Unit 9 Gases
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Transcript Unit 9 Gases
Chemistry I Honors
Unit 9: Gases
Objectives #1-3: Introduction to the
Kinetic Theory of Gases
I. The Kinetic Theory
A. Assumptions of the Kinetic Theory
Gases are composed of tiny particles
that are arranged far apart from each
other
Gases are composed of individual
atoms such as in the element neon or
molecules such as in the element
oxygen
Objectives #1-3: Introduction to the
Kinetic Theory of Gases
Collisions that may occur between gas
particles are elastic with no net loss of
energy
Gas particles are in constant, random
motion
No attractive forces exist between gas
particles
The temperature of a gas depends on the
average kinetic energy of the particles
(video clip Ch. 3 “Common Gas Properties”)
Objectives #1-3: Introduction to the
Kinetic Theory of Gases
II. The Kinetic Theory and its Implications on
the Properties of Gases
The temperature of a gas is directly related
to its kinetic energy
The temperature at which no kinetic energy is
present is called absolute zero; this
temperature is 0 on the Kelvin scale and -273
on the Celsius scale
Gases are able to expand freely due to their
random motion and the lack of attractive
forces between the particles of a gas
Gas particles are tiny and can move past each
other with ease due to the lack of attractive
forces between gas particles
Objectives #1-3: Introduction to the
Kinetic Theory of Gases
The density of gases is generally less than
other substances due to the ability of gases
to move freely
Gases can be compressed due to the great
distances between gas particles
The ability of gases to spread out
spontaneously or diffuse is due to the rapid
motion of gas particles; particles with less
mass and faster velocities spread out at a
faster rate than heavier, slower gas particles
(This is why you smell cookies baking! )
Objectives #1-3: Introduction to the
Kinetic Theory of Gases
Due to their small size, gas particles can
be made to pass through small openings;
this characteristic of effusion depends on
the velocity and molar mass of the gas
particles present; the rate of effusion is
directly proportional to the velocity of the
gas particles and inversely proportional to
the molar mass of the gas particles
When gas particles collide against a
surface they exhibit pressure; which is
defined as the force per unit area
Objectives #4-9 : Properties of Gases
Common Units of Pressure:
Unit
Symbol
Standard
Value
Pascal/Kilopascal Pa/kPa
101325/101.3
Millimeters of Hg mm Hg
760
Atmospheres
atm
1
Torr
torr
760
Objectives #4-9: Properties of Gases
Pressure Conversion Examples:
• Convert 2.5 atm to mmHg
2.5 atm x 760 mmHg = 1900 mmHg
1 atm
• Convert 300. Pa to atm
300. Pa x 1 atm_____ = .00296 atm
101325 Pa
Objectives #4-9: Properties of Gases
Dalton’s Law of Partial Pressure
The total pressure of a mixture of gases is
equal to the sum of the partial pressures of
each of the individual gases in the mixture
PT = P1 + P2 + P3 + ….
This concept must be kept in mind when gases
are collected over water in the laboratory;
the vapor pressure of water must be
subtracted from the measured pressure of
the gas in order to obtain the true pressure
of the gas being collected
Pgas = Patm - Pwater
Objectives #4-9: Properties of Gases
The vapor pressure due to water
increases with increasing temperature
(see attached chart)
Molar Volume
at standard temperature and pressure
(STP), which are defined as O o C and 1
atm, 1 mole of any gas occupies 22.4 L;
this is referred to as molar volume
*Interpreting Vapor Pressure Charts
Examples:
1. Which substance has the highest vapor pressure?
2. Which substance has the lowest vapor pressure?
3. Boiling occurs when the vapor pressure of a
substance equals the external atmospheric
pressure. At what external pressure will ethanol
boil at 40oC?
4. A substance is volatile when it readily evaporates.
Which substance is the most volatile?
5. Which substance has the strongest intermolecular
forces?
6. Which substance has the weakest intermolecular
forces?
Objectives #4-9: Properties of Gases
Relationships Among Gas Characteristics
A. Amount of Gas vs. Pressure (assumes
temperature and volume are held
constant)
What is happening at the particle level:
Objectives #4-9: Properties of Gases
As the amount of gas in a container
increases, the pressure increases
This illustrates a direct relationship
between the amt. of the gas and the
pressure of the gas
Objectives #4-9: Properties of Gases
B. Pressure vs. Volume of Gases (assumes
constant temperature)
What is happening at the particle level:
Objectives #4-9: Properties of Gases
As the pressure of a fixed amount of gas
increases, its volume decreases
This illustrates an inverse relationship
Objectives #4-9: Properties of Gases
C. Temperature vs. Volume (assumes
constant pressure)
What is happening at the particle level:
Objectives #4-9: Properties of Gases
As the Kelvin temperature of a fixed amount of gas
increases, its volume increases
This illustrates a direct relationship:
Objectives #4-9: Properties of
Gases
D. Temperature vs. Pressure (assuming
constant volume)
What is happening at the particle level:
Objectives #4-9: Properties of Gases
As the Kelvin temperature of a fixed
amount of gas increases, its pressure
increases
This illustrates a direct relationship
Objectives #4-9: Properties of
Gases
Ideal vs. Real Gases
Ideal gases always follow the kinetic
theory under any conditions
Real gas particles do have attractive
forces among each other
Real gases no longer act as ideal gases
under conditions of high pressure and
extremely low temperature
Objective #10 Solving problems Involving the
Gas Laws
*the Gas Laws
*the Gas Laws are mathematical formulas
based on the relationships discussed in the
previous section of notes
P1V1 / T1 = P2V2 / T2 (Combined Gas Law)
PV = nRT (Ideal Gas Law)
(video clip “Ch. 4 Ideal Gas Law)
I. Combined Gas Law
*examples:
1. To what pressure must a gas be
compressed to get it into a 9.00 L tank
if it occupies 90.0 L at 1.00 atm?
Can a variable be eliminated?
P1V1 = P2V2
P1V1 / V2 = P2
(1.00 atm) (90.0 L) / 9.00 L = P2
10.0 atm = P2
2. A container with a movable piston
contains .89 L of methane gas at
100.50C. If the temperature of the gas
drops to 11.3oC, what is the new volume
of the gas?
K = C + 273
= 100.5oC + 273
= 374 K
K = C + 273
= 11.3oC + 273
= 284 K
Can a variable be eliminated?
V1 / T1 = V2 / T2
V1T2 = T1V2
V1T2 / T1 = V2
(.89 L) (284 K) / 374 K = V2
.676 L = V2
3. A sample of gas occupies a volume of
5.0 L at a pressure of 650. torr and a
temperature of 24oC . We want to put
the gas in a 100. ml container that can
only withstand a pressure of 3.0 atm.
What temperature must be maintained
so that the container does not explode?
5.0 L = 5000 ml
650. torr = .855 atm
24oC = 297 K
P1V1 / T1 = P2V2 / T2
Can a variable be eliminated?
P1V1T2 = T1P2V2
T2 = T1P2V2 / P1V1
T2 = T1P2V2 / P1V1
T2 = (297 K) (3.0 atm) (100. ml) /
= (.855 atm) (5000 ml)
= 21 K
4. A sample of gas occupies 2.00 L at
STP. What volume will it occupy at 27oC
and 200. mmHg?
P1V1 / T1 = P2V2 / T2
Can a variable be eliminated?
P1V1T2 = T1P2V2
P1V1T2 / T1P2 = V2
(760 mm Hg) (2.00 L) (300 K) /
(273 K) (200. mm Hg) = V2
8.35 L = V2
II.Additional Problems
*Recall that at STP conditions, 1 mole of
any gas = 22.4 L
*examples:
1. Calculate the volume of .55000 moles
of gas at STP.
.55000 moles X 22.4 L / 1 mole = 12.320 L
2. Calculate the moles of gas contained in
350 L of gas at STP.
350 L X 1 mole / 22.4 L = 16 moles
3. Calculate the mass in grams of 3.50 L
of chlorine gas.
3.50 L X 71.0 g / 22.4 L = 11.1 g
Objective #11 The Ideal Gas Law and Gas
Stoichiometry
PV=nRT
*ideal gas constant:
R = PV / nt
R = ( 1 atm) (22.4 L) / (1 mole) (273K)
= .0821 L . atm. / mole.K
*examples:
1. What is the temperature of a .65 L
sample of fluorine gas at 620. torr which
contains 1.3 mol fluorine?
T = PV / nR
*examples:
1. What is the temperature of a .65 L
sample of fluorine gas at 620. torr
which contains 1.3 mol fluorine?
T = PV / nR
= (620 torr / 760 torr) (.65 L) /
(1.3 mol) (.0821 L.atm / mol.K)
= 5.0 K
2. A 25.0 gram sample of argon gas is placed
inside a container with a volume of 10.0 L
at a temperature of 65oC. What is the
pressure of argon at this temperature?
PV=nRT
P = nRT / V
= (25.0 g / 39.9 g) (.0821) (338 K) /
10.0 L
= 1.74 atm.
*there are two types of gas stochiometry
problems:
at STP
non STP
*examples:
1. Calculate the volume of hydrogen gas
that can be produced from the
reaction of 5.00 g of zinc reacted in an
excess of hydrochloric acid. Assume
STP conditions.
Zn + 2HCl --›
H2 + ZnCl2
5.00 g Zn X 1 mole Zn / 65.4 g Zn X
1 mole H2 / 1 mole Zn X 22.4 L / 1 mole H2
= 1.71 L
2. Calculate the volume in liters of oxygen
gas that can be produced from the
decomposition of 3.50 X 1024 formula
units of potassium chlorate. Assume STP
conditions.
2KClO3 --› 2KCl + 3O2
3.50 X 1024 formula units KClO3 X
1 mole KClO3 / 6.02 X 1023 f. units X
3 mole O2 / 2 mole KClO3 X 22.4 L / 1 mole =
195 L
3. Calculate the volume of hydrogen
produced at 1.50 atm and 19oC by the
reaction of 26.5 g of calcium metal
with excess water. The vapor pressure
of water is 16.5 mmHg.
Ca + 2H2O --›
H2 + Ca(OH)2
*use amount of given reactant and
stoichiometry to determine moles of gas
desired in problem:
26.5 g Ca X 1 mole Ca / 40.1 g Ca X
1 mole H2 / 1 mole Ca = .661 mole H2
*determine net pressure of gas:
16.5 mmHg X 1atm/760 mmHg = .0217 atm
1.50 atm - .0217 atm = 1.48 atm
*use moles of gas found in ideal gas law to
calculate volume of gas:
PV=nRT
V = nRT / P
= (.661 moles) (.0821) (292 K) / 1.48 atm
= 10.7 L
4. Calculate the volume of chlorine gas
produced at 1.25 atm at 25oC from the
reaction of 5.00 g of sodium chloride
and an excess of fluorine.
2NaCl + F2 --› 2NaF + Cl2
5.00 g NaCl X 1 mole NaCl / 58.5 g NaCl X
1 mole Cl2 / 2 mole NaCl = .0427 mole Cl2
PV = nRT
V = nRT / P
= (.0427 mole) (.0821) (298 K) /1.25 atm
= .836 L