Chapter-9

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Gaseous State
Chapter 9
9.1 The Behavior of Gases
Characteristics of Gas
• Gases consist of Particles that are
relatively far apart
• Gas particles move very rapidly
• Gas particles have little interaction until
they collide.
• Gases expand to fill the available space.
Description of Gases
• Gases can be described in various physical
properties.
–
–
–
–
–
Pressure
Temperature
Density
Volume
Number of Particles
• Changing one physical parameter will change
one or more others.
Pressure
• Pressure (P) = Force/area
• There are several units for Pressure.
– Pounds/Sq. Inch (psi)
– 1 Atmosphere (atm) = 760 mm Hg = 760 torr
– 1 atm = 101,325 Pa (pascal)
– 1 atm = 14.7 psi
EXAMPLE
• Calculate the amount of pressure a 105 lbs.
woman exerts by stepping down on a stiletto
heel (1/16 sq inch)? Compare this to an
Elephant (12,000 lbs., 16 sq. in. foot)?
PROBLEM
• The depths of the ocean can exceed 380
atm of pressure. How many pascals is
this? How many psi?
9.2 Factors That Affect the
Properties of Gases
Boyle’s Law
• For a given mass of gas at a constant
temperature, volume varies inversely with
pressure.
π‘π‘œπ‘›π‘ π‘‘π‘Žπ‘›π‘‘
𝑉=
𝑃
Implications of Boyle’s Law
𝑉1 =
π‘π‘œπ‘›π‘ π‘‘π‘Žπ‘›π‘‘
𝑃1
𝑉1 𝑃1 = π‘π‘œπ‘›π‘ π‘‘π‘Žπ‘›π‘‘
𝑉2 =
π‘π‘œπ‘›π‘ π‘‘π‘Žπ‘›π‘‘
𝑃2
𝑉2 𝑃2 = π‘π‘œπ‘›π‘ π‘‘π‘Žπ‘›π‘‘
Therefore
𝑉1 𝑃1 = 𝑉2 𝑃2
Ideal Gases
• Any gas that behaves according to the
predicted linear relationships.
– Pressure/Volume
– Volume/Temperature
PROBLEM
• Given an ideal gas held at constant mass and
temperature, what would be the volume of the
gas if it was originally 125 ml, and its pressure is
dropped from 1.40 atm to 0.500 atm?
Charles’s Law
• For a given mass of a gas at constant
pressure, volume is directly proportional to
temperature on an absolute (Kelvin) scale.
𝑉 = π‘π‘œπ‘›π‘ π‘‘π‘Žπ‘›π‘‘ × π‘‡
Implications of Charles’s Law
𝑉1 = π‘π‘œπ‘›π‘ π‘‘π‘Žπ‘›π‘‘ × π‘‡1 𝑉2 = π‘π‘œπ‘›π‘ π‘‘π‘Žπ‘›π‘‘ × π‘‡2
𝑉1
𝑇1
= π‘π‘œπ‘›π‘ π‘‘π‘Žπ‘›π‘‘
𝑉2
𝑇2
= π‘π‘œπ‘›π‘ π‘‘π‘Žπ‘›π‘‘
Therefore
𝑉1 𝑉2
=
𝑇1 𝑇2
Kelvin Scale
• So far we have measured temperature in
degrees Celsius, a relative scale.
• Absolute temperature is measure in Kelvin
(K, note no degree sign is used), this scale
is related to the absolute motion of atoms.
0 K = -273.15 °C
Therefore Temp °C +273.15 = Temp K
PROBLEM
• Calculate the following temperatures in
Kelvin.
– Boiling Point of Water
– Freezing Point of Water
PROBLEM
• Given an ideal gas held at constant mass and
pressure, what would be the temperature of the
gas if it was originally at 125 °C, and its volume
is dropped from 1.00 L to 0.500 L?
Combined Gas Law
• Combining Boyle’s and Charles’s Law
results in the following equation.
𝑃1 𝑉1
𝑃2 𝑉2
=
𝑇1
𝑇2
PROBLEM
• A sample of Hydrogen gas occupies 1.25 L
at 80.0 °C and 2.75 atm. What would its
volume be at 185 °C and 5.00 atm?
Guy-Lussac and Avagadro
• Guy-Lussac’s Law stated that combined
gases always reacted in small whole
number ratios (sound familiar?)
• Avagadro hypothesized that the volume
occupied by a gas is proportional to the
number of gas particles and hence the
moles of gas.
Avagadro’s Hypothesis
• At a given pressure and temperature, equal
volumes of gas contain equal numbers of
moles.
• Standard Temperature and Pressure (STP)
are defined as 273.15 K and 1 atm, at which
1.00 mole of gas occupies 22.414 L, defined
as molar volume
PROBLEM
• What volume would 1.00 mole of gas
occupy at 37°C and 1.00 atm?
9.3 The Ideal Gas Law
Review of the Relationships
• Boyle’s Law: (C is a constant)
𝑉=
𝐢
𝑃
at constant n, T
• Charles’s Law:
𝑉 = 𝐢𝑇 at constant n, P
• Avagadro’s Hypothesis:
𝑉 = 𝐢𝑛 at constant P, T
Combine the Equations
• Assume the constants combined equal
another constant we will call the Ideal Gas
Constant, R
𝑅𝑛𝑇
𝑉=
𝑃
• Rearrange this into the Ideal Gas Law
𝑷𝑽 = 𝒏𝑹𝑻
So What is the Value for R?
• Assuming a standard gas you can
calculate the value of R
𝑃𝑉 = 𝑛𝑅𝑇 becomes 𝑅 =
𝑃𝑉
𝑛𝑇
For a Standard gas 1.000 mole at 273.15 K
and 1.000 atm has a volume of 22.414 L
𝑅=
(1.00 π‘Žπ‘‘π‘š)×(22.414 𝐿)
(1.00 π‘šπ‘œπ‘™)×(273.15 𝐾)
= 0.08206
πΏβˆ™π‘Žπ‘‘π‘š
π‘šπ‘œπ‘™βˆ™πΎ
PROBLEM
• Calculate the Ideal Gas Constant if
pressure is measured in torr
PROBLEM
The volume of an oxygen cylinder is 2.025 L. At a
temperature of 32 °C the cylinder has a pressure
of 945 torr. How many moles of Oxygen gas are in
the cylinder?
PROBLEM
• Given the amount of oxygen in the
previous question, what is its mass?
• What is its density?
Partial Pressures
• Dalton’s Law of Partial Pressure: Gases
in a mixture behave independently and
exert they same pressure as they would
alone.
π‘ƒπ‘šπ‘–π‘₯π‘‘π‘’π‘Ÿπ‘’ = 𝑃𝐴 + 𝑃𝐡 +𝑃𝐢 …
PROBLEM
Given a 2.500 L container at 45.00°C, what would
be the total pressure in the container from a
mixture of 2.100 moles gaseous sulfur dioxide and
1.800 moles hydrogen sulfide?
9.5 Gases and Chemical
Reactions
The return of Stoichiometry
• You can use the gas laws to solve
stoichiometry problems by converting
values between states of matter.
Example
• A sample of hydrogen gas has a volume of 8.00 L at a
pressure of 5.00 atm and a temperature of 25 °C. What
volume of gaseous water is produced in the following
reaction at 150.0 °C and 0.947 atm, if all the hydrogen
gas reacts with copper (II) oxide
CuO(s) + H2(g) οƒ  Cu(s) + H2O(g)
ALGORITHM
• mol of Hydrogen (PV=nRT)
• mol of Water (formula)
• V of Water (PV=nRT)
PROBLEM
How many grams of calcium carbonate is formed if 3.45 L
of carbon dioxide, measured at 45.0 C and 1.37 atm, react
with excess CaO?
CaO(s) + CO2(g) οƒ  CaCO3(S)
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