Lecture 19 Overview Ch. 4-5 - Rutgers Physics & Astronomy

Lecture 19 Overview Ch. 4-5 - Rutgers Physics & Astronomy

Lecture 19 Overview Ch. 4-5 List of topics 1. 2. 3. 4. Heat engines Phase transformations of pure substances, Clausius-Clapeyron Eq. van der Waals gases Thermodynamic potentials, chemical reactions P e 1 QC QH 2 1 TH 3 V1 V2 Problem 1 (heat engine) Working substance for the cycle shown in the Figure is 1 mole of ideal gas. Find the efficiency of this heat engine (in terms of TH and TC). TH Q 12 C P dT C P TH TC 0 12 TC TC TC V 23 Q 23 CV dT CV TC TH 0 TH

V1 31 QH Q12 V1 TC Q 31 W31 PdV R TC ln R TC ln V2 TH V2 QC Q23 Q31 T f RTH TC R TC ln C 2 TH e 1 f 1 RTH TC 2 CV f R 2 0 f C P CV R 1 R 2 f TH TC1 TC ln TC 1 2 TH f 1 TH TC 2

Problem (vdW) V, P1 U vdW V, P2 An insulating membrane (does not conduct heat) divides an insulated tank into two equal volumes. Each volume contains one mole of the same van der Waals gas (the constants a and b are known). The pressure in volume I is P1, in volume II P2 . The piston has been removed. Find the pressure, explain your reasoning. N 2a f N 2a U ideal Nk BT V 2 V 2 2N a f N 2a f N 2a f Nk BT1 Nk BT2 2 N k BT f 2 V 2 V 2 2V 2 2N a 1 P P1 P2 2 2V 2 Nb 2V 2

2 Nk BT f 1 T f T1 T2 2 Isothermal Process for a vdW gas U Q W Vf Vf U vdW T 1 1 N a V V f i 2 V f Nb Nk BT N 2 a 1 2 1 W PdV 2 dV Nk BT ln N a V V V Nb V V Nb f i Vi Vi i V Nb Q1 2 Nk BTH ln 2 V1 Nb

Latent heat, dS for phase transformations 10 kg of water at 200C is converted to ice at - 100C by being put in contact with a reservoir at - 100C. This process takes place at constant pressure and the heat capacities at constant pressure of water and ice are 4180 and 2090 J/kgK respectively. The heat of fusion of ice is 3.34105 J/kg. (a) Calculate the heat absorbed by the cold reservoir. (b) Calculate the change in entropy of the closed system reservoir + water/ice. The conversion consists of three processes: (a) water at 200C water at 00C; (b) water at 00C ice at 00C; (c) ice at 00C ice at -100C: (a) the heat absorbed by the cold reservoir Qres 10 kg 4180 20 3.34 105 2090 10 J/kg 4.385 106 J (b) the change in entropy of the sub-system water/ice: 273 K McPwater dT 273 cooling of water S a McP ln 2955 J/K T 293 293 K MLvap 10 kg 3.34 105 J/kg Sb 12,234 J/K forming ice T0 273K 263 K ice Mc 263 ice P dT cooling of ice S c McP ln 780 J/K T 273 273 K Qres 4.385 106 J

S res 16,673 J/K The increase of entropy of the reservoir: Tres 263 K The total change of entropy of the whole system: S S a Sb S c S res 704 J/K Vapor equation 1. 2. For Hydrogen (H2) near its triple point (Ttr=14K), the latent heat of vaporization Lvap=1.01 kJ/mol. The liquid density is 71 kgm-3, the solid density is 81 kgm-3, and the melting temperature is given by Tm =13.99+P/3.3, where Tm and P measured in K and MPa respectively. Compute the latent heat of sublimation. Lvap where P = 90 MPa . P P exp 0 The vapor pressure equation for H2: 0 RT Compute the slope of the vapor pressure curve (dP/dT) for the solid H2 near the triple point, assuming that the H2 vapor can be treated as an ideal gas. 1. Near the triple point: P liquid solid Ptr Lmelt gas Ttr Ttr VL VS dT / dP 2 10 3 kg/mol 2 10 3 kg/mol 14 3 3 81 kg/m

71 kg/m 162 J 6 1/3.3 10 Lvap Ttr SG S L T Lmelt Ttr S L S S Lsub Ttr SG S S Lsub Ttr SG S S Lvap Lmelt 1010 162 J/mol 1172 J/mol Vapor equation (cont.) 2. At the solid-gas phase boundary: dP Lsub L sub dT Ttr VG VS TtrVG Assuming that the H2 vapor can be treated as an ideal gas VG PtrVG RTtr RTtr RTtr 8.3 J/K mol 14 K Ptr P0 exp Lvap / RTtr 9 107 Pa exp - 1010 J/ 8.3 J/K mol 14 K 7.69 10-3 m 3 / mol dP Lsub 1172 J/mol 1.09 10 4 Pa/K 3 3 dT TtrVG 14 K 7.69 10 m / mol Phase transformations The pressure-temperature phase diagram of carbon is shown below. For simplicity, assume that the molar volumes of graphite and diamond are independent of temperature and pressure at 5.310-6 and 3.410-6 m3, respectively. 1 kbar = 108 N/m2 .

(a) Determine the latent heat per mole of transformation at T = 1000 K. (b) Sketch the graph of Gibbs free energy of carbon at the constant temperature T = 2000 K as a function of pressure between P = 50 and 70 kbar. Mark the transition pressure. Explain the changes, if any, of the slope of G. (c) Sketch the graph of entropy per mole of the material at the constant pressure P = 90 kbar as a function of temperature between 4000 and 5500 K. State any assumptions you make and explain your graph. (a) dP L dT TV dP dT 90 108 Pa 48.9kJ 1000 K 5.3 10 m 3.4 10 m 3500 K (b) L TV G. 6 3 dG T , N 6 3 SdT VdP dN T , N VdP the slope of G changes at the phase transition because of the change in volume

63 kbar P Phase transformations, cont. (c) S S 4000 L Tmelt 5500 T C.-C. Equation The vapor pressure of solid ammonia is given by the relation: ln P 23.03 3754 / T where units mm of Hg, T absolute temperature. The vapor pressure of liquid ammonia is given by the relation: ln P 19.5 3063 / T (a) (b) (c) (a) What is the temperature of triple point? Compute the latent heat of vaporization at the triple point. (Assume that the vapor can be treated as an ideal gas, and the density of vapor is negligibly small compared to that of the liquid) The latent heat of sublimation at the triple point is 31.4 kJ/mol. What is the latent heat of melting at the triple point? Ttr from the equation 23.03 3754 / Ttr 19.5 3063 / Ttr Ttr = 195 K ln P 19.5 3063 / T dP / dT 3063P / T 2 dP L L The Clausius-Clapeyron eq. gives L 3063PVG / T 3063R 25.4 kJ/mol dT TV TVG (b) From the equation for liq. ammonia (c) Denote Sg, Sl, and Ss as the entropy for vapor, liquid and solid at triple point.

The latent heat of vaporization - Ttr S g Sl The latent heat of sublimation - Ttr S g S s The latent heat of melting - Ttr S L S S Ttr SG S S Ttr SG S L 31.4 kJ/mol - 25.4 kJ/mol 6 kJ/mol C-C eq., phase transformation) A cylinder closed with a piston is filled with the saturated water vapor at T = 1000C. The vapor is heated up by 10C, and, at the same time, the piston is moved to prevent condensation and to keep the vapor saturated (the system is moving along the coexistence curve). Find the relative change in the vapor volume, V/V. Assume that the vapor is an ideal gas, the latent heat of vaporization Lvap=40.7 kJ/mol, and the vapor density is negligible in comparison with the water density. The Clausius-Clapeyron equation for the coexistence curve liquid-gas: For an ideal gas: Lvap Lvap Lvap P P T TVG T T VG VL TVG PiVi Pf V f P V For a small temperature changes : Ti Tf P V Lvap T Lvap T V P 40.7 kJ/mol 1 K

0.035 3.5% 2 2 V P TVP RT 8.3 J/mol K 373 K phase transformations The triple point for water corresponds to Ttr=0.010C and Ptr =0.006 bar. At T~Ttr,, the latent heat of melting is 335 kJ/kg, and the slope of the solid-vapor phase coexistence curve is dP/dT=50 Pa/K . Assume that water vapor behaves as an ideal gas near its triple point. Find the latent heat of vaporization. Along the solid-gas phase equilibrium curve: Lsub Ttr Lmelt Ttr Lvap Ttr Lsub L dP sub dT T v gas vsolid Tv gas - the latent heat of sublimation at the triple point dP Lvap Ttr Lsub Ttr Lmelt Ttr Ttr vgas Lmelt Ttr dT sub 3 nRTtr dP 2 10 g / kg 8.3 J / K mol 3 Ttr 50 Pa / K 335 10 J / kg Lmelt Ttr 273K 2 Ptr dT sub 18 g / mol 6 10 Pa

2863 103 J / kg 335 103 J / kg 2529 103 J / kg Chem. equilibrium Consider the following equilibrium at 500 K: CO(g) + 2 H2 (g) CH3OH(g) The equilibrium concentrations are: [CO] = 0.0911 M, [H2] = 0.0822 M, [CH3OH] = 0.00892 M, what is the value of the equilibrium constant? Is G positive or negative when the reactants are transformed into products? K CH 3OH CO H 2 2 0.00892 14.5 2 0.0911 0.0822 Since the value of the equilibrium constant is greater than one, G <0. Consider the reaction: N2(g) + 3H2 (g) 2NH3 (g) Keq = 0.5 at 400 K Suppose we mix the following initial concentrations: nNH 3 1 M nN 2 1 M nH 2 1 M In which direction will the reaction go? NH 3 2 1 2 1 K eq N 2 H 2 3 11 3 Thus, the reaction will go to the left (some ammonia molecules will be transformed into hydrogen and nitrogen). Problem (chem. equilibrium) Consider the following reaction: CO(g) + H2O(g) CO2 (g) + H2 (g) Keq = 23.8 at 600 K If the initial amounts of CO and H2O were both 0.100 M, what will be the amounts of each reactant and product at equilibrium? H2O

CO start change finish CO2 H2 0.1 0.1 0 0 -x -x +x +x 0.1-x 0.1 - x x x We start with n0CO and n0H20 moles of the reacting gases and define as the yield x the number of moles of CO2 and H2 that the reaction will produce at equilibrium: The mass action law requires: n 0 H2 x2 x2 K 2 0 x nCO 2 x 0.1 x This is a quadratic equation with respect to x : Thus, in equilibrium,

nH 2 0.083 M x K 0.1 x 0.1 K x 0.083 1 K nCO 2 0.083 M nH 2 0 0.017 M nCO 0.017 M Chem. equilibrium Consider the following reaction: A (g) + 2B (g) C (g) +3D (g), where A, B, C, D are some molecules. In equilibrium, nA 1 M nB 1 M nC 1 M nD 1 M (a) Calculate the equilibrium constant and the standard Gibbs energy for this reaction at a given temperature. (b) To the above equilibrium system we add an extra 1M of A. What are the new equilibrium concentrations of each component? (a) The mass action law: 3 C D K 1 2 A B G 0 1 K exp k BT G 0 0 (b) A B C D

initial 2 1 1 1 change -x - 2x +x + 3x final 2-x 1 - 2x 1+x 1+3x nA 2 M - x 1 x 1 3x 3 2 x 1 2 x 2 nB 1 M - 2x nC 1 M x nD 1 M 3x 1

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