Ch06lecture 150104200613-conversion-gate01

Copyright © 2011 Pearson Canada Inc.General Chemistry: Chapter 1Slide 1 of 41
PHILIP DUTTON
UNIVERSITY OF WINDSOR
DEPARTMENT OF CHEMISTRY AND
BIOCHEMISTRY
TENTH EDITION
GENERAL CHEMISTRY
Principles and Modern Applications
PETRUCCI HERRING MADURA BISSONNETTE
Gases 6

Gases

The gaseous state of three halogens (group 17)
FIGURE 6-1

The Concept of Pressure
FIGURE 6-2
P (Pa) =
Area (m2
)
Force (N)
Force (N) = g (m/s2
) x m (kg)
Force
Pressure

Liquid Pressure
FIGURE 6-3
liquid pressure is directly proportional to the liquid density and the height of the liquid column
P (Pa) =
A
F =
A
W =
A
g x m =
A
g x V x δ =
A
g x h x A x δ = g x h x δ

Measurement of atmospheric pressure with a mercury barometer
FIGURE 6-4
Standard Atmospheric Pressure
1.00 atm, 101.325 kPa, 1.01325 bar, 760 torr, ~760 mm Hg

Measurement of gas pressure with an open-end manometer
FIGURE 6-5

Relationship between gas volume and pressure – Boyle’s Law
FIGURE 6-6
PV = constantP α
1
V
6-2 Simple Gas Laws

6-2 CONCEPT ASSESSMENT
A 50.0 L cylinder contains nitrogen gas at a pressure of 21.5 atm. The
contents of the cylinder are emptied into an evacuated tank of unknown
volume. If the final pressure in the tank is 1.55 atm, then what is the
volume of the tank?
(a)(21.5/1.55) x 50.0 L (b) (1.55/21.5) x 50.0 L
(b)(c) v21./(1.55 x 50.0) L (d) (1.55/(21.5 x 50.0) L

6-2 CONCEPT ASSESSMENT (CONTINUED)
Use Boyle’s Law
P1V1 = P2V2 V2 =
P1V1
P2
therefore (a) is the answer
Calculation shows that Vtank = 644 L

Gas volume as a function of temperature
FIGURE 6-7
V = b TV α T

Standard Temperature and Pressure
Gas properties depend on conditions.
IUPAC defines standard conditions of temperature
and pressure (STP).
P = 1 Bar = 105
Pa
T = 0°C = 273.15 K

Avogadro’s Law
Gay-Lussac 1808
Small volumes of gases react in the ratio of small whole
numbers.
Avogadro 1811
At a fixed temperature and pressure, the volume of a
gas is directly proportional to the amount of gas.

Molar volume of a gas visualized
FIGURE 6-9
At STP
1 mol gas = 22.711 L gas
At fixed T and P
V ∝ n or V = c n

Formation of Water – actual observation and Avogadro’s hypothesis
FIGURE 6-8

6-3 Combining the Gas Laws:
The Ideal Gas Equation
and the General Gas Equation
Boyle’s law V ∝ 1/P
Charles’s law V ∝ T
Avogadro’s law V ∝ n
V ∝
nT
P

The Ideal Gas Equation
R =
PV
nT
PV = nRT

Applying the ideal gas equation
Copyright © 2011 Pearson Canada Inc.Slide 18 of 41 General Chemistry: Chapter 6

The General Gas Equation
R = =
P2V2
n2T2
P1V1
n1T1
=
P2
T2
P1
T1
If we hold the amount and volume constant:

Using the Gas Laws

6-4 Applications of the Ideal Gas Equation
PV = nRT and n =
m
M
PV =
m
M
RT
M =
m
PV
RT
Molar Mass Determination

Gas Density
δ =
m
V
PV =
m
M
RT
MP
RTV
m
= δ =
KEEP IN MIND
that gas densities are
typically much smaller than
those of liquids and solids.
Gas densities are usually
expressed in grams per liter
rather than grams per
milliliter.

6-5 Gases in Chemical Reactions
Stoichiometric factors relate gas quantities to
quantities of other reactants or products.
Ideal gas equation relates the amount of a gas to
volume, temperature and pressure.
Law of Combining Volumes can be developed using
the gas law.

6-6 Mixtures of Gases
Partial pressure
Each component of a gas mixture exerts a
pressure that it would exert if it were in the
container alone.
• Gas laws apply to mixtures of gases.
• Simplest approach is to use ntotal, but....

Dalton’s law of partial pressures illustrated
FIGURE 6-12
The total pressure of a mixture of gases is the sum of
the partial pressures of the components of the mixture.

Partial Pressure
Ptot = Pa + Pb +…
Va = naRT/Ptot and Vtot = Va + Vb+…
Va
Vtot
naRT/Ptot
ntotRT/Ptot
= =
na
ntot
Pa
Ptot
naRT/Vtot
ntotRT/Vtot
= =
na
ntot
na
ntot
= χaRecall

Collecting a gas over water
FIGURE 6-13
Ptot = Pbar = Pgas + PH2O

Visualizing Molecular Motion
FIGURE 6-14
6-7 Kinetic Molecular Theory of Gases
• Particles are point masses in constant,
random, straight line motion.
• Particles are separated by great
distances.
• Collisions are rapid and elastic.
• No force between particles.
• Total energy remains constant.

Derivation of Boyle’s Law
ek ∝
1
2
µ υ2
Translational kinetic energy,
Frequency of collisions,
Impulse or momentum transfer,
Pressure proportional to impulse
times frequency
v ∝ υ
Ν
ς
I ∝ µ υ
2
mu
V
N
P ∝
PV = c

Pressure and Molecular Speed
2
um
V
N
3
1
P =
FIGURE 6-15
2
u=
um is the modal speed
uav is the simple average
urms
Three dimensional systems lead to:

Pressure
M
3RT
u
uM3RT
umRT3
um
3
1
PV
rms
2
2
A
2
A
=
=
=
=
N
N
Assume one mole:
PV=RT so:
NAm = M:
Rearrange:

Distribution of Molecular Speeds – the effect of mass and temperature
FIGURE 6-16
M
3RT
urms =

Distribution of molecular speeds – an experimental determination
FIGURE 6-17

Temperature
(T)
R
2
3
e
e
3
2
RT
)um
2
1
(
3
2
um
3
1
PV
A
k
k
22
A
N
N
NN
A
A
=
=
==Modify:
PV=RT so:
Solve for ek:
Average kinetic energy is directly proportional to temperature!

6-8 Gas Properties Relating to the
Kinetic-Molecular Theory
Diffusion
Net rate is proportional to
molecular speed.
Effusion
A related phenomenon.

Graham’s Law
rate of effusion of A
rate of effusion of B
=
(υρµ σ)Α
(υρµ σ)Β
=
3ΡΤ/ΜΑ
3ΡΤ/ΜΒ
=
ΜΒ
ΜΑ
Only for gases at low pressure (natural escape, not a jet).
Tiny orifice (no collisions)
Does not apply to diffusion.
Ratio used can be:
• Rate of effusion (as above)
• Molecular speeds
• Effusion times
• Distances traveled by molecules
• Amounts of gas effused.

Intermolecular forces of attraction
FIGURE 6-22
6-9 Nonideal (Real) Gases
Compressibility factor
PV/nRT = 1 for ideal gas.
Deviations for real gases.
PV/nRT > 1 - molecular
volume is significant.
PV/nRT < 1 – intermolecular
forces of attraction.

The behaviour of real gases – compressibility factor as a function of
pressure at 0°C
FIGURE 6-20
Real Gases

van der Waals Equation
P +
n2
a
V2
V – nb = nRT
The van der Waals equation reproduces the observed
behavior of gases with moderate accuracy. It is most
accurate for gases comprising approximately spherical
molecules that have small dipole moments.

End of Chapter Questions
A problem is like a knot in a ball of wool:
If you pull hard on any loop:
The knot will only tighten.
The solution (undoing the knot) will not be achieved.
If you pull lightly on one loop and then another:
You gradually loosen the knot.
As more loops are loosened it becomes easier to undo
the subsequent ones.
Don’t pull too hard on any one piece of
information in your problem, it tightens.

Ch06lecture 150104200613-conversion-gate01

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