/T = Const
- Adiabatic: T 2 /T 1= (V 1 /V 2 ) -1
- P 1 V 1 = P 2 V 2 ; C P/C V=
- C P C V= R
13. Sheet (1)
- A sample containing of 1.0 mole of argon expands isothermally at 0 C from 22.4 dm 3to 44.8 dm 3 .Calculate Q, W, U, and H if the gas expands:
- reversibly
- against a constant external pressure equal to the final pressure of the gas
- freely
14. Sheet (1)
- 2. The constant pressure heat capacity of a sample of a prefect gas was found to vary with temperature according to the expression:
- C P / (J/K) = 20.17 + 0.3665T
- Calculate Q, W, U, and H when the temperature is raised from 25C to 200C:
- (a) at constant pressure,
- (b) at constant volume.
15. Sheet (1)
- 3. An ideal gas undergoes the following sequences of mechanically reversible processes in a closed system;
- a) From an initial state 70C and 1 bar, it is compressed adiabatically to 150 C.
- b) It is then cooled from 150 to 70C at constant pressure.
- c) Finally, it is expanded isothermally to its original state.
- Calculate W, Q, U, and H of each of the three processes and for the entire cycle. Take Cp = (5/2)R and Cv = (3/2) R
16. Second Law of Thermodynamics
- Direction of energy :
- A spontaneous process is a process in which the final state is more probable than the initial state.
- Every system which is left to itself will, on average, change toward a system of maximum entropy.
17. Second Law of Thermodynamics
- Entropy definition:
- A measure of disorder of the system.
- S is equal to the heat Q it absorbs, divided by T.
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 19. Sheet 1
- 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.
20. Third Law
- Law of zero entropy
- S (at any T) = S (at 298 K) +
21. Sheet 1
- 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+ (0.110 6 ) /T 2. Knowing that the standard entropy for water equal 69.91 J/mol K and heat of vaporization Hv =44.016 kJ/mol.
22. Thermodynamic equilibrium
- S is a measure of equilibrium during reversible adiabatic process or isentropic process.
- To define equilibrium in process other than isentropic process we use free energies.
23.
- Gibbs free energy
- G = H TS
- H = total energy
- TS = unavailable energy
- dG = VdP SdT = fn ( P, T)
- G = 0 at constant T and P
- G = -ve(The process will be spontaneous)
24.
- Isothermal Process:
- dG = VdPG = RT ln(P 2 /P 1 )
- Isobaric process:
- dG= -SdTG= - SdT
- Isochoric Process:
- G = VdP - - SdT
25.
- Helmholtzfree energy
- A = U TS
- U = Total energy
- TS = unavailable energy
- dA = -PdV SdT
- A = 0 at constant V and T
- G = H TS = U + PV- TS = A + PV
- G =A + (PV)
26. Sheet 1
- 6. Calculate G, A, and S for each of the following processes:
- a) Reversible melting of 36 gm of ice at 1 atm and 0C.
- b) Reversible vaporization 39 gm of C 6 H 6at its normal boiling point of 80
27. Sheet 1
- 7. A sample consisting of 1 mole of argon is taken through the following cyclic process:
- a) Isobaric expansion at 1 atm from 25 to 50 liters.
- b) Isochoric cooling at 50 liters from 1 to 0.5 atm.
- c) Isothermal compression to the initial state.
- Calculate U, H,S G, A for each step and for the cycle. Argon may be considered ideal gas with C v =3/2R, C p =5/2R
28. Thank you