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Do Problem 40 in Chapter 1 PowerPoint Presentation
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Do Problem 40 in Chapter 1

Do Problem 40 in Chapter 1

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Do Problem 40 in Chapter 1

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  1. Do Problem 40 in Chapter 1 Pid= RT/Vm Vm,id= RT/P Pvdw = RT/(Vm – b) – a/Vm2 Add Vid into VdW equation. Adjust Vm to change P value until target P reached. Have class do handout Assign problem 20 and 35 for Friday

  2. What is the composition of the Universe? Thermodynamics – The study of the interactions between matter and energy. The Laws of Thermodynamics 1. Conservation of Energy or … “You can’t Win” 2. The Entropy of the Universe is Increasing or … “You can’t break even” 3. A Pure Substance has 0 Entropy at 0K

  3. Matter & energy energy Open System adiabatic no heat transfer Closed System isolated system Isothermal (T) Surroundings Isobaric (P) Isochoric (V)

  4. ENERGY OF SYSTEM internal energy kinetic energy (½mv2) potential energy depends on the force field applied to the system. E.g. gravity, electrostatic, etc. E = K + V + U U = Utr + Uvib + Urot + Uel + Uint + Unuc Assuming K and V are not changing …… DE = DU

  5. Thermodynamic Properties Intensive Extensive P T n V U H, S, G isobaric isothermal isochoric V/n or Vm U/n = Um, etc

  6. The 1st Law of Thermodynamics DU = q + w work (w) heat (q) Conservation of Energy – The energy of the universe is constant. The energy of a system can change if an equal amount of energy is transferred to/from the surroundings in the form of heat or work. For a given change in U, the values of q and w will depend on how the process is carried out; q & w are not state functions. State FunctionA property of a system that is indicative of the state of the system at the current time. P, V, T, U, H, G, S Change of State – DT, DU, DP, etc. …. The values of DT, DU, etc. are independent of path. Equation of State – PV = nRT; G = H –TS; indicates the relationship between the values of various state functions with respect to other state functions

  7. Work (w) Time (t) s Volts (E) WA-1 or J s-1 A-1 current (I) Ampere (A) work = force applied over a distance extension work dw = Fxdx and w = ∫ Fxdx gravitational work F = ma = mg (constant) dw = mg dh or w = mgh electrical work w = EIt

  8. Work derivations dw = Fxdx ….. w =  dw =  Fxdxif Fx is cst … = Fx (x2 - x1) 2.1 dw = Fxdx: for gas P = F/A (area) & F = PA dw = PA dx = P  Adx = P dV dw = - P dV(- due to sign convention) expansion w is (-) compression w is (+)

  9. PV (expansion) work system must allow for DV. P = F/A & Fx = PA

  10. PV (expansion) work system must allow for DV. Fx = PA w = - Fx dx = - PA dx = - P dV

  11. Reversible vs. Irreversible Pext = Pf = constant

  12. Reversible vs. Irreversible Pext is not constant

  13. w = -RT ln (2) = -1729 J Work for reversible, isothermal, IG expansion n = 1.00 w = -  P dV = - nRTdV/V = -nRTln (V2/V1) PV Isotherm P (Pa) V (m3)

  14. w = -P (V2 - V1) = -1247 J Work for irreversible, isothermal, IG expansion n = 1.00 w = -  P dV = - P  dV = -P (V2 – V1) P (Pa) V (m3) 2.4

  15. Solid/liquid volume changes are small compared to gas therefore work is much less. Work - Solid/liquid examples w = - ∫ P dVapplies universally – not just to gases however, do not try to substitute P = nRT/V for solids/liquids! Work done during phase changes (e.g. - side walk cracks from Duluth winters) See water data on page 37 How much work is done when 18.0 g of water freezes at 0 ºC? density: ice = 0.915 g ml-1, water = 0.9999 g ml-1 W = -0.17 J V (m3) = 1/(g ml-1) • 1 x 10-6 m3/cm3 • #g Vm (m3) = 1/(g ml-1) • 1 x 10-6 m3/cm3 • FW g/mol How much work is done heating 18.0 g of water from 0→100 ºC? density: water (0ºC) = 0.9999 g ml-1, water (100ºC) = 0.9585 g ml-1 W = 0.079 J

  16. Special case – Expansion into a vacuum an irreversible process What is Pext? What is w? w = -P • (V2 – V1) = 0 Assign 2.14 and 2.16

  17. 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 c = specific heat: e.g. cal g-1oC-1

  18. Heat Capacity CP = dqP/dT (J K-1) or CP,m (J K-1 mol-1) CV = dqV/dT (J K-1) or CV,m (J K-1 mol-1) substance CP (J mol-1 K-1) Cu 24.4 H2O 75.9 Fe 24.8 Pb 32.9 Cg 8.5 Cd 6.2 What would make the most efficient frying pan? How much heat does it take to raise a 10carat diamond (1 carat = 0.2 g) from 298K – 500K?

  19. Isothermal change of state for an ideal gas T causes these motions so if DT = 0 they don’t contribute to DU 0 for chemical process 0 for IG DU = DUtr + DUvib + DUrot + DUel + DUint + DUnuc 0 if no l absorption and DT = 0 DU = 0 any isoT, IG process q = DU – w = -w

  20. Isothermal change of state for an ideal gas T causes these motions so if DT = 0 they don’t contribute to DU 0 for chemical process 0 for IG DU = DUtr + DUvib + DUrot + DUel + DUint + DUnuc 0 if no l absorption and DT = 0 DU = 0 any isoT, IG process q = DU – w = -w

  21. Heat Capacity - CV,m and DU CV,m = dqv/dT J K-1mol-1 FromDUsys = q + w ……. DeriveDU = qV assume that only expansion work is possible assume constant volume DU = qV (constant V heat) Sub in dU for dq in CV,m expression: CV= (dU/dT)V Multiply both sides by dT: dU = CVdT Integrate both sides: ∫dU = DU = ∫CVdT • Assume CV is constant over T: • DU = CV∫dT = CV (T2 – T1) = CVDT • This is independent of how the process is carried out.

  22. Equipartition Theorem (CM) For a collection of particles at thermal equilibrium the average contribution of each ‘degree of freedom’ to the total energy is 1/2kT. Accurate for translation, Very good for rotations, Poor for vibrations. Utr contributes 3/2kT toward U for IG (per particle) since 3 dimensions of motion Utrcontributes 3/2 RT toward Umfor IG(dU/dT) = CV,m = 3/2 R Monatomic IG (no rotations/vibrations) e.g. He, Ne, Ar… CV,m = 1.5R = 12.47 J mol-1 K-1

  23. Enthalpy (H) Begin with 1st Law DU = q + w (or dU = dq + dw) cst P and expansion work only By definition H = U + PV or dH = dU + d(PV) dH= dqP – P dV + P dV + V dP = dqP DH = qP U is cst V heat (qV) while H is cst P heat (qP) CP = dqP/dT = (dH/dT)P anddH = CPdT DH =  dH=  CPdT = CPDT if CPis independent of T If CP,m = f(T) = a + bT + c/T2 then DH = qP=  CPdT…. =  (a + bT + c/T2)dT

  24. Calculating q, w, DH and DU? melting vs. boiling (n = 1) Properties of Water How can we calculate DU? Is melting/boiling a Cst V or Cst P process? What are easiest variable(s) to determine from given data? How can we calculate work? w = -P (V2 – V1)

  25. How different are DH and DU? Solids or Liquids H = U + PV & dH= dU+ d(PV) since solids/liquids are not very compressible D(PV) ~ 0 (very small) & DH DU Ideal Gas H = U + PV & PV = nRT show DH = DU + Dn(g)RT

  26. Calculations Find q, w, DU, and DH for ……. Isobaric heating/cooling of solids, liquids, or gases data needed: CP,m (or CV,mfor gases) and density Phase changes - data needed: DHtr - density IG changes of state with specified conditions …. isothermal (reversible/irreversible) adiabatic (reversible)

  27. Starting Points for w, q, DU, & DH calculations Regardless of condition …. Work w = -  P dV Internal energy DU =  CV dT DU = q + w DH = DU + D(PV) enthalpy DH =  CP dT CP,m = CV,m + R (ideal gas)

  28. DU =  CVdT w = -  P dV DU = q + w DH =  CPdT 2.14 Gas – V1 = 377 ml to V2 = 119 ml with constant P = 1550 torrwhile removing 124.0 J of heat. DU = q + wq = -124.0 P (Pa) = 1550/760 • 101325 = 2.07 x 105 w = -∫ P dV = - P (V2 – V1) = - 2.07 x 105 (0.000119 – 0.000377) = + 53.4 J DU = 53.4 – 124.0 = -70.6 J What is DH? DH = DU + P • DV -70.6 1550/760 • 101325 -0.000258 = -124 = qP.

  29. DU =  CVdT w = -  P dV DU = q + w DH =  CPdT DH = DU + D(PV) • 2.16 IG n = 1.000 mol V1 = 1.0 L to V2 = 10.0 L T = 298K • a. reversibly b. irreversibly against cst P = 1.00 atm • w = -∫ P dV = - nRTln(V2/V1) = -1.000 • 8.314 • 298 • ln(10): wrev= -5700 J • w = -∫ P dV = - P (V2 – V1) = -101325 • (0.010 – 0.0010): wirr= -912 J • You get more work output when the expansion is done reversibly. Calculate DH, DU, and q for each part above? DH = 0 , DU = 0, and q = +5700 J (reversible) q = +912 J (irreversible)

  30. DU =  CVdT w = -  P dV DU = q + w DH =  CPdT DH = DU + D(PV) Note that CP,m - CV,m = R • 2.16 IG n = 1.000 mol V1 = 1.0 L to V2 = 10.0 L T = 298K • a. reversibly b. irreversibly against cst P = 1.00 atm • w = -∫ P dV = - nRTln(V2/V1) = -1.000 • 8.314 • 298 • ln(10): wrev= -5700 J • w = -∫ P dV = - P (V2 – V1) = -101325 • (0.010 – 0.0010): wirr= -912 J For n = 1, V = 0.00100 m3, T = 298 K — gas is He with CV,m = 12.5 J mol-1 K-1 Gas is heated at constant volume to 400K. Using IG law and equations above…. Calculate P1 and P2, w, DU and DH P1 = 2.48 x 106 Pa P2 = 3.33 x 106 Pa w = 0 DU = 1272 J What is CP,m? DH = 2122 J = 20.8 J mol-1 K-1.

  31. Reversible, Adiabatic IG Process q = 0 w = DU = - PdV = CVdT note that P, V, & T are all changing. Equate above two expressions & solve to get ...... - RT/VmdV= CV,mdT ln (V1/V2)R/Cv,m = ln (T2/T1) - R dV/Vm= CV,mdT/T (V1/V2)R/Cv,m= T2/T1 - R/CV,mln(V2/V1) = ln(T2/T1) T2/T1 = (V1/V2)R/Cv,mor P1V1 g= P2V2 g(g= CP,m/CV,m) DH = DU + D(PV)

  32. PV graph T2/T1 = (V1/V2)R/Cv 1 mole of an IG at P = 2.00 bar and T = 300. K is expanded adiabatically to a final volume of 0.0250 m3. What is w, q, DU, and DH. CV.m =15.0 J mol-1 K-1 isothermal P (Pa) adiabatic T↓ V (m3)

  33. CP,m – CV,m = R for an ideal gas (proof) Define:DH = n CP,mDT , DU = n CV,mDT let n = 1, rearrange & sub into above CP,m – CV,m = (DH/DT) - (DU/DT) substitute DH = DU + D (PV) = {DU + (D(PV)}/DT) - (DU/DT) sub D (PV) = R DT (for n = 1) = (DU + R DT - DU)/DT) CP,m – CV,m = R Internal Pressure CP - CV = [(dU/dV)T+ P](dV/dT)P

  34. Joule Experiment: measure DT as gas expands into vacuum: mJ = (dT/dV)Ufind (dU/dV)T for real gas Adiabatic Walls ― q = 0 but w = 0 also since Pext = 0 (dT/dU)V(dU/dV)T(dV/dT)U = -1 & (dU/dV)T = -mJCV Result: DT was too small to measure with Joule’s apparatus

  35. Joule-Thomson Experiment: (dT/dP)H = mJT P1 P2 P1, V1 T1 (dT/dH)P(dH/dP)T(dP/dT)H = -1 (dH/dP)T = -mJTCP P1 P2, V2 T2 P2 Adiabatic Walls: Measure T change JT throttling is used to liquefy gases when mJT> 0 ( since dP < 0) Inversion temperature: mJT > 0 below inv. T

  36. Inversion temperature: mJT > 0 below inv. T JT throttling is used to liquefy gases when mJT> 0 ( since dP < 0)

  37. Dependent on bonds in molecules Bond Energy – Uel difference between energy of AOs & MOs Separated atoms Energy bond length atom separation U = Utr + Uvib + Urot + Uel + Uint + Unuc DUrx & DHrx

  38. Standard Enthalpy of Reaction DHo298 = Sini Hom,298,i= Sini Hom,298,i (products) - Sini Hom,298,i (reactants) CH4 + 2O2 CO2 + 2H2O nCO2 = 1; nH2O = 2 nO2 = -2; nCH4 = -1 DHo298 = H˚CO2 + 2H˚H2O - H˚CH4– 2H˚O2 Can’t know H˚i U = Utr + Uvib + Urot + Uel + Uint+ Unuc Replace with DH˚f,298,i

  39. DHo298 = Si viDHof,298,i DHof,298,i = Standard Heat of Formation The standard heat of formation of any element in its ‘natural’ state is 0. This eliminates any concern over Unuc, and focuses on energy differences due to bond formation. e.g. C(gr)+ O2(g) → CO2(g) This can be experimentally determined using a bomb calorimeter, and is the value listed in your thermodynamic tables = -393.5 kJ mol-1.

  40. Bomb Calorimeter 1. Reaction heats water – DT measured Power Supply w = EIt experiment step #1 DU = 0 P + K 25 + DToC R + K 25oC DUrx (298) Uel P + K 25oC experiment step #2 desired information 2. Cool to original T1 Requirements: ~ 100% Products exothermic no side reactions 3. T1→ T2;w = EIt = qrx = DUrx ignite

  41. Conversion from DUrx to DHrx DHrx = DUrx + D(PV) Only gases contribute significantly to D(PV) Let D(PV) = DngRT ……. DHrx = DUrx + DngRT Examples: C(gr)+ O2(g)→ CO2(g)Dn = 0 CO(g)+ ½ O2(g) → CO2(g)Dn= -½

  42. Cg + ½ O2(g)→ COg This can’t be done in a bomb calorimeter Cg + O2(g)→ CO2(g) COg + ½ O2(g) → CO2 These can be done in a bomb calorimeter DH˚rx = DH˚f,CO2 - DH˚f,CO DH˚f,CO = DH˚f,CO2 - DH˚rx Using similar procedures the values for DH˚f has been determined for most common compounds and the results can be found in standard thermodynamic tables.

  43. For a reaction at T  298K ……. dH = ∫ CP dT ~ CPDT DHT – DH298 = ∫298TDCP,rx dT DCPrx = SiniDCPof,298,i DHT = DH298 + DCP,rx (T – 298) Note: you must keep units consistent: convert DCP,rx to kJ

  44. Ethanol combustion at 298 K at 500 K DHT = DH298 + DCP,rx (T – 298) Note: you must keep units consistent: convert DCP,rx to kJ

  45. AVG Bond Energy in kJ mol-1 AVG Bond Energy in kJ mol-1 C – C 344 C = C 615 H – H 436 C – Cgr 717 C = O 725 O – H 463 C – H 415 C = N 615 N - H C – O 350 N = N 418 C – N 292 O = O 498 O – O 143 C – F 441 N – N 159 C – Cl 328 C ≡ C 812 Cl – Cl 243 C – Br 276 C ≡ N 890 F – F 158 C – S 259 N ≡ N 946 If DHf,m,i is not in table ….. DHrx,298 can be estimated from tables of averaged bond energies. DHº298 = Si -ni• bond energy (note sign is opposite) Ethanol combustion Errors  as resonance energy  - particularly aromatic cpds