第一章气体及溶液 (Gas and Solution)

第一章 气体及溶液 (Gas and Solution)

Outline 1.1 Gas 1.1.1 Properties of Gases 1.1.2 Gas Laws 1.1.3 Ideal Gas Law 1.1.4 Gas Mixtures & Partial Pressures 1.1.5 Nonideal Behavior: Real Gases 1.2 Solution 1.2.1 Units of concentration

1.1 Gas 1.1.1 Properties of Gases  Chemists describe gases: P (Pressure) V (Volume) T (Temperature (Kelvins)) n (amount (mol))

 Pressure Unit: 1 atm = 760 mm Hg = 101.325 kPa = 1.01325  105 Pa Pa (Pascal): SI unit 1 Pa = 1 newton / m2 (1 bar = 1  105 Pa = 0.9869 atm)

1.1.2 Gas Laws:  Based on experimental basis in 17th & 18th centuries 1.1.2.1 Compressibility of Gases: Boyle’s Law  e. g. bicycle pump  Boyle’s Law:V 1/P(at constant n & T)  PV = CB  P1V1 = P2V2(at constant n & T)

Example 1.1 A sample of gaseous nitrogen in a 65.0-L automobile air bag has a pressure of 745 mm Hg. If this sample is transferred to a 25.0-L bag with the same temperature as before, what is the pressure of the gas in the new bag? Solution: P1 = 745 mm Hg, V1 = 65.0 L P2 = ?, V2 = 25.0 L P1V1 = P2V2 P2 = P1V1 / V2 = (745 mm Hg)(65.0 L) / 25.0 L = 1940 mm Hg#

1.1.2.2 Effect of Temperature on Gas Volume: Charles’s Law VT(at constant n & P)  V / T = CC  V1 / T1 = V2 / T2 (at constant n & P)

Example 1.2 Suppose you have a sample of CO2 in a gas-tight syringe. The gas volume is 25.0 mL at 20.0 C. What is the final volume of the gas if you hold the syringe in your hand to raise its temperature to 37 C? Solution: V1 = 25.0 mL, T1 = 20.0 + 273.2 = 293.2 K V2 = ?, T2 = 37.0 + 273.2 = 310.2 K V1 / T1 = V2 / T2 V2 = T2V1 / T1 = (310.2 K)(25.0 mL) / 293.2 K = 26.4 mL#

1.1.2.3 Combining Boyle’s & Charles’s Laws: General Gas Law  P1V1 / T1 = P2V2 / T2(at constant n)

Example 1.3 Helium-filled balloons are used to carry scientific instruments high into the atmosphere. Suppose a balloon is launched when the temperature is 22.5 C and the barometric pressure is 754 mm Hg. If the balloon’s volume is 4.19  103 L (and no helium escapes from the balloon), what will the volume be at a height of 20 miles, where the pressure is 76.0 mm Hg and the temperature is 33.0 C? Solution: V1 = 4.19  103 L, P1 = 754 mm Hg, T1 = 295.7 K V2 = ? , P2 = 76.0 mm Hg,T2 = 240.2 K P1V1 / T1 = P2V2 / T2 V2 = (T2 / P2)/ (P1V1 / T1) = 3.38  104 L#

1.1.2.4 Avogadro’s hypothesis (假设) Vn(at constant T & P)

Example 1.4 Ammonia can be made directly from the elements: N2(g) + 3 H2(g)  2 NH3(g). If you begin with 15.0 L of H2(g), what volume of N2(g) is required for complete reaction (both gases being at the same T and P)? What is the theoretical yield of NH3, in liters, under the same conditions? Solution: N2(g) + 3 H2(g)  2 NH3(g) Vn(at constant T & P)  V (N2 required) = (15.0 L) / 3  1 = 5.00 L#  V (NH3 produced) = (15.0 L) / 3  2 = 10.0 L#

1.1.3 Ideal Gas Law  Boyle’s Law:V 1/P(at constant n & T)  Charles’s Law:VT(at constant n & P)  Avogadro’s hypothesis: Vn(at constant T & P)  V nT / P  V= R(nT / P)  PV=nRT R (摩尔气体常数) is determined by experiments

At 1 atm & 273.15 K (0 C), 1 mole gas occupies 22.414 L  R = PV / nT = (1 atm)(22.414 L) / [(1 mol)(273.15 K)] = 0.082 (LatmK1mol1) We can also use other units of pressure to get R  R = 8.315 kPaLK1mol1 (= 8.315 JK1mol1) = 0.08315 barLK1mol1

Example 1.5 The nitrogen gas in an automobile air bag, with a volume of 65 L, exerts a pressure of 829 mm Hg at 25 C. What amount of N2 gas (in moles) is in the air bag? Solution: V1 = 65 L, P = 829 mm Hg (1.09 atm), T = 298.2 K, n = ? PV=nRT n = PV/ RT = (1.09 atm)(65 L) / [(0.082)(298.2 K)] = 2.9 mol#

1.1.4 Gas Mixtures & Partial Pressures  Air you breathe: a mixture of N2, O2, Ar, CO2,…  Atomspheric pressure = Sum of the pressures exerted by each gas = Sum of partial pressures of each gas  Dalton’s law of partial pressures: Ptotal = P1 + P2 + P3 + 

 Consider a mixture of three iedal gases, A, B, and C, is contained in a given volume (V) at a given temperature (T): The pressure exerted by each gas: PAV= nART PBV= nBRT PCV= nCRT Ptotal= PA+PB+PC = nA(RT/V) + nB(RT/V) + nC(RT/V) = (nA + nB + nC)(RT/V)  Ptotal= ntotal(RT/V)

 mole fraction(莫尔分数): X XA = nA / (nA + nB + nC) = nA / ntotal  PA= XA(Ptotal)

Example 1.6 Halothane, C2HBrClF3, is a nonflammable, nonexplosive, and nonirritating gas that is commonly used as an inhalation anesthetic (麻醉剂). Suppose you mix 15.0 g of halothane vapor with 23.5 g of oxygen gas. If the total pressure of the mixture is 855 mm Hg, what is the partial pressure of each gas? Solution: Step 1. Calculate mole fractions: Moles of C2HBrClF3 = 15.0 / 197.4 = 0.0760 mol Moles of O2 = 23.5 / 32.00 = 0.734 mol Mole fraction of C2HBrClF3 = 0.0760 / (0.0760 + 0.734) = 0.0938 Xoxygen = 1  0.0938 = 0.9062

Solution (continued): Setp 2. Calculate partial pressures: Phalothane = (Xhalothane)(Ptotal)  Phalothane = 0.0938  855 mm Hg = 80.2 mm Hg#  Poxygen =855  80.2 = 775 mm Hg#

1.1.5 Nonideal Behavior: Real Gases  Real gases: molecular volume, intermolecular forces  van der Waals equation: observed pressure observed V = Videal [P +a(n / V)2][V bn] =nRT correction for correction for intermolecular forces molecular volume (a and b are experimentally determined constants)

1.2 Solution A solution (溶液) is a homogeneous mixture of two or more substances in a single phase.  solvent (溶剂) and solute (溶质) Two cases: (a) solid + solvent (b) liquid + liquid

1.2.1 Units of concentration (A) Weight percentage (质量百分比浓度) Weight % A = mass of A / total mass of the mixture x 100% e.g. The alcohol-water mixture has 46.1 g of ethanol and 162 g of water, so the weight % of alcohol is: 46.1 / (46.1 + 162) x 100% = 22.2%

(B) Molality (m) (质量莫尔浓度) Molality of solute (m) = amount of solute (mol) / kilograms of solvent

(C) Mole fraction (XA) (莫尔分数浓度) Mole fraction of A (XA) = nA / (nA + nB + nC + ) e. g. A solution contains 1.00 mol (46.1 g) of ethanol (C2H5OH) in 9.00 mol (162 g) of water.  the mole fraction of alcohol: Xethanol = 1.00 / (1.00 + 9.00) = 0.100  the mole fraction of water: Xwater = 9.00 / (1.00 + 9.00) = 0.900 Note Xethanol + Xwater = 1.000

(D) Molar concentration (Molarity; 溶质的量浓度; M) Molar concentration of solute A = amount of A (mol) / volume of solution (L) Note: 1L = 1 dm3 Unit: mol·dm3;mol·L; M

Problem 1. A 3.0-L bulb containing He at 145 mm Hg is connected by a valve to a 2.0-L bulb containing Ar at 355 mm Hg. Calculate the partial pressure of each gas and the total pressure after the valve between the flasks is opened. Before mixing After mixing Valve open He V = 3.0 L P = 145 mm Hg He + Ar He + Ar Ar V = 2.0 L P = 355 mm Hg

Problem 2. Chlorine trifluoride, ClF3, is a valuable reagent because it can be used to convert metal oxides to metal fluorides: 6 NiO(s) + 4 ClF3(g)  6 NiF2(s) + 2 Cl2(g) + 3 O2(g) (a) What mass of NiO will react with ClF3 gas if the gas has a pressure of 250 mm Hg at 20ºC in a 2.5-L flask? (b) If the ClF3 described in part (a) is completely consumed, what are the partial pressures of Cl2 and O2 in the 2.5-L flask at 20ºC (in millimeters of mercury)? What is the total pressure in the flask?

Problem 3. Concentrated aqueous ammonia has a molarity of 14.8 (mole/L) and a density of 0.9 g/cm3. What is the molality of the solution? Calculate the mole fraction and weight percentage of NH3.

第一章气体及溶液 (Gas and Solution)