Lecture 17 Overview Ch. 4-5 List of topics Heat engines (and refrigerators) Thermodynamic potentials Phase transformations of pure substances and Clausius-Clapeyron Equation Phase transformations of binary mixture van der Waals gases Format of the second midterm: four problems with multiple questions. Total: 100 points. Only textbook and cheat-sheets (handwritten!) are allowed. No homeworks and lecture notes. DO NOT forget to bring your calculator!

Problem 1 (heat engine) TC P TH V V1 V2 1  2 2  3 3  1 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). 1  2 2  3 3  1

Problem2 (heat engine) 1  2 2  3 3  1 Working substance for the cycle shown in the Figure is 1 mole of van der Waals gas. Find the efficiency of this heat engine (in terms of T1 and T2). 12: adiabatic, 23: isothermal, 31: isochoric. 1  2 2  3 3  1

Problem2 (heat engine) (cont) Working substance for the cycle shown in the Figure is 1 mole of van der Waals gas. Find the efficiency of this heat engine (in terms of TH and TC). 12: adiabatic, 23: isothermal, 31: isochoric.

Problem (vdW) 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. V, P1 V, P2

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/kg·K respectively. The latent heat of ice is 3.34·105 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 (b) the change in entropy of the sub-system “water/ice”: cooling of water forming ice cooling of ice The increase of entropy of the reservoir: The total change of entropy of the whole system:

Vapor equation For Hydrogen (H2) near its triple point (Ttr=14K), the latent heat of vaporization Lvap=1.01 kJ/mol. The liquid density is 71 kg·m-3, the solid density is 81 kg·m-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. The vapor pressure equation for H2: where P0 = 90 MPa . 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 gas Ttr T

Vapor equation (cont.) 2. At the solid-gas phase boundary: Assuming that the H2 vapor can be treated as an ideal gas

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: - the latent heat of sublimation at the triple point

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 5.3×10-6 and 3.4×10-6 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) (b) G. the slope of G changes at the phase transition because of the change in volume 63 kbar P

Phase transformations, cont. 4000 5500 T

phase transformation of binary mixture The phase diagram of a solution of B in A, at a pressure of 1 bar, is shown in the Figure (TA>TB). The upper bounding curve (the dew-point curve) of the two-phase is T=TA-(TA-TB)  xB2. The lower bounding curve (the bubble-point curve) is T=TA-(TA-TB)  (2- xB)  xB. An enclosed chamber containing equal mole numbers of A and B in vapor phase is brought to its condensing temperature at the dew-point curve. What is the composition of the first drop of condensed liquid?

phase transformation of binary mixture (cont) Does condensation tends to increase or decrease the mole fraction of B in the remaining vapor?