Unformatted text preview: Chapter 8 Objectives
8.1. Explain how gas molecules differ from liquid and solid molecules.
molecules.
8.2. Outline the basic relationships between pressure, volume,
temperature, and the number of moles observed with gases.
gases.
8.3. Include a discussion of how the relationships in 8.2 above can be
can
applied experimentally to identify unknown gases using density
and molecular mass.
8.4. Provide an overview of the Kinetic Theory of Gases and include a
brief discussion of average kinetic energy and molecular speed.
molecular
8.5. Differentiate between the processes of diffusion and effusion.
8.5.
effusion.
Introduce Graham’s Law.
Graham’ Law. 8/12/2009 Zumdahl Chapter 5 1 The Person Behind the Science Evangelista Torricelli (16081647)
Highlights – In 1641, moved to Florence to assist the
astronomer Galileo
– Designed first barometer Barometer
P = – It was Galileo who suggested that
Evangelista Torricelli use mercury in his
vacuum experiments
– Torricelli filled a fourfoot long glass
fourtube with mercury and inverted the tube
into a dish
Moments in a Life
– Succeeded Galileo as professor of
mathematics at the University of Pisa – Asteroid (7437) Torricelli named in
his honor
8/12/2009 Zumdahl Chapter 5 2 1 Units of
Pressure Pascal (Pa) 8/12/2009 Zumdahl Chapter 5 Boyle’s Law
1662 Charles’ Law
1787 V∝P 1 V∝T 3 P1V1 = P2V2
at a fixed T V1 / V2 = T1 / T2
at a fixed P Avogadro V∝n 1811 n = no. of moles V ∝ n T P1 at a fixed T & P Proportionality constant found
to be independent of gas identity Zumdahl Chapter 5
4
The above relationships work well up to pressures above 1 atm 8/12/2009 2 Boyle’s Experiments The product of the pressure and volume,
PV, of a sample of gas is a constant at a
PV,
constant temperature: at fixed T and n
8/12/2009 Zumdahl Chapter 5 5 Sample Problem Using Boyle’s Law
The long 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
is
3. The temperature of the air trapped inside does not change.
cm
Compute the pressure inside the pump. (at fixed T & n) (1.02 atm)(1131 cm3) = P2(517 cm3)
8/12/2009 Zumdahl Chapter 5 6 3 Charles’ Law: The Effect of Temperature on Gas Volume
Charles’ For all gases, Charles’ data extrapolate
gases, Charles’
to −273.15 oC in the limit where V → 0
(Fundamental limit for T)
T) (a)
(b) 8/12/2009 Zumdahl Chapter 5 Pa = Pb
na > n b 7 The Absolute (Kelvin) Temperature Scale
According to Charles’ Data:
Charles’ where α = (273.15 oC)1 independent of gas
identity = coefficient of thermal expansion
and V0 ≡ volume when t = 0.
The Kelvin (K) temperature scale: T(K) ≡ 273.15 + t(C)
Gas volume is proportional to T(K):
(fixed P & n)
8/12/2009 Zumdahl Chapter 5 8 4 The Ideal Gas Equation of State
From the work of Boyle, Charles, and Avogadro, we know that V
is proportional to nT/P and that the proportionality constant is
nT/
universal, i.e., the same for all gases:
universal, PV = nRT (R is the universal ideal gas constant) R = 8/12/2009 Zumdahl Chapter 5 9 Sample Problem
At one point during its ascent, a weather balloon filled with
helium at a volume of 1.0 × 104 L at 1.00 atm and 30oC reaches an
altitude z at which the temperature is −10 oC, yet the balloon
volume is unchanged. What is the pressure at altitude z? n1 = n2 PV1 P2V2
1
=
n1T1 n2T2
1.00
atm
30 ° C = V1 = V2 x atm
− 10 ° C ▬► 1.00 atm ▬►
303.15K = x atm
263.15K x = 0.868 atm
8/12/2009 Zumdahl Chapter 5 10 5 What is the Value for R?
One mole of an ideal gas at T = 0 oC and P = 1atm is found to
occupy a volume of 22.414 L. R= (1atm)(22.414L)
PV
=
= 0.08206 L atm mol1 K1
(1.00 mol)(273.15 K)
nT (101.325x 103 N m2 ) (22.414 x 103 m3 )
R=
= 8.3145 J mol1 K1
(1.00 mol) (273.15K) 8/12/2009 Zumdahl Chapter 5 11 Physical Properties of an Ideal Gas
In an ideal gas, the molecules are far enough apart
(on average) that
1. The potential energy of interaction between gas
molecules is negligibly small.
2. The volume occupied by the gas molecules is a
negligible fraction of the total volume. 8/12/2009 Zumdahl Chapter 5 12 6 Sample Problem
What mass of hydrogen gas is needed to fill a weather balloon
to a volume of 10,000. L at 1.00 atm and 30.0 °C? Strategy
1. Use PV = nRT
2. Find the number of moles
3. Convert moles to mass using moles = mass
divided by molar mass 8/12/2009 Zumdahl Chapter 5 13 Sample Problem (con’t)
What mass of hydrogen gas is needed to fill a weather balloon
to a volume of 10,000. L at 1.00 atm and 30.0 °C? n= PV
RT (1.00 atm) (1x10 4 L)
(0.08206 Latmmol 1 K 1 )( 3 03.15K) n= m
⇒ m = nM
M 8/12/2009 n = 402 moles m grams H2 = 402 mol x 2.015 g H2
1 mol H2 = 810 g H2
Zumdahl Chapter 5 14 7 Deviations from Ideal Gas Behavior
One mole of an ideal gas at T = 0 oC and P = 1atm
is found to occupy a volume of 22.414 L. 8/12/2009 Zumdahl Chapter 5 15 Gas Density and Molar Mass
Rearrange PV = nRT
PV = 8/12/2009 m
RT
M → Zumdahl Chapter 5 m
P
M
=
V RT
m
P
=d=
M
V
RT 16 8 Sample Problem
Calculate the density of gaseous hydrogen at a pressure of
1.32 atm and a temperature of 45oC. 1.32 atm
2 g H2
x
−1
−1
(0.082 L atm mol K )(233.15K) 1mol H 2 8/12/2009 = d = 0.142 g/L Zumdahl Chapter 5 17 Sample Problem
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
fluorine
occupies 7.40 L at STP (P = 1.00 atm and T = 273.15 K). Determine the
(P
approximate molar mass of the fluorocarbon and give its molecular formula.
molecular formula.
⎛ 1mol C ⎞
⎜
⎟
n C = 7.94 g C x ⎜
⎟ = 0.661mol C
⎝ 12 g C ⎠
⎛ 1mol F ⎞
n F = 37.66 g F x ⎜
⎜ 19 g F ⎟ = 1.982 mol F
⎟
⎝
⎠ M =d } Empirical Formula is m
RT where
d=
P
V −1 −1
⎛ 45.60g ⎞⎛ 0.082 L atm mol K x 273K ⎞ ~
⎜
⎟
⎟⎜
M =⎜
⎟
1atm
⎝ 7.40L ⎠⎝
⎠ 140 g/mol ~ 2 MCF3 Therefore, molecular formula is
8/12/2009 Zumdahl Chapter 5 18 9 Mixtures of Gases
Dalton’s Law of Partial Pressures
The pressure exerted by a mixture of ideal gases is the sum of the
the
pressures that each one would exert if it occupied the container
alone. P1 = n RT
n RT
n1RT
, P3 = 3
, P2 = 2
V
V
V Ptotal = P1 + P2 + P3 = (n1 + n 2 + n 3 ) RT
V ⎛ RT ⎞
Ptotal = n Total ⎜
⎟
⎝ V ⎠
8/12/2009 Zumdahl Chapter 5 19 Mole Fractions and Partial Pressures
The mole fraction of a component in a mixture is defined as the
number of moles of the component divided by the total number of
moles present. Mole Fraction of A = X A
XA = nA
nA
=
n tot n A + n B + ... + n N For ideal gases:
PA V = n A RT
Ptot V = n tot RT
divide equations
PA V n A RT
P
n
n
=
or A = A or PA = A Ptot
Ptot V n tot RT
Ptot n tot
n tot
8/12/2009 Zumdahl Chapter 5 →
20 10 Sample Problem
A solid hydrocarbon is burned in air in a closed container, producing a mixture
producing
of gases having a total pressure of 3.34 atm. Analysis of the mixture shows it to
mixture
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
fraction
pressure of carbon dioxide in this mixture. n tot = n H 2 O + n CO 2 + n O 2 + n N 2
n tot =
X CO2 0.34 0.792 0.288 3.790
+
+
+
= 0.1809
18
44
32
28
n CO 2
0.018
=
=
= 0.0995
n tot
0.1809 PCO 2 = X CO 2 Ptot = 0.0995 x 3.34 atm = 0.332 atm 8/12/2009 Zumdahl Chapter 5 21 Mixtures of Gases
Dalton’s Law of Partial Pressures
The pressure exerted by a mixture of ideal gases is the sum of the
the
pressures that each one would exert if it occupied the container
alone. P1 = n RT
n RT
n1RT
, P3 = 3
, P2 = 2
V
V
V Ptotal = P1 + P2 + P3 = (n1 + n 2 + n 3 ) RT
V ⎛ RT ⎞
Ptotal = n Total ⎜
⎟
⎝ V ⎠
8/12/2009 Zumdahl Chapter 5 22 11 Mole Fractions and Partial Pressures
The mole fraction of a component in a mixture is defined as the
number of moles of the component divided by the total number of
moles present. Mole Fraction of A = X A
XA = nA
nA
=
n tot n A + n B + ... + n N For ideal gases:
PA V = n A RT
Ptot V = n tot RT
divide equations
PA V n A RT
P
n
n
=
or A = A or PA = A Ptot
Ptot V n tot RT
Ptot n tot
n tot
8/12/2009 → Zumdahl Chapter 5 23 Sample Problem
A solid hydrocarbon is burned in air in a closed container, producing a mixture
producing
of gases having a total pressure of 3.34 atm. Analysis of the mixture shows it to
mixture
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
fraction
pressure of carbon dioxide in this mixture. n tot = n H 2 O + n CO 2 + n O 2 + n N 2
n tot =
X CO2 0.34 0.792 0.288 3.790
+
+
+
= 0.1809
18
44
32
28
n CO 2
0.018
=
=
= 0.0995
n tot
0.1809 PCO 2 = X CO 2 Ptot = 0.0995 x 3.34 atm = 0.332 atm 8/12/2009 Zumdahl Chapter 5 24 12 The Kinetic Molecular Theory of Gases
The Ideal Gas Law is an empirical relationship based on
experimental observations (Boyle, Charles, & Avogadro).
The Kinetic Molecular Theory is a simple theoretical model
that attempts to explain the behavior of gases. 8/12/2009 Zumdahl Chapter 5 25 The Kinetic Molecular Theory of Gases
1. The molecules of an ideal gas are constantly moving in random directions
with a distribution of speeds. Collisions of the particles with the walls of the
speeds.
container are the cause of the pressure exerted by the gas. The gas molecules
occupy a negligibly small fraction of the total volume.
2. 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
collisions,
velocities. The gases are assumed to neither attract or repel each other. The
velocities.
The
collisions of the molecules with each other and with the walls of the container
of
are elastic, i.e., no energy is lost during a collision. Essentially, the gas
elastic,
molecules behave like tiny billiard balls undergoing ceaseless random motion.
random motion.
3. The average kinetic energy of a collection of gas particles is found to be
directly proportional to the temperature of the gas in degrees Kelvin.
Kelvin. See pages 351 − 357 in text for derivation of key results
8/12/2009 Zumdahl Chapter 5 26 13 Pressure ∝ (impulse per collision) x (rate of collisions with walls) impulse per collision ∝
rate of collisions with walls ∝ rate of collisions with walls ∝ =>
8/12/2009 Zumdahl Chapter 5 P ∝ (m × u) × (N/V) × u 27 => But, the molecules of a gas are
But,
constantly moving in random
directions with a distribution of speeds.
speeds. u2 ≡
8/12/2009 Zumdahl Chapter 5 Meansquare speed
Meanof all molecules
28 14 Temperature and Molecular Motion
As shown in the text derivation, the proportionality constant
―
―
―
relating PV to Nmu2 is 1/3, i.e., PV = Nmu2/3 = nNAmu2/3
Therefore, PV/n = RT = = ideal gas law molar
kinetic
energy 8/12/2009 Zumdahl Chapter 5 29 The Meaning of Temperature u 2 = 3RT
M The temperature of a gas in degrees K is a measure
of the random motions of the gas atoms.
Higher T => greater random motion
8/12/2009 Zumdahl Chapter 5 30 15 MaxwellBoltzmann Distribution of Speeds
N2 f(u) = 4π (m/2πkBT)3/2 u2 exp(−mu2/2kBT)
f(u)
4π (m/2π
exp(−
kB ≡ R / NA = 1.38 x 10−23 J K−1 molecule−1 Temperature is a measure of the average kinetic energy of
molecules when their speeds have a MaxwellBoltzmann
Maxwelldistribution. i.e., when the molecules are at thermal equilibrium.
distribution.
equilibrium.
8/12/2009 Zumdahl Chapter 5 31 Distributions of Molecular Speeds
u2 = uavg = 3RT
M 8RT
π
M
ump : uavg : urms = 1.000 : 1.128: 1.225 8/12/2009 Zumdahl Chapter 5 32 16 Sample Problem
At a certain temperature, the rootmeansquarespeed of the
root mean squaremolecules of hydrogen in a sample of gas is 1055 m s1. Compute
the rootmean square speed of molecules of oxygen at the same
roottemperature.
temperature. Strategy
1. Find T for the H2 gas with a urms = 1055 m s1
2. Find urms of O2 at the same temperature 8/12/2009 Zumdahl Chapter 5 33 Strategy
1. Find T for the H2 gas with a urms = 1055 m s1 u rm s (u
T H 2
rm s = = u )
(u 2 =
H 2
rm s 3 R T
M = 2 3 R T
M ) 2 M H 2 3 R = (1055 m s1)2 (0.002016 kg mol1) / 3 (8.315 J K1 mol1) 8/12/2009 Zumdahl Chapter 5 34 17 Strategy
2. Find urms of O2 at the same temperature
At a constant temperature, urms is proportional to M−1/2. Therefore, urms for O2 =
= 1055 m s1 x
urms for O2 = 264.8 m s1 8/12/2009 Zumdahl Chapter 5 35 Gaseous Diffusion and Effusion
Diffusion: Mixing of gases via their ceaseless random motions
Distance diffused by NH
uavg(NH3)
____________3 ~ _____
Distance diffused by HCl = MHCl
____ 1/2 (M )
NH3 = 1.5 uavg(HCl)
(HCl) approximately
correct Effusion: Rate of passage of a gas through a tiny orifice in a chamber
Rate of effusion of gas 1
uavg(1)
__________________ = ______ =
Rate of effusion of gas 2 uavg(2) (M )
M2
___ 1/2 1 Graham’s Law of Effusion
Graham’
8/12/2009 Zumdahl Chapter 5 36 18 Collisions of Gas Molecules with Container Walls
See Text, pages 361363 for details
ZA ≡ no. of collisions per unit time of gas molecules
with an area A of container wall 1.00 atm of O2 at T = 27 oC makes 2.72 x 1023
collisions per second per cm2 of container surface
8/12/2009 Zumdahl Chapter 5 37 Intermolecular Collisions
See Text, pages 363−365 for details
363− Z ≡ no. of collisions per unit time between gas molecules
Z = (d ≡ molecular diameter) Z−1 = mean time between collisions
λ ≡ = mean distance traveled between collisions λ = uavg Z−1 =
For O2 at T = 27 oC and P = 1 atm:
atm: Z 8/12/2009x 109 collisions/s
= 4.0 Z −1 Chapter 5
Zumdahl = 2.5 x 10−10 s 7
λ ~ 1 x 10−38m 19 Real Gases
Ideal Gas behavior is generally observed under
conditions of low pressure and high temperature. For an ideal gas, PV/nRT = 1 at all pressures.
PV/nRT
For real gases, attractive forces dominate at intermediate pressures
pressures
8/12/2009
Zumdahl Chapter 5
39
and repulsive forces dominate at high pressures Van der Waals Equation of State
Johannes van der Waals (1873)
– Correction to ideal gas equation of state for
attractive forces in gases (and liquids)
– Correction to ideal gas equation of state to
account for volume occupied by the molecules 8/12/2009 Zumdahl Chapter 5 40 20 The Person Behind the Science Johannes van der Waals (18371923)
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
8/12/2009 Zumdahl Chapter 5 41 Chapter 5: Gases
5.1.
5.2.
5.3.
5.4.
5.5.
5.6.
5.7.
5.8.
5.9.
5.10.
5.11. 8/12/2009 Early Experiments
The Gas Laws of Boyle, Charles and Avogadro
The Ideal Gas Law
Gas Stoichiometry
Dalton’s Laws of Partial Pressure
Dalton’
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 Zumdahl Chapter 5 42 21 ...
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 Bottomley
 pH, Charles’ Law, Zumdahl Chapter

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