1. Results of Midterm 2 0 102030405060708090 points # of students GradePoints A85-100 B+B+ 75-84 B60-74 C+C+ 50-59 C30-49 D,F<30 100 # of students Midterm 1

2. Problem 1 (heat engine) 1 – 2 2 – 3 3 – 1 T1T1 P T2T2 1 2 3 V (20) The heat engine uses an ideal gas as its working substance. 1-2 is an isochoric process, 2-3 – adiabatic, 1-3 – isothermal. T 1 and T 2 are given. Find  Q for each process and the efficiency of the heat engine in terms of T 1 and T 2 (  =1+2/f, C V =(f/2)Nk B ). (10) Which way of increasing the efficiency of the Carnot heat engine is better – to increase the temperature of the hot reservoir by  T (  T <<T H, T C ) or to decrease the temperature of the cold reservoir by the same  T? Explain.

3. Problem 2 vdW (20) Two insulated tanks with volumes V 1 and V 2 are connected by a valve. Each tank contains one mole of the same monatomic van der Waals gas (the constants a and b are known). Initially, when the valve was closed, both gases were at the same temperature T i. The valve is open and the system reaches its equilibrium state. Find the equilibrium temperature T f. Is T f higher or lower than T i ? For which value of V 1 /V 2 you expect T f to be equal to T i? the temperature will decrease V 1, T i V 2, T i if

4. Problem 3 (phase transformations) (25) The pressure of the saturated water vapor at T = 0.01 0 C is 0.006 bar (these T and P correspond to the triple point of water). The latent heat of the solid-liquid transformation at 0.01 0 C is 335 kJ/kg, the latent heat of the liquid-gas transformation is 2500 kJ/kg. Find the pressure of the saturated water vapor at T = -1 0 C. Along the solid-gas phase equilibrium curve: - the latent heat of sublimation at the triple point - we neglected the volume of solid and expressed the gas volume using the ideal gas law. n – the number of moles in 1 kg.

5. Problem 4 (chem. equilibrium) (25) Consider the following reaction at T = 298K and P = 1bar: H 2 CO 3 (aq)  HCO 3 -- (aq) + H + (aq) (a) Using the data in the Table on p.404, calculate  G for this reaction, and the equilibrium constant K. (b) If the initial amount of H 2 CO 3 is 1 mole, what will be the amounts of each reactant and product at equilibrium? The mass action law: Thus, in equilibrium, H 2 CO 3 HCO 3 -- H+H+ initial100 change-x+ x final1-xxx H 2 CO 3 HCO 3 -- H+H+  H (kJ) -699.65-691.990 S (J/K)187.491.20  G (kJ) -623.08-586.770 To obtain the value of  G for the reaction, subtract  G of the reactants from  G of the products: