Chapter 10: Gases

Overview Pressure Standard Conditions Gas Laws Barometer & Atmospheric Pressure Standard Conditions Gas Laws Boyle’s Law Charles’ Law Avogadro’s Law Ideal Gas Law

Gas Laws under Two Conditions Gas Densities Darlton’s Law of Partial Pressure Kinetic Molecular Theory Molecular Effusion/Diffusion Graham’s Law Deviation from Ideality

Characteristics Solids Liquids Gases have own shape and volume particles close together with strong interaction Liquids have own volume but assume shape of container particles farther apart but have moderate interaction Gases assume shape and volume of container particles far apart with little/no interaction highly compressible

Pressure P = F/A Barometer Standard Pressure Force in Newtons Area in m2 Barometer P in N/m2 = Pascal unit 1 x 105 N/m2 = 1 x 105 Pa or 100 kPa Standard Pressure 1 atm = 760 mm Hg = 1.01325 x 105 Pa = 101.325 kPa (or torr)

force of the atmosphere force of the column when atmospheric force equals the force of the column the atmospheric pressure is measured as “h”

Gas Laws Boyle’s Law Charles’ Law Avogadro’s Law P µ 1/V constant T, n volume increases as pressure decreases Charles’ Law V µ T constant P, n volume increases as temperature increases Avogadro’s Law V µ n constant P, T volume increases as moles of gas (n) increases

Ideal Gas Law combines all gas laws PV = nRT R = 0.0821 L-atm mol-K any volumes must be in liters any temperatures must be in kelvin any pressures must be in atmospheres STP or SC -- standard temperature/pressure P = 1 atm (same as 760 mm Hg) T = 273 K (same as 0° C)

Problem 10. 3: A flashbulb contains 2. 4 x 10 -4 mol of O2 gas at 1 Problem 10.3: A flashbulb contains 2.4 x 10 -4 mol of O2 gas at 1.9 atm and 19°C . What is the volulme? PV = nRT or V = nRT P V = 2.4x10 -4 mol x 0.0821 L-atm x 292 K mol-K 1.9 atm V = 3.0 x 10 -3 L or 3.0 mL or 3.0 cm3

Gas Laws Under Two Conditions P1V1 = P2V2 T1 T2 Problem 10.4: Pressure in a tank is kept at 2.20 atm. When the temp. is -15°C the volume is 28,500 ft3. What is the volume is the temp. is 31°C P1 = P2 = 2.20 atm T1 = 258 K T2 = 304 K V1 = 28,500 ft3 V2 = P1 V1 T2 P2 T1 V2 = 28,500 ft3 x 304 K = 258 K 33,600 ft3

Gas Densities n = P from PV = nRT V RT d = PMM RT n = moles x g/mol = g = d = PMM V L L RT d = PMM RT (atm)g mol L atm ( K) mol K

Dalton’s Law of Partial Pressures total pressure of a mixture = sum of each partial pressure PT = P1 + P2 + P3 . . . . each partial pressure = the pressure each gas would have if it were alone P1 = n1RT P2 = n2RT P3 = n3RT V1 V2 V3 PT = n1RT + n2RT + n3RT = (n1 + n2 + n3) RT V1 V2 V3 V volumes are the same

P1 = n1 therefore P1 = n1 PT PT nT nT n1 = X1 mole fraction nT P1 = X1 PT

Kinetic Molecular Theory Gases consist of particles in constant, random motion Volume of gas particles is negligible Attractive and repulsive forces are negligible Average kinetic energy is proportional to temperature Collisions are elastic

molecular speed u = root mean square speed or speed of molecule with average kinetic energy R is the gas constant (8.314 J/mol-K), T is temp. in K & MM is molar mass What is the rms speed of an He atom at 25°C? u = (3 x 8.314 kg-m2/s2-mol-K x 298 K)1/2 ( 4.00 x 10 -3 kg/mol ) u = 1.36 x 103 m/s

Effusion/Diffusion small molecules will effuse/diffuse faster than large molecules effusion diffusion

Graham’s Law where r is rate of speed & MM is the molar mass Problem 10.14: Calculate the ratio of the effusion rates of N2 and O2. rN2 = 1.07 rO2

Deviation from Ideality Occurs at very high pressure or very low temperature Correction due to volume ideal law assumes molecules have no volume for molecules which are far apart, this is a good assumption must correct for the volume of the molecules themselves

Correction due to attraction of molecules ideal law assumes the molecules have no attraction to each other for molecules which are far apart, this is a good assumption must correct for actual attraction of molecules correction for molecular volume correction for molecular attraction