Download CHEM1310 Lecture - School of Chemistry and Biochemistry

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
Chemistry in the Atmosphere
States of Matter
Solid
Liquid
Gas
We start with gases because they are simpler
than the others.
5/24/2017
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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
Boyle’s Law
P1V1 = P2V2
V x P = const
(fixed T,n)
1662
Charles’ Law
V / T = const
(fixed P,n)
1787
Avogadro
1811
V1 / V2 = T1 / T2
V / n = const
n = number of moles
(fixed P,T)
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,
273°C
T=-
Charles’ Law: T vs V
The Absolute Temperature Scale
V = Vo
(
t
1+
273.15oC
)
Kelvin temperature scale
T (Kelvin) = 273.15 + t (Celsius)
Gas volume is proportional to Temp
Charles’ Law: The Effect of Temperature on
Gas Volume
V vs T
V1 / V2 = T1 / T2
(at a fixed pressure and for a fixed amount of gas)
Avogadro’s law (1811)
V = an
n= number of moles of gas
a = proportionality constant
For a gas at constant temperature and pressure the volume
is directly proportional to the number of moles of gas.
Boyle’s Law
V = kP
Charles’ Law
-1
V = bT
P1V1 = P2V2
(at a fixed temperature)
V1 / V2 = T1 / T2
(at a fixed pressure)
Avogadro
V = an
(at a fixed pressure
and temperature)
n = number of moles
V=
-1
nRTP
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 .
P1V1 P2V2

n1T1 n2T2
n1 = n2
V1 = V2
P1
P2

T1
T2
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
PV = nRT
What is R, universal gas constant?
the R is independent of the particular gas studied
PV
(1atm)(22.414L)

R
nT (1.00 mol)(273.15 K)
R  0.082057 L atm mol K
-1
-1
(101.325 x 103 N m-2 ) (22.414 x 10-3 m3 )
R
(1.00 mol) (273.15K)
1
R  8.3145 N m mol K
R  8.3145 J mol 1 K 1
1
PV = nRT
ideal gas law constants
R  0.082057 L atm mol K
-1
1
R  8.3145 J mol K
1
-1
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 ̊ C?
1) Use PV = nRT; n=PV/RT.
2) Find the number of moles.
3) Use the atomic weight to find the mass.
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 ̊ C?
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
Gas Stoichiometry
Use volumes to determine stoichiometry.
The volume of a gas is easier to measure than
the mass.
Gas Density and Molar Mass
PV  nRT
m
PV  RT
M
Rearrange
m
P

M
V RT
m
P
d
M
V
RT
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.
Example
Fluorocarbons 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.
RT
M d
P
m
d
V
1 1
 45.60g  0.082 L atm mol K x 273K 



1atm
 7.40L 

M  138 g mol 1
 1mol C 
  0.661 mol C  0.661 mol  1 part C
n C  7.94 g C x 
 12 g C 
 1mol F 
  1.982 mol F  0.661 mol  3 parts F
n F  37.66 g F x 
 19 g F 
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.
n 3 RT
n1RT
n 2 RT
P1 
, P2 
, P3 
V
V
V
RT
Ptotal  P1  P2  P3  n1  n 2  n 3 
V
 RT 
Ptotal  n Total 

 V 
Mole Fractions and Partial Pressures
The 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.
Mole Fraction of A  X A
nA
nA
XA 

n tot n A  n B  ...  n N
PA V  n A RT
Ptot V  n tot RT
PA  X A Ptot
divide equations
PA V n A RT
PA
nA
nA

or

or PA 
Ptot
Ptot V n tot RT
Ptot n tot
n tot
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.
n tot  n H 2O  n CO 2  n O 2  n N 2
0.34 0.792 0.288 3.790
n tot 



 0.1809
18
44
32
28
n CO 2
0.018
X CO2 

 0.0995
n tot 0.1809
PCO 2  X CO 2 Ptot  0.0995 x 3.34 atm  0.332 atm
2NH4ClO4 (s) → N2(g) + Cl2 (g) + 2O2 (g) + 4 H2 (g)
The Kinetic Molecular Theory of Gases
• 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
1. 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 and Molecular Motion
Pressure  (impulse per collision) x
(frequency of collisions with the
walls)
• frequency of collisions
 speed of molecules (u)
• impulse per collision
 momentum (m × u)
• frequency of collisions
 number of molecules per unit
volume (N/V)
P  (m × u) × [(N/V) × u]
Pressure and Molecular Motion
P  (m × u) × [(N/V) ×
u]
PV  Nmu2
Correction: The molecules have a distribution of speeds.
PV  Nmu2
Mean-square speed of all molecules =
u2
Pressure and Molecular Motion
PV  Nmu2
Make some substitutions
1. 1/2m u2 = 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)
The Kinetic Molecular Theory of Gases
PV
2
= RT = (KE)ave
3
n
or
3
(KE)ave = RT
2
3RT
u 
M
2
Speed Distribution
Temperature is a measure of the average kinetic
energy of gas molecules.
Velocity Distributions
Distribution of Molecular Speeds
3RT
u 
M
2
u rms
3RT
 u 
M
2
8RT
uavg 
πM
ump 
2RT
M
ump : uavg : urms = 1.000 : 1.128: 1.225
Example
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
O2
u
rms 
= 1055 ms-1
u  M

T
H2
rms
2
H2
3R
2. Find urms of O2 at the same
temperature
u Orms2
 u 2H2 M H2
3R 
 3R


M O2
u Orms2 
H2 about 4 times velocity of O2
u
O2
rms
3RT
M O2



u 2H2 M H2
M O2
(1005) 2 (2)

 264.8ms 1
32
Gaseous Diffusion and Effusion
Diffusion: mixing of
Gases
e.g., NH3 and HCl
Effusion: rate of passage of a gas through a tiny orifice in a
chamber.
3RT
Rate of Eff A u Arms
 B
Rate of Eff B u rms


3RT
MA
3RT
MB
MB
 enrichment factor
MA
u rms  u 2 
M
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.
MB
enrichment factor 
MA
Effussion of N 2
32  (6 x 19)

 2.28
Effusion of SF6
2 x 14
# molecules of N 2
X
2.28 =
=
# molecules of SF6 100 molecules SF6
X = 228 molecules of N 2
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
PV  nRT
n 2

 Pobs  a(V )  V  nRT   nRT


• 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”
The Person Behind the Science
Lord Kelvin (William Thomson) 1824-1907
“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
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=gxd xh