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