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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
Elements that exist as gases at 250C and 1 atmosphere
• 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
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 = 6894.757 Pa Barometer
Pressure = ForceArea
(force = mass x acceleration)
Pressure = Force/AreaSI Units
Force = mass x accelerationForce = kg-m/s2 = Newton Pressure = Newton/m2 = Pascal
Customary UnitsPressure = atm, torr, mmHg, bar, psiRelate SI to customary1.01325 X 105 Pascal = 1 atm = 760 torr
bar = 105 Pascal
PRESSURE Units and Measurement
Sea level 1 atm
4 miles 0.5 atm
10 miles 0.2 atm
PRESSUREMercury Barometer
Figure 5.4
As P (h) increases V decreases
Boyle’s Law
P ∝ 1/V
P x V = constant
P1 x V1 = P2 x V2
Boyle’s Law
Constant temperatureConstant amount of gas
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
V1 = 946 mL
P2 = ?
V2 = 154 mL
P2 = P1 x V1
V2
726 mmHg x 946 mL154 mL
= = 4460 mmHg
As T increases V increases
Charles’ LawDefinition of Temperature
V = V0 + at
V = aT
0 273.15Temperature(K)
Variation of gas volume with temperatureat constant pressure.
V ∝ T
V = constant x T
V1/T1 = V2/T2 T (K) = t (0C) + 273.15
Charles’ & Gay-Lussac’s
Law
Temperature must bein Kelvin
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 = 3.20 L
T1 = 398.15 K
V2 = 1.54 L
T2 = ?
T2 = V2 x T1
V1
1.54 L x 398.15 K3.20 L
= = 192 K
V1/T1 = V2/T2
T1 = 125 (0C) + 273.15 (K) = 398.15 K
Equal volumes of gases contain the same number of molecules at constant T,P
22.414 L of any gas contains 6.022 X 1023 atoms (or molecules) at 0 ℃ and 1 atm.
Avogadro’s Hypothesis
Avogadro’s Law
V number of moles (n)
V = constant x n
V1/n1 = V2/n2
Constant temperatureConstant pressure
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
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)1P
V ∝ nT
P
V = constant x = RnT
P
nT
PR is the gas constant
PV = nRT
The conditions 0 0C and 1 atm are called standard temperature and pressure (STP).
PV = nRT
R = PVnT
=(1 atm)(22.414L)
(1 mol)(273.15 K)
R = 0.082057 L • atm / (mol • K) = 8.3145 J / (mol • K)
Experiments show that at STP, 1 mole of an ideal gas occupies 22.414 L.
Q. What is the volume (in liters) occupied by 49.8 g of HCl at STP?
PV = nRT
V = nRTP
T = 0 0C = 273.15 K
P = 1 atm
n = 49.8 g x 1 mol HCl36.45 g HCl
= 1.37 mol
V =1 atm
1.37 mol x 0.0821 x 273.15 KL•atmmol•K
V = 30.6 L
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
nRV
= PT
= constant
P1
T1
P2
T2
=
P1 = 1.20 atm
T1 = 291 K
P2 = ?
T2 = 358 K
P2 = P1 x T2
T1
= 1.20 atm x 358 K291 K
= 1.48 atm
Density (d) Calculations
d = mV =
PMRT
m is the mass of the gas in g
M is the molar mass of the gas
Molar Mass (M ) of a Gaseous Substance
dRTP
M = d is the density of the gas in g/L
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?
dRTP
M = d = mV
4.65 g2.10 L
= = 2.21 g
L
M =2.21
g
L
1 atm
x 0.0821 x 300.15 KL•atmmol•K
M = 54.6 g/mol
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) 6CO2 (g) + 6H2O (l)
g C6H12O6 mol C6H12O6 mol CO2 V CO2
5.60 g C6H12O6
1 mol C6H12O6
180 g C6H12O6
x6 mol CO2
1 mol C6H12O6
x = 0.187 mol CO2
V = nRT
P
0.187 mol x 0.0821 x 310.15 KL•atmmol•K
1.00 atm= = 4.76 L
Dalton’s Law of Partial Pressures
V and T are
constant
P1 P2 Ptotal = P1 + P2
Consider a case in which two gases, A and B, are in a container of volume V.
PA = nART
V
PB = nBRT
V
nA is the number of moles of A
nB is the number of moles of B
PT = PA + PB XA = nA
nA + nB
XB = nB
nA + nB
PA = XA PT PB = XB PT
Pi = Xi PT mole fraction (Xi) = ni
nT
Q. A sample of natural gas contains 8.24 moles of CH4, 0.421 moles of C2H6, and 0.116 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
Xpropane = 0.116
8.24 + 0.421 + 0.116
PT = 1.37 atm
= 0.0132
Ppropane = Xpropane x 1.37 atm = 0.0181 atm
2KClO3 (s) 2KCl (s) + 3O2 (g)
Bottle full of oxygen gas and water vapor
PT = PO + PH O2 2
Chemistry in Action:
Scuba Diving and the Gas Laws
P V
Depth (ft) Pressure (atm)
0 1
33 2
66 3
Air Bag Chemistry
Air Bag Chemistry
Crash Sensor
Air Bag
Nitrogen Gas
Inflator
Air Bag Chemistry
On ignition:
2 NaN3 2Na + 3N2
Secondary reactions: 10 Na + 2 KNO3 K2O + 5 Na2O + N2
K2O + Na2O + SiO2 K2Na2SiO4
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) = 0.00125 g/L (273°C)d(N2,liq) = 0.808 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)
Collisions of Gas Particles
Collisions of Gas Particles
The gas molecules in the container are in random motion
압력과 기체분자 운동
• 운동량 변화 ∆ p = mvx –(mvx) = 2mvx
• 분자의 왕복시간 ∆ t = 2a/vx
• 분자 한 개가 면 A 에 미치는 힘 f = ∆p/ ∆t= 2mvx/(2a/vx)= mvx
2/a
• N 개의 분자에 의해 면 A 에 가해지는 압력• P = Σf/a2 = m(vx1
2 + vx22 + ... + vxN
2)/a3
=Nmvx2/V
PV=Nmvx2
압력과 기체분자 운동• 공간 상에서 무작위로 움직이는 분자는 특별한 방향에 대한
선호도가 없을 것이므로
분자당 평균운동에너지
Gas Pressure in the Kinetic Theory
)2
1(
3
2
3
13
1
3
222
22
22222222
vmNvNmvNmPV
vv
vvvvvvvv
x
x
xxxxzyx
2
2
1vmKE
운동에너지와 온도
• 이상 기체방정식
• 기체에 열이 가해지면 기체의 내부에너지와 온도가 증가 !
2 21 2
3
1( )
3 2PV Nmv N mv
Boltzmann constant
Mean Kinetic Energy per particle
nRTPV
J/K1038.1
2
3
2
1
23
2
AB
B
N
Rk
TkvmKE
P = gas pressure, V=volume = mean square velocity
N = number of gas moleculesn = mole number of gas molecules = N/NA
NA = Avogadro’s numberm = mass of the gas molecule
M = molar mass of the gas molecule
2v
기체분자운동이론(Kinetic Molecular Theory for Gases)
22
3
1
3
1vnMvNmPV
N = number of molecules, k = Boltzmann constant, J K-1,n = number of moles = N/NA,
R = gas constant, 0.08204 l atm mol-1 K-1
T = temperature, P = gas pressure,
V = volume,NA = Avogadro’s number
PV = NkT = nRT
실험관찰에 의하면 (Ideal Gas Equation)
M
RT
m
Tkvu
m
Tkv
Brms
B
33
3
2
2
Maxwell-Boltzmann Distribution for Molecular Speeds
f(u) = speed distribution function, m = molecular mass,k = Boltzmann constant,T = temperature, u = speed
)2
exp(42
22
23
kT
muπu
πkT
mf(u)
/
u~ u+du 사이의 속도를 가진 분자들의 수
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 distributionof molecular velocities v ~ v+∆v 사이의 분자개수 ∆ N 측정가능 !!
http://leifi.physik.uni-muenchen.de/web_ph12/originalarbeiten/stern/molekularstr.htm
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.
M
RTu
M
RTu
M
RTu
rms
mp
3
8
2
Apparatus for studying molecular speed distribution
5.7
Chopper method gravitation method
The distribution of speedsfor nitrogen gas molecules
at three different temperatures
The distribution of speedsof three different gases
at the same temperature
M
RTurms
3
Chemistry in Action: Super Cold Atoms
Gaseous Na Atoms1.7 × 10-7 K
Bose-Einstein Condensate1995, NIST Image
Gaseous Diffusion/Effusion
Diffusion of Ammonia and HCl
Effusion enrichment of UF6
Gas diffusion is the gradual mixing of molecules of one gas with molecules of another by virtue of their kinetic properties.
NH3
17 g/molHCl
36 g/mol
NH4Cl
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
Deviations from Ideal Behavior
1 mole of ideal gas
PV = nRT
n = PVRT
= 1.0
Repulsive Forces
Attractive Forces
Effect of intermolecular forces on the pressure exerted by a gas.
Real Gas
Has volumeHas interaction(usually attractive)
Van der Waals equationnonideal gas
P + (V – nb) = nRTan2
V2( )}
correctedpressure
}
correctedvolume
Gasa [(L2 · atm)/mol2]
b [10-2
L/mol]
He 0.03412 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
Selected Values for a and b for the van der Waals Equation
Hydrogen Economy & Fuel Cell
Chlorine Destroys Ozone
but is not consumed in the process
Ozone Depletion
Crutzen Molina RowlandHolland (The Netherlands)
Max-Planck-Institute for Chemistry
Mainz, Germany
1933 -
USA (Mexico)
Department of Earth,Atmospheric
and Planetary Sciences and
Department of Chemistry,MIT
Cambridge, MA, USA1943 -
USA
Department of Chemistry, University of California
Irvine, CA, USA
1927 -
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 :
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(Thot))× 풍선의 부피 (V) ×g외부 공기의 무게 = 공기의 밀도 (ρ(Tcold))× 풍선의 부피 (V) ×g
공기의 밀도 ρ(T) = w/V = Mp/RT
부피 (m3)600 ~ 2800 ~ 17,000
Tcold ~ 20 ℃Thot ~ 100 ℃부력 ~ 700kg
Lake Nyos, the killer Lake of Cameroon
Clarke, T., Taming Africa’s killer lake, Nature, V. 409, p. 554-555, February 2001.
Lake Nyos, the killer Lake
Clarke, T., Taming Africa’s killer lake, Nature, V. 409, p. 554-555, February 2001.
Test of the de-gassing operation in 1995