Transcript Unit 9 Powerpoint

Chemistry I Honors
Unit 9: Gases
Objectives #1-4: 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 #1-4: 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
Objectives #1-4: 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
Objectives #1-4: 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 #1-4: 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
Diffusion v Effusion
Objectives #5-8: Relationships Among
Gas Characteristics
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 #5-8: Relationships among
Gas Characteristics
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 #5-8: Relationships among
Gas Characteristics
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
Interpreting Vapor Pressure Charts
Objectives #5-8: Relationships among
Gas Characteristics
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
Relationships Among Gas Characteristics
Objectives #5-8: Relationships among
Gas Characteristics
Relationships Among Gas Characteristics
A. Amount of Gas vs. Pressure (assumes
temperature and volume are held
constant)
 What is happening at the particle level:
How can you tell the
can is full?
Gases expand to fit the
space, so increasing
the particles makes the
can heavier !
Objectives #5-8: Relationships among
Gas Characteristics
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

Empty can =
less pressure,
so the can is
lighter weight
Objectives #5-8: Relationships Among
Gas Characteristics
B. Pressure vs. Volume of Gases (assumes
constant temperature)
 What is happening at the particle level:
Why are
balloons “overfilled” with
helium?
Objectives #5-8: Relationships among
Gas Characteristics
 As the pressure of a fixed amount of gas
increases, its volume decreases
 This illustrates an inverse relationship
Objectives #5-8: Relationships among
Gas Characteristics
C. Temperature vs. Volume (assumes
constant pressure)
 What is happening at the particle level:
Why does tires tend to
look like they are low in air
on a cold winter morning?
Cold night temperatures
result in slower movement
of the gas particles, so the
volume of the gas SEEMS to
decrease. As the car moves,
friction of the tire on the
road increases the temp of
the air in the tire.
Objectives #5-8: Relationships Among
Gas Characteristics
 As the Kelvin temperature of a fixed amount of gas
increases, its volume increases
 This illustrates a direct relationship:
Objectives #5-8: Relationships
Among Gas Characteristics
D. Temperature vs. Pressure (assuming
constant volume)
 What is happening at the particle level:
Keeping & transporting
unopened pop cans in the
car in the summer is NOT
advisable. Why ?
Increasing the temperature
in the car increases the
pressure in the can. If the
pressure of the gas is
greater than the pressure
exerted by the wall of the
can, the can will explode!
Objectives #5-8: Relationships Among
Gas Characteristics
As the Kelvin temperature of a fixed
amount of gas increases, its pressure
increases
This illustrates a direct relationship
Objectives #5-8: Relationships
Among Gas Characteristics
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 #9 Solving Problems Involving the
Gas Laws
*The Gas Laws are mathematical formulas
based on the relationships discussed in the
previous section of notes…
Combined Gas Law: P1V1 = P2V2
T1
T2
**Note that all temps must be in KELVINS!!
Ideal Gas Law: (P)(V) = (n)(R)(T)
I. Combined Gas Law
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?
Step 1: Can a variable be eliminated?
Yes; temperature is not a factor…
Step 2: Substitute & Solve…
P1V1 = P2V2
P2 = P1V1 = (1.00 atm) (90.0 L)
V2
9.00 L
P2 = 10.0 atm
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?
Step 1: Convert temps!!
K = C + 273
= 100.5oC + 273
= 374 K, AND
K = C + 273
= 11.3oC + 273
= 284 K
Step 2: Can a variable be eliminated?
Yes; pressure is not a factor…
Step 3: Substitute and solve…
V1 = V2
T1 T2
(V1)(T2) = (T1)(V2)
V1T2 = V2
T1
(.89 L) (284 K) = V2
374 K
.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?
Step 1: Standardize units!!
5.0 L = 5000 ml
650. torr = .855 atm
24oC = 297 K
Step 2: Can a variable be eliminated?
No, all components are used…
Step 3: Substitute and solve…
P1V1 = P2V2
T1
T2
P1V1T2 = T1P2V2
T2 = T1P2V2 = (297 K)(3.0 atm)(100. ml)
P1V1
(.855 atm) (5000 ml)
T2 = 21 K
4. A sample of gas occupies 2.00 L at STP.
What volume will it occupy at 27oC and 200. mm
Hg?
P1V1 = P2V2
T1
T2
Can a variable be eliminated?
(P1)(V1)(T2) = (T1)(P2)(V2)
P1V1T2 = V2
T1P2
(760 mm Hg) (2.00 L) (300 K) = V2
(273 K) (200. mm Hg)
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 = 12.320 L
1 mole
2. Calculate the moles of gas contained in
350 L of gas at STP.
350 L x 1 mole = 16 moles
22.4 L
3. Calculate the mass in grams of 3.50 L
of chlorine gas.
3.50 L x 71.0 g = 11.1 g
22.4 L
Objective #11: The Ideal Gas Law and Gas
Stoichiometry
Ideal Gas Law: (P)(V)=(n)(R)(T)
R = ideal gas constant, so…
R = (P)(V) = (1 atm) (22.4 L) = .0821 L . atm.
(n)(t)
(1 mole) (273K)
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
(620 torr/760 torr/atm = 0.82 atm )
nR
= (0.82atm ) (.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.
Gas Stoichiometry
There are two types of gas stochiometry
problems:
1) at STP (Standard Temp & Pressure)
2) non STP
Gas Stoichmetry: 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 + 2 HCl  H2 + ZnCl2
5.00 g Zn x 1 mole Zn x 1 mole H2 x 22.4 L H2
65.4 g Zn 1 mole Zn 1 mole H2
= 1.71 L H2
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.
2 KClO3
 2 KCl
+
3 O2
3.50 x 1024 formula units KClO3 x
1 mole KClO3
x 3 mole O2
x 22.4 L
6.02 X 1023 f. units
2 mole KClO3
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. (Ignore the vapor
pressure of water)
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
*use moles of gas found in ideal gas law to
calculate volume of gas:
PV=nRT
V = nRT / P
= (.661 moles) (.0821) (292 K)
1.50 L
= 10.6 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.
2 NaCl + F2  2 NaF + Cl2
First, find the moles of Cl2…
5.00 g NaCl x 1 mole NaCl x 1 mole Cl2
58.5 g NaCl
2 mole NaCl
= .0427 mole Cl2
Now, use the ideal gas relationship…
PV = nRT
V = nRT
P
= (.0427 mole) (.0821) (298 K)
1.25 atm
= .836 L