5 Physical Characteristics of Gases Gases assume the volume and shape of their containers.Gases are the most compressible state of matter.Gases will mix evenly and completely when confined to the same container.Gases have much lower densities than liquids and solids.
6 (force = mass x acceleration) AreaBarometerPressure =(force = mass x acceleration)Units of Pressure1 pascal (Pa) = 1 N/m21 atm = 760 mmHg = 760 torr1 atm = 101,325 Pabar = 105 Papsi = lb/in2 = Pa
7 PRESSURE Units and Measurement Pressure = Force/AreaSI UnitsForce = mass x accelerationForce = kg-m/s2 = Newton Pressure = Newton/m2 = PascalCustomary UnitsPressure = atm, torr, mmHg, bar, psiRelate SI to customaryX 105 Pascal = 1 atm = 760 torrbar = 105 Pascal
13 Boyle’s Law P ∝ 1/V Constant temperature P x V = constant Constant amount of gasP1 x V1 = P2 x V2
14 P1 x V1 = P2 x V2 P1 = 726 mmHg P2 = ? V1 = 946 mL V2 = 154 mL P1 x V1 Q. A sample of chlorine gas occupies a volume of 946 mL at a pressure of 726 mmHg. What is the pressure of the gas (in mmHg) if the volume is reduced at constant temperature to 154 mL?P1 x V1 = P2 x V2P1 = 726 mmHgP2 = ?V1 = 946 mLV2 = 154 mLP1 x V1V2726 mmHg x 946 mL154 mL=P2 == 4460 mmHg
16 Charles’ Law Definition of Temperature V = V atV = aTTemperature(K)
17 Variation of gas volume with temperature at constant pressure. Charles’ & Gay-Lussac’s LawV ∝ TTemperature must bein KelvinV = constant x TV1/T1 = V2/T2T (K) = t (0C)
18 Q. A sample of carbon monoxide gas occupies 3. 20 L at 125 0C Q. A sample of carbon monoxide gas occupies 3.20 L at 125 0C. At what temperature will the gas occupy a volume of 1.54 L if the pressure remains constant?V1/T1 = V2/T2V1 = 3.20 LV2 = 1.54 LT1 = KT2 = ?T1 = 125 (0C) (K) = KV2 x T1V11.54 L x K3.20 L=T2 == 192 K
19 Avogadro’s Hypothesis Equal volumes of gases contain the same number of molecules at constant T,PL of any gascontains X 1023atoms (or molecules)at 0 ℃ and 1 atm.
20 Avogadro’s Law V a number of moles (n) V = constant x n V1/n1 = V2/n2 Constant temperatureConstant pressureV = constant x nV1/n1 = V2/n2
21 4NH3 + 5O2 → 4NO + 6H2O 1 mole NH3 → 1 mole NO At constant T and P Q. Ammonia burns in oxygen to form nitric oxide (NO) and water vapor. How many volumes of NO are obtained from one volume of ammonia at the same temperature and pressure?4NH3 + 5O2 → 4NO + 6H2O1 mole NH3 → 1 mole NOAt constant T and P1 volume NH3 → 1 volume NO
25 Ideal Gas Equation 1 Boyle’s law: V ∝ (at constant n and T) P Charles’ law: V ∝ T (at constant n and P)Avogadro’s law: V ∝ n (at constant P and T)V ∝nTPV = constant x = RnTPR is the gas constantPV = nRT
26 R = 0.082057 L • atm / (mol • K) = 8.3145 J / (mol • K) PV = nRT PV The conditions 0 0C and 1 atm are called standard temperature and pressure (STP).Experiments show that at STP, 1 mole of an ideal gas occupies L.PV = nRTR =PVnT=(1 atm)(22.414L)(1 mol)( K)R = L • atm / (mol • K)= J / (mol • K)
27 PV = nRT nRT V = P 1.37 mol x 0.0821 x 273.15 K V = 1 atm V = 30.6 L Q. What is the volume (in liters) occupied by 49.8 g of HCl at STP?T = 0 0C = KP = 1 atmPV = nRTn = 49.8 g x1 mol HCl36.45 g HCl= 1.37 molV =nRTPV =1 atm1.37 mol x x KL•atmmol•KV = 30.6 L
28 PV = nRT n, V and R are constant nR V = P T = constant P1 T1 P2 T2 = Q. Argon is an inert gas used in lightbulbs to retard the vaporization of the filament. A certain lightbulb containing argon at 1.20 atm and 18 0C is heated to 85 0C at constant volume. What is the final pressure of argon in the lightbulb (in atm)?PV = nRTn, V and R are constantnRV=PT= constantP1 = 1.20 atmT1 = 291 KP2 = ?T2 = 358 KP1T1P2T2=P2 = P1 xT2T1= 1.20 atm x358 K291 K= 1.48 atm
29 d is the density of the gas in g/L Density (d) Calculationsm is the mass of the gas in gmV=PMRTd =M is the molar mass of the gasMolar Mass (M ) of a Gaseous SubstancedRTPM =d is the density of the gas in g/L
30 dRT P M = d = m V = = 2.21 1 atm x 0.0821 x 300.15 K M = M = Q. A 2.10-L vessel contains 4.65 g of a gas at 1.00 atm and 27.00C. What is the molar mass of the gas?dRTPM =d =mV4.65 g2.10 L== 2.21gL2.21gL1 atmx x KL•atmmol•KM =M =54.6 g/mol
31 C6H12O6 (s) + 6O2 (g) 6CO2 (g) + 6H2O (l) Gas StoichiometryQ. What is the volume of CO2 produced at 370 C and 1.00 atm when 5.60 g of glucose are used up in the reaction?C6H12O6 (s) + 6O2 (g) CO2 (g) + 6H2O (l)g C6H12O mol C6H12O mol CO V CO25.60 g C6H12O61 mol C6H12O6180 g C6H12O6x6 mol CO2= mol CO2V =nRTP0.187 mol x x KL•atmmol•K1.00 atm== 4.76 L
32 Dalton’s Law of Partial Pressures V and T are constantP1P2Ptotal = P1 + P2
33 PA = nART V PB = nBRT V XA = nA nA + nB XB = nB nA + nB PT = PA + PB Consider a case in which two gases, A and B, are in a container of volume V.PA =nARTVnA is the number of moles of APB =nBRTVnB is the number of moles of BXA =nAnA + nBXB =nBnA + nBPT = PA + PBPA = XA PTPB = XB PTPi = Xi PTmole fraction (Xi) =ninT
34 Ppropane = Xpropane x 1.37 atm Q. A sample of natural gas contains 8.24 moles of CH4, moles of C2H6, and moles of C3H8. If the total pressure of the gases is 1.37 atm, what is the partial pressure of propane (C3H8)?Pi = Xi PTPT = 1.37 atm0.116Xpropane ==Ppropane = Xpropane x 1.37 atm= atm
35 PT = PO + PH O 2KClO3 (s) 2KCl (s) + 3O2 (g) Bottle full of oxygen gas and water vapor2KClO3 (s) KCl (s) + 3O2 (g)PT = PO + PH O2
39 Air Bag ChemistryCrash SensorAir BagNitrogen GasInflator
40 Air Bag Chemistry 2 NaN3 2Na + 3N2 On ignition: Secondary reactions: 10 Na + 2 KNO3 K2O + 5 Na2O + N2K2O + Na2O + SiO2 K2Na2SiO4
41 Kinetic Molecular Theory of Gases 1. A gas is composed of molecules that are separated from each other by distances far greater than their own dimensions. The molecules can be considered to be points; that is, they possess mass but have negligible volume.d(N2,g) = g/L (273°C)d(N2,liq) = g/mL (-195.8°C)2. Gas molecules are in constant motion in random directions. Collisions among molecules are perfectly elastic.3. Gas molecules exert neither attractive nor repulsive forces on one another.(no interaction)
44 The gas molecules in the container are in random motion
45 압력과 기체분자 운동 운동량 변화 ∆p = mvx –(mvx) = 2mvx 분자의 왕복시간 ∆t = 2a/vx f = ∆p/ ∆t= 2mvx/(2a/vx)= mvx2/aN개의 분자에 의해 면 A에 가해지는 압력P = Σf/a2 = m(vx12 + vx vxN2)/a3=Nmvx2/VPV=Nmvx2
46 압력과 기체분자 운동 공간 상에서 무작위로 움직이는 분자는 특별한 방향에 대한 선호도가 없을 것이므로 Gas Pressure in the Kinetic Theory분자당 평균운동에너지
47 운동에너지와 온도 이상 기체방정식 기체에 열이 가해지면 기체의 내부에너지와 온도가 증가! Mean Kinetic Energy per particleBoltzmann constant
48 기체분자운동이론 (Kinetic Molecular Theory for Gases) P = gas pressure, V=volume= mean square velocityN = number of gas moleculesn = mole number of gas molecules = N/NANA = Avogadro’s numberm = mass of the gas moleculeM = molar mass of the gas molecule
49 실험관찰에 의하면 (Ideal Gas Equation) PV = NkT = nRTN = number of molecules,k = Boltzmann constant, J K-1,n = number of moles = N/NA,R = gas constant, l atm mol-1 K-1T = temperature,P = gas pressure,V = volume,NA = Avogadro’s number
50 Maxwell-Boltzmann Distribution for Molecular Speeds u~ u+du 사이의 속도를 가진 분자들의 수f(u) = speed distribution function,m = molecular mass,k = Boltzmann constant,T = temperature, u = speed
51 분자속도와 에너지 분포 (Molecular velocity and energy distribution) 슈테른(Otto Stern)의 분자속도 측정노벨물리학상 수상자(1943, w/ Gerlach)Schematic diagram of a stern type experiment for determining the distributionof molecular velocities v ~ v+∆v 사이의 분자개수 ∆N측정가능!!
52 Molecular speed Distribution of N2 gas Maxwell Boltzmann distribution of molecular speeds in nitrogen gas at two temperatures. The ordinate is the fractional number of molecules per unit speed interval in (km/s)-1.
53 Apparatus for studying molecular speed distribution Chopper methodgravitation method5.7
54 The distribution of speeds of three different gases at the same temperatureThe distribution of speedsfor nitrogen gas moleculesat three different temperatures
55 Chemistry in Action: Super Cold Atoms Gaseous Na Atoms1.7 × 10-7 KBose-Einstein Condensate1995, NIST Image
56 Gaseous Diffusion/Effusion Diffusion of Ammonia and HClEffusion enrichment of UF6
57 Gas diffusion is the gradual mixing of molecules of one gas with molecules of another by virtue of their kinetic properties.NH4ClNH317 g/molHCl36 g/mol
58 Kinetic-Molecular Theory for Gaseous Behavior Principal Issues (잠재적 문제점)Negligible Volume and No interactionHold only at low P, high T; for dilute gasesElastic CollisionsOnly in Neutonian mechanics is the reverse of an event as likely as the event itself.In the real world you cannot “unscramble” eggs because of entropy effects resulting from large ensembles of molecules
59 Deviations from Ideal Behavior 1 mole of ideal gasRepulsive ForcesPV = nRTn =PVRT= 1.0Attractive Forces
60 Real GasEffect of intermolecular forces on the pressure exerted by a gas.Has volumeHas interaction(usually attractive)
61 Selected Values for a and b for the van der Waals Equation Gasa [(L2 · atm)/mol2]b [10-2 L/mol]He2.370Ne0.2111.71Ar1.343.22Kr2.323.98Xe4.192.66H20.24442.661N21.3903.913O21.3603.183CO23.5924.267CH42.254.28CCl420.413.8C2H24.3905.136Cl26.4935.622C4H1014.4712.26C8H1837.3223.68Van der Waals equationnonideal gasP (V – nb) = nRTan2V2()}correctedpressure}correctedvolume
65 Molina Rowland Crutzen Holland (The Netherlands) USA USA (Mexico)Department of Earth,Atmosphericand Planetary Sciences andDepartment of Chemistry,MITCambridge, MA, USA1943 -Holland (The Netherlands)Max-Planck-Institute for ChemistryMainz, Germany1933 -USADepartment of Chemistry,University of CaliforniaIrvine, CA, USA1927 -
67 Science of Hot Air Balloon 부력 = 외부 공기의 무게 – 내부 기체의 무게내부 기체의 무게= 공기의 밀도(ρ(Thot))× 풍선의 부피(V) ×g외부 공기의 무게= 공기의 밀도(ρ(Tcold))× 풍선의 부피(V) ×g공기의 밀도ρ(T) = w/V = Mp/RT2008 National Balloon Classic in Indianola, Iowa.Tcold ~ 20 ℃Thot ~ 100 ℃부력 ~ 700kg부피(m3)600 ~ 2800 ~ 17,000
68 Lake Nyos, the killer Lake of Cameroon 1986년 8월 이산화탄소가 호수에서 떠올라왔다. 계곡을 타고 수십 킬로미터를 가면서 수천마리의 가축과 1700여명을 사망하게 하였다.이들 중 상당수는 잠들어 있었다.Clarke, T., Taming Africa’s killer lake, Nature, V. 409, p , February 2001.
69 Lake Nyos, the killer Lake Test of the de-gassing operation in 1995Clarke, T., Taming Africa’s killer lake, Nature, V. 409, p , February 2001.