Slide 1Gases SubstancesPhysical Characteristics of GasesGases exert PressurePressure UnitsGas LawsGas Laws Cont’dGas Laws Cont’dGas ConstantIdeal Gas LawClicker: 10.1Standard Temp. & Press.Slide 13Combined Gas LawSample ProblemClicker: 10.2Molar Mass & Ideal Gas Law (Exp. 10)Density & Ideal Gas LawSample ProblemsSample ProblemsClicker 16.3:Gases in Chemical Rxns (Stoichiometry)Slide 23Sample ProblemsSlide 25Gas MixturesSample ProblemRelating Partial & Total PressureSample ProblemCollecting Gas over Water (Exp. 9)Sample ProblemAppendix BSample Problem1GasesChapter 10Chemistry the Central Scienceby: Brown, Lemay, Bursten, Murphy & WoodwardPresented by: Dr. Stacey Gulde2Gases SubstancesAir: Mixture of many gassesCommon gases: Usually have low MwNoble gasesDiatomicsOzone (O3)Molecular compounds•Ex: CO, CO2, N2OSolids or liquids can be forced to exist as gases called vaporsEx. H2O3Physical Characteristicsof Gases1. Automatically take the shape & volume ofits containerMolecules are farther apart in a gas than either a solid or liquid•Lots of empty space b/t molecules2. Highly compressibleAble to mush gases into smaller volumes3. Always make homogeneous mixtures (uniform)4Pressure (P) – force exerted by molecules in motionGravity pulls atmospheric gases to Earth’s surface•Exerting 14.7 psi (lbs/in2)-Perspective: Car tires usually 27 – 32 psi•Felt uniformlyHow come we aren’t crushed under thispressure?•Pressure inside our bodiesequalizes this pressureGases exert PressureWhat if No insidePressure?AreaForceP 5Pressure UnitsSI Unit = Pascal (Pa) Other common units:Standard atmospheres (atm) = pressure at sea level & 0C•1 atm = 1 x105 PaMillimeter of mercury (mmHg) = from barometer•Also known as torr (torricelli)-1 mmHg = 1 torr-1 atm = 760mmHgExact valuesP=pressureg=gravityd=density of liquidh=height of liquid21mN21smkgdhP g6Gas Laws4 variables describe gases:1. Pressure (P)2. Volume (V)3. Temperature (T)4. Amount (n) = # of moles1. Bolye’s Law: P & VIf pressure increases, volume decreasesP  1/V (@ constant n and T)•PV = constant7Gas LawsCont’d2. Charles’s Law: V & TIf increase temp, volume increases•Think hot air balloonV  T (@ constant n and P)•V=T(constant)-Temp must be in Kelvin8Gas LawsCont’d3. Avogadro’s Law: V & nNOTE: 1st proposed by Louis Gay-Lussac,but better interpreted by Amedeo AvogadroIf increase moles, volume increases•Think blowing up a balloonV  n (@constant T and P)•V=n(constant)9Gas ConstantWhat is the constant in each Gas law?1. Boyle’s: PV= constant2. Charles's: V=T(constant)3. Avogadro’s: V=n(constant)Ideal (universal) gas constant = RHow do you know which to use?•Depends on the unitsPV = RV = TRV = nRRewritten:KmolJKmolatm L8.314R 0.08206 Ror10Ideal Gas LawCombines all 3 gas laws into 1:Required units for this equation:P = atmV = Ln = molesT = K•TK=TC + 273nRTPV Clicker: 10.1Which value of the gas constant should always be used with the Ideal gas law?1112Standard Temp. & Press.Standard Temperature & Pressure (STP) – special set of conditionsTemperature = 0CPressure = 1.00 atmIf have 1.00 mole of gas at STP what is the volume?Standard molar volume =22.4 LnRTPV (1.00atm)V(1.00mol))(0.08206KmolatmL(273K)13Sample ProblemIf 12.3g of steam is confined to 18.0 mL, what is the temp (C) if the pressure is 1.75x106 torr?P = 1.75x106 torrT = ?C V = 18.0 mL12.3 g H2O12.3 g H2O= 0.683mols H2O mols LTK= TC + 273= 466 C18.0 mL= 0.0180L atm1.75x106 torr= 2.30x103atm739 = TC + 273 atm2.30x103L) (0.0180 KmolatmL08206.0 mols 0.683nRTPV KmolatmL0.08206 RnRPVT T739KT OH 18.02gOH 1mol221000mL1L760torr1atm14Combined Gas Law – shows relationships for the same gas at different conditionsUse Ideal gas equation & make a proportion •Solve for what’s changing & set that equal to each other •GENERIC example:Combined Gas LawRnTPV1111TnVP2222TnVPnRTPV 15Sample ProblemA 5.50 L volume of gas has a pressure of 0.950 atm at 25C. What is the volume if the gas is as STP?V = 5.50 LP = 0.950 atmT = 25C 298 KP2 = 1.00 atmT2 = 0C 273 KV2 = ?(cross multiply)(1 gas, 2 conditions)Solve for what’s changing Katm 982V2nRTPV 222111TVPTVP  K 298L 50.5atm 0.950  KLatm 425.14264.79LV2  K 273V1.00atm216Clicker: 10.2The pressure in a natural-gas tank is maintained at 2.20 atm. On a day when the temperature is -15C, the volume of gas in the tank is 3.25x103 m3. What is the volume of the same quantity of gas on a day when the temperature is 31C?What equation do you use?A. PV=nRTB. C. D. 222111TVPTVP22221111TnVPTnVP2211TVTV17Molar Mass & Ideal Gas Law(Exp. 10)Can find Molar mass (Mm) of unknown gas:Substitute into Eq:Solve for Mm:nRTPV Mmmn MmmRTPV molgramsMm nmMm  UNITS nforsolve PVmRTMm18Density & Ideal Gas LawCan also determine the density of a gas:Close look at Mm eq:Solve for d:PVmRTMmVmdPRTdMm  RTPMmd  d SubstituteSample Problems1. What is the density of CO2 at 745 torr and 65C?19d=?P=745 torrT= 65C 338 K atm745 torr760 torr1 atm=0.980 atmMm = 12.01+2(16.00) =44.01g/mol RTPMmd  d molg01.44 atm980.0 KmolatmL08206.0 K338Lgd 55.1Sample Problems2. A gas is collected from a smoke stack at 741 torr, 35C into a container that has a volume of 2.0L and an evacuated mass of 200.00g. After filling, the container weights 204.93g. What is the molar mass of the gas?20P=741atmT=35CV= 2.0L 308KEmpty = 200.00gFilled = 204.93gGas mass = 4.93g atm741 torr760 torr1 atm=0.9745atmMm = ? L0.2PVmRTMm Mm g93.4 atm975.0 KmolatmL08206.0 K308molgMm 64Clicker 16.3:Which of the following pollutants could be our mystery gas from the previous problem?A. COB. CO2C. NO2D. SO22122Gases in Chemical Rxns(Stoichiometry)Use mol:mol ratio to relate information between reactants & productsNote: to use gas law, compound MUST BE a GAS!Sample Problems1. Air bags in cars are inflated by the decomposition of