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 Pressure is the amount of force applied per area. 22,000 lbs!!! (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 1 N 1 m2 STP (standard T & P) 273 K 1 atm 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 which 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) Gas pressure is caused by collisions with the container walls. 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 particles is NEGLIGIBLE, compared to total volume of container. Ideally: Vgas = Vcontainer Vgas = Vcontainer – Vparticles In a 1.000 L container, gas only expands into 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)

(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 = R = P ↑ , V ↓ T ↑ , V ↑ T ↑ , P ↑ given on exam P1V1 n1T1 P2V2 n2T2 = R = (changes in P,V,T,n) constant (initial) (final) (inversely proportional) (directly proportional) P ↑ , V ↓ T ↑ , V ↑ T ↑ , P ↑ n ↑ , V ↑

PV = nRT Ideal-Gas Changes PV = nRT V1 The pressure on a 411 mL sample of gas is decreased from 812 torr to 790 torr. How will volume of the gas be affected? P2 V2 = ? P1 PV = nRT (812)(0.411) = nRT (790) (?) = nRT V increased

PV = nRT Ideal-Gas Changes PV = nRT 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. How will the temperature in be affected? T1 V2 T2 = ? PV = nRT (1.20)(10.0) = nR(298) (1.20)(8.83) = nR(?) T decreased =

PV = nRT Ideal-Gas Changes PV = nRT O2  O3 3 2 n1 V1 A 12.2 L sample of 0.500 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 PV = nRT HW p. 432 #1, 23, 26 (1.00)(12.2) = (0.500) R (298) (1.00) (?) = (?) R (298) (1.00) V = (0.335) R (298) O2  O3 3 2 V = 8.19 L 0.500 mol O2 x 2 mol O3 = 3 mol O2 0.333 mol O3 n2 n1 n2 = ?

PV = nRT Ideal-Gas Calc’s 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 atoms of He? n = 0.268 mol He

PV = nRT Ideal-Gas Calc’s PV = nRT m T P V M = ? 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

PV = nRT Ideal-Gas Calc’s PV = nRT HW p. 438 #92, 29, 46, 35, 38 T P V @STP A gas at 25.0oC and 1.20 atm occupies a volume of 4.28 L. What volume will it occupy at STP? R = 0.08206 L∙atm mol∙K PV = nRT (1.20 atm)(4.28 L) = n (0.08206)(298 K) n = 0.210 mol (1.00)V = (0.210)(0.08206)(273) V = 4.70 L

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 The 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 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 9.85 g BaCO3 x 1 mol BaCO3 x 197.34 g BaCO3 1 mol CO2 x 1 mol BaCO3 22.4 L CO2 = 1 mol CO2

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) (would be 89.6 L if at STP) = ____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 the % 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 KEAr = KEHe (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 SO2 HW p.437 #8, 74, 76a

“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) (high KE) (high Vtotal) WHY? (weak IMAFs) (tiny Vparticles) (negligible) (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

Slides #34-37 demonstrate calculation of gas changes using the combined gas law. His is no longer required on the new AP Exam

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