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