Chapter 10 Gases Lecture 1Characteristics-- Vol. is equal to Vol. of the container it is in-- Highly compressible-- 2 or more gases ALWAYS make a HOMOgenous mixture(Defining feature of GAS NO INTERMOLECULARFORCES)Gases have no interactionsLiquids are together but still movingSolids can move but not flowing (locked in place)10.2 PressurePressure = Force/AreaPressure exerted when molecule hits side of containerUnits-- Atmosphere (atm) 1 atm= 760 mmHg-- mmHg Torr-- Kilopascals (kpa)-- Pounds per square inch (PSI) not used in science10.3 Ideal Gas Laws4 variables used to define state of a gas--Temp.--Pressure--Vol.--# of mol of gasThese characteristics arise b/c of gas molecules are far apart & have NO intermolecular forces b/w themSee HW 10.1All related quantitatively by Ideal Gas EquationPV= nRTR= gas constant=0.082 (L-atm/mol K)3 Special Cases of Ideal Gas Law1.-- The P-V relationship: Boyles Law (constant n, T)>PV=nRT PV= constantV= constant-- As you compress Vol. pressure become greater.(P & V are inversely related)2. -- The T- V relationship Charles Law (Constant n, PPV=nRTV= (nR/P)T3. -- Quantity (n)—V relationship: Avagadro’s Law (Constant P,T)PV=nRTV=n(RT/P)10.4 The Ideal Gas Equation1st introduce standard temp. and Pressure (STP)P= 1 atmT=0cCalculate the Vol. of 1 mol of gas at STP(1atm)V= (1mol)(0.082 L-atm/mol-K)(0+273 K)V=22.4L*Doesn’t matter what gas 1 mol of ANY gas will have 22.4 L at STP*See HW 10.3nRT is a constant(nR/P) is a constant(RT/P) is a constant10.4 ContinuedRelating the Ideal Gas Equation and the Gas LawA gas at 15c and 182 mL is reduced to 82.0 mL while P is held constant. What is the final Temp?Temperature has to decrease to SLOW down interactions when VOL is REDUCEDPV1=nRT1 V1/T1=nR/PPV2=nRT2 V2/T2=nR/PIn general P1V1/T1= P2V2/T210.5 Further application ofIdeal Gas Equation, Gas Density and Molar mass (See Lab This Week)--PV=nRT n(MW/V=P(MW)/RT--n/V= P/RT(multiply both sides by Molecular Weight (MW))D=m/v n(mw)/V(mol)(g/mol)/L=g/L=dnR/P is a constantnR/P is a constantSo V1/T1=V2/T20.182L/15+273 = 0.082L/T2T2=130KSEE HW 10.5, 10.6, 10.7Chapter 10.5 Lecture 2Vol of Gases in Chemical Reactionsex.Airbags(chunk of NaN3 w/ igniter is a solid becomes Nitrogen gas)2NaN3(s)2Na+N2(g) -exothermic reaction, it gives off heat-once it cools down the vol. decreasesCh. 10.6 Gas Mixtures and Partial Pressure Dalton’s Law of Partial Pressure>The total pressure of a mixture of gases equals the sum of the pressures of each gaswould exert if it were present alone.PT=P1+P2PT=n1(RT/V)+n2(RT/V)ConstantPT=(RT/V) (n1+n2)Partial Pressure and Mole FractionP1/PT= [n1(RT/V)]/[nT(RT/V)]P1/PT=n1/nTP1=(n1/nT)PTP1=X1PTA mixture of gases contains 0.445 mol of N2, 0.982 mol of H2 and 0.479 mol of NH3.If the total pressure of the mixture is 2.35 atm what is he partial pressure of N2?nT=1.906 PT=2.35 PN2=(.0445/1.906)(2.35atm)Pn2=.05499 atm*TOTAL Pressure inside of container is equal to the pressure exerted by the gases in the container plus the pressure of the molecules of the water inside the containerIf an airbag has a vol. of 36L filled by N2(g) at a pressure of 1.15 atm. & temp. of 26c how many grams of NaN3 must be decomposed?PV=nn2RT(1.15)(36)=n(0.0821L-atm/molK)(26c)Symbol for Mole fraction =x1See HW 10.10Chapter 10.6 Cont.Collecting Gases over waterIn the case of collecting gases over water you need to include the PH2O in your calculationA sample of KCLO3 is decomposed producing O2 which is collected over H2O2KCLO3(s) KCL(s)+2O2(g)The vol. of the O2(g) collected is 250 ml at 26c and 765 Torr total pressure. How many moles of O2 are collected?PT=PO2+PH2O765=PO2+25.2mmHg nO2=PO2V/RTPO2=740mmHg(740mmHg)(1atm/760mmHg)(.250L)/[(0.0821L-atm/mol-K)(299K)]NO2=9.92E-3mol O2Chapter 10.7Kinetic Molecular Theory (KMT)Why does the Ideal Gas Equation work?1. Gases consist of large numbers of molecules all moving in random motion2. Volume of gas molecules is negligible (insignificant) compared to the volume of the container3. Attractive and repulsive forces b/w gas molecules is negligible (randommovement)4. Energy can be transferred b/w gases molecules during collisions, but theaverage kinetic energy of all gas molecules does NOT change if temperature isheld constant.5. Average Kinetic energy of molecules is proportional to temperature b/c gas molecules have more velocity at high temperatures.faster moving gas molecules KE=12mV2slower moving gas moleculesTemperature dictates how fast molecules moveApplication of Volume increase at constant temperature1.V =constantpressure Boyles LawAt constant temperature gas molecules move at same velocity and same KE Aszxs vol. INCREASES they have a longer way to go to reach container walls so Pressure DECREASESApplications of KMT to Gas LawsEffect of Vol. increases at constant temperature, Boyles Law (prev. lecture)Effect of Temp. increase at constant Vol.As you increase temp. KE of gas molecules increases because their velocity increasesso they hit the container wall more often and with GREATER KE resulting in higher pressure.10.9 Real Gases: Deviations from Ideal Behavior>Volume of gas is nor always negligible (seen at low vol, high pressure or high number of moles of gas.>Attractive forces are not always negligible (seen at low vol, high pressure, low temp. when gas molecules are moving slow enough to feel attractions. (induce dipole)Chapter 13 Properties of SolutionsThe Solution process—Why do soutions form? Because it is energetically favorable to do so!In general solutions form when the attractive forces b/w solute and solvent are comparable to those between solute and solute and solvent and solvent.ex. Why does NaCl dissolve in H2O? There are 3 types of interactions>solute—solute Hsolute –requires energy to break NaCl ionic bondENDOthermicSince temp is the same, KE is the same, interaction is the same but because the numberof interactions b/w the molecules and the wall of the container are lower pressure is lower.SEE HWK 10.11 10.13KE stays the same>solvent—solvent Hsolvent –requires energy to break H bond of H2O ENDOthermic>solute—solvent Hmix –EXOthermic gain energy from ion dipole b/w Na+, Cl- and H2OOverall energy of solution process∆ Hsolution=∆ Hsolute+∆ Hsolvent+∆ HmixIn general>H is greater than Hsolute + Hsolvent a solution will form>H is less than Hsolute + Hsolvent a solution will NOT form>H is a