Chapter 2 The First Law: the concepts

1. Chapter 2 The First Law: the concepts Thermodynamics: the study of thetransformations of energy, enables us to discuss all these mattersquantitatively and to make useful prediction. Conservation of energy in the universe. Universe = System + Surrounding System: isthe part of the worldin which we have a special interest. Surrounding: is where we make our measurement.

2. The basic concepts: Open system: Matter and energycan be transferred between system and surrounding. Close system: only energycan be transferred. Isolated system: can exchangeneither energy no matterwith its surrounding.

3. 2.1 Work , Heat, and Energy: Internal energy change = Work + Heat (measurable) Work: is done when an object is movedagainst an opposing force. W > 0: when work is done on the (isolated) system by the surrounding. W < 0: when the system does work to the surrounding. Energy of a system is its capacity to do work. Heat: the energy of a systemchanges as a result of temperature difference. Heat transfer direction isdependent on temperature. q > 0: when the system absorbs the heat from surroundings. (Endothermic) q < 0: when the system releases the heat to surrounding. (Exothermic)

4. Diathermic: A boundary that does permit energy transfer as heat. Adiabatic: A boundary that does not permit energy transfer as heat.

5. Consider a reaction (e.g. vaporization of water) occurs in the system: In anadiabatic container: (a). An endothermic reaction; T of the system decreases. (b). An exothermic reaction ; T raises. In andiathermic system: (c). An endothermic reaction, T = constant, q is adsorbed from surrounding. (d). An exothermic reaction, T = constant, q is released to surrounding.

6. Molecular interpretation 21. In molecular terms, heat is the transfer of energy thatmakes use of chaotic motion. Chaotic motion of molecules is called thermal motion. In a hotter system (or surrounding), the thermal motion is more vigorous. Work is the transfer of energy thatmakes use of organized motion. Work: energy transfer making use of theorganized motionof particles. Heat: energy transfer making use of thethermal motion. Work and heat can coexist in a process or chemical reaction.

7. When a system heats its surrounding, molecules of the system stimulate the thermal motion of the surroundings. When a system does work, it stimulates orderly motion in the surroundings. .

8. 2.2 The Internal Energy: In thermodynamics, the total energy of a system called its internal energy, U. Internal energy =kinetic energy + potential energy ΔU = Uf – Ui : the change in internal energy The internal energy is astate function, that its value depends only on the current state of the system and isindependent of pathway. The internal energy is anextensive property. (Extensive property changes with the amount of substance) The unit of the internal energy: joule (J) Molar internal energy: kJ/mol

9. Molecular interpretation 2.2 A molecule has a certain number of degree of freedom. Equipartitiion theorem of classical mechanics: For a collection of particles at thermal equilibrium at a temperature, the average energy of each quadratic contribution (degree of freedom) to the energy is same and equal to ½ kT. For the case of a monatomic perfect gas at a temperature T, Ek = ½ mvx2 + ½ mvy2 + ½ mvz2 (translation energy) (no potential energy exists in the perfect gases) Um = Um(0) + 3/2 RT Um(0): is the molar internal energy at T = 0.

10. When the gas of polyatomic molecules (nonlinear moleucle), the rotation (Rx, Ry, Rz) energy is an additional contribution of 3/2RT to the internal energy: Um = Um(0) + 3RT A linear molecule can rotate only around two axes, so it has two rotational modes of motion, each contributing ½ kT to the internal energy. Um = Um(0) + 5/2RT

11. The internal energy of a system may be changedether by doing work on the system or by heating it. Heat and work areequivalent ways of changing a system’s internal energy. ΔU = q + w In acquisitive convention, w > 0; q > 0 if energy is transfer to the system as work or heat w < 0; q < 0 if energy is lost from the system as work or heat. In scientific representations,sign + quantity.  If a system isolated from its surroundings, then no change in internal energy takes place. First Law of Thermodynamics: The internal energy of an isolated system is constant.

12. The mechanical definition of heat In adiathermic system,ΔU is the same as in adiabatic, but we might find that thework must do is not the sameas that in adiabatic. The difference is definedas the heat adsorbed by the system in the process: Path I: adiabatic process ΔU = q + w = Wad Path II : diathermic process q = U – w = Wad – w Exp. Wad = 42 kJ; w = 50 kJ q = wad – w = – 8 kJ Path I Path II (W dia) (W ad)

13. 2.3 Expansion work: In an infinitesimal change: dU = dq + dw In physics: dw = – Fdz (against the opposing force) For expansion work, dw = – pex dV When the volume changes from Vi to Vf W =–  pex dV vf vi

14. Other types of work: non-expansion work or additional work Work = intensive factor (pressure) x extensive factor (volume)

15. (b). Free expansion: System expands to a vacuum. pex = 0 ; w = 0. (C). Expansion against constant pressure: W = –  pex dV = – pex (Vf – Vi) = – pexΔV A p, V-graph used to compute expansion work is called anindicator diagram.

16. (d). Reversible expansion: A reversible change: is a change that can bereversed by an infinitesimal modification of a variable. Equilibrium: if an infinitesimal changes in the conditions inopposite directions results in opposite change in its state. In the view of the work (reverse): dw = – pex dV = – pdV W = – pex dV

17. (e). Isothermal reversible expansion: An isothermal reversible expansion of the perfect gas: PV = nRT Wrev = – pex dV = –nRT ln (Vf/Vi) | Wrev| > | Wone-step| Themaximum workavailable from a system operating between specified initial and final states and passing along a specified path is obtained when the change takes placereversibly. Vf Vi

18. One-step expansion (膨脹) Pex = 1/4 P1 Work = – Pex –ΔV = – ¼ P1 (4V1 – V1) = – 3/4 P1V1 P1; V1 ¼ P1; 4V1

19. Infinite-step (無限多步)expansion: The external pressure is always almost exactly equal to the pressure of the gas. Reversible process：P～Pex P = Pex+ΔP (P  0) ∣Work∣＝∫Pex dV P ～ PexPex ～ P＝ RT/V∣W∞∣＝∣Wrev∣＝∫nRT/V dV

20. When Pex = 0, such expansion of gas is called free expansion (自由膨脹):

21. 2.4 Heat transaction: dU = dq + dwexp + dwe At constant V (dwexp = 0) and no other kind worked produced (dwe = 0). dU = dqv  U = qv (a). Calorimeter: The most common device for measuring ΔU is the adiabatic bomb calorimeter. Constant-volume container.

22. No net loss of heat from the calorimeter, the calorimeter is adiabatic. For a calorimeter: q = CΔT; C: calorimeter constant (b) Heat capacity: The internal energy of a substance increases when its temperature is raised. The slope of tangent to the curve at any temperature is called the heat capacity of the system. Cv = (U/T)v At constant volume

23. Heat capacity areextensive properties. However,the molar heat capacity at constant volume is intensive. Unit: kJmol-1K-1 Specific heat capacity: Unit: kJK-1g-1 In general, internal energies depend on the temperature and decrease at low temperature. For perfect gas: Cv is T-independent Cv = 3/2 R (monatomic molecules) Cv = 3 R (polyatomic molecules)

24. At constant volume: dU = Cv dT If Cv is T-independent, ΔU = CvΔT For measuring ΔU, heat supplied at constant volume. qv = CvΔT = ΔU 2.5 Enthalpy: When the system is free to change its volume,dU  dq. The heat supplied at constant p is equal the change in another thermodynamic property, enthalpy H.

25. At constant p, some of energy supplied as heat may escape into the surrounding as work. H = U + PV H is a state function. dH = dq (at pex= constant, no additional work) dH = dU + pdV + Vdp = dq – pdV + pdV + Vdp = dq + vdP At constant p: dH = dqp ΔH = qp

26. (b). The measurement of an enthalpy change An enthalpy change can be measured calorimetrically by monitoringthe temperature changeat constant p. (isobar calorimeter) Because solids and liquids have small molar volume, Hm = Um + pVm≈ Um ΔHm = ΔUm (solid and liquid) For the perfect gas,H = U + pV = U + nRT The change of enthalpy in a reaction that produce or consumes gas is: ΔH = ΔU + Δ(nRT)

27. (c). The variation of enthalpy with temperature: H = f (T, p ,V) The enthalpy increase with the temperature. Heat capacity at constant pressure (Cp): theslopeof the tangent to a plot of enthalpy against T. Cp = (H/T)p Cp,m: heat capacity per mole of material. dH = CpdT (at constant p) If Cp is a constant:ΔH = CpΔT at constant pressure

28. A common approximate empirical expression: Cp,m = a + bT + c/T2 H(T2) – H(T1) = a (T2 – T1) + ½ b(T22–T12) – c(1/T2 – 1/T1)

29. The relationship between heat capacities: q = ΔU – w = ΔU + pexdV At constant V, w = 0; At constant P, w = –pexdV qp > qv ;Cp > Cv For the perfect gas: Cp – Cv = nR For 1 mole perfect gas: Cp – Cv = R

30. I 2.1 Differential Scanning Calorimeter (DSC) A DSC measures the energy transferred as heat to or from a sample at a constant pressure during a physical or chemical change. A DSC consists of two small compartments that are heated electrically at a constant rate. T = T0 + T

31. To maintain the same temperature in both compartments, excess energy is transferred as heat to or from the sample during the process. If no physical or chemical change occurs, qp = CpT (Cp independent of temperature) The chemical or physical process requires the transfer of qp + qp,ex qp,ex: excess energy transferred as heat to attain the same change in temperature qp + qp,ex = (Cp + Cp,ex)T  Cp,ex = qp,ex/ T = qp,ex/t = Pex/ Pex = the excess electrical power necessary to equalize the temperature

32. A DSC trace, also call a thermogram, consists of a plot of Pex(Cp,ex) against T. The enthalpy change associated with the process is H = T1Cp,ex dT 2.6 Adiabatic Changes: Adiabatic process： A process in which no energy as heat flows into or out of the system. q＝0 ; ΔE＝q ＋ W＝ W； For an ideal gas: dE ＝ nCvdT ； W＝ – Pex dV T2

33. For a reversible, adiabatic expansion－compression of an ideal gas. dE＝nCvdT＝ –Pex dV＝ –nRT / V dV Cv / T dT＝ –R / V dV Form T1 to T2 (infinitesimal changes) Cv∫1/ T dT＝ –R∫1/ V dV; Cv㏑T2 / T1＝ – R㏑V2 / V1 T2 / T1＝ (V1 / V2)r-1； r＝Cp / Cv T1V1r-1 ＝ T2V2r-1 PV＝nRT ； T2 / T1＝ P2V2 / P1V1 P1V1r＝ P2V2r

34. For isothermal expansion (At const. T，ΔE＝0) P1V1＝ P2V2＝ constant For adiabatic expansion (q＝0，T will change) P1V1r＝ P2V2r＝constant Cp/Cv > 1

35. Thermochemistry The study of theheat produced or requiredby chemical reactions. Internal energy = Kinetic energy (T) + Potential energy Potential energy = Physical potential + Chemical potential Physical potential is from intermolecular interaction. Chemical potential is ascribed to intramolecular chemical bond. For chemical reactions (Thermochemistry): Vessel + content (chemicals): system In general, chemical reaction takes place at constant p qp = ΔH

36. At constant pressure: An endothermicprocess (q > 0) hasΔH > 0. Anexothermic process (q < 0)is one for whichΔH < 0. 2.7 Standard enthalpy changes: H(T, P, n…….) Thestandard stateof a substance at a specified temperature is its pure format 1 bar. Thestandard enthalpy change for a reactionor a physical process is the difference between the enthalpy of the products instandard statesand the enthalpy of the reactants in their standard states, all at the same specified temperature.

37. Standard enthalpy of vaporization: H2O(l) H2O(g) ΔvapHØ (373K) = + 40.66 kJ/mol Standard enthalpy may be reportedfor any temperature. The conventional temperature =298.15 K (a). Enthalpy of physical change: Standard enthalpy of transition (a change of physical state): ΔtrsHØ

38. Enthalpy is a state function: (path-independent) ΔsubHØ = ΔfusHØ + ΔvapHØ (at same T) ΔHØ(AB) = – ΔHØ(BA)

40. (b). Enthalpies of Chemical Change: Standard reaction enthalpy, ΔrHØ, is the change in enthalpy when reactants in their standard states change to products in their standard states. CH4(g) + 2 O2 (g) CO2(g) + 2H2O(l) ΔrHØ = – 890 kJ/mol Thermochemical equation: a chemical equation + standard reaction enthalpy 2 A + B 3C + D ; ΔrHØ = ? ΔrHØ = ∑ v HmØ – ∑ v HmØ The general form: ΔrHØ = ∑ vJHmØ(J) ; vJ: stoichiometric number; HmØ(J): the standard molar enthalpy of a species J at standard temperature. reactants products

41. Somestandard reaction enthalpies have special names and a particular significant. The standard enthalpy of combustion cHØ. C6H12O6(s) + 6O2 6CO2(g) + 6 H2O(l) cHØ = – 2808 kJmol-1

42. I2.2 Food and Energy Reserve: The thermodynamically properties of fuels and foods are commonly discussed in term of their specific enthalpy (cH/M). A typical 18 – 20 year old man requires a daily input of about 12 MJ; a woman needs about 9.0 MJ.

43. Glucose has a specific enthalpy of 16 kJg-1. Digestible carbohydrates has a specific enthalpy of 17 kJg-1. Indigestible cellulose helps to move digestion products through the intestine. Fats (long-chain esters) have a specific enthalpy of 38 kJg-1. The specific enthalpyHydrocarbon oils used as fuel is 48 kJg-1. In Arctic(北極) species, the stored fat acts as layer of insulation. In desert species, the fat is also a source of water, one of its oxidation products. Proteins are also used as a source of energy, but their components (amino acids) are often used to construct other protein instead. Radiation is one means of discarding heat; another is evaporation and energy demands of vaporization of water (2.4 kJg-1). Vigorous excises  1 – 2 dm3 of perspired water per hour  2.4 – 5.0 MJh-1.

44. (c). Hess’s law: The standard enthalpy of an overall reaction is thesumof the standard enthalpies of the individual reactionsinto which a reaction may be divided. C(s) + 2 H2(g)CH4(g)ΔHf0 = ? C(s) + O2(g)CO2(g) ΔH10 = -394 kJ 2 H2(g) + O2(g) H2O(l) ΔH20= -572 kJ CH4(g)+ 2O2(g) CO2(g) + H2O(l) ΔH30 = -891 kJ ΔH0 = -ΔH30 + ΔH10+ ΔH20 = -75 kJ The importance of Hess’s law is that information about a reaction of interest, which may be difficult to determined directly, can assembled from information on other reactions.

45. Illustration: N2 + 3 H2 2 NH3 v(N2) = –1 ; v(H2) =–3; v(NH3) = +2 ΔrHØ = 2HmØ(NH3) – {(HmØ(N2) + 3 HmØ(H2)} However, how to have the value of HmØ(NH3) ???

46. 2.8 Standard Enthalpies of formation: Standard enthalpy of formation ΔfHØ: the standard reaction enthalpy for the formation of the compound from its elementsin their reference states. Reference state of an element: its most stable state at the specified temperature and 1 bar. 6C(s, graphite) + 3 H2(g) C6H6(l) ΔH = + 49.0 kJ/mol The standard enthalpy offormation of liquid benzeneat 298 K is + 49.0 kJ/mol. The standard enthalpies of formation of elements in their reference states are zero at all temperatures because they are the enthalpies of such “null” reactions as N2(g)  N2(g).

47. Definitions of Standard State: 1. Gas : 1 atm (or 1 bar = 105 Pa). 2. In solution, concentration = 1.0 M at 1 atm. 3. A pure substance in condensed state, pure liquid or solid. 4. Element: it exists (most stable) under conditions of 1 atm and the temperature of interest (25 oC). Notice the physical state or the structure of reactant and product: ΔHof of H2O(g)  ΔHof of H2O (l) ΔHof of C(graphite) = 0 ΔHof of C(diamond) = 2 kJ/mol ΔHof of many compounds are listed in Table 2.5 and 2.6

48. Enthalpies of formation (Hof) are always given per mole of product with the product in its standard state.

49. (a). The reaction enthalpy in terms of enthalpies of formation: Conceptually, we can regard a reaction as proceeding by decomposing the reactants into their elements and then forming those elements into the products. ΔrHØ = ∑ vJΔHfØ(J) Δ HfØ(J): the standard enthalpy of formation of a species J at interested temperature.

50. Reference (standard) state  Set up a scale for measuring thermodynamic properties. Hof of element at standard state  0 Reactants Elements Products Hof (reactants ) Hof = 0 Hof (products)