第一章 气体及溶液 (Gas and Solution)

Outline 1.1 Gas Properties of Gases Gas Laws Ideal Gas Law Gas Mixtures & Partial Pressures Nonideal Behavior: Real Gases 1.2 Solution Units of concentration

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

 Pressure Unit: 1 atm = 760 mm Hg = kPa =  10 5 Pa Pa (Pascal): SI unit 1 Pa = 1 newton / m 2 (1 bar = 1  10 5 Pa = atm)

1.1.2 Gas Laws:  Based on experimental basis in 17th & 18th centuries Compressibility of Gases: Boyle’s Law  e. g. bicycle pump  Boyle’s Law: V  1/P (at constant n & T)  PV = C B  P 1 V 1 = P 2 V 2 (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: P 1 = 745 mm Hg, V 1 = 65.0 L P 2 = ?, V 2 = 25.0 L P 1 V 1 = P 2 V 2 P 2 = P 1 V 1 / V 2 = (745 mm Hg)(65.0 L) / 25.0 L = 1940 mm Hg #

Effect of Temperature on Gas Volume: Charles’s Law V  T (at constant n & P)  V / T = C C  V 1 / T 1 = V 2 / T 2 (at constant n & P)

Example 1.2 Suppose you have a sample of CO 2 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: V 1 = 25.0 mL, T 1 = = K V 2 = ?, T 2 = = K V 1 / T 1 = V 2 / T 2 V 2 = T 2 V 1 / T 1 = (310.2 K)(25.0 mL) / K = 26.4 mL #

Combining Boyle’s & Charles’s Laws: General Gas Law  P 1 V 1 / T 1 = P 2 V 2 / T 2 (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  10 3 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: V 1 = 4.19  10 3 L, P 1 = 754 mm Hg, T 1 = K V 2 = ?, P 2 = 76.0 mm Hg, T 2 = K P 1 V 1 / T 1 = P 2 V 2 / T 2 V 2 = (T 2 / P 2 )/ (P 1 V 1 / T 1 ) = 3.38  10 4 L #

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

Example 1.4 Ammonia can be made directly from the elements: N 2 (g) + 3 H 2 (g)  2 NH 3 (g). If you begin with 15.0 L of H 2 (g), what volume of N 2 (g) is required for complete reaction (both gases being at the same T and P)? What is the theoretical yield of NH 3, in liters, under the same conditions? Solution: N 2 (g) + 3 H 2 (g)  2 NH 3 (g) V  n (at constant T & P)  V (N 2 required) = (15.0 L) / 3  1 = 5.00 L #  V (NH 3 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 & K (0  C), 1 mole gas occupies L  R = PV / nT = (1 atm)( L) / [(1 mol)( K)] = (L  atm  K  1  mol  1 ) We can also use other units of pressure to get R  R = kPa  L  K  1  mol  1 (= J  K  1  mol  1 ) = 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 N 2 gas (in moles) is in the air bag? Solution: V 1 = 65 L, P = 829 mm Hg (1.09 atm), T = 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 N 2, O 2, Ar, CO 2,…  Atomspheric pressure = Sum of the pressures exerted by each gas = Sum of partial pressures of each gas  Dalton’s law of partial pressures: P total = P 1 + P 2 + P 3 + 

 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: P A V = n A RT P B V = n B RT P C V = n C RT P total = P A +P B +P C = n A (RT/V) + n B (RT/V) + n C (RT/V) = (n A + n B + n C )(RT/V)  P total = n total (RT/V)

 mole fraction ( 莫尔分数 ): X X A = n A / (n A + n B + n C ) = n A / n total  P A = X A (P total )

Example 1.6 Halothane, C 2 HBrClF 3, 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 C 2 HBrClF 3 = 15.0 / = mol Moles of O 2 = 23.5 / = mol Mole fraction of C 2 HBrClF 3 = / ( ) = X oxygen = 1  =

Solution (continued): Setp 2. Calculate partial pressures: P halothane = (X halothane )(P total )  P halothane =  855 mm Hg = 80.2 mm Hg #  P oxygen =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 = V ideal [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 / ( ) 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 (X A ) ( 莫尔分数浓度 ) Mole fraction of A (X A ) = n A / (n A + n B + n C +  ) e. g. A solution contains 1.00 mol (46.1 g) of ethanol (C 2 H 5 OH) in 9.00 mol (162 g) of water.  the mole fraction of alcohol: X ethanol = 1.00 / ( ) =  the mole fraction of water: X water = 9.00 / ( ) = Note X ethanol + X water = 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 dm 3 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 mixingAfter mixing He V = 3.0 L P = 145 mm Hg Ar V = 2.0 L P = 355 mm Hg Valve open He + Ar

Problem 2. Chlorine trifluoride, ClF 3, is a valuable reagent because it can be used to convert metal oxides to metal fluorides: 6 NiO(s) + 4 ClF 3 (g)  6 NiF 2 (s) + 2 Cl 2 (g) + 3 O 2 (g) (a) What mass of NiO will react with ClF 3 gas if the gas has a pressure of 250 mm Hg at 20ºC in a 2.5-L flask? (b) If the ClF 3 described in part (a) is completely consumed, what are the partial pressures of Cl 2 and O 2 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/cm 3. What is the molality of the solution? Calculate the mole fraction and weight percentage of NH 3.