Transcript V. Two More Gas Laws

Ch. 10 & 11 - Gases
V. Two More Laws
(p. 322-325, 351-355)
Read these pages first!
The Behavior of Real Gases




The molar volume is not constant as is expected for
ideal gases.
These deviations due to an attraction between some
molecules.
Finite molar molecular volume.
For compounds that deviate from ideality the van der
Waals equation is used:
2a 

n
P +
(V - nb) = nRT

2 
V 


where a and b are constants that are characteristic of
the gas.
Applicable at high pressures and low temperatures.
John A. Schreifels
Chemistry 211
2
The Kinetic Theory – Molecular Theory
of Gases

Microscopic view of gases is called the
kinetic theory of gases and assumes that
• Gas is collection of molecules (atoms) in
continuous random motion.
• The molecules are infinitely small point-like
particles that move in straight lines until they
collide with something.
• Gas molecules do not influence each other except
during collision.
• All collisions are elastic; the total kinetic energy is
constant at constant T.
• Average kinetic energy is proportional to T.
John A. Schreifels
Chemistry 211
3
Gases and Gas Pressure




They form homogeneous solutions. All gases dissolve in each
other.
• Gases are compressible.
• Large molar volume.
Barometer usually mercury column in tube; mm Hg is a
measure of pressure.
Manometer tube of liquid connected to enclosed container
makes it possible to measure pressure inside the container.
Pressure
• One of the most important of the measured quantities for
gases
• defined as the force/area P = f/area.
• Pressure has traditionally been measured in units relating to
the height of the Hg and is thus expressed as mm Hg = 1
Torr.
John A. Schreifels
Chemistry 211
4
Gas Pressure




Pressure is directly proportional to the height of the column in a
barometer or manometer.
F mg dVg
P 

A
A
A
 dgh
Mercury often used but other low density liquids are used for low
pressure changes:
P = dHgghHg = doilghoil or dHghHg = doilhoil.
E.g. Water is sometimes used to determine pressure; determine
the height of water if the barometer pressure was 750 mmHg.
The density of Hg = 13.596 g/cm3 and 1.00 g/cm3 respectively.
Solution:
hH2O  hHg 
John A. Schreifels
Chemistry 211
dHg
dH2O
 750 mmHg 
 10197 mmH 2O
13.596 g / cm3
1.00 g / cm3
5
B. Dalton’s Law
 The
total pressure of a mixture
of gases equals the sum of the
partial pressures of the
individual gases.
Ptotal = P1 + P2 + ...
When a H2 gas is
collected by water
displacement, the gas in
the collection bottle is
actually a mixture of H2
and water vapor.
B. Dalton’s Law
 Hydrogen
gas is collected over water at
22.5°C. Find the pressure of the dry
gas if the atmospheric pressure is 94.4
The totalkPa.
pressure in the collection bottle is equal to atmospheric
pressure and is a mixture of H2 and water vapor.
GIVEN:
PH2 = ?
Ptotal = 94.4 kPa
PH2O = 2.72 kPa
Look up water-vapor pressure
on p.899 for 22.5°C.
WORK:
Ptotal = PH2 + PH2O
94.4 kPa = PH2 + 2.72 kPa
PH2 = 91.7 kPa
Sig Figs: Round to least number
of decimal places.
B. Dalton’s Law

A gas is collected over water at a temp of
35.0°C when the barometric pressure is 742.0
torr. What is the partial pressure of the dry gas?
The total pressure in the collection bottle is equal to barometric
pressure and is a mixture of the “gas” and water vapor.
GIVEN:
Pgas = ?
Ptotal = 742.0 torr
PH2O = 42.2 torr
Look up water-vapor pressure
on p.899 for 35.0°C.
WORK:
Ptotal = Pgas + PH2O
742.0 torr = PH2 + 42.2 torr
Pgas = 699.8 torr
Sig Figs: Round to least number
of decimal places.
C. Graham’s Law
 Diffusion
• Spreading of gas molecules
throughout a container until
evenly distributed.
 Effusion
• Passing of gas molecules
through a tiny opening in a
container
C. Graham’s Law
 Speed
of diffusion/effusion
• Kinetic energy is determined by
the temperature of the gas.
• At the same temp & KE, heavier
molecules move more slowly.
 Larger m  smaller v
KE =
2
½mv
C. Graham’s Law
 Graham’s
Law
• Rate of diffusion of a gas is
inversely related to the square root
of its molar mass.
• The equation shows the ratio of
Gas A’s speed to Gas B’s speed.
vA

vB
mB
mA
C. Graham’s Law
 Determine
the relative rate of diffusion
for krypton and bromine.
The first gas is “Gas A” and the second gas is “Gas B”.
Relative rate mean find the ratio “vA/vB”.
vA

vB
v Kr

v Br2
m Br2
m Kr
mB
mA
159.80g/mol
 1.381

83.80g/mol
Kr diffuses 1.381 times faster than Br2.
C. Graham’s Law

vA

vB
A molecule of oxygen gas has an average
speed of 12.3 m/s at a given temp and pressure.
What is the average speed of hydrogen
molecules at the same conditions?
mB
mA
vH 2
12.3 m/s

32.00 g/mol
2.02 g/mol
vH 2
vH 2
vO2

mO2
mH 2
Put the gas with
the unknown
speed as
“Gas A”.
12.3 m/s
 3.980
vH2  49.0m/s
C. Graham’s Law

An unknown gas diffuses 4.0 times faster than
O2. Find its molar mass.
The first gas is “Gas A” and the second gas is “Gas B”.
The ratio “vA/vB” is 4.0.
vA

vB
vA

v O2
mB
mA
mO2
mA
Square both
sides to get rid
of the square
root sign.

32.00 g/mol
g/mol 
32.00
 4.0 



m
A
A


32.00 g/mol
16 
mA
32.00 g/mol
mA 
 2.0 g/mol
16
2
TEAM PRACTICE!
 Work
the following problems in your
book. Check your work using the
answers provided in the margin.
• p. 324
 SAMPLE PROBLEM 10-6
 PRACTICE 1 & 2
• p. 355
 SAMPLE PROBLEM 11-10
 PRACTICE 1, 2, & 3