Presentation Transcript

1. Find q, w, DU, and DH for 5.0 grams H2O at 20°C heated to become gas at 100°C and P = 0.200 atm. n = 5.0 g ÷ 18.0 g mol-1 = 0.28 mols Heat water from 20ºC to 100ºC. (DH = q = CP• DT) w = - ∫ P dV = - P • (V2 – V1) V2= 5.0 ÷ .9584 • 1 x 10-6 = 5.2 x 10-6 m3 V1= 5.0 ÷ .9982 • 1 x 10-6 = 5.0 x 10-6 m3 DU = DH - D(PV) = q + w Phase change from liquid to vapor at 100 ºC. find Vgas 3. Expansion of gas. isoT IG expansion
2. For a reaction at T  298K ……. dH= ∫ CPdT ~ CPDT so …. dDHrx= ∫ DCP,rxdT ~ DCP,rxDT so …. DHT – DH298 = ∫298TDCP,rxdT DCPrx = SiniDCPof,298,i DHT = DH298 + DCP,rx (T – 298) Note: you must keep units consistent: convert DCP,rx to kJ
3. DHo298 = SiniDHof,298,i Ethanol combustion C2H6Ol + 3O2→ 2CO2 + 3H2Og at 298 K -1235 J DHT= DH298 + DCP,rx (T – 298) keep units consistent: convert DCP,rx to kJ DCP,rx= -29.05 J mol-1 K-1 = -0.02905 kJ mol-1 K-1 at 500 K -1229 J
4. Early 1800’s – heat is a substance (caloric) 1824 – Carnot – Did work on heat engines that led to 2nd Law and the concept of entropy Died very young ……. 1850s – William Thomson (Lord Kelvin) – Planck and Clausius Statements of 2nd Law of Thermodynamics based on Carnot’s work
5. Spontaneity – Will a process occur? Spontaneous: dropped glass breaks gas expands into vacuum (or area of reduced P) ice melts at room T sugar dissolves in water diamonds change into graphite at 1 atm. P? Not spontaneous: pieces of broken glass reassemble and jump to table gas molecules all rush to one half of flask volume water freezes at room T oil dissolves in water water reacts to form oxygen and hydrogen gas
6. Heat (q) A measure of thermal energy transfer that can be measured by the change in T of the system. Tc Tf Th Tf q = m1c1(Th-Tf) = m2c2(Tf-Tc) q Clausius - It is impossible for a system to undergo a cyclic process for which the sole effect is the flow of heat into the system from a cold reservoir and the flow of an equivalent amount of heat out of the system into a hot reservoir.
7. HEAT ENGINE What limits engine output? IG – reversible, isoT expansion Hot Reservoir Engine IG qh -w e = work out ÷ heat in = -w/qh Perfect if qh = -w & e = 1 qh
8. CYCLIC ENGINE IG w(-) qh(+) IG cyclic so …. DU = 0 qc(-) H C e = -w/(qh) = (qh + qc)/qh
9. 332560 332560 CARNOT CYCLE ― reversible heat engine (IG fluid) StateT (K)V (m3)P (Pa) 1 400 .0010 2 400 .0020 3 300 4 300 1 400 .0010 1. Isothermal expansion at TH 166280 4. Adiabatic Compression TCTH achieves original P, V, & T 2. Adiabatic Expansion TH TC .00308 81002 3. Isothermal Compression .00154 162003 For n = 0.10 moles IG T2/T1 = (V1/V2)R/Cv CV = 12.47 (R/CV ~ 0.667) What is net work of one cycle?
10. 332560 332560 CARNOT CYCLE ― reversible heat engine (IG fluid) StateT (K)V (m3)P (Pa) 1 400 .0010 2 400 .0020 3 300 4 300 1 400 .0010 1. Isothermal expansion at TH 166280 4. Adiabatic Compression TCTH achieves original P, V, & T 2. Adiabatic Expansion TH TC .00308 81002 3. Isothermal Compression .00154 162003 For n = 0.10 moles IG T2/T1 = (V1/V2)R/Cv CV = 12.47 (R/CV ~ 0.667) What is net work of one cycle?
11. Carnot Cycle 1 4 P Pa 2 3 V m3
12. Carnot Cycle – heat engine – IG working fluid CV,m= 12.47 Is this an efficient process? n = 0.1 e = 58/231 = 0.25 CP,m = CV,m + R w = -  P dV DU = q + w DU =  CVdT DH = DU + D(PV) DH =  CPdT Iso T expansion adiabatic expansion Iso T compression adiabatic compression
13. Carnot Cycle dU = dq + dw = dq - PdV dU = CVdT = dqrev -nRTdV/V  T & ∫ & 0 =&  CV dT/T = dqrev/T - nR dV/V 0 0 0 dqrev/T is a State Function dqrev/T = dS S = entropy Entropy (S) is a state function such that dS = dqrev/T
14. Efficiency for Carnot Engineoperating between Th = 400K & Tc = 300K e = -w/qh = (qh + qc)/qh = 1 + qc/qh e = -(-58)/231 or 1 + -173/231 = 0.25 Because qrev/T is a state function ……… qh/Th + qc/Tc = 0 … qh/Th = -qc/Tc or - Tc/Th= qc/qh& erev= 1 - Tc/Th erev = 1 - 300/400 = 0.25 The theoretical efficiency of a reversible heat engine can be predicted simply by knowing the temperature of the hot and cold reservoirs.
15. The 2nd Law of Thermodynamics Esuper Kelvin-Planck - It is impossible for a system to undergo a cyclic process for which the sole effect is the flow of heat into the system from a hot reservoir and the performance of an equivalent amount of work on the surroundings. erev (1 - Tc/Th) > eirrev No cyclic engine can have an efficiency of 1. TH qh qh win backwards Work in Erev wnet qc,super qc,rev TC
16. Reversible process DSuniv = DSsys + DSsurr = ∫ dqrev,sys/T + ∫ dqrev,surr/T = 0 Heat (qrev) surr Reversible Process dSsys = dq/T For a reversible process in closed system DSuniv = 0. But – no ‘real’ system can ever be truly reversible!
17. Irreversible process DSuniv = DSsys + DSsurr = ∫ dqrev,sys/T + ∫ dqrev,surr/T > 0 surr = Universe Heat (qrev) Irreversible Process dSsys = dq/T For an irreversible process in closed system DSuniv > 0. The Entropy of the Universe is increasing DSuniv = DSsys + DSsurr > 0
18. DSuniv = DSsys + DSsurr > 0 dSsurr = dqrev/T Heat (qrev) surr Irreversble Process qirrev dSsys> dq/T otherwise violates 2nd Law the system process acts as a superengine. -qsys (irrev) = qsurr (rev) & |DSsys | < DSsurr& DSuniv > 0. For a spontaneous, irreversible process in closed system, DSuniv > 0. Closed dS≥ dq/T or dq – TdS≤ 0 A spontaneous process in a closed system need not have a positive change in entropy!
19. TH qh Refrigerator DSsys < 0 but … DSsurr> |DSsys| > 0 win Erev qc TC
20. 1st Law – The energy of isolated system is constant 2ndLaw The entropy of an isolated system increases in the course of a spontaneous change. Or ….. entropy is a measure of disorder. An isolated system tends to become more disorderd.
21. System isolated e.g. universe closed open 2nd Law 1st Law DE = 0 DS ≥ 0 dq – TdS≤ 0 DU = q + w Entropy (S) is a state function such that …. dS = dqrev/T Clausius - It is impossible for a system to undergo a cyclic process for which the sole effect is the flow of heat into the system from a cold reservoir and the flow of an equivalent amount of heat out of the system into a hot reservoir. Kelvin-Planck - It is impossible for a system to undergo a cyclic process for which the sole effect is the flow of heat into the system from a hot reservoir and the performance of an equivalent amount of work on the surroundings.
22. wrev= -RT ln (2) = -1729 J qrev = +1729 J W at 300K, isothermal, IG expansion DSsurr = -1250/300 = -4.17 J Spontaneous Irrev Process DSsys = +1729/300 = + 5.76 J DSsys > q/T (4.17 J) dSsys> dq/T PV Isotherm P (Pa) wirrev= -100000 (0.0125) = -1250 J q = +1250 J surr DSuniv = -4.17 + 5.76 = 1.59 J V (m3)
23. wrev= -RT ln(0.5) = 1729 J qrev = +1729 J W at 300K, isothermal, IG compression DSsurr = +2500/300 = +8.33 J Spontaneous Irrev Process DSsys = -1729/300 = - 5.76 J DSsys > q/T (-8.33 J) dSsys> dq/T PV Isotherm P (Pa) wirrev= -200000 (0.0125) = +2500 J q = -2500 J surr DSuniv = +8.33 - 5.76 = 1.59 J V (m3)
24. The Entropy of the Universe is increasing DSuniv = DSsys + DSsurr > 0 dSsurr = dq/T surr Spontaneous – Irrev Process qirrev dSsys> dq/T For a spontaneous, irreversible process in closed system, DSuniv > 0. Closed dS≥ dq/T or dq – TdS≤ 0 A spontaneous process in a closed system need not have a positive change in entropy!
25. Spontaneity criteria – Will a process occur? Isolated system: DSsys > 0 dq – TdSsys≤ 0 Closed system: DSuniv > 0 How do you calculate DS for an Irreversible Processes? Devise a reversible process and calculate DS for this Spontaneous (isolated): a hot body transfers heat to a colder body an ideal gas expands into vacuum (or area of reduced P)
26. Tc Tf Th Tf Irreversible heating qrev = CPdT DS = 24.4 • (ln 323/273) + 24.4 • ln (323/373) = DS = 4.10 + -3.51 = 0.59 JK-1 q (Cu: CP = 24.4 J mol-1 K-1), 1 mole: Th = 100°C, Tc = 0°C, so Tf = 50°C DS =  dqrev/T =  CP dT/T = CP ln (T2/T1) (sum for both blocks)
27. Joule Experiment - IG expansion into vacuum Adiabatic so q = 0. Isolated so DS must be > 0. Is it reversible? How would you go about finding DS? n = 1, T = 298K, V1 = 0.001m3, V2 = 0.002 m3 qrev = nRTln(V2/V1) & DS = qrev/T = nRln(V2/V1) > 0 DS = 1 • 8.314 • ln 2 = 5.76 JK-1?
28. Ideal gas change of state devise reversible process to carry out change Pi – V2 – T1 P1 – V1 – T1 q = -w = nRT ∫ dV/V DS = q/T = nRln (V2/V1) 1 mole of an IG is at 3L and 300K undergoes a change to 0.5 L and 370K. What is DS for this change? q = DU = nCV ∫ dT DS = q/T = nCVln (T2/T1) q = DU = nCV ∫ dT DS = q/T = nCVln (T2/T1) q = -w = nRT ∫ dV/V DS = q/T = nRln (V2/V1) P2 – V2 – T2 DS = nR∫dV/V + nCV,m∫ dT/T isoT expansion cst V heating Pi – V1 – T2
29. Phase changes (reversible) Will melting ice have a + or – entropy change? qP(rev) = DHtr DS = ∫ qP/T = DHtr/T What is DS for the melting of 1kg of ice at 0ºC? DHmelt = 6007 J mol-1 & CP = 38.1 (ice) and 75.4 (water) J mol-1 K-1 What is DS for the melting of 1kg of ice at -10ºC? 0°C 0°C -10°C 1 2 3 -10°C Devise reversible process: 1) cst P heating to 0ºC 2) melt at 0ºC 3) cst P cooling to -10ºC Is DS (+)? Is the process spontaneous?
30. Entropy of Mixing IG’s= 2 separate IG expansion problemsConstant T & PA = PB DSa = qrev/T = -w/T = naR  dV/V = naR ln (V/Va) = -naR ln (na/n) = -naR ln ca DS = -naR lnca - nbR lncb DS = -naR ln(V/Va) - nbR ln(V/Vb)
31. Summary – Find DS for …… Heating (solid/liquid): DS = T1T2CPdT/T Phase Ds at nmp or nbp: DS = DHtr/T IG changes of state: DS = nCV,mln (T2/T1) + nRln (V2/V1) mixing: DS =Si - niRln (ci)
32. 1/36 = 1/6  1/6 6/36 = 1/6
33. Entropy and Probability Ω= 1 Ω= 1 + 16 + 36 + 16 + 1 = 70 S = f(p) for two events p1&2= p1p2 h(p1&2) = f(p1) + g(p2) = h(p1p2) ln(x) + ln(y) = ln(xy) f & g must be ln functions “The impossibility of an uncompensated decrease of entropy seems to be reduced to an improbability.” Boltzmann S = k ln(p) = k lnΩ
34. What is Entropy? Work (w) = ordered motion vs. heat (q) = disordered motion dS = dqrev/T and DS = if dqrev/T = J K-1 Boltzmann distribution and Entropy Ni/N0 = exp(DE/kT)= exp(DS/k) At low T the energy of a system is more ordered The same amount of heat (q) will have a larger effect on the order of the system than it will at high T.
35. nonmetal: CP = CV = aT3 metal: CP = CV = aT3 + bT CP vs. T for SO2 Cst P Heating (solid/liquid): dqP = DH = ∫ CPdT DS = T1T2CPdT/T
36. The Third Law (Nernst) “The 3rd Law was based on empirical data regarding the changes in DG for any process as T approaches 0K. However, this led to conclusions regarding DS …… “The entropy of any pure, perfect crystal is zero at 0K.” “The entropy change accompanying any physical or chemical transformation  0 as T  0 provided all the substances involved are perfectly ordered.” Si= 0 at 0K (i represents any pure substance) if S0,m = 0 then ..... Si,m,T= 0TCPdT/T “If a substance is a solid at 298K the above equation and the 3rd Law allow the determination of the ‘Conventional’ entropy of that substance at 298K. If the substance is a liquid or gas, the entropy of phase changes has to be included ….
37. nonmetal: CP = CV = aT3 metal: CP = CV = aT3 + bT CP vs. T for SO2 Conventional Entropy Tables DSom,i =  0TnmpDCPo/T dT + DHfus/T(if liquid) + TnmpTnbpDCPo/T dT + DHvap/T (if gas)+  Tnbp298DCPo/T dT
38. DHfus/T DHvap/T CP/T vs. T for SO2 Integral = DS
39. DHoT = SiniDHof,T,i DSoT = SiniSom,T,i Methane burns at 298K. CH4(g) + 2O2(g) → CO2(g) + 2H2O(l) Methane burns at 298K. CH4(g) + 2O2(g) → CO2(g) + 2H2O(g)
40. Criteria for Spontaneity isolated dS > 0 Closed dS ≥ dq/T or dq – TdS ≤ 0 Cst V,T: dU – TdS ≤ 0 let A = U - TS dA ≤ 0 Cst P,T: dH – TdS ≤ 0 let G = H - TS dG ≤ 0
41. Free Energy A = U - TS Helmholtz Free Energy dA = dU – TdS = dq + dw – TdS if rev then dw = dwmax & dS = dq/T dA = dwmax A called ‘work’ function It is equivalent to the maximum work a system can perform A is minimized for any irrev process in a closed system at cst V & T.
42. Free Energy G = H - TS Gibbs Free Energy dG = dH – TdS = dU + PdV – TdS = dq + dw + PdV– TdS if rev then dw = dwrev & dS = dq/T dG = dwrev – dwexp = dwnonexp dG is often referred to as the maximum ‘useful’ work G is minimized for any irrev process in a closed system at cst P & T.
43. Measuring DG Heating and Cooling: DG is undefined for temperature changes. DG can only be determined for constant T processes. However, the change in DG for a process with change in T can be determined. 2.Isothermal processes: DG = DH – T DS 3.Phase changes at nmp or nbp: DG = DH – T DS sub DStr = DHtr/T DG = DHtr – T DHtr/T = 0 At the nmp or nbp two phases are in equilibrium and DGtr = 0
44. Measuring DG 4. Chemical Reactions: DG298= SiniDGºf,m,i (DG298 can be determined from thermodynamic table data) or…. DG298 = DH298 – 298 • DS298 5. Chemical Reactions (not at 298): Find DHT and DST using DCP values inthermodynamic tables Then…. DGT = DHT – TDST DHºT=DHo298 + DCPºdT DSºT=DSo298 + DCPºdT/T
45. H2(g) + ½ O2(g)→ H2O(l) Df DG˚f = -285.83 – 298(-0.163343) = -237.15 DS˚f = 0.06991 - .130684 – ½ 0.205138 = -0.163343 DG˚f -285.83 – 298(0.06991) = -306.66
46. Glucose + 6O2 6CO2 + H2O D D
47. 0˚C 0˚C What is DG for conversion of supercooled water at -10˚C to ice at -10˚C? (n = 1) -10˚C You can’t measure DG for constant P heating/cooling! -10˚C DH = 1  75.4  10 + 1  -6007 + 1  38.07  -10 = -6312 DS = 75.4  ln(273/263) – 6007/273 + 38.07  ln(263/273) = -20.61 DG = -6312 – 263(-20.61) = -892 J
48. Free Energy of Mixing IG’sDG = DH – TDS = 0 - TDS = -TDS DS = -naR ln(V/Va) - nbR ln(V/Vb) DS = -naR lnca - nbR lncb DG = naRT ln(V/Va) - nbRT ln(V/Vb) DG = naRTlnca - nbRTlncb
49. dU = dq + dw (if closed, rev & wexp) dU = dqrev - PdV dqrev/T = dS & dqrev = TdS dU = TdS - PdV Since U is a state function, this also applies to irreversible processes. This is the “fundamental equation”
50. 1. dU = TdS - PdV (derive from 1st) 2. H = U + PV 3. A = U - TS 4. G = H - TS 5. CV = (dU/dT)V= T(dS/dT)V 6. CP = (dH/dT)P= T(dS/dT)P Gibbs Equations 1. dU = TdS - PdV (derive from 1st) 2. dH = TdS + VdP 3. dA = -SdT - PdV 4. dG = -SdT + VdP
51. 1. dU = TdS - PdV U (S,V) 2. dH = TdS + VdP H (S,P) 3. dA = -SdT – PdV A (T,V) 4. dG = -SdT + VdP G (T,P) e.g. dG = (dG/dT)P dT + (dG/dP)T dP (dG/dT)P = -S and (dG/dP)T = V T = (dU/dS)V -P = (dU/dV)S T = (dH/dS)P V = (dH/dP)S -S = (dA/dT)V -P = (dA/dV)T
52. Euler Reciprocity Relations Given z = z (x,y) dz = (dz/dx)Y dx + (dz/dy)x dy let M = (dz/dx)Y & N = (dz/dy)x then..... (dM/dy)x = (dN/dx)Y dU = TdS – PdV ……. (dT/dV)S = -(dP/dS)V
53. 1. dU = TdS - PdV (derive from 1st) 2. dH = TdS + VdP 3. dA = -SdT - PdV 4. dG = -SdT + VdP (dT/dV)S = -(dP/dS)V (dT/dP)S = (dV/dS)P Maxwell Relations (dS/dV)T = (dP/dT)V (dS/dP)T = -(dV/dT)P
54. Starting Points dU = TdS - PdV (derive from 1st) dH = TdS + VdP dA = -SdT – PdV dG = -SdT + VdP CV = (dU/dT)V= T(dS/dT)V CP = (dH/dT)P= T(dS/dT)P a = 1/V (dV/dT)P k = -1/V (dV/dP)T a/k = (dP/dT)V
55. desired relationships .......... (dS/dT)P (dS/dP)T(dS/dT)V (dS/dV)T (dH/dT)P (dH/dP)T (dG/dT)P (dG/dP)T (dU/dV)T(dU/dT)V (dA/dT)V (dA/dV)T some of the relationships are fairly easy.... (dS/dT)P = CP/T (dH/dT)P = CP (dS/dT)V = CV/T (dU/dT)V = CV CV = (dU/dT)V= T(dS/dT)V & CP = (dH/dT)P= T(dS/dT)P a = 1/V (dV/dT)P k = -1/V (dV/dP)T a/k = (dP/dT)V
56. desired relationships .......... (dS/dT)P(dS/dP)T (dS/dT)V (dS/dV)T (dH/dT)P(dH/dP)T (dG/dT)P (dG/dP)T (dU/dV)T (dU/dT)V (dA/dT)V (dA/dV)T ÷ by dP at cst T dH = TdS + VdP (dH/dP)T = T(dS/dP)T + V from Maxwell... (dS/dP)T = -(dV/dT)P = - aV ....... (dH/dP)T = -TVa + V a = 1/V (dV/dT)P k = -1/V (dV/dP)T a/k = (dP/dT)V
57. Internal P = (dU/dV)T dU = TdS - PdV dVT (dU/dV)T = T(dS/dV)T - P apply Euler/Maxwell from dA = -SdT - PdV (dU/dV)T = T(dP/dT)V – P gas (dU/dV)T = T(a/k)T - Psol/liq for an IG show that(dU/dV)T = 0 a = 1/V (dV/dT)P k = -1/V (dV/dP)T a/k = (dP/dT)V
58. Hydrophobic Interactions C3H8(aq) C3H8(l) DH298 ~ +8 kJ mol-1 DS298 ~ +80 J mol-1 DG298 = 8 - 298  0.080 = -16 kJ mol-1 Hydrophobic effects are due to the entropy increase of the solvent which must be more ordered when it is in the presence of nonpolar solutes compared to when it is surrounded by other water molecules.
59. The effect of P on Gibbs energy or (dG/dP)T (dG/dP)T = V & dG = V dP DG =  V dP ~ V (P2 – P1) for solids/liquids = nRT ln (Pf/Pi) for an IG For an IG …. DG = Gm(P) – Gm(P°) = RT ln (P/P°) … Gm(P) = Gºm + RT ln (P/Pº) ideal gas
60. dG = -SdT + VdP This Gibb’s equation assumes that there is no change in the amount of substance in the closed system. However, if you open the lid and throw in some substance, G will change. It will also change if there is a chemical reaction or phase changes that how much of a particular substance is present. This concept is handled by expanding the Gibb’s equation above to …… dG = -SdT + VdP + Si (dG/dni)T,P,n´ Chemical potential (mi) = (dG/dni)T,P,n´ Gm(P) = Gºm + RT ln (P/Pº) ideal gas or … mi(P) = mi(P˚) + RT ln(Pi/P˚)
61. dG = -SdT + VdP + Si (dG/dni)T,P,n´ What can cause a change in ni? 1) Chemical Reactions … Chapter 4 2) Phase changes … Chapter 5 dG = -SdT + VdP + Si,a (dG/dnia)T,P a represents phase (e.g. s, l or g)? or multiple crystalline forms of the solid … … liquid crystal …. … superconducting … … transition to chaos …