World of Chemistry—Chapter 13 Gases
I.
II.
Price
Section 13.1—Pressure
A. A gas will fill any container in which it is placed
B. The air in earth’s atmosphere are pressing down on you with a
little less than 14.69 pounds of pressure on every square inch
of your body
C. Atmospheric pressure is measured with a barometer
D. Units of pressure
1. 760 millimeters of Mercury (mm Hg)
760 torr (short for Torricelli)
1 standard atmosphere (atm)
14.69 pounds per square inch (psi)
101,325 pascals (Pa)
All of these amounts are EQUAL!
Section 13.2—Pressure and Volume: Boyle’s Law
A. Robert Boyle (1627-1691) showed that pressure multiplied by
volume equals a constant—this is Boyle’s Law
PV = k
B. We say that pressure and volume are inversely related meaning
that as one goes up the other goes down and vice versa
C. Therefore, if we double the pressure we cut the volume in half.
(this relationship holds true only if the amount of gas and its temperature
remain the same)
D. Calculations involving Boyle’s Law
1. Boyle’s Law allows us to make calculations involving
changing pressure and volume like example 13.2
2. P(initial) x V(initial) = k = P(final) x V(final)
OR
P(initial) x V(initial) = P(final) x V(final)
We can use this equation to solve problems
Example 13.2
If a 1.5 ℓ sample of gas has a pressure of 56 torr, and the
pressure is changed to 150 torr what is the new volume?
P1 x V1 = P2 x V2
(1.5ℓ) x (56torr) = (?ℓ) x (150torr)
84 = (?ℓ) x (150torr)
84 = ?ℓ
150
0.56 = ℓ
III.
Section 13.3—Volume and Temperature: Charles’s Law
A. Jacques Charles (1746-1823) experimented with hot air
balloons and discovered that as the temperature of a gas
increases, its volume increases as well—volume and
temperature are directly proportional
B. Charles’s work led to an inference about absolute zero
C. Charles’s Law is written as
V = b (a constant)
T
Where V is the volume in liters and T is the temperature in Kelvins
D. We can manipulate the above equation, just as with Boyle’s law,
for initial volume and temperature to determine final volume
and temperature
E. The final equation looks like this:
or, simply
V1 = b = V2
T1
T2
V1 = V2
T1 T2
F. We can use Charles’s Law to solve problems like example 13.4
on p.409
Example 13.4
A 2 liter sample of air is collected at 298K and then cooled to 278
K. The pressure is held constant at 1.0 atm. Calculate the volume
of the air at 278K.
Charles’s Law tells us that
V1 = V2
T1 T2
are given we get this equation:
2ℓ
298K
=
we then multiply both sides by 278K
by simply substituting what we
?
278K
278K x 2ℓ = ?
x 278K
298K 278K
solve this like a conversion factor
278K x 2ℓ = ?
298K
IV.
we find that our new volume is 1.87ℓ
Section 13.4—Volume and Moles: Avagadro’s Law
A. Similar to Boyle’s and Charles’s laws we find a direct
relationship between volume and number of moles…that is, if
we double the number of moles we double the volume at a
constant pressure and temperature
B. We write Avagadro’s Law like this:
V=a
n
where V is volume in liters and n is the number of moles and a is a constant
C. We can manipulate this equation for initial and final
measurements
V1 = V2
n1
V.
n2
D. We can use Avagadro’s Law to solve problems like example 13.7
Section 13.5 The Ideal Gas Laws
A. You may have noticed some similarities in the past 3 sections…
B. We can combine all of these relationships to derive the Ideal
Gas Law
PV = nRT
where P is pressure, V is volume, n is number of moles, R is the
ideal gas constant 0.08206 L atm/K mol
and T is the temperature
C. We can solve many Ideal Gas Law problems using this equation
D. The Ideal Gas Law is often written as
P1V1 = P2V2
T1
T2
This is known as the Ideal Gas Law Equation
VI.
Section 13.6—Dalton’s Law of Partial Pressure
A. 1803 John Dalton (remember Dalton’s atomic theory? p.52)
discovered that “For a mixture of gases in a container, the
total pressure exerted is the sum of the partial pressures of
the gases present.”
B. This became known as Dalton’s Law of Partial Pressures
C. Mathematically, Dalton’s Law is written as
P(total) = P1 + P2 + P3
D. We can use the ideal gas law to figure out our partial pressures
and, hence, the total pressure of a mixture of gases in a single
container.
P(total) = n(total) x RT
V
VII.
STP
A. What is the volume of 1 mole of an ideal gas at 273K (0oC) and
1 atm?
B. By using the ideal gas law we can see that
V= (1.0 mole) (0.08206)(273K) = 22.4 Liters
1 atm
C. This is known as STP (Standard Temperature and Volume) and
is used as a shortcut in many gas law problems.
D. In Summary: Standard Temperature and Volume
P = 1.0atm
V = 22.4 ℓ
n = 1.0 mol
R = 0.08206
T = 273K