Sheets Page 1 Lecture 22 Lecture 22: Gases 1 Read: BLB 10.1–10.5 HW: BLB 10.2,19a,b,23,26,30,39,41,45,49 Sup 10:1–6 Know: • gases & gas laws • PV=nRT(!!!!) • partial pressures • density & molecular mass Exam #2: “NO SCORE”? Not a problem. Please talk with Mike Joyce in 210 Whitmore Final Exam: Monday, May 4 @ 12:20; MUST register on elion for a final exam conflict or overload by March 8. See http://www.registrar.psu.edu/exams/exam_overload.cfm http://www.psu.edu/dus/handbook/exam.html#conflict Need help?? Get help!! TAs in CRC (211 Whitmore) and SI—hours on Chem 110 website; my office hours (Mon 12:30-2 & Tues 10:30-12 in 324 Chem Bldg [or 326 Chem])Sheets Page 2 Lecture 22 Itʼs a gas, gas, gas gas molecules: • are constantly • are far apart (10 times as far as they are big; occupy < 0.1% of the total volume) • move in lines from collision to collision • move at higher temperatures • move at lower temperatures as a result: • expand in whatever volume is available • have pressure (collisions with walls) • mix completely with one another • can be easily compressed • when cooled, will eventually condense (compression may be necessary) • gases are described in terms of pressure (P), temperature (T), volume (V), & # moles (n) • all gases behave similarly at low pressure • gases will mix in all proportions with other gases to form mixturesSheets Page 3 Lecture 22 Pressure • pressure: force/unit area force: kg•m•s–2, Newton (N) unit area: m2 • SI unit for pressure: 1 N•m–2 = 1 Pa (Pascal) • standard atmospheric pressure**: 1 atm ≡ 1.013 × 105 Pa 1 atm = 760 torr (or mm Hg) [1 atm = 14.7 lb/in2] **real atmospheric pressure varies with altitude, temperature and weather • how big is one atm? 10 tons / m2! or 600 lbs/headSheets Page 4 Lecture 22 Measuring pressure • barometer used to measure Patm by forces; (Torricelli, early 1600ʼs) • measure P in terms of height of Hg 1 atm ≡ 760 torr “=” 760 mm Hg [know this conversion!] [factoid: 760 mm Hg = 10 m H2O]Sheets Page 5 Lecture 22 The ideal gas law • state of gas described by: P, V, T in K (degrees Kelvin), & n • STP: standard temperature (273.15 K) & pressure (1 atm) NOTE: STP for gases is not the same as standard state conditions used for ΔH!! (more in Chap 5) • absolute temperature in Kelvin (K): °C + 273.15 = K • Avagadro: V ∝ n V/n = constant* (P,T fixed) *proportionality constant is the same for ALL gases aVnSheets Page 6 Lecture 22 The ideal gas law (cont.) • Boyle: V ∝ 1/P PV = constant (T,n fixed) • Charles: V ∝ T V/T = constant (P,n fixed) note: T in absolute T (K)! • All of these laws can be described by one equation: PV = nRT • R = gas constant; units of R are very, very important! R = 0.08206L atmmol K R = 8.314Jmol K R =1.987calmol KaVT (°C)-273.15°C = 0 Ka1 atm 2 atm 4 atmaVPaV1/PSheets Page 7 Lecture 22 Example: • for problems: given known quantities, solve for the unknown What is the volume (V) occupied by 1.00 mole of gas at exactly 0°C and 1 atm (STP)? V = ?P = 1atmn = 1molT = 0° + 273 = 273KR =0.08206L atmmol KPV = nRTSheets Page 8 Lecture 22 Example: The gas in a 750 mL vessel at 105 atm and 27°C is expanded into a vessel of 54.5 L and –10°C. What is the final pressure? V1= 750mL1L1000mL!"#$%&= 0.75LP1= 105atmT1= 27° + 273 = 300KV2= 54.5LP2= ?T2= !10° + 273 = 263KPV = nRT so P1V1= nRT1n is not changing & R is a constant, so...nR =P1V1T1 & nR =P2V2T2Sheets Page 9 Lecture 22 Density and molecular weight Example: can derive d = PM/RT from PV=nRT where d = density M = molar mass You should derive this on your own! (see BLB pp 406–407) Calculate the average molar mass of dry air if it has a density of 1.17 g/L at 21°C and 740.0 torr. d = 1.17g / LP = 740.0torr( )1atm760torr!"#$%&= 0.974atmT = 21° + 273 = 294KM = ?d = PM / RTSheets Page 10 Lecture 22 Example: What is the density of oxygen, in grams per liter, at 25°C and 0.850 atm? d = ? need g / LP = 0.850atmT = 25° + 273 = 298KM = 32g / molSheets Page 11 Lecture 22 Daltonʼs law of partial pressures • partial pressure: the pressure a gas would have if it was the only gas in the container • Daltonʼs law of partial pressures: total pressure is equal to the sum of partial pressures • mole fraction: ratio of na/ntot; dimensionless Xa = na/ntot note: mole fractions must sum to 1 Ptot= P1+ P2+ P3+ ...P =nRTVPtot=n1RTV+n2RTV+n3RTV+ ...Ptot= n1+ n2+ n3+ ...( )RTVntot= n1+ n2+ n3+ ...Ptot=ntotRTVX1= n1/ ntotP1=n1RTVP1=n1ntotPtotP1=n1ntotntotRTV!"#$%&P1= X1PtotSheets Page 12 Lecture 22 Example: Mix 5 moles of CO2, 2 moles of N2 and 1 mole of Cl2 in a 40 L container at 0°C. (a) What is the pressure in the container? (b) What is the partial pressure of each gas in the container? (c) What volume does each gas occupy in the container? nCO 2= 5molnN 2= 2molnCl 2= 1molntot= 8molXCO 2=XN 2=XCl 2=check: X!= 1.000[ ]V = 40LT2= 0° + 273 = 273KPtot= ?Ptot=ntotRTVPtot=therefore, PCO 2=PN 2=PCl 2=check: P!= Ptot= 4.48atm[ ]Sheets Page 13 Lecture 22 Example: 3.0 L of He at 5.0 atm and 25°C are combined with 4.5 L of Ne at 2.0 atm and 25°C in a 10 L vessel at constant temperature. What is the partial pressure of the He in the 10 L vessel? on to next page… He (initial): VHe= 3LPHe= 5.0atmT = 25° + 273 = 298KnHe=PHeVHeRTnHe=Ne (initial): VNe= 4.5LPNe= 2.0atmT = 25° + 273 = 298KnNe=PNeVNeRTnNe=ntot= nHe+ nNentot=therefore, XHe=XNe=note: do not need XNe for problemcheck: X!= 1.000[ ]Sheets Page 14 Lecture 22 Ptot in 10L vessel: (final)Vtot= 10LPtot= ??ntot= 0.981molT = 25° + 273 = 298KPtot=ntotRTVtotPtot=Now we need to find PHe(final)PHe=Sheets Page 15 Lecture 22 Before next class: Read: BLB 10.6–9 HW: BLB 10:5,8,59,61,71,75,77,82,83,84 Sup 10:7–15 Know: • kinetic-molecular theory • effusion & diffusion • real gases (van der Waals) Answers: p. 7: 22.4 L p. 8: Pfinal = 1.27 atm p. 9: 29.0 g/mol p. 10: 1.11 g/L p. 12: (a) Ptot = 4.48 atm; (b) PCO2 = 2.80 atm, PN2 = 1.12 atm, PCl2 = 0.56 atm; (c) 40 L p. 13: 1.5
View Full Document
Unlocking...