Thermo - Part I - Chemistry [PDF]

Thermo - Part I Extra Problems to Try Problem 1: 50.0 mL dilute AgNO3 is added to a solution with OH– ions in a coffee cup calorimeter. Ag2O precipitates and the temperature of the liquid goes from 23.78 to 25.19°C. Assume that the mixture has the same specific heat as water. Calculate "qsurr" for a mass of 150. g. Is the reaction exothermic or endothermic? Solution: qSurroundings = mass × specific heat × change in temp qSurroundings = 150. g × 4.184 J/g⋅ºC × (25.19 - 23.78 ºC) qSurroundings = 885 Joules Rxn exothermic (T of mixture rose; ∴rxn released heat to its surroundings).

Problem 2: When ice at 0°C melts to liquid H2O, it absorbs 334 J per gram. The specific heat of water is 4.184 J g– 1 deg–1. Suppose the heat needed to melt a 35.0g ice cube is absorbed from water (0.210 kg) in a calorimeter at 21.0°C. What is the final water temperature? [This is the kind of calculation that all experienced party givers do before throwing a party!] Solution: Another heat transfer problem! heat associated with (ice melting + ice water warming) = heat associated with water cooling (massICE × heat to melt ice) + (massICE H2O × specific heat H2O × ΔT) = –(massCal H2O × specific heat H2O × ΔT) (35.0 g × 334 J/g) + (35.0 g × 4.184 J/g⋅deg × {Tf – 0ºC})= – (210 g × 4.184 J/g⋅deg × {Tf – 21.0ºC}) 11690 + 146.44Tf = –878.64Tf + 18451.44 1025.08Tf = 6761.44 Tf = 6.6ºC PROBLEM on page 13: When an ideal gas expands at constant temperature (i.e., isothermal expansion), ΔU is zero. Consider 1.00 L of an ideal gas initially at 9.00 atm and 15°C. (a) Calculate q and w if the gas expands isothermally against a pressure of 1.00 atm. (b) Calculate q and w if the gas expands isothermally first against a pressure of 3.00 atm, and then against 1.00 atm. (c) Calculate q and w if the gas expands isothermally first against a pressure of 3.00 atm, then 2.00 atm and then against 1.00 atm. (d) Comment on the results.

Solution: a) V1 = 1.00 L, P1 = 9.00 atm, P2 = 1.00 atm, find V2 using P1V1 = P2V2 V2 = (9.00 atm × 1.00 L)/1.00 atm = 9.00 L w = −PΔV = −1.00 atm (9.00 L - 1.00 L) = −8.00 atm⋅L Conversion factor: 1 atm⋅L = 101.325 Joules ∴ w = −8.00 atm⋅L × 101.325 Joules/ atm⋅L = −811 Joules = − 0.811 kJ ΔU = q + w = 0; q = –w = +0.811 kJ b) V1 = 1.00 L, P1 = 9.00 atm, P2 = 3.00 atm; using P1V1 = P2V2, V2 = 3.00 L w1 = −PΔV = −3.00 atm × (3.00 L − 1.00 L) = −6.00 atm⋅L w1 = −6.00 atm⋅L × 101.325 Joules/ atm⋅L = −0.608 kJ V2 = 3.00 L, P2 = 3.00 atm, P3 = 1.00 atm; using P2V2 = P3V3, V3 = 9.00 L w2 = −1.00 atm⋅L × (9.00 L - 3.00 L) = −6.00 atm⋅L w2 = −6.00 atm⋅L × 101.325 Joules/ atm⋅L = −0.608 kJ wTotal = w1 + w2 = −0.608 kJ + −0.608 kJ = −1.216 kJ qTotal = –wTotal = +1.216 kJ c) First expansion is the same as the above, i.e., V2 = 3.00 L & w1 = −0.608 kJ V2 = 3.00 L, P2 = 3.00 atm, P3 = 2.00 atm; using P2V2 = P3V3, V3 = 4.50 L w2 = −PΔV = −2.00 atm × (4.50 L − 3.00 L) = −3.00 atm⋅L w2 = −3.00 atm⋅L × 101.325 Joules/ atm⋅L = −0.304 kJ V3 = 4.50 L, P3 = 2.00 atm, P4 = 1.00 atm; using P2V2 = P3V3, V3 = 9.00 L w2 = −1.00 atm⋅L × (9.00 L - 4.50 L) = −4.50 atm⋅L w2 = −4.50 atm⋅L × 101.325 Joules/ atm⋅L = −0.456 kJ wTotal = w1 + w2 + w2 = −0.608 kJ + (−0.304 kJ) + (−0.456 kJ) = −1.368 kJ qTotal = –wTotal = +1.368 kJ d) The conclusion is that you would get the maximum amount of work done on the surroundings by doing the expansion in an infinite number of very small steps.