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Unformatted text preview: Discussion Section Problems; 1) Suppose a lake has a surface area of 10000m2. water at the surface of the lake is warmed to a temperature of 298K by heat from the sun. water at the bottom of the lake is 20 degrees lower in temperature. It is proposed that a heat engine be placed in the lake that will exploit this temperature gradient to produce work. It is claimed that this engine can produce energy at an average rate of 1 MWatt (i.e. 106 Watts). Assuming the average solar flux into the lake (i.e. the rate of energy transfer from the sun into the lake) is 500 Watts m-2, does this engine violate: 1) the First Law of Thermodynamics; 2) the Second Law of Thermodynamics; 3) both the First and Second Law; 4) neither the First or the Second Law. Explain your answer. Watt=Js-1. Solution: The total solar flux into the lake is Flux = 10000m2 500Wm!2 = 5"106W = 5MWatts ( )( ) Thermal efficiency = TH ! TL 20 = = 0.067 TH 298 Maximum Theoretical Output = 0.067 5"106W = 0.34MW ( )( ) The claim is the engine can output on average 1MWatt. This does not violate the First Law because the average output of energy does not exceed the solar flux. But the proposed output assumes a thermal efficiency of 20%, which is more than three times the thermal efficiency designated by the Second Law. It violates the Second Law.
2) Supercooled water is liquid water that has been cooled below its normal freezing point. This state is thermodynamically unstable and converts to ice irreversibly. If two moles of supercooled water are transformed into ice at T=263.0. K and at a constant pressure of 1.00 atm., calculate the enthalpy change H and entropy change S of the water. For water at T=273.0 K, Hfusion=-6.010 kJ mol-1. Calculate also the entropy change of the surroundings and the universe
o For liquid water the constant pressure heat capacity is CP = 75.3 JK -1mol -1. For ice the o constant pressure heat capacity is CP = 36.2 JK -1mol -1. Assume all heat capacities are constant between 263.0.K and 273.0 K. Assume the surroundings are at T=263K. Solution: To address this problem we exploit the fact that the entropy is a state function and set up a threestep pathway. Because S is a state function, S is path independent. Therefore, S for the transformation H2O (l, 263 K) H2O (s, 263 K) is equal to S for the transformation H2O (l, 263 K) H2O (l, 273 K) H2O (s, 273 K) H2O (s, 263 K) because the intial and final state are identical for the two pathways. We use the second pathway to calculate S for the process. H 2 O(l , 263 K) H 2O(s, 263 K) H 2 O(l , 273 K) H 2O(s, 273 K) H = H sys = nCP ( H 2 O, l ) T273- 263 + nH fusion + nCP ( H 2O, s ) T263- 273 = nH fusion + nCP T = -12.020kJ + 2 ( 75.3 J K -1 - 36.2 J K -1 ) ( 273.0 K - 263.0 K ) = -12.020 103 J + ( 78.2 J K -1 ) (10.0K ) = -12.020 103 J + 7.82 102 J = -11.238 103 J 273.0 2H fusion 263.0 S sys = nCP ( H 2 O, l ) ln + nCP ( H 2O, s ) ln + 273K 263.0 273.0 12.020 103 J = 2 ( 75.3J K -1 ) ln (1.038 ) - + 2 (36.2JK -1 ) ln ( 0.9633) 2.730 102 K = 5.62JK -1 - 44.03JK -1 - 2.70JK -1 = -41.11JK -1 11.238 103 J = 42.7 JK -1 K 263K Suniverse = S sys + S surr = -41.1JK -1 + 42.7 JK -1 = 1.6 JK -1 S surr =
3) The standard entropies of CO, CO2, and O2 at 298.15 K are
o S298.15 ( CO, g ) = 197.67 J K -1 mol -1 o S298.15 ( CO2 , g ) = 213.74 J K -1 mol -1 o S298.15 ( O2 , g ) = 205.138 J K -1 mol-1 The constant pressure heat capacities for of CO, CO2, and O2 are given by
o CP (CO, g ) = 29.14J K -1 mol-1 o CP (CO2 , g ) = 37.13J K -1 mol-1 o CP (O2 , g ) = 29.36J K -1 mol-1 Calculate S o for the reaction CO(g) + 1/2 O2(g) CO2(g) at 475 K. Solution 475 o o o -1 -1 -1 -1 -1 -1 S475 ( CO, g ) = S298 ( CO, g ) + CP ln = 197.67 J K mol + ( 29.14 JK mol ) ( 0.47 ) = 211.26 JK mol 298 475 o o o -1 -1 -1 -1 -1 -1 S475 ( CO2 , g ) = S298.15 ( CO 2 , g ) + CP ln = 213.74 J K mol + ( 37.13JK mol ) ( 0.47 ) = 231.19J K mol 298 475 o o o -1 -1 -1 -1 -1 -1 S475 ( O2 , g ) = S298.15 ( O2 , g ) + CP ln = 205.138 J K mol + ( 29.36 JK mol ) ( 0.47 ) = 218.94 JK mol 298 o o o Srxn = S 475 ( CO 2 , g ) - 1 S 475 ( O 2 , g ) - S 475 ( CO, g ) 2 = (1mol ) ( 231.19J K -1 mol-1 ) - ( 1 ) ( 218.94 JK -1mol -1 ) - (1mol ) ( 211.26 JK -1mol -1 ) 2 = -89.54 JK -1
4) The procedure for determination of standard Gibbs free energies of formation is illustrated with this example. Urea, N2H4CO, is a waste product from protein metabolism. Urea may be synthesized from its constituent elements as follows: C ( gr ) + N2 ( g ) + 2H 2 ( g ) + 1 O2 ( g ) N2 H 4CO ( s ) 2 Calculate G o for urea using standard entropies. For urea H o = -332.96kJmol -1 f f C(gr) N2(g) O2(g) H2(g) urea o -1 -1 5.69 191.49 205.03 130.59 104.60 S JK mol ( ) !S !f = 1mol 104.60JK "1mol "1 " 1mol 5.69JK "1mol "1 " 1mol 191.49JK "1mol "1 ( )( ) ( )( ) ( )( ) " 2mol 130.59JK "1mol "1 " (0.5mol)(205.03) = "456.28JK "1 #!G !f = !H !f " T !S !f = "332960J " 298.15K "456.28JK "1 = "196920J ( )( ) ( )( ) ...
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This note was uploaded on 01/18/2012 for the course CHEM 152A taught by Professor Chiu during the Fall '12 term at University of Washington.
- Fall '12