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Work, Heat and Internal Energy: The First Law. System – the specific part of the universe of interest to us Surroundings – the part of the universe not.

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Presentation on theme: "Work, Heat and Internal Energy: The First Law. System – the specific part of the universe of interest to us Surroundings – the part of the universe not."— Presentation transcript:

1 Work, Heat and Internal Energy: The First Law

2 System – the specific part of the universe of interest to us Surroundings – the part of the universe not contained in the system

3 3 types of Systems open system – exchanges mass and energy closed system – exchanges energy but no mass isolated system – no exchange of either mass or energy

4 Open system Closed System corkinsulation Isolated System

5 State of a system the system is in a definite state when each of its properties has a definite value. Change in state initial state final state Path initial and final states intermediate states

6 Process reversible or irreversible transformation Cyclic transformation begins and ends at the same state variables.

7 Isothermal dT = 0 Isochoric dV = 0 Isobaric dP = 0

8 Work (w) any quantity that flows across the system’s boundary and is completely convertible into the lifting of a mass in the surroundings. How much work was done? Unit of work = J = 1 kg m/s 2

9

10 A single-stage expansion process State 1 State 2 Piston (T, P 1, V 1 ) mass (m) Piston (T, P 2, V 2 ) mass (m) Direction of piston h2h2 h1h1

11 The work done in the surroundings w surr = P ext  V The work done by the system w sys = - w surr = - P ext  V For an infinitesimal volume change dw sys = - P ext dV

12 If the system is in equilibrium F sys = -F ext P = P ext For a simple system d w rev = - P dV

13 Ideal gas as the working fluid.

14 For an isothermal process (ideal gas as working fluid)

15 dw irr = -P ext dV for a constant external pressure

16 Heat - the quantity that flows across the boundary of the system during a change in state due to temperature difference between system and surroundings HOT to COLD (never the other way around)!!!

17 Measured by determining the temperature change of some known object C - the heat capacity of the system.

18 Integrate the infinitesimal heat flow

19 Exothermic - system to surroundings Endothermic – surroundings to system surroundings system heat

20 Heat flows during phase changes - latent heats Latent heat of vapourisation Latent heat of fusion

21 Subject our system to a cyclic transformation

22 The following would be true for an exact differential

23 The infinitesimal change in the internal energy  For a general process

24 In general, we write U as a function of T and V

25 Examine the first partial derivative

26 Define the constant volume heat capacity, C V

27 For a system undergoing an isochoric temperature change  For a macroscopic system

28 Examine the second partial derivative

29 A T 1, V m,1, P 1 B Stirrer Valve Thermal insulation FF OO C O C O 50 40 30 20 10 0 20 30 40 50 120 100 80 0 20 40 60 40

30 The partial derivative is known as the Joule coefficient,  J.

31 The change in the internal energy under isothermal conditions is related to the Joule Coefficient

32 For an adiabatic process, q = 0!! The first law becomes

33 For an ideal gas undergoing a reversible, adiabatic process

34 Defining the enthalpy of the system Re-examine the piston with the weight on top Piston (T, P, V) mass (m)

35 The first law n Integrating

36 Define the enthalpy of the system, H

37 In general, we write H as a function of T and P

38 Examine the first partial derivative

39 Define the constant pressure heat capacity, C P

40 For a system undergoing an isobaric temperature change  For a macroscopic system

41 For an ideal gas In general

42 Examine the second partial derivative

43 Porous Plug Thermal insulation T 1, P 1, V m,1 T 2, P 2, V m,2 FF OO C O C O 50 40 30 20 10 0 20 30 40 50 120 100 80 0 20 40 60 40 FF OO C O C O 50 40 30 20 10 0 20 30 40 50 120 100 80 0 20 40 60 40

44 The partial derivative is known as the Joule-Thomson coefficient,  JT.

45 The change in the enthalpy under constant pressure conditions is related to the Joule-Thomson Coefficient

46 The shorthand form for a chemical reaction  J = chemical formula for substance J J = stoichiometric coefficient for J

47 The enthalpy change for a chemical reaction H m [J] = molar enthalpies of substance J n J = number of moles of J in the reaction

48 Reaction beginning and ending with equilibrium or metastable states Note – Initial and final states have the same temperature and pressure!

49 We note that 1 mole of a reaction occurs if

50 A reaction that begins and ends with all substances in their standard states The degree sign, either  or  P = 1.00 bar [aqueous species] = 1.00 mol/ kg T = temperature of interest (in data tables - 25  C or 298 K).

51 We note that for 1 mole of a reaction under standard conditions

52 A "chemical thermodynamic reference point." For CO and CO 2 C (s) + O 2 (g)  CO 2 (g) C (s) + ½ O 2 (g)  CO (g)

53 The formation reaction 1 mole of a compound constituent elements stable state of aggregation at that temperature. Formation of 1.00 mole of Na 2 SO 3 (s) 2 Na(s) + S(s) + 3/2 O 2 (g)  Na 2 SO 3 (s) ‘Formation enthalpy of Na 2 SO 3 (s)’,  f H°[Na 2 SO 3 (s)]

54  f H° is a measurable quantity! Compare CO (g) with CO 2 (g) C (s) + 1/2 O 2 (g)  CO (g)  f H° [CO(g)] = -110.5 kJ/mole C (s) + O 2 (g)  CO 2 (g)  f H° [CO 2 (g)] = - 393.5 kJ/mole

55 Formation enthalpies - thermodynamic reference point! H o m [J] =  f H  [J] H m  [elements] = 0 kJ / mole. Use the tabulated values of the formation enthalpies

56 The enthalpy change for a given reaction is calculated from the formation enthalpies as Notes Reverse a reaction Multiply a reaction by an integer

57 A calorimeter - device containing water and/or another substance with a known heat capacity Calorimeters – either truly or approximately adiabatic systems

58  U = q v.

59  H = q p

60 The enthalpy and the internal energy both represent quantities of heat.  U = q v.  H = q p. Relate the two state functions using the following relationship  U =  H -  PV

61 Enthalpy of solution Enthalpy of dilution Enthalpy of fusion Enthalpy of vapourisation

62  sol H - heat absorbed or released when a quantity of solute is dissolved in fixed amount of solvent  sol H = H m (sol’n) – H m (component) H(component) = H m (solid) + H m (solvent) Two definitions Standard Limiting

63 For the process, HCl (aq, 6 M)  HCl (aq, 1 M). The Enthalpy of dilution of the acid.  dil H = H m (sol’n 2) – H m (sol’n,1)

64 Differentiate the reaction enthalpy with temperature

65  r C  p - the heat capacity change for the reaction

66 Examine a chemical reaction. C (s) + O 2 (g)  CO 2 (g)  U = U[CO 2 (g)] – U[C(s)] – U[O 2 (g)] Note -  r H  = -393.5 kJ/mole

67 Use tabulated values of formation enthalpies to obtain  r H°. May also estimate reaction enthalpies using an indirect method.

68 Hess’s Law – the enthalpy change for a given reaction is the same whether the reaction occurs in a single step or in many steps.

69 Examine the following reactions H 2 (g)  H (g) + H (g)  U° = 433.9 kJ Cl 2 (g)  Cl (g) + Cl (g)  U° = 239.5 kJ Bond dissociation energies. Enthalpy changes are designated D (H- H) and D (Cl-Cl).

70 CO 2 (g)  C (g) + 2 O (g)  U  = 740 kJ  H of this reaction D(C=O) What about dissociating methane into C + 4 H’s? CH 4 (g)  C(g) + 4 H(g)  U° = 1640 kJ 4 C-H bonds in CH 4  D (C-H)  410 kJ/mol

71 Note: all chemical reactions involve the breaking and reforming of chemical bonds Bonds break - we add energy. Bonds form - energy is released.  r U°   D(bonds broken) -  D(bonds formed)

72 These are close but not quite exact. Why? The bond energies we use are averaged bond energies ! This is a good approximation for reactions involving diatomic species. Can only use the above procedure for GAS PHASE REACTIONS ONLY!!!


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