Download Chapter 12 Gases

Gases
Chapter 12
Properties of Gases
 Compressibility
 Mass
 Volume (fill container)
 Exert Pressure
 Diffuse through other gases
 Low Density
Kinetic Molecular Theory (KMT)
1.
2.
3.
4.
5.
6.
All gases are matter (have mass and take up space).
These particles are in constant, rapid and random motion.
All collisions are perfectly elastic. (no energy is lost in the
collision)
The force of gas particle collisions on the walls of the
container creates pressure; however, gas particles do not
exert force on each other.
At a given temperature, all gas particles have the same
amount of kinetic energy (temperature).
The distance between gas particles is very large.
Gas Pressure
Gas pressure results from the force of colliding
particles on a given area.
The metric unit of force is the newton (similar to the
ounce in the English system).
A newton of force acting on a square meter is called a
pascal.
Normal atmospheric pressure at sea level is
101.3 kPa.
Measuring Gases
 In order to describe a gas, we use four variables to make
predictions about the behavior of the particles.
Amount of a gas (n) expressed in moles
2. Amount of space the gas takes up, volume (V), as measured in
liters. *note: this is the volume of the container
3. Measure of kinetic energy, temperature (T), in degrees Kelvin
1.
1.
4.
*note: K = C + 273
Measure of the force exerted by the gas particles on the walls
of the container, pressure (P)
Kinetic Energy and Temperature
Temperature is an indication of kinetic energy.
T1
T2
T2 > T1
Speed of Particles
The Kelvin Temperature Scale
Kinetic Energy of Molecules
0
(Absolute Zero Scale)
K = oC + 273
-273oC
-300
0
-200
73
-100
173
0
100 oCelcius scale
273
373 Kelvin scale
Pressure
 Pressure is the measure of the amount of force of an object
per unit area.
 Newton / m2 = Pascal
 Atmospheric Pressure is the amount of pressure exerted by
the entire atmosphere on the surface of the Earth.
 A barometer is an instrument that measures atmospheric
pressure
 A manometer is an instrument that measures the pressure of an
enclosed gas
Mercury Barometer
Manometer
STP

Pressure and temperature changes affect gas
volume.

To compare experimental results, scientist convert
their results to standard temperature and pressure
(STP)
 Standard Conditions:
Standard Temperature = 0oC (273 K)
Standard Pressure = 101.3 kPa (1atm) {760 torr}
Pressure Units
 Kilopascals (kPa)
 Atmospheres (atm)
1 atm = 101.3 kPa
 Torricellis (torr)
1 torr = 1 mm Hg
760 torr = 101.3 kPa
(other units are listed on page 420)
Pressure Units
Atmosphere (atm)
Millimeter Hg (mmHg)
Pascal (Pa)
Pounds per square inch (psi)
Kilopascal (kPa)
 1 atm = Standard Pressure
 1 atm = 760 mmHg
 1 atm = 101,325 Pa
 1 atm = 14.7 psi
 1 atm = 101.3 kPa
Conversions
 Covert using Dimensional Analysis
 Information you have x conversion factor (unknown/known)
 Convert 665 mmHg to kPa
 665 mmHg x 101.3 kPa = 88.6 kPa
760 mmHg
 Classwork: page 421 #1-4
 Homework: page 422 #1-12
Boyle’s Law
Pressure and Volume are inversely proportional at a
given temperature and number of molecules.
VP = k (a constant value)
Pressure Effects on Volume
V  1/P
Inverse Proportionality
Boyle’s Law continued
 V 1P 1 = k 1
and
V 2P 2 = k 2
since k1 = k2 (that’s why its called a
constant)
V 1P 1 = V 2P 2
This is the mathematical
expression for Boyle’s Law
A Boyle’s Law Problem
A gas in a 242 cm3 container exerts a pressure of
87.6 kPa.
What volume would the gas occupy at standard
atmospheric pressure?
Identify Variables
V1 =
242 cm3
V2 =
?
P1 =
87.6 kPa
P2 = 101.3 kPa
Two Methods of Solving
1. Algebraic:
V1P1 = V2P2
V2 = V1P1
P2
V2 = 242 cm3 x 87.6 kPa
101.3 kPa
V2 = 209 cm3
Another Method
2. Logic and reason:
The new volume (V2) depends upon the old
volume (V1) multiplied by a pressure change.
87.6
V2 = 242 cm3 x _______
101.3
Since the pressure is increasing, the volume should
decrease.
Multiply by a fraction less than 1.
Charles’s Law
At a given pressure and number of molecules, Volume
and Temperature are directly proportional.
V = k
T
Two Temperature Scales
b
Volume
y = mx + b
V = kT + b
V = kT + 0
V = kT
0
b
-300
0
-200
73
-100
173
0
100 oCelcius scale
273
373 Kelvin scale
Charles’s Law Continued
 V1 = k1
T1
V2 = k2
T2
V 1 = V2
T1 T2
and
since k1 = k2
This is the mathematical
expression for Charles’s Law
A Charles’s Law Problem
A 225 cm3 volume of gas is collected at 58oC.
What volume would the gas occupy at standard
temperature?
Identify Variables
V1 =
V2 =
225 cm3
?
T1 =
58oC + 273 = 331 K
T2 =
0oC + 273 = 273 K
Two Methods of Solving
1. Algebraic:
V1 / T1 = V2 / T2
V2 = V1T2
T1
V2 = 225 cm3 x 273 K
331 K
V2 = 186 cm3
The Other Method
2. Logic and reason:
The new volume (V2) depends upon the old
volume (V1) multiplied by a temperature change.
273
V2 = 225 cm3 x _______
331
Since the temperature is decreasing, the volume
should decrease.
Multiply by a fraction less than 1.
Gay-Lussac’s Law
P1 = k1
T1
and
P2 = k2
T2
since k1 = k2
P1 = P2
T1 T2
This is the mathematical expression for
Gay-Lussac’s Law
The Combined Gas Law
 In most laboratory situations, both temperature and pressure
change.
 The Combined Gas Law is used to calculate the affects of
these changes.
A Combination of Gas Laws
Boyle’s Law:
V  1/P
Charles’s Law:
V  T

Gay-Lussac’s Law:
PT
V  T/P
To change a proportionality to an equality…
multiply by a proportionality constant.
V = kT
P
V1P1 = V2P2
VP = k
T1
T2
T
V 1P1 = V 2P 2
T1
T2
Solve for each variable
V2 = V1P1T2 / T1P2
T2 =
V2P2T1 / V1P1
P2 =
V1P1T2 / T1V2
Sample Problem
What volume would 955 cm3 of a gas measured at 58
oC and 108.0 kPa occupy at 76 oC and 123.0 kPa?
3
955
cm
V1 =
V2 = ?
T1 = 331 K
T2 = 349 K
P1 =
P2 =
108.0 kPa
123.0 kPa
Method #1
V1P1 / T1 = V2P2 / T2
V2 = V1P1T2 / (T1P2)
V2 = (955 cm3) (108.0 kPa) (349 K)
(331 K) (123.0 kPa)
V2 = 884 cm3
Method #2
108
349 x --------V2 = 955 cm3 x --------331
temperature increases
volume increases
pressure increases
volume decreases
V2 = 884 cm3
123
Gas Law Variables
P
V
T
When T is constant P V
When P is constant T V
When V is constant T P
Dalton’s Law of Partial Pressures
 Gases are often collected by water displacement.
 These gases contain water vapor.
 To find the pressure of the dry gas alone, Dalton’s Law of
Partial Pressures is used.
 The total pressure exerted by a mixture of gases is equal to the sum of
the partial pressures of each gas in the mixture.
Dalton’s Law of Partial Pressures
 Gases are often collected by water displacement.
 These gases contain water vapor.
 To find the pressure of the dry gas alone, Dalton’s Law of
Partial Pressures is used.
 The total pressure exerted by a mixture of gases is equal
to the sum of the partial pressures of each gas in the
mixture.
The Mathematical Expression of
Dalton’s Law
PT = P1 + P2 + …. Pn
Example:
Gas A = 1.0 atm
Gas B = 1.5 atm
Gas C = 0.5 atm
PT = 3.0 atm
The Pressure of a “Dry” Gas
When a gas is collected over water, the total
pressure is the result of the dry gas plus the
water vapor pressure.
PT = Pgas + Pvapor
To determine the pressure of the dry gas alone,
subtract the water vapor pressure from the total
pressure. ( P. 401 )
Pgas = PT - Pvapor
Diffusion
 If a bottle of ammonia were opened at the front of the room,
the odor could soon be detected at the back of the room.
Diffusion is the random
movement of particles through
another substance.
Effusion is gas under pressure
escaping through a small
opening.
Diffusion
Kinetic Energy
The equation for finding the kinetic energy of a particle:
K.E. = ½ mv2
m = mass of the particle
v = velocity (speed) of the particle
Kinetic Energy
K.E. = ½ mv2
#@*~!!*
~##~@!&
Graham’s Law
 If gases A and B are at the same temperature…..
____
____
K.E.A = K.E.B
 ½ mAv2A = ½ mBv2B


mAv2A = mBv2B
Collecting terms…
v2A = mB
v 2B
mA
vA =
vB

mB
mA
This is the
mathematical
expression for
Graham’s Law
The Meaning of Graham’s Law
“The ratio of the velocities of two gases is equal to the
square root of the inverse ratios of their masses.”
mAv2A = mBv2B
if:
mA > mB
then:
vB > vA
(The lightest gas always travels faster)
See practice problems on page 438
A Graham’s Law Experiment
HCl(g) + NH3(g)  NH4Cl(s)
HCl
v = d/t
vA = dA/tA
vB = dB/tB
tA = t B
vA / vB = dA / dB
(experimental)
vA / vB = √mB / mA
(accepted)
% error:
| exp. – accepted | x 100
accepted
NH3
Ideal Gases
 The gas laws we have studied only apply to ideal gases.
 There are no ideal gases.
Properties of Ideal Gases

Ideal gases have:
1. Point mass
(Particles have mass but occupy no volume)
2. No mutual attractions
(No matter how close or how slow the
particles are, they do not attract each other)
Ideal Gas Laws Work
 Real gas particles occupy negligible volume compared to the
total volume occupied by the gas…
(unless under very high pressures)
Real Gases
Ideal Gas Laws Work
•Real gas particles occupy negligible volume
compared to the total volume occupied by
the gas…
(unless under very high pressures)
•Real gas particles have such weak
intermolecular attraction forces that they do
not affectively attract each other…
(unless at very low temperatures)
Real Gas Laws
 Ideal gas laws work for real gases except at very high pressures or at
very low temperatures.
 Why not use real gas laws that work all the time?
Ideal gas law:
PV = KT
Solve for V
Real gas law:
[(P + n2a)/V2] [V – nb] = KT
Solve for V
The Real Gas Law Solution
V = -(P + n2a)  (P + n2a)2 – 4(KT)(P +n2a)(nb)
2KT
Manometer Problems
A Closed Arm Manometer
= ? kPa
165 torr x 101.3 kPa
760 torr
= 22.0 kPa
Manometer Problems
An Open Arm Manometer
The gas exerts 120 torr
more pressure than the
atmosphere.
Gas pressure =
872 torr
872 torr x 101.3 kPa =
760 torr
116 kPa