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Gases Chapter 5.

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Presentation on theme: "Gases Chapter 5."— 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 250C and 1 atmosphere

4

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)
Area Barometer Pressure = (force = mass x acceleration) Units of Pressure 1 pascal (Pa) = 1 N/m2 1 atm = 760 mmHg = 760 torr 1 atm = 101,325 Pa bar = 105 Pa psi = lb/in2 = Pa

7 PRESSURE Units and Measurement
Pressure = Force/Area SI Units Force = mass x acceleration Force = kg-m/s2 = Newton Pressure = Newton/m2 = Pascal Customary Units Pressure = atm, torr, mmHg, bar, psi Relate SI to customary X 105 Pascal = 1 atm = 760 torr bar = 105 Pascal

8 10 miles 0.2 atm 4 miles 0.5 atm Sea level 1 atm

9 PRESSURE Mercury Barometer

10 Figure 5.4

11 As P (h) increases V decreases

12 Boyle’s Law

13 Boyle’s Law P ∝ 1/V Constant temperature P x V = constant
Constant amount of gas P1 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 V2 P1 = 726 mmHg P2 = ? V1 = 946 mL V2 = 154 mL P1 x V1 V2 726 mmHg x 946 mL 154 mL = P2 = = 4460 mmHg

15 As T increases V increases

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

17 Variation of gas volume with temperature at constant pressure.
Charles’ & Gay-Lussac’s Law V ∝ T Temperature must be in Kelvin V = constant x T V1/T1 = V2/T2 T (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/T2 V1 = 3.20 L V2 = 1.54 L T1 = K T2 = ? T1 = 125 (0C) (K) = K V2 x T1 V1 1.54 L x K 3.20 L = T2 = = 192 K

19 Avogadro’s Hypothesis
Equal volumes of gases contain the same number of molecules at constant T,P L of any gas contains X 1023 atoms (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 temperature Constant pressure V = constant x n V1/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 + 6H2O 1 mole NH3 → 1 mole NO At constant T and P 1 volume NH3 → 1 volume NO

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23

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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 ∝ nT P V = constant x = R nT P R is the gas constant PV = 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 = nRT R = PV nT = (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 = K P = 1 atm PV = nRT n = 49.8 g x 1 mol HCl 36.45 g HCl = 1.37 mol V = nRT P V = 1 atm 1.37 mol x x K L•atm mol•K V = 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 = nRT n, V and R are constant nR V = P T = constant P1 = 1.20 atm T1 = 291 K P2 = ? T2 = 358 K P1 T1 P2 T2 = P2 = P1 x T2 T1 = 1.20 atm x 358 K 291 K = 1.48 atm

29 d is the density of the gas in g/L
Density (d) Calculations m is the mass of the gas in g m V = PM RT d = 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 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? dRT P M = d = m V 4.65 g 2.10 L = = 2.21 g L 2.21 g L 1 atm x x K L•atm mol•K M = M = 54.6 g/mol

31 C6H12O6 (s) + 6O2 (g) 6CO2 (g) + 6H2O (l)
Gas Stoichiometry Q. 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 CO2 5.60 g C6H12O6 1 mol C6H12O6 180 g C6H12O6 x 6 mol CO2 = mol CO2 V = nRT P 0.187 mol x x K L•atm mol•K 1.00 atm = = 4.76 L

32 Dalton’s Law of Partial Pressures
V and T are constant P1 P2 Ptotal = 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 = nART V nA is the number of moles of A PB = nBRT V nB is the number of moles of B XA = nA nA + nB XB = nB nA + nB PT = PA + PB PA = XA PT PB = XB PT Pi = Xi PT mole fraction (Xi) = ni nT

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 PT PT = 1.37 atm 0.116 Xpropane = = 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 vapor 2KClO3 (s) KCl (s) + 3O2 (g) PT = PO + PH O 2

36

37 Scuba Diving and the Gas Laws
Chemistry in Action: Scuba Diving and the Gas Laws Depth (ft) Pressure (atm) 1 33 2 66 3 P V

38 Air Bag Chemistry

39 Air Bag Chemistry Crash Sensor Air Bag Nitrogen Gas Inflator

40 Air Bag Chemistry 2 NaN3 2Na + 3N2 On ignition: Secondary reactions:
10 Na + 2 KNO3 K2O + 5 Na2O + N2 K2O + 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)

42 Collisions of Gas Particles

43 Collisions of Gas Particles

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/a N개의 분자에 의해 면 A에 가해지는 압력 P = Σf/a2 = m(vx12 + vx vxN2)/a3 =Nmvx2/V PV=Nmvx2

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

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

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

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

50 Maxwell-Boltzmann Distribution for Molecular Speeds
u~ u+du 사이의 속도를 가진 분자들의 수 f(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 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 ∆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 method gravitation method 5.7

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

55 Chemistry in Action: Super Cold Atoms
Gaseous Na Atoms 1.7 × 10-7 K Bose-Einstein Condensate 1995, NIST Image

56 Gaseous Diffusion/Effusion
Diffusion of Ammonia and HCl Effusion 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. NH4Cl NH3 17 g/mol HCl 36 g/mol

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 Repulsive Forces PV = nRT n = PV RT = 1.0 Attractive Forces

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

61 Selected Values for a and b for the van der Waals Equation
Gas a [(L2 · atm)/mol2] b [10-2 L/mol] He 2.370 Ne 0.211 1.71 Ar 1.34 3.22 Kr 2.32 3.98 Xe 4.19 2.66 H2 0.2444 2.661 N2 1.390 3.913 O2 1.360 3.183 CO2 3.592 4.267 CH4 2.25 4.28 CCl4 20.4 13.8 C2H2 4.390 5.136 Cl2 6.493 5.622 C4H10 14.47 12.26 C8H18 37.32 23.68 Van der Waals equation nonideal gas P (V – nb) = nRT an2 V2 ( ) } corrected pressure } corrected volume

62 Hydrogen Economy & Fuel Cell

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

64

65 Molina Rowland Crutzen Holland (The Netherlands) USA
USA (Mexico) Department of Earth, Atmospheric and Planetary Sciences and Department of Chemistry, MIT Cambridge, MA, USA 1943 - Holland (The Netherlands) Max-Planck-Institute for Chemistry Mainz, Germany 1933 - USA Department of Chemistry, University of California Irvine, CA, USA 1927 -

66 Hindenberg Crew: 40 to 61 Capacity: 50-72 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
부력 = 외부 공기의 무게 – 내부 기체의 무게 내부 기체의 무게 = 공기의 밀도(ρ(Thot))× 풍선의 부피(V) ×g 외부 공기의 무게 = 공기의 밀도(ρ(Tcold))× 풍선의 부피(V) ×g 공기의 밀도 ρ(T) = w/V = Mp/RT 2008 National Balloon Classic in Indianola, Iowa. Tcold ~ 20 ℃ Thot ~ 100 ℃ 부력 ~ 700kg 부피(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 1995 Clarke, T., Taming Africa’s killer lake, Nature, V. 409, p , February 2001.


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