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Chapter 8 Gases. The Gas Laws of Boyle, Charles and Avogadro The Ideal Gas Law Gas Stoichiometry Dalton’s Laws of Partial Pressure The Kinetic Molecular Theory of Gases Effusion and Diffusion Collisions of Gas Particles with the Container Walls Intermolecular Collisions Real Gases - PowerPoint PPT Presentation
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Chapter 8Gases
The Gas Laws of Boyle, Charles and Avogadro
The Ideal Gas Law
Gas Stoichiometry
Dalton’s Laws of Partial Pressure
The Kinetic Molecular Theory of Gases
Effusion and Diffusion
Collisions of Gas Particles with the Container Walls
Intermolecular Collisions
Real Gases
Chemistry in the Atmosphere
04/21/23 2
States of Matter
SolidLiquid Gas
We start with gases because they are simpler than the others.
Pressure (force/area, Pa=N/m2):
A pressure of 101.325 kPa is need to raise the column of Hg 76 cm (760 mm).
“standard pressure”
760 mm Hg = 760 torr = 1 atm = 101.325 kPa
V1 / V2 = T1 / T2 (fixed P,n)
P1V1 = P2V2 (fixed T,n)
Boyle’s Law
Charles’ Law
V x P = const
V / T = const
V / n = const (fixed P,T)Avogadro
1662
1787
1811 n = number of moles
Boyle’s Law: Pressure and Volume
The product of the pressure and volume, PV, of a sample of gas is a constant at a constant temperature:
PV = k = Constant
(fixed T,n)
Boyle’s Law: The Effect of Pressure on Gas Volume
Example
The cylinder of a bicycle pump has a volume of 1131 cm3 and is filled with air at a pressure of 1.02 atm. The outlet valve is sealed shut, and the pump handle is pushed down until the volume of the air is 517 cm3. The temperature of the air trapped inside does not change. Compute the pressure inside the pump.
Charles’ Law: T vs V
At constant pressure, the volume of a sample of gas is a linear function of its temperature.
V = bT
T(°C) =273°C[(V/Vo)]When V=0, T=-273°C
Charles’ Law: T vs V
The Absolute Temperature Scale
Kelvin temperature scale
T (Kelvin) = 273.15 + t (Celsius)
Gas volume is proportional to Temp
V = Vo ( 1 + ) t 273.15oC
Charles’ Law: The Effect of Temperature on Gas Volume
V1 / V2 = T1 / T2 (at a fixed pressure and for a fixed amount of gas)
V vs T
Avogadro’s law (1811)
V = an
n= number of moles of gas
a = proportionality constantFor a gas at constant temperature and pressure the volume
is directly proportional to the number of moles of gas.
V1 / V2 = T1 / T2 (at a fixed pressure)
P1V1 = P2V2 (at a fixed temperature)
Boyle’s Law
Charles’ Law
V = kP -1
V = bT
V = an (at a fixed pressure and temperature)Avogadro
V = nRTP-1
n = number of moles
PV = nRT
ideal gas law
an empirical law
Example
At some point during its ascent, a sealed weather balloon initially filled with helium at a fixed volume of 1.0 x 104 L at 1.00 atm and 30oC reaches an altitude at which the temperature is -10oC yet the volume is unchanged. Calculate the pressure at that altitude .
n1 = n2
V1 = V222
22
11
11
Tn
VP
Tn
VP=
2
2
1
1
T
P
T
P=
P2 = P1T2/T1 = (1 atm)(263K)/(303K)
STP (Standard Temperature and Pressure)
For 1 mole of a perfect gas at O°C (273K)
(i.e., 32.0 g of O2; 28.0 g N2; 2.02 g H2)
nRT = 22.4 L atm = PV
At 1 atm, V = 22.4 L
STP = standard temperature and pressure
= 273 K (0o C) and 1 atm
The Ideal Gas Law
What is R, universal gas constant?
the R is independent of the particular gas studied
11
11
K mol J 8.3145R
K mol m N 8.3145R−−
−−
=
=
PVR
nT= (1atm)(22.414L)
(1.00 mol)(273.15 K)=
-1-1 K mol atm L 0.082057 R =
(273.15K) mol) (1.00
)m 10 x (22.414 )m N 10 x (101.325 R
3-3-23
=
PV = nRT
PV = nRT
ideal gas law constants
11 K mol J 8.3145R −−=
-1-1 K mol atm L 0.082057 R =
Example
What mass of Hydrogen gas is needed to fill a weather balloon to a volume of 10,000 L, 1.00 atm and 30 8 C?
1) Use PV = nRT; n=PV/RT.
2) Find the number of moles.
3) Use the atomic weight to find the mass.
n = PV/RT =
(1 atm) (10,000 L) (293 K)-1 (0.082 L atm mol-1 K-1)-1
= 416 mol
(416 mol)(1.0 g mol-1) = 416 g
Example
What mass of Hydrogen gas is needed to fill a weather balloon to a volume of 10,000 L, 1.00 atm and 30 8 C?
Use volumes to determine stoichiometry.
Gas Stoichiometry
The volume of a gas is easier to measure than the mass.
Gas Density and Molar Mass
MRT
Pd
V
m
MRT
P
V
m
Rearrange
==
=RT
M
mPV
nRTPV
=
=
Gas Density and Molar Mass
Example
Calculate the density of gaseous hydrogen at a pressure of 1.32 atm and a temperature of -45oC.
Example
Fluorocarbons are compounds containing fluorine and carbon. A 45.6 g sample of a gaseous fluorocarbon contains 7.94 g of carbon and 37.7 g of fluorine and occupies 7.40 L at STP (P = 1.00 atm and T = 273 K). Determine the molecular weight of the fluorocarbon and give its molecular formula.
ExampleFluorocarbons are compounds of fluorine and carbon. A 45.60 g sample of a gaseous fluorocarbon contains 7.94 g of carbon and 37.66 g of fluorine and occupies 7.40 L at STP (P = 1.00 atm and T = 273.15 K). Determine the approximate molar mass of the fluorocarbon and give its molecular formula.
1
11
mol g 138M
1atm
273K x Kmol atm L 0.082
7.40L
45.60g
−
−−
=
⎟⎟⎠
⎞⎜⎜⎝
⎛⎟⎠
⎞⎜⎝
⎛
P
RTdM =
F mol 1.982F g 19
F 1mol x F g 37.66n
C mol 0.661C g 12
C 1mol x C g 7.94n
F
C
=⎟⎟⎠
⎞⎜⎜⎝
⎛=
=⎟⎟⎠
⎞⎜⎜⎝
⎛= Cpart 1mol 0.661 =÷
F parts 3mol 0.661 =÷
Vm
d =
Mixtures of Gases
Dalton’s Law of Partial Pressures
The total pressure of a mixture of gases equals the sum of the partial pressures of the individual gases.
( )
⎟⎠
⎞⎜⎝
⎛=
++=++=
===
V
RTn P
V
RTnnnPPP P
V
RTnP ,
V
RTnP ,
V
RTnP
Totaltotal
321321total
33
22
11
Mole Fractions and Partial PressuresThe mole fraction of a component in a mixture is define as the number of moles of the components that are in the mixture divided by the total number of moles present.
NBA
A
tot
AA
A
n...nn
n
n
nX
X A ofFraction Mole
+++==
=
tottot
AA
tot
A
tot
A
tot
A
tot
A
tottot
AA
Pn
nPor
n
n
P
Por
RTn
RTn
VP
VP
equations divide
RTnVP
RTnVP
===
==
totAA PXP =
Example
A solid hydrocarbon is burned in air in a closed container, producing a mixture of gases having a total pressure of 3.34 atm. Analysis of the mixture shows it to contain 0.340 g of water vapor, 0.792 g of carbon dioxide, 0.288 g of oxygen, 3.790 g of nitrogen, and no other gases. Calculate the mole fraction and partial pressure of carbon dioxide in this mixture.
atm 0.332atm 3.34 x 0.0995PXP
0.09950.1809
0.018
n
nX
0.180928
3.790
32
0.288
44
0.792
18
0.34n
n n n nn
totCOCO
tot
COCO2
tot
NOCOOHtot
22
2
2222
===
===
=+++=
+++=
2NH4ClO4 (s) → N2(g) + Cl2 (g) + 2O2 (g) + 4 H2 (g)
• The Ideal Gas Law is an empirical relationship based on experimental observations.– Boyle, Charles and Avogadro.
• Kinetic Molecular Theory is a simple model that attempts to explain the behavior of gases.
The Kinetic Molecular Theory of Gases
The Kinetic Molecular Theory of Gases1. A pure gas consists of a large number of identical molecules separated by distances that are large compared with their size. The volumes of the individual particles can be assumed to be negligible (zero).
2. The molecules of a gas are constantly moving in random directions with a distribution of speeds. The collisions of the particles with the walls of the container are the cause of the pressure exerted by the gas.
3. The molecules of a gas exert no forces on one another except during collisions, so that between collisions they move in straight lines with constant velocities. The gases are assumed to neither attract or repel each other. The collisions of the molecules with each other and with the walls of the container are elastic; no energy is lost during a collision.
4. The average kinetic energy of a collection of gas particles is assumed to be directly proportional to the Kelvin temperature of the gas.
Pressure (impulse per collision) x (frequency of collisions with the walls)
• impulse per collision
momentum (m × u)
• frequency of collisions
number of molecules per unit volume (N/V)
• frequency of collisions
speed of molecules (u)
P (m × u) × [(N/V) × u]
Pressure and Molecular Motion
P (m × u) × [(N/V) × u]
PV Nmu2
PV Nmu2
Correction: The molecules have a distribution of speeds.
Mean-square speed of all molecules = u2
Pressure and Molecular Motion
Pressure and Molecular Motion
PV Nmu2
1. 1/2m = kinetic energy (KEave) of one molecule.
2. KE is proportional to T (KEave= RT)3. Divide by 3 (3 dimensions)4. N = nNa (molecules = moles x molecules/mole)
u2
Make some substitutions
The Kinetic Molecular Theory of Gases
ave
ave
or
PV 2 = RT = (KE)3n
3(KE) = RT2
M
3RTu2 =
Speed Distribution
Temperature is a measure of the average kinetic energy of gas molecules.
Velocity Distributions
Distribution of Molecular Speeds
M
3RTu2 =
M
3RTuu 2
rms ==
M
2RTump =
M
8RTuavg
∂=
ump : uavg : urms = 1.000 : 1.128: 1.225
At a certain speed, the root-mean-square-speed of the molecules of hydrogen in a sample of gas is 1055 ms-1. Compute the root-mean square speed of molecules of oxygen at the same temperature.
Strategy
1. Find T for the H2 gas with a urms = 1055 ms-1
2. Find urms of O2 at the same temperature
( )2
2
2
3
Hrms Hu M
TR
=
2
2
2 2
2
2
2 22
2
2
Orms
O
2H H
Orms
O
2H HO
rmsO
2O 1rms
3RTu
M
u M3R
3Ru
M
u Mu
M
(1005) (2)u 264.8ms
32−
=
⎛ ⎞⎜ ⎟⎜ ⎟⎝ ⎠=
=
= =
H2 about 4 times velocity of O2
Example
Gaseous Diffusion and EffusionDiffusion: mixing of Gases
Effusion: rate of passage of a gas through a tiny orifice in a chamber.
ArmsBrms
A
B
B
A
uRate of Eff A
Rate of Eff B u
3RTM
3RTM
Menrichment factor
M
=
=
= =
M
3RTuu 2
rms ==
e.g., NH3 and HCl
Example
A gas mixture contains equal numbers of molecules of N2 and SF6. A small portion of it is passed through a gaseous diffusion apparatus. Calculate how many molecules of N2 are present in the product of gas for every 100 molecules of SF6.
B
A
Menrichment factor
M=
2
6
2
# molecules of N X2.28 = =
# molecules of SF6 100 molecules SF
X = 228 molecules of N
2
6
Effussion of N 32 (6 x 19)2.28
Effusion of SF 2 x 14
+= =
Real Gases
• Ideal Gas behavior is generally conditions of low pressure and high temperature
PV = nRT
PV = 1.0
nRT
Real Gases
• Kinetic Molecular Theory model– assumed no interactions between gas
particles and – no volume for the gas particles
• 1873 Johannes van der Waals– Correction for attractive forces in gases (and
liquids)– Correction for volume of the molecules
Pcorrected Vcorrected = nRT
The Person Behind the Science
Johannes van der Waals (1837-1923)
Highlights– 1873 first to realize the
necessity of taking into account the volumes of molecules and
– intermolecular forces (now generally called "van der Waals forces") in establishing the relationship between the pressure, volume and temperature of gases and liquids.
Moments in a Life– 1910 awarded Nobel Prize in
Physics ( )2( )obs
PV nRT
nP a V nRT nRT
V
=
⎡ ⎤+ − =⎢ ⎥⎣ ⎦
• Significant Figures– Zeros that follow the last non-zero digit sometimes are
counted.– E.g., for 700 g, the zeros may or not be significant.– They may present solely to position the decimal point – But also may be intended to convey the precision of the
measurement.– The uncertainty in the measurement is on the order of
+/- 1 g or +/- 10g or perhaps +/- 100 g– It is impossible to tell which without further information.
– If you need 2 sig figs and want to write “40” use either• Four zero decimal point “40.” or• “4.0 x 10+1”
“When you can measure what you are speaking about and express it in numbers, you know something about it; but when you cannot measure it, when you cannot express it in numbers, you knowledge is of a meager and unsatisfactory kind; it may be the beginning of knowledge but you have scarcely, in your thoughts advanced to the stage of science, whatever the matter may be.”
Lecture to the Institution
of Civil Engineers,
3 May 1883
The Person Behind the Science
Lord Kelvin (William Thomson) 1824-1907
The Person Behind the Science
Evangelista Torricelli (1608-1647)
Highlights– In 1641, moved to Florence to
assist the astronomer Galileo. – Designed first barometer– It was Galileo that suggested
Evangelista Torricelli use mercury in his vacuum experiments.
– Torricelli filled a four-foot long glass tube with mercury and inverted the tube into a dish.
Moments in a Life– Succeed Galileo as professor of
mathematics in the University of Pisa.
– Asteroid (7437) Torricelli named in his honor
Barometer
P = g x d x h
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