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Gases Chapter 5. 기체 기체상태의 물질 기체의 압력 기체 법칙 보일의 법칙, 샤를의 법칙, 아보가드로의 법칙 분압 기체분자운동론 온도와 에너지, 확산 실제기체 및 반데르발스 상태방정식.

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Presentation on theme: "Gases Chapter 5. 기체 기체상태의 물질 기체의 압력 기체 법칙 보일의 법칙, 샤를의 법칙, 아보가드로의 법칙 분압 기체분자운동론 온도와 에너지, 확산 실제기체 및 반데르발스 상태방정식."— Presentation transcript:

1 Gases Chapter 5

2 기체 기체상태의 물질 기체의 압력 기체 법칙 보일의 법칙, 샤를의 법칙, 아보가드로의 법칙 분압 기체분자운동론 온도와 에너지, 확산 실제기체 및 반데르발스 상태방정식

3 Elements that exist as gases at 25 0 C and 1 atmosphere

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5 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. Physical Characteristics of Gases

6 Units of Pressure 1 pascal (Pa) = 1 N/m 2 1 atm = 760 mmHg = 760 torr 1 atm = 101,325 Pa bar = 10 5 Pa psi = lb/in 2 = Pa Barometer Pressure = Force Area (force = mass x acceleration)

7 Pressure = Force/Area SI Units Force = mass x acceleration Force = kg-m/s 2 = Newton Pressure = Newton/m 2 = Pascal Customary Units Pressure = atm, torr, mmHg, bar, psi Relate SI to customary X 10 5 Pascal = 1 atm = 760 torr bar = 10 5 Pascal PRESSURE Units and Measurement

8 Sea level1 atm 4 miles0.5 atm 10 miles0.2 atm

9 PRESSURE Mercury Barometer

10 Figure 5.4

11 As P (h) increases V decreases

12 Boyle’s Law

13 P ∝ 1/V P x V = constant P 1 x V 1 = P 2 x V 2 Boyle’s Law Constant temperature Constant amount of gas

14 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? P 1 x V 1 = P 2 x V 2 P 1 = 726 mmHg V 1 = 946 mL P 2 = ? V 2 = 154 mL P 2 = P 1 x V 1 V2V2 726 mmHg x 946 mL 154 mL = = 4460 mmHg

15 As T increasesV increases

16 Charles’ Law Definition of Temperature V = V 0 + at V = aT Temperature(K)

17 Variation of gas volume with temperature at constant pressure. V ∝ TV ∝ T V = constant x T V 1 /T 1 = V 2 /T 2 T (K) = t ( 0 C) Charles’ & Gay-Lussac’s Law Temperature must be in Kelvin

18 Q. A sample of carbon monoxide gas occupies 3.20 L at C. At what temperature will the gas occupy a volume of 1.54 L if the pressure remains constant? V 1 = 3.20 L T 1 = K V 2 = 1.54 L T 2 = ? T 2 = V 2 x T 1 V1V L x K 3.20 L = = 192 K V 1 /T 1 = V 2 /T 2 T 1 = 125 ( 0 C) (K) = K

19 Equal volumes of gases contain the same number of molecules at constant T,P L of any gas contains X atoms (or molecules) at 0 ℃ and 1 atm. Avogadro’s Hypothesis

20 Avogadro’s Law V  number of moles (n) V = constant x n V 1 /n 1 = V 2 /n 2 Constant temperature Constant pressure

21 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? 4NH 3 + 5O 2 → 4NO + 6H 2 O 1 mole NH 3 → 1 mole NO At constant T and P 1 volume NH 3 → 1 volume NO

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25 Ideal Gas Equation Charles’ law: V ∝ T  (at constant n and P) Avogadro’s law: V ∝  n  (at constant P and T) Boyle’s law: V ∝  (at constant n and T) 1 P V ∝V ∝ nT P V = constant x = R nT P P R is the gas constant PV = nRT

26 The conditions 0 0 C and 1 atm are called standard temperature and pressure (STP). PV = nRT R = PV nT = (1 atm)(22.414L) (1 mol)( K) R = L atm / (mol K) = J / (mol K) Experiments show that at STP, 1 mole of an ideal gas occupies L.

27 Q. What is the volume (in liters) occupied by 49.8 g of HCl at STP? PV = nRT V = nRT P T = 0 0 C = K P = 1 atm n = 49.8 g x 1 mol HCl g HCl = 1.37 mol V = 1 atm 1.37 mol x x K Latm molK V = 30.6 L

28 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 0 C is heated to 85 0 C at constant volume. What is the final pressure of argon in the lightbulb (in atm)? PV = nRT n, V and R are constant nR V = P T = constant P1P1 T1T1 P2P2 T2T2 = P 1 = 1.20 atm T 1 = 291 K P 2 = ? T 2 = 358 K P 2 = P 1 x T2T2 T1T1 = 1.20 atm x 358 K 291 K = 1.48 atm

29 Density (d) Calculations d = m V = PMPM RT m is the mass of the gas in g M is the molar mass of the gas Molar Mass ( M ) of a Gaseous Substance dRT P M = d is the density of the gas in g/L

30 Q. A 2.10-L vessel contains 4.65 g of a gas at 1.00 atm and C. What is the molar mass of the gas? dRT P M = d = m V 4.65 g 2.10 L = = 2.21 g L M = 2.21 g L 1 atm x x K Latm molK M = 54.6 g/mol

31 Gas Stoichiometry Q. What is the volume of CO 2 produced at 37 0 C and 1.00 atm when 5.60 g of glucose are used up in the reaction? C 6 H 12 O 6 (s) + 6O 2 (g) 6CO 2 (g) + 6H 2 O (l) g C 6 H 12 O 6 mol C 6 H 12 O 6 mol CO 2 V CO g C 6 H 12 O 6 1 mol C 6 H 12 O g C 6 H 12 O 6 x 6 mol CO 2 1 mol C 6 H 12 O 6 x = mol CO 2 V = nRT P mol x x K Latm molK 1.00 atm == 4.76 L

32 Dalton’s Law of Partial Pressures V and T are constant P1P1 P2P2 P total = P 1 + P 2

33 Consider a case in which two gases, A and B, are in a container of volume V. P A = n A RT V P B = n B RT V n A is the number of moles of A n B is the number of moles of B P T = P A + P B X A = nAnA n A + n B X B = nBnB n A + n B P A = X A P T P B = X B P T P i = X i P T mole fraction (X i ) = nini nTnT

34 Q. A sample of natural gas contains 8.24 moles of CH 4, moles of C 2 H 6, and moles of C 3 H 8. If the total pressure of the gases is 1.37 atm, what is the partial pressure of propane (C 3 H 8 )? P i = X i P T X propane = P T = 1.37 atm = P propane = X propane x 1.37 atm= atm

35 2KClO 3 (s) 2KCl (s) + 3O 2 (g) Bottle full of oxygen gas and water vapor P T = P O + P H O 22

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37 Chemistry in Action: Scuba Diving and the Gas Laws PV Depth (ft)Pressure (atm)

38 Air Bag Chemistry

39 Crash Sensor Air Bag Nitrogen Gas Inflator

40 Air Bag Chemistry On ignition: 2 NaN 3  2Na + 3N 2 Secondary reactions: 10 Na + 2 KNO 3  K 2 O + 5 Na 2 O + N 2 K 2 O + Na 2 O + SiO 2  K 2 Na 2 SiO 4

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(N 2,g) = g/L (273°C) d(N 2,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)

42 Collisions of Gas Particles

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44 The gas molecules in the container are in random motion

45 압력과 기체분자 운동 운동량 변화 ∆p = mv x –(mv x ) = 2mv x 분자의 왕복시간 ∆t = 2a/v x 분자 한 개가 면 A 에 미치는 힘 f = ∆p/ ∆t= 2mv x /(2a/v x )= mv x 2 /a N 개의 분자에 의해 면 A 에 가해지는 압력 P = Σf/a 2 = m(v x1 2 + v x v xN 2 )/a 3 =Nmv x 2 /V PV=Nmv x 2

46 압력과 기체분자 운동 공간 상에서 무작위로 움직이는 분자는 특별한 방향에 대한 선호도가 없을 것이므로 분자당 평균운동에너지 Gas Pressure in the Kinetic Theory

47 운동에너지와 온도 이상 기체방정식 기체에 열이 가해지면 기체의 내부에너지와 온도가 증가 ! Boltzmann constant Mean Kinetic Energy per particle

48 P = gas pressure, V=volume = mean square velocity N = number of gas molecules n = mole number of gas molecules = N/N A N A = Avogadro’s number m = mass of the gas molecule M = molar mass of the gas molecule 기체분자운동이론 (Kinetic Molecular Theory for Gases)

49 N = number of molecules, k = Boltzmann constant, J K -1, n = number of moles = N/N A, R = gas constant, l atm mol -1 K -1 T = temperature, P = gas pressure, V = volume, N A = Avogadro’s number PV = NkT = nRT 실험관찰에 의하면 (Ideal Gas Equation)

50 Maxwell-Boltzmann Distribution for Molecular Speeds f(u) = speed distribution function, m = molecular mass, k = Boltzmann constant, T = temperature, u = speed u~ u+du 사이의 속도를 가진 분자들의 수

51 분자속도와 에너지 분포 (Molecular velocity and energy distribution) 슈테른 (Otto Stern) 의 분자속도 측정 노벨물리학상 수상자 (1943, w/ Gerlach) Schematic diagram of a stern type experiment for determining the distribution of molecular velocities  v ~ v+∆v 사이의 분자개수 ∆N 측정가능 !!

52 Molecular speed Distribution of N 2 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 5.7 Chopper methodgravitation method

54 The distribution of speeds for nitrogen gas molecules at three different temperatures The distribution of speeds of three different gases at the same temperature

55 Chemistry in Action: Super Cold Atoms Gaseous Na Atoms 1.7 × K Bose-Einstein Condensate 1995, NIST Image

56 Gaseous Diffusion/Effusion Diffusion of Ammonia and HCl Effusion enrichment of UF 6

57 Gas diffusion is the gradual mixing of molecules of one gas with molecules of another by virtue of their kinetic properties. NH 3 17 g/mol HCl 36 g/mol NH 4 Cl

58 Kinetic-Molecular Theory for Gaseous Behavior Principal Issues ( 잠재적 문제점 ) –Negligible Volume and No interaction Hold only at low P, high T; for dilute gases –Elastic Collisions Only 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 gas PV = nRT n = PV RT = 1.0 Repulsive Forces Attractive Forces

60 Effect of intermolecular forces on the pressure exerted by a gas. Real Gas Has volume Has interaction(usually attractive)

61 Van der Waals equation nonideal gas P + (V – nb) = nRT an 2 V2V2 () } corrected pressure } corrected volume Gas a [(L 2 · atm)/mol 2 ] b [10 -2 L/mol] He Ne Ar Kr Xe H2H N2N O2O CO CH CCl C2H2C2H Cl C 4 H C 8 H Selected Values for a and b for the van der Waals Equation

62 Hydrogen Economy & Fuel Cell

63 Chlorine Destroys Ozone but is not consumed in the process Ozone Depletion

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65 Crutzen Molina Rowland Holland (The Netherlands) Max-Planck-Institute for Chemistry Mainz, Germany USA (Mexico) Department of Earth, Atmospheric and Planetary Sciences and Department of Chemistry, MIT Cambridge, MA, USA USA Department of Chemistry, University of California Irvine, CA, USA

66 Hindenberg Crew: 40 to 61 Capacity: passengers Length: 245 m, Diameter: 41 m Volume: 200,000 m³ Powerplant: 4 × Daimler-Benz diesel engines, 890 kW (1,200 hp) each 기체 : 수소 ( 헬륨으로 계획되었으나, 수소로 대체 ) 부력 :

67 Science of Hot Air Balloon 2008 National Balloon Classic in Indianola, Iowa. 부력 = 외부 공기의 무게 – 내부 기체의 무게 내부 기체의 무게 = 공기의 밀도 (ρ(T hot ))× 풍선의 부피 (V) ×g 외부 공기의 무게 = 공기의 밀도 (ρ(T cold ))× 풍선의 부피 (V) ×g 공기의 밀도 ρ(T) = w/V = Mp/RT 부피 (m 3 ) 600 ~ 2800 ~ 17,000 T cold ~ 20 ℃ T hot ~ 100 ℃ 부력 ~ 700kg

68 Lake Nyos, the killer Lake of Cameroon Clarke, T., Taming Africa’s killer lake, Nature, V. 409, p , February 2001.

69 Lake Nyos, the killer Lake Clarke, T., Taming Africa’s killer lake, Nature, V. 409, p , February Test of the de-gassing operation in 1995


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