principles of chemistry i chem 1211 chapter 6
DESCRIPTION
PRINCIPLES OF CHEMISTRY I CHEM 1211 CHAPTER 6. DR. AUGUSTINE OFORI AGYEMAN Assistant professor of chemistry Department of natural sciences Clayton state university. CHAPTER 6 THE GASEOUS STATE. CHARACTERISTICS OF GASES. - Gases are composed of nonmetals - PowerPoint PPT PresentationTRANSCRIPT
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PRINCIPLES OF CHEMISTRY I
CHEM 1211
CHAPTER 6
DR. AUGUSTINE OFORI AGYEMANAssistant professor of chemistryDepartment of natural sciences
Clayton state university
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CHAPTER 6
THE GASEOUS STATE
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CHARACTERISTICS OF GASES
- Gases are composed of nonmetals
- Gases form homogeneous mixtures of each other
Air mixture of 78% N2
21% O2
0.9% Arother substances (CO2, H2, Ne, Kr, He)
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Gases at Ordinary Temperature and Pressure
Noble gases (monatomic gases)He, Ne, Ar, Kr, Xe
Diatomic GasesH2, N2, O2, F2, Cl2
Other Common Gasespropane, ammonia, carbon dioxide,
hydrogen sulfide, methane, carbon monoxide, sulfur dioxide
CHARACTERISTICS OF GASES
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PRESSURE
- Gases exert pressure on any surface they come in contact with
- Pressure is force applied per unit area
A
FP
A = area (m2)F = force (newton, N = kg-m/s2)P = pressure (N/m2 = pascal, Pa)
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F = m x a
m = mass (kg)
a = acceleration (m/s2)
PRESSURE
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THE GAS LAWS
Four Variables Define the Physical State of any GasAmount (mole)
Temperature (K)Volume (L)
Pressure (bar, Pa, mm Hg, torr, atm, psi)
1 bar = 105 Pa1 atm = 760 mmHg
= 760 torr = 1.01325 x 105 Pa
= 101.325 kPa = 14.7 psi
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mm Hg: millimeters mercuryatm: atmosphere (atmospheric pressure = 1atm)
Pa: Pascalpsi: pound per square inch (Ib/in2)
Pressure Instrumentsbarometers, manometers, gauges
760, 700, 650 mm Hg- Considered to have 3 significant figures
THE GAS LAWS
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BOYLE’S LAW
- The volume of a fixed amount of a gas is inversely proportional to the pressure applied to the gas
if the temperature is kept constant
PV = constant
P1V1 = P2V2
- P1 and V1 are the pressure and volume of a gas at an initial set of conditions
- P2 and V2 are the pressure and volume of the same gas under a new set of conditions
- The temperature and amount of gas remain constant
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A sample of N2 gas occupies a volume of 3.0 L at 6.0 atm pressure . What is the new pressure if the gas is
allowed to expand to 4.8 L at constant temperature?
P1 = 6.0 atm V1 = 3.0 LP2 = ? V2 = 4.8 L
P1V1 = P2V2
(6.0 atm)(3.0 L) = (P2)(4.8 L)
P2 = 3.8 atm
BOYLE’S LAW
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CHARLES’S LAW
- The volume of a fixed amount of gas is directly proportional to its absolute temperature
if the pressure is kept constant
- V1 and T1 are the volume and absolute temperature of a gas at an initial set of conditions
- V2 and T2 are the volume and absolute temperature of the same gas under a new set of conditions
- The pressure and amount of gas remain constant
constantT
V
2
2
1
1
T
V
T
V
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A sample of Ar gas occupies a volume of 1.2 L at 125 oCand a pressure of 1.0 atm. What is the new temperature, in Celsius, if the volume of the gas is decreased to 1.0 L at the
same pressure?
V1 = 1.2 L T1 = 125 oC = 398 K
V2 = 1.0 L T2 = ?
2T
L1.0
K398
L1.2
T2 = 332 K = 59 oC
CHARLES’S LAW
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AVOGADRO’S LAW
- The volume of a gas maintained at constant temperature and pressure is directly proportional to the
number of moles of the gas
n = number of moles of a gas
constantn
V
2
2
1
1
n
V
n
V
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AVOGADRO’S LAW
Avogadro’s Hypothesis
- Equal volumes of gases at the same temperature and pressure contain equal numbers of molecules
At Standard Temperature and Pressure (STP)1 mol of any gas (= 6.022 x 1023 molecules)
occupies a volume of 22.41 L
Conditions of STPTemperature = 0 oC = 273 K = 32 oF
Pressure = 1.00 atm (101.325 kPa or 100 kPa)
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THE IDEAL GAS LAW
PV = nRT
Considering all three gas laws
V α 1/P V α T V α n
P
nTαV
P
nTRV
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R is the ideal gas constant
= 0.08206 L-atm/mol.K
= 8.314 J/mol-K
= 8.314 m3-Pa/mol-K
= 1.987 cal/mol-K
= 62.36 L-torr/mol-K
THE IDEAL GAS LAW
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RELATING THE GAS LAWS
PV = nRT
If n is constant
constantT
PV
2
22
1
11
T
VP
T
VP
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A 1.00-L container is filled with 0.500 mole of CO gas at35.0 oC. Calculate the pressure, in atmospheres, exerted
by the gas in the container
PV = nRT
P = ? V = 1.00 L n = 0.500 molT = 35.0 oC = 308 K R = 0.08206 atm.L/mol.K
(P)(1.00 L) = (0.500 mol)(0.08206 atm.L/mol.K)(308 K)
P = 12.6 atm
RELATING THE GAS LAWS
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A balloon filled with helium initially has a volume of 1.00 x 106 L at 25 oC and a pressure of 752 mm Hg.Determine the volume of the balloon after a certain
time when it encounters a temperature of -33 oC and a pressure of 75.0 mm Hg
P1 = 752 mm Hg V1 = 1.00 x 106 L T1 = 25 oC = 298 K P2 = 75.0 mm Hg V2 = ? T2 = -33 oC = 240 K
L10x8.08K)Hg)(298mm(75.0
K)L)(24010xHg)(1.00mm(752
TP
TVPV 6
6
12
2112
RELATING THE GAS LAWS
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GAS DENSITIES AND MOLAR MASS
- Density (d) = mass/volume (m/V)
- Rearrange the ideal gas equation and multiply both sides by molar mass (M) to give
RT
MP
V
Mn
M x n = mass (m) and m/V = d
RT
MPd
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Calculate the density of carbon dioxide gas at 0.45 atm and 252 K
K)K)(252-L/mol-atm(0.08206
atm) 5g/mol)(0.4 (44.01d
Density (d) = 0.96 g/L
GAS DENSITIES AND MOLAR MASS
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Calculate the average molar mass of dry air if its density is 1.17 g/L at 21 oC and 740.0 torr
Molar mass = 29.0 g/mol
P
dRTM
torr)(740.0
K) K)(294-torr/mol-L g/L)(62.36 (1.17M
GAS DENSITIES AND MOLAR MASS
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REACTION STOICHIOMETRY
Consider the following reaction
4Al(s) + 3O2(g) → 2Al2O3(s)
Calculate the mass of aluminum that would react completely with 2.00 L of pure oxygen gas at STP
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1 mol O2 at STP = 22.4 L
REACTION STOICHIOMETRY
22
222 Omol0.0893
)O L (22.4
)O mol )(1O L (2.00O moles
Almol0.119)O mol (3
Al) mol )(4O mol (0.0893Al moles
2
2
Alg21.3Al) mol (1
Al) g Al)(26.98 mol (0.119Al mass
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- At the same conditionsvolumes of gases combine in the same proportions as
the coefficients of the chemical equation
Example3H2(g) + N2(g) → 2NH3(g)
This implies that
- 3 mol H2 reacts with 1 mol N2 to produce 2 mol NH3
- 3 L H2 reacts with 1 L N2 to produce 2 L NH3
REACTION STOICHIOMETRY
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DALTON’S LAW OF PARTIAL PRESSURES
- The total pressure exerted by a mixture of gases is the sum of the partial pressures of the individual gases present
- The partial pressure is the pressure that a gas in a mixture of gases would exert if it were present alone under the same
conditions
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- Ptotal is the total pressure of a gaseous mixture
- P1, P2, P3,…. are the partial pressures of the individual gases
- ntotal is the total number of moles of a gaseous mixture
- n1, n2, n3,….are the number of moles of the individual gases
V
RTn
V
RT......nnn......PPPP total321321total
DALTON’S LAW OF PARTIAL PRESSURES
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The total pressure by a mixture of He, Ne, and Ar gases is3.50 atm. Find the partial pressure of Ar if the partial pressures
of He and Ne are 0.50 atm and 0.75 atm, respectively
Ptotal = 3.50 atmP1 = 0.50 atmP2 = 0.75 atm
P3 = ?
Ptotal = P1 + P2 + P3
P3 = Ptotal - (P1 + P2) = 3.50 atm - (0.50 atm + 0.75 atm) = 2.25 atm
DALTON’S LAW OF PARTIAL PRESSURES
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For gas 1 in a mixture of gases with n1 molesMole fraction (x1) = n1/ntotal
total1totaltotal
11 PxP
n
nP
O2 is 21% of airMole fraction of O2 (x1) = 0.21
Total atmospheric pressure = 1 atmPartial pressure of O2 = P1 = (0.21)(1 atm) = 0.21 atm
= (0.21)(760 mm Hg) = 160 mm Hg
DALTON’S LAW OF PARTIAL PRESSURES
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EXPERIMENT TO COLLECT GAS OVER WATER
- Magnesium reacts with HCl to produce hydrogen gas according to the following equation
Mg(s) + 2HCl(aq) → MgCl2(aq) + H2(g)
- The H2 gas is collected in a gas collection tube initially filled with water and inverted in a water pan
- The volume of H2 gas is measured by raising or lowering the tube such that the water levels in the tube and pan are the same
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For this condition
The pressure inside the tube is equal to the atmospheric pressure outside
Ptotal = Pgas + Pwater
Ptotal = total atmospheric pressure
Pgas = pressure of gas in tube
Pwater = vapor pressure of water at experimental temperature
EXPERIMENT TO COLLECT GAS OVER WATER
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- Gases consist of large numbers of small molecules that are in continuous random motion
- Attractive and repulsive forces between gas molecules are negligible
- The combined volume of all the molecules of a gas is negligible relative to the total volume in which the
gas is contained
KINETIC MOLECULAR THEORY OF GASES
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- Collisions between molecules are perfectly elastic (the average kinetic energy of the colliding molecules does
not change at constant temperature)
- The average kinetic energy of the molecules is proportional to the absolute temperature (molecules of all gases have the
same average kinetic energy at a given temperature)
KINETIC MOLECULAR THEORY OF GASES
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AVERAGE KINETIC ENERGY
2mu2
1ε
ε = average kinetic energy
m = mass of an individual molecule
u = root-mean-square (rms) velocity at a given temperature
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M
3RTu
AVERAGE KINETIC ENERGY
Root-Mean-Square Velocity
- Square-root of the average of the squares
- The lower the molar mass of a gas the higher the u
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Diffusion - The spread of one substance throughout space or
throughout a another substance
Effusion - Escape of gas molecules through a tiny hole
into an evacuated space
GRAHAMS LAW OF EFFUSION
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GRAHAMS LAW OF EFFUSION
- r1 and r2 are the effusion rates of gases 1 and 2 respectively- M1 and M2 are the molar masses of gases 1 and 2 respectively
- The rate of effusion of a gas is inversely proportional to the square root of its molar mass
1
2
2
1
M
M
r
r
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- The rate of effusion of a gas is directly proportional to the rms velocity of the molecules
1
2
2
1
2
1
2
1
M
M
3RT/M
3RT/M
u
u
r
r
GRAHAMS LAW OF EFFUSION
- The time it takes for a gas to effuse is inversely proportional to the rate of effusion
1
2
1
2
2
1
M
M
t
t
r
r
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It took 4.5 minutes for 1.0 L helium to effuse through a porousbarrier. How long will it take for 1.0 L Cl2 gas to effuse under
identical conditions?
g/mol 4.00
g/mol 70.90
minutes 4.5
t 2
t2 = 19 minutes
GRAHAMS LAW OF EFFUSION
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REAL GASES
Mean Free Path
- The average distance traveled by a molecule between collisions
- Real Gases deviate from ideal gas behavior at high pressures
- Deviation is very small at low pressures (usually below 10 atm)
- Deviation increases with decreasing temperature
- Deviation is significant near the temperature at which a given gas changes to the liquid state
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- Molecules of real gases have finite volumes and attract one another
- van der Waals equation
nRTnb)(VV
anP
2
2
REAL GASES
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a = constant (measure of how strongly gas molecules attract each other
b = constant (measure of the small but finite volume occupied by gas molecules)
n2a/V2 accounts for the attractive forces and adjusts pressure upwards
nb accounts for the finite volume occupied by molecules
REAL GASES
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Calculate the pressure exerted by 0.2500 mol N2 gas in a 5.000 L container at 22.5 oC using both the ideal gas law
and the van der Waals equation. Compare the results.
PV = nRT
P = nRT/V
= (0.2500 mol)(0.08206 L-atm/mol-K)(295.5 K)/(5.000 L)
= 1.212 atm
REAL GASES
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nRTnb)(VV
anP
2
2
atm1.211P
)5.295)(08206.0)(2500.0(
)]0391.0)(2500.0([(5.000)(5.000)
(1.39)(0.2500)P
2
2
REAL GASES