Unit 6 (Chp 10): Gases Chemistry, The Central Science, 10th edition Theodore L. Brown; H. Eugene LeMay, Jr.; and Bruce E. Bursten Unit 6 (Chp 10): Gases John Bookstaver St. Charles Community College St. Peters, MO  2006, Prentice Hall, Inc.

Characteristics of Gases Unlike liquids and solids, they Expand to fill their containers. (indefinite volume) Are highly compressible. Have extremely low densities.

Pressure F P = A Atmospheric Pressure: weight of air per area 22,000 lbs!!! Pressure is the amount of force applied per area. (per sq. meter) P = F A Atmospheric Pressure: weight of air per area

Pressure Units (at sea level) STP (standard T & P) 273 K 1 atm empty space (a vacuum) Units (at sea level) 1 atm = 760 mmHg 760 torr 101.3 kPa h 760 mm STP (standard T & P) 273 K 1 atm 1 N 1 m2 Atmospheric Pressure (weight of air) 657 760 1) 657 mmHg to atm 2) 830 torr to atm 3) 0.59 atm to torr 830 760 0.59 x 760

Kinetic-Molecular Theory KMT is a model that explains the Properties (P, V, T, n) and Behavior (motion, energy, speed, collisions) of gases.

5 Parts of Kinetic-Molecular Theory Gas particles are in constant random motion. 2) Collisions with the walls of the container cause gas pressure. P = F A Collisions are elastic (no KE lost). (ideally) P = F A P = F A

5 Parts of Kinetic-Molecular Theory 3) Attractive forces (IMAFs) are negligible. IMAFs 4) Volume of gas particles is negligible. (compared to total volume of container) Ideally: Vgas = Vcontainer Vgas = Vcontainer – Vparticles In a 1.000 L container, gas fills about 0.999 L of volume 1.000 L container has 1.000 L of gas (negligible)

5 Parts of Kinetic-Molecular Theory Average KE of gas particles is… …directly proportional to Kelvin temp. (K not oC) no negative temp’s, no negative energies, no negative volumes, etc.) KEavg α T video clip – Describing the invisible properties of gas (TEDEd) (3 min) video clip

(inversely proportional) Boyle’s Law (P & V) P ↑ , V ↓ (inversely proportional) 10 L 5 L 1 atm 2 atm

(directly proportional) Charles’ Law (V & T) T ↑ , V ↑ (directly proportional) 60 L 30 L how absolute zero was estimated 300 K 150 K

(directly proportional) Lussac’s Law (P & T) T ↑ , P ↑ (directly proportional) 500 kPa 100 kPa 200 kPa 300 K 600 K

(directly proportional) Avogadro's Hypothesis At the same ___ & ___, equal _________ of gas must contain equal _________. T P volumes moles (n) (particles) All at: P = 1 atm T = 25oC V = 1.0 L O2 He CO2 Avogadro's Law (V & n) n ↑ , V ↑ 2 mol He 1 mol He add gas (directly proportional)

PV = nRT Ideal Gas Law PV = R nT R = 0.08206 L∙atm/mol∙K So far we’ve seen that V  1/P (Boyle’s law) V  T (Charles’s law) P  T (Lussac’s law) V  n (Avogadro’s law) constant PV nT = R (all gases same ratio) R = 0.08206 L∙atm/mol∙K ideal gas constant: given on exam PV = nRT Ideal Gas Law NO Units of : mL , mmHg , kPa , grams , °C

PV = nRT P1V1 n1T1 P2V2 n2T2 P1V1 n1T1 P2V2 n2T2 = R = = P ↑ , V ↓ given on exam NOT given on exam P1V1 n1T1 P2V2 n2T2 P1V1 n1T1 P2V2 n2T2 = R = = constant (initial) (final) (changes in P,V,T,n) (inversely proportional) (directly proportional) P ↑ , V ↓ T ↑ , V ↑ T ↑ , P ↑ n ↑ , V ↑

PV = nRT Ideal-Gas Changes P1V1 n1T1 P2V2 n2T2 = PV = R nT V1 The pressure on a 411 mL sample of gas is decreased from 812 torr to 790 torr. What will be the new volume of the gas? P2 P1 V2 = ? P1V1 n1T1 P2V2 n2T2 = 422 mL = (812)(411) = (790) V2 PV nT (812)(411) (790) = V2 = R

PV = nRT Ideal-Gas Changes P1V1 n1T1 P2V2 n2T2 = PV = R nT V1 A 10.0 L sample of a gas is collected at 25oC and then cooled to a new volume of 8.83 L while the pressure remains at 1.20 atm. What is the final temperature in oC? T1 V2 P1V1 n1T1 P2V2 n2T2 T2 = 263 K = ? = –10 oC = (8.83)(298) (10.0) T2 = (10.0) (298) (8.83) T2 = PV nT T2 (10.0) = (8.83)(298) = R

PV = nRT Ideal-Gas Changes P1V1 n1T1 P2V2 n2T2 = 3 O2  O3 2 PV nT = R HW p. 432 # 1, 23, 89, 26 V1 n1 A 13.1 L sample of 0.502 moles of O2 is held under conditions of 1.00 atm and 25.0oC. If all of the O2 is converted to Ozone (O3), what will be the volume of O3? V2 = 8.73 L P1V1 n1T1 P2V2 n2T2 n2 = ? (13.1) (0.502) V2 . (0.335) = = 3 O2  O3 2 n2 PV nT 0.502 mol O2 x 2 mol O3 = 3 mol O2 0.335 mol O3 = R n1

How many molecules of He? Ideal-Gas Equation PV = nRT V P T A 5.00 L He balloon has 1.20 atm at 0.00oC. How many moles of He gas are in the balloon? n PV = nRT R = 0.08206 L∙atm∙mol–1∙K–1 (1.20 atm)(5.00 L) = n (0.08206)(273 K) (1.20)(5.00) (0.08206)(273) = n How many molecules of He? n = 0.268 mol He

PV = nRT Ideal-Gas Equation PV = nRT HW p. 438 #92, 29, 46, 38, 35 m T A sample of aluminum chloride gas weighing 0.0500 g at 350.oC and 760 mmHg of pressure occupies a volume of 19.2 mL. Calculate the Molar Mass of the gas. PV = nRT R = 0.08206 L∙atm mol∙K (1.00 atm)(0.0192 L) = n (0.08206)(623 K) m M (given on exam) n = 0.000376 mol n = __0.0500 g_ 0.000376 mol M = so… m n Molar Mass grams mole M = = AlCl3 = 133 g/mol

PVm = nRT Molar Volume: Vm = _____ 22.4 L 1 mol the ______ of ______ of any gas at ____. volume 1 mole STP PVm = nRT (1.00 atm) Vm = (1 mol )(0.08206)(273 K) The volume of 1 mole of any gas at STP will be: Vm = _____ 22.4 L 1 mol but… ONLY at STP!!!

Gas Stoich with Molar Volume Assume reactions below occurred at STP. Calculate the mass of NH4CI reacted with Ca(OH)2 to produce 11.6 L of NH3(g) . 2 NH4Cl + Ca(OH)2  2 NH3 + CaCI2 + 2 H2O Calculate the volume of CO2 gas produced when 9.85 g of BaCO3 is completely decomposed. BaCO3(s)  BaO(s) + CO2(g) 11.6 L NH3 x 1 mol NH3 x 22.4 L NH3 2 mol NH4Cl x 2 mol NH3 53.49 g NH4Cl = 1 mol NH4Cl 27.7 g 1.12 L

KClO3(s)  KClO(s) + O2(g) NOT at STP PV = nRT use… What volume of O2 gas is produced from 490 g KClO3 at 298 K and 1.06 atm? KClO3(s)  KClO(s) + O2(g) 1 mol KClO3 1 mol O2 490 g KClO3 x x = 4.00 mol O2 122.55 g KClO3 1 mol KClO3 (1.06 atm) V = (4.00 mol )(0.08206)(298 K) = ____L O2 92.3 L O2 Molar Mass of KClO3 is 122.55 g/mol

Dalton’s Law of Partial Pressures Ptotal = PA + PB + PC + … Mole Fraction (XA) HW p. 436 #63 64 66 moles of A XA = WS 6b #1-4 total moles PA = Ptotal x XA The mole fraction (XA) is like a % of total moles that is A, but without the % or x 100.

= Ptotal = PH2O + Pgas Pgas = Ptotal – PH2O equalize water level inside & outside Ptot Patm = Ptotal = PH2O + Pgas When one collects a gas over water, there is water vapor mixed in with the gas. To find only the pressure of the gas, one must subtract the water vapor pressure from the total pressure. Pgas = Ptotal – PH2O

PV = nRT Ptotal = PH2O + Pgas Calculate the mass of 0.641 L of H2 gas collected over water at 21.0oC with a total pressure of 750. torr. The vapor pressure of water at 21.0oC is 20.0 torr. 750. = 20.0 + PH2 nH2 = 0.0255 mol PH2 = 730. torr mH2 = 0.0514 g PH2 = 730/760 = 0.961 atm (0.961)(0.641) = nH2 (0.08206)(294)

Study the models below. What can be said quantitatively about the molecular speed (v) of a gas in relation to its molar mass (M) and its temperature (T)? ↑ M , ↓ v ↑ T , ↑ v

KE = ½ mv2 WHY? ↑ M , ↓ v ↑ T , ↑ v (given on exam) Compare the molecular speed (v) of these gases: 1) at 25.0 oC (i) Helium (ii) Oxygen (O2) 2) at 50.0 oC (i) Helium (ii) Oxygen (O2) Does the data support your conclusions from the models on the previous slide about effects of T and M on v ? 1360 m/s 482 m/s 1420 m/s 502 m/s WHY? ↑ M , ↓ v ↑ T , ↑ v

Distributions of Molecular Speed Temp (K) & KEavg are directly proportional T α KEavg (KMT) average molecular speed (v) KE = ½ mv2 (given on exam) Therefore: T & v are __________ proportional directly ↑ T , ↑ v

½ mv2 = ½ mv2 ↑ M , ↓ v ↓ M , ↑ v Speed vs. Molar Mass KE1 = ½ m1v12 Gases at the same Temp, have the same _____. KEavg KE = ½ mv2 Ar: M = 40 g/mol He: M = 4.0 g/mol KE1 = ½ m1v12 KE2 = ½ m2v22 KEHe = KEAr (at same T) ½ mv2 = ½ mv2 M & v are __________ proportional inversely ↑ M , ↓ v ↓ M , ↑ v

KE = ½ mv2 Effusion Diffusion ↑T, ↑v ↑M,↓v KE = ½ mv2 ½ mv2 = ½ mv2 escape of gas particles through a tiny hole spread of gas particles throughout a space ↑M,↓v ½ mv2 = ½ mv2 64 g/mol (SO2) 16 g/mol (CH4) CH4 _____ is __ times faster than _____. 2 HW p.437 #8,74,76a SO2

Real (non-Ideal) Gases In the real world, the behavior of gases only conforms to the ideal-gas equation under “ideal” conditions. (usually) PV = nRT (ONLY under ideal conditions) Ideal: (High T) (Low P) (weaker IMAFs) (high KE) (high Vtotal) (negligible) WHY? Non-Ideal: (Low T) (High P) (stronger IMAFs) (low KE) (low Vtotal) (not negligible)

Ideal Gas vs Non-Ideal Gas T: ↑ P: ↓ NON T: ↓ P: ↑ more KE/speed weaker IMAFs more avg. dist. less KE/speed stronger IMAFs less avg. dist. HW p.437 #81,82,83 IMAFs (ideal P) (ideal V) n2a V2 ) (V ) = nRT (P + − nb “observed” V too high b/c size of particles not negligible compared to total volume “observed” P too low b/c attractive forces not negligible, collisions less frequent and of less force Vgas = Vcontainer – Vparticles

n2a V2 ) (V ) = nRT (P + − nb Van der Waal Equation: a & b are two constants. a: measures how strongly the gas particles are attached together. b: a measure of finite volume occupied by the molecules. a & b generally increase with increasing molecular mass ?? Larger, massive molecules have large volumes and tend to have greater IMF. n2a V2 ) (V ) = nRT (P + − nb