Chapter 9 Gases: Their Properties and Behavior 1 of 4 Ch. 9 Gases ...

Chapter 9 Gases: Their Properties and Behavior 1 of 4Ch. 9 Gases: Their Properties and Behavior9.1 Gases and Gas Pressuregases expand to fill a container.if mix of several compounds or elemental gases – homogeneous mixture (solution).atoms or molecules – very far apart from each other mostly empty space.density of water vapor: 0.803 g/Ldensity of water: 1000 g/L liquid more than 1000 times as dense as the gas…pressure (P) =Force (F)Area (A)Pressure at sea level called 1 atmosphere (atm)1 atmosphere of pressure pushes a column of mercury up 760 mm (or water 10.6 meters – 34.8 feet)9.2 Gas LawsBoyle’s LawIf pressure increases, the volume of a gas decreases.V α 1 Por V = 1 Px constant or PV = constantCharles Law:If T increases, average velocity of atoms/molecules of gas increase – greater force from collisions – If thepressure stays constant, the volume will increase.V α T or V = T x constant orVT = constantNOTE: For Charles Law, T must be in an absolute temperature scale. For SI units: KelvinAvogadro’s law:If the number of gas atoms/molecules increased, volume increases (if P constant).VV α n or V = n x constant orn = constantwhere n = moles of gas9.3 The Ideal Gas LawCombining all 3 relations:V α nV α T

Chapter 9 Gases: Their Properties and Behavior 2 of 4V α 1 PV α n TPor V = n TPR => PV = nRT called the “Ideal Gas Equation” (know)where R is called the gas constant; Given on exams: R = 0.08206liter atmmol KExample: What is the volume of 1.00 mol of gas under standard state conditions (0 o C, 1 atm)?PV = nRTV = nRTPneed T in K: 0 + 273 = 273 KV = 1.00 mol 0.08206 liter atm mol-1 K -1 273 K1.00 atmcalled the molar volume of a gas.All gases, 1 atm, 273 K, 1 mol: occupy 22.4 L= 22.4 LCan use ideal gas equation to calculate changes in variables given new conditions.Example: Suppose that a balloon has a volume of 10.2 liters at a pressure of 1.00 atm and a temperature of25.2 o C. The balloon is submerged in water to a depth where the pressure is 1.30 atm. The volume hasdecreased to 7.69 liters. What is the temperature of the water.PV = nRT; n, R are constant, so rearrange ideal gas equation to put all constants on one side and allvariables on the other.initial conditions:P 1 V 1T 1= nRfinal conditions:P 2 V 2T 2= nR and:rearrange to solve for unknown:P 1 V 1T 1= P 2V 2T 2T 2 = P 2V 2 T 1P 1 V 1T 1 = 25.0 + 273.15 = 298.1 5 KT 2 = 1.30 atm x 7.69 l x 298.1 5 o C1 atm x 10.2 l= 292.2 K (19.1 o C)Use units!

Chapter 9 Gases: Their Properties and Behavior 3 of 49.4 Stoichiometry Relationships With GasesGases and rxn eqns:Knowing P, V, T, can calculate the moles of gas for use with a rxn equation.Example: How many g of KClO 3 need to be decomposed to make 50.0 l of O 2 ? P= 1 atm, T = 273 K.2 KClO 3 (s) 2 KCl (s) + 3 O 2 (g)V O2 n O2 n KClO3 m KClO3PV = nRT => n = PVRT = 1.00 atm x 50.0 l0.08206 l atm mol –1 K –1 x 273.15 Kn = 2.23 mol O 22.23 mol O 2 ⎝ ⎜⎛ 2 mol KClO 33 mol O ⎠ ⎟⎞2⎝ ⎜⎛ 122.6 g1 mol KClO ⎠ ⎟⎞ = 182 g3The ideal gas law applies to all gases, therefore addition of 2 different gases together, would expectpressures to be additive.9.5 Gas Mixtures and Partial PressuresDalton’s Law of partial pressures (9.5)P total = P 1 + P 2 + P 3 +…mole fraction = X =n 1n totalP 1 = n 1n totalP total = X P totalExample: What is the mole fraction of O 2 in air. Assume P = 0.998 atm and P O2 = 0.209 atm.P O2 = X P total => X = P O2 0.209 atm=P total 0.998 atm = 0.2099.6 Kinetic Molecular Theory– explain all the observed laws.Postulates:1. Gases consist of molecule/atoms in continuous random motion2. The volume occupied by molecules negligible compared to vessel.3. Attractive and repulsive forces between species are negligible.4. Energy can be transferred by collisions, but total E is conserved, so average E constant.5. Average KE proportional to absolute T.

Chapter 9 Gases: Their Properties and Behavior 4 of 4Pressure is due to collision or species on walls of vessel. Decreasing the volume by half, doubles thenumber of collisions.Doubling the number of particles doubles the number of collisions, doubling the pressure.Doubling the T, doubles the average kinetic energy, doubling the frequency of the collisions, doubling thepressure.9.8 The Behavior of Real GasesReal gases DO deviate from the ideal gas law.At high pressures – gases occupy a significant fraction of the vessel volume and forces of attraction morelikely – so greater deviations from ideal behavior.At low temperatures – gases have lower kinetic energies relative to attractive forces – so less ideal behavior.