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

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

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

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

2)Collisions with the walls of the container cause gas pressure. P = FAFA FAFA FAFA 1)Gas particles are in constant random motion. 5 Parts of Kinetic-Molecular Theory Collisions are elastic (no KE lost). (ideally)

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

5) Average KE of gas particles is… …directly proportional to Kelvin temp. ( K not o C ) no negative temp’s, no negative energies, no negative volumes, etc.) video clip KE avg α T 5 Parts of Kinetic-Molecular Theory

Boyle’s Law (P & V) 1 atm2 atm 10 L 5 L (inversely proportional) P ↑, V ↓

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

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

Avogadro's Hypothesis At the same ___ & ___, equal _________ of gas must contain equal _________. CO 2 He O2O2 1 mol He 2 mol He n ↑, V ↑ Avogadro's Law (V & n) All at: P = 1 atm T = 25 o C V = 1.0 L moles (n) add gas volumes TP (particles) (directly proportional)

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

(changes in P,V,T,n) P ↑, V ↓T ↑, V ↑ T ↑, P ↑ n ↑, V ↑ (directly proportional)(inversely proportional) NOT given on exam PV = nRT = R= R constant given on exam P2V2n2T2P2V2n2T2 P1V1n1T1P1V1n1T1 == P1V1n1T1P1V1n1T1 P2V2n2T2P2V2n2T2 (initial)(final)

1.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? = 422 mL P1P1 P2P2 V2V2 V1V1 P2V2n2T2P2V2n2T2 P1V1n1T1P1V1n1T1 = = ? (812)(411) = (790) V 2 (812)(411) (790) = V2V2 Ideal-Gas Changes PV = nRT PV nT = R

2.A 10.0 L sample of a gas is collected at 25 o C 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 o C? V1V1 T1T1 V2V2 T2T2 P2V2n2T2P2V2n2T2 P1V1n1T1P1V1n1T1 = = ? (10.0) (298) = (8.83) T2T2 T 2 (10.0) = (8.83)(298) (10.0) = T2T2 Ideal-Gas Changes = 263 K = –10 o C PV = nRT PV nT = R

3.A 13.1 L sample of moles of O 2 is held under conditions of 1.00 atm and 25.0 o C. If all of the O 2 is converted to Ozone (O 3 ), what will be the volume of O 3 ? = 8.73 L n1n1 n 2 = ? V2V2 V1V1 O 2  O mol O 2 x 2 mol O 3 = 3 mol O mol O 3 n2n2 n1n1 P2V2n2T2P2V2n2T2 P1V1n1T1P1V1n1T1 = (13.1) (0.502) = V 2. (0.335) Ideal-Gas Changes PV = nRT HW p. 432 # 1, 23, 89, 26 PV nT = R

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

Ideal-Gas Equation PV = nRT Molar Mass 2.A sample of aluminum chloride gas weighing g at 350. o C and 760 mmHg of pressure occupies a volume of 19.2 mL. Calculate the Molar Mass of the gas. 133 g/mol (1.00 atm)( L) = n ( )(623 K) n = mol PV = nRT R = L∙atm mol∙K grams mole M = __ g_ mol = AlCl 3 = HW p. 438 #92, 29, 46, 38, 35 n = mMmM M = mnmn so… (given on exam) m T P V M = ?

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

Gas Stoich with Molar Volume Assume reactions below occurred at STP. 1.Calculate the mass of NH 4 CI reacted with Ca(OH) 2 to produce 11.6 L of NH 3 (g). 2 NH 4 Cl + Ca(OH) 2  2 NH 3 + CaCI H 2 O 2.Calculate the volume of CO 2 gas produced when 9.85 g of BaCO 3 is completely decomposed. BaCO 3 (s)  BaO (s) + CO 2 (g) 27.7 g 1.12 L 11.6 L NH 3 x 1 mol NH 3 x 22.4 L NH 3 2 mol NH 4 Cl x 2 mol NH g NH 4 Cl = 1 mol NH 4 Cl

3.What volume of O 2 gas is produced from 490 g KClO 3 at 298 K and 1.06 atm? KClO 3 (s)  KClO (s) + O 2 (g) = ____L O g KClO 3 x g KClO 3 mol KClO 3 x mol O = L O 2 NOT at STP Molar Mass of KClO 3 is g/mol PV = nRT use… 4.00 (1.06 atm) V = (4.00 mol )( )(298 K)

Mole Fraction (X A ) P A = P total x X A X A = moles of A total moles The mole fraction (X A ) is like a % of total moles that is A, but without the % or x 100. WS 6b #1-4 P total = P A + P B + P C + … Dalton’s Law of Partial Pressures HW p. 436 #

P total = P H 2 O + P gas W hen 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. P gas = P total – P H 2 O equalize water level inside & outside =

1.Calculate the mass of L of H 2 gas collected over water at 21.0 o C with a total pressure of 750. torr. The vapor pressure of water at 21.0 o C is 20.0 torr. P total = P H 2 O + P gas PV = nRT 750. = P H 2 P H 2 = 730. torr P H 2 = 730/760 = atm (0.961)(0.641) = n H 2 ( )(294) n H 2 = mol m H 2 = g

↑ T, ↑ v ↑ M, ↓ v 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)?

Compare the molecular speed (v) of these gases: 1) at 25.0 o C(i) Helium (ii) Oxygen (O 2 ) 2) at 50.0 o C(i) Helium (ii) Oxygen (O 2 ) 3)Does the data support your conclusions from the models on the previous slide about effects of T and M on v ? ↑ T, ↑ v ↑ M, ↓ v 1360 m/s482 m/s 1420 m/s502 m/s KE = ½ mv 2 (given on exam) WHY?

Distributions of Molecular Speed average molecular speed (v) KE = ½ mv 2 (KMT) (given on exam) Therefore: T & v are __________ proportional directly Temp (K) & KE avg are directly proportional T α KE avg ↑ T, ↑ v

Gases at the same Temp, have the same _____. Speed vs. Molar Mass KE avg ½ m v 2 = ½ m v 2 KE 1 = ½ m 1 v 1 2 Ar: M = 40 g/mol He: M = 4.0 g/mol KE = ½ mv 2 (at same T) KE He = KE Ar KE 2 = ½ m 2 v 2 2 ↑ M, ↓ v ↓ M, ↑ v M & v are __________ proportional inversely

Effusion escape of gas particles through a tiny hole spread of gas particles throughout a space Diffusion ↑ T, ↑ v ↑M,↓v↑M,↓v KE = ½ mv 2 64 g/mol (SO 2 ) 16 g/mol (CH 4 ) _____ is __ times faster than _____. CH 4 SO 2 2 KE = ½ m v 2 ½ m v 2 = ½ m v 2 HW p.437 #8,74,76a

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

Ideal Gas vs Non-Ideal Gas ) (V ) = nRT(P(P “observed” P too low b/c attractive forces not negligible, collisions less frequent and of less force “observed” V too high b/c size of particles not negligible compared to total volume (ideal P) (ideal V) NON T: ↓ P: ↑ IDEAL T: ↑ P: ↓ more KE/speed weaker IMAFs more avg. dist. less KE/speed stronger IMAFs less avg. dist. IMAFs V gas = V container n2aV2n2aV2 − nb + HW p.437 #81,82,83 – V particles