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128 Chapter 6 Thermochemistry = Tcoffee2) = 120.0 X 4.18 J X Rearrange to solve for Tf. 9.0 = 120.0 9.0 g = 120.0 g Tf + 10083.1 g Check: The units (°C) are correct. The temperature is closer to the original coffee temperature because the mass of coffee is so much larger than the ice mass. 6.125 KE = 1 For an ideal gas, v = = = KE₁ = 3 RT₂ RT₁ = At constant V, = so = At constant P, = q + = qp PAV = PAV. Because PV = nRT, for one mole of an ideal gas at constant P, PAV = RAT; so = q + = qp - PAV = PAV = AH RAT. This gives 2 3 RAT = RAT or = 5 = so = 5 6.127 q = AH = 454 1 X mol = 1025.405 kJ = 1030 kJ and W = -PAV. Assume that P = 1 atm (exactly) and V = VG VL, where VL = 454 X 0.9998 1 mL X 1000 = 0.4540908 L and PV = nRT. Rearrange to solve for = = 771.1545 L 1 atm V = VG VL = 771.1545 L = 770.7004 and so = PAV = 1.0 atm X = = -7.81 X 10⁴ J = -78.1 kJ Finally, E = q + = 78.0719515 kJ = = 950 kJ. 6.129 + + = AH = 1303 kJ/mol and W = -PAV. Assume that P = 1 atm (exactly) and = VG because the gas volumes are so much larger than the liquid volumes. The change in the number of moles of gas = 8 mol 25 2 mol = -4.5 mol and PAV = AnRT. Rearrange to atm -4.5 X 0.08206 solve for = = = 110.0425 L and so P 1 atm = -PAV = 1.0 atm X X = +11147.30J = + = +11kJ Finally, = q + = -1303kJ + = = 1292 kJ. Conceptual Problems 6.131 (d) Only one answer is possible: = 6.133 (a) At constant P, = q + W = qp + = AH + w; W = = q. 6.135 The aluminum cylinder will be cooler after 1 hour because it has a lower heat capacity than does water (less heat needs to be pulled out for every °C temperature change). Copyright © 2017 Pearson Education, Inc.