Gases
description
Transcript of Gases
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Chapter 12
Gases
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Properties of GasesCompressibilityMassVolume (fill container)Exert PressureDiffuse through other gasesLow Density
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Kinetic Molecular Theory (KMT)1. All gases are matter (have mass and take up
space).2. These particles are in constant, rapid and
random motion.3. All collisions are perfectly elastic. (no energy is
lost in the collision)4. The force of gas particle collisions on the walls of
the container creates pressure; however, gas particles do not exert force on each other.
5. At a given temperature, all gas particles have the same amount of kinetic energy (temperature).
6. The distance between gas particles is very large.
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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.
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Measuring GasesIn order to describe a gas, we use four variables
to make predictions about the behavior of the particles.1. Amount of a gas (n) expressed in moles2. 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. *note: K = C + 273
4. Measure of the force exerted by the gas particles on the walls of the container, pressure (P)
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Kinetic Energy and Temperature
Temperature is an indication of kinetic energy.
Fra
ctio
n of
Par
ticl
es
Speed of Particles
T1
T2 T2 > T1
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Kelvin scale
The Kelvin Temperature Scale
(Absolute Zero Scale)
-300 -200 -100 0 100 oCelcius scaleKin
etic
Ene
rgy
of M
olec
ules
0
-273oC
0 73 173 273 373
K = oC + 273
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PressurePressure 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 pressureA manometer is an instrument that measures
the pressure of an enclosed gas
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Mercury Barometer
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Manometer
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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}
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Pressure UnitsKilopascals (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)
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Pressure Units
Atmosphere (atm)Millimeter Hg (mmHg)
Pascal (Pa)Pounds per square inch
(psi)Kilopascal (kPa)
1 atm = Standard Pressure
1 atm = 760 mmHg1 atm = 101,325 Pa1 atm = 14.7 psi1 atm = 101.3 kPa
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ConversionsCovert using Dimensional AnalysisInformation 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 #1-4Homework: page 422 #1-12
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Boyle’s Law
Pressure and Volume are inversely proportional at a given temperature and number of molecules.
VP = k (a constant value)
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Pressure Effects on Volume
V 1/P
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Inverse Proportionality
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Boyle’s Law continuedV1P1 = k1
and V2P2 = k2
since k1 = k2 (that’s why its called a constant)
V1P1 = V2P2
This is the mathematical expression for Boyle’s Law
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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 =
V2 =
P1 =
P2 =
242 cm3
?
87.6 kPa
101.3 kPa
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Two Methods of Solving
1. Algebraic:
V1P1 = V2P2
V2 = V1P1
P2
209 cm3
V2 = 242 cm3 x 87.6 kPa101.3 kPa
V2 =
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Another Method2. Logic and reason:
The new volume (V2) depends upon the old volume (V1) multiplied by a pressure change.
V2 = 242 cm3 x _______
Since the pressure is increasing, the volume should decrease.
Multiply by a fraction less than 1.
87.6
101.3
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Charles’s LawAt a given pressure and number of
molecules, Volume and Temperature are directly proportional.
V = kT
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Two Temperature Scales
-300 -200 -100 0 100 oCelcius scale
0 73 173 273 373 Kelvin scale
0
Vol
ume
y = mx + b
V = kT + bV = kT + 0
V = kT
b
b
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Charles’s Law Continued
V1 = k1
T1
and V2 = k2
T2
since k1 = k2
V1 = V2
T1 T2This is the mathematical expression for Charles’s Law
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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 =
T1 =
T2 =
225 cm3
?
58oC + 273 = 331 K
0oC + 273 = 273 K
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Two Methods of Solving
1. Algebraic:
V1 / T1 = V2 / T2
V2 = V1T2
T1
V2 = 225 cm3 x 273 K 331 K
V2 = 186 cm3
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The Other Method
2. Logic and reason:
The new volume (V2) depends upon the old volume (V1) multiplied by a temperature change.
V2 = 225 cm3 x _______
Since the temperature is decreasing, the volume should decrease.
Multiply by a fraction less than 1.
273
331
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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
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The Combined Gas LawIn most laboratory situations, both
temperature and pressure change.
The Combined Gas Law is used to calculate the affects of these changes.
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A Combination of Gas Laws
Boyle’s Law:V 1/P
Charles’s Law:V T
V T / P
To change a proportionality to an equality… multiply by a proportionality constant.
V = kT PVP = k T
V1P1 = V2P2
T1 T2
Gay-Lussac’s Law: P T
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V1P1 = V2P2
T1 T2 Solve for each variable
V2 = V1P1T2 / T1P2
T2 = V2P2T1 / V1P1
P2 = V1P1T2 / T1V2
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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?
V1 =
V2 =
T1 =
T2 =
P1 =
P2 =
955 cm3
?
331 K
349 K
108.0 kPa
123.0 kPa
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Method #1V1P1 / T1 = V2P2 / T2
V2 = V1P1T2 / (T1P2)
V2 = (955 cm3) (108.0 kPa) (349 K)
(331 K) (123.0 kPa)
V2 = 884 cm3
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Method #2
V2 = 955 cm3 x --------- x ---------
temperature increasesvolume increases
V2 = 884 cm3
349
331
pressure increases
volume decreases
123
108
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Gas Law Variables
P
T
V
When T is constant P VWhen P is constant T V
When V is constant T P
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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.
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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.
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The Mathematical Expression of Dalton’s Law
PT = P1 + P2 + …. Pn
Example:
Gas A = 1.0 atmGas B = 1.5 atmGas C = 0.5 atm
PT =3.0 atm
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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
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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.
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Diffusion
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Kinetic Energy
The equation for finding the kinetic energy of a particle:
K.E. = ½ mv2
m = mass of the particlev = velocity (speed) of the particle
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Kinetic Energy
K.E. = ½ mv2
#@*~!!*~##~@!&
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Graham’s LawIf gases A and B are at the same
temperature…..
____ ____ K.E.A = K.E.B
½ mAv2A = ½ mBv2
B
mAv2A = mBv2
B
Collecting terms…
v2A = mB
v2B mA
vA = mB
vB mA
This is the mathematical expression for Graham’s Law
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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 = mBv2
B
if: mA > mB
then: vB > vA
(The lightest gas always travels faster)See practice problems on page 438
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A Graham’s Law Experiment
HCl(g) + NH3(g) NH4Cl(s)
HCl NH3
v = d/t
vA = dA/tA
vB = dB/tB
tA = tB
vA / vB = dA / dB
(experimental)
vA / vB = √mB / mA
(accepted)
% error:
| exp. – accepted |accepted
x 100
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Ideal Gases
The gas laws we have studied only apply to ideal gases.
There are no ideal gases.
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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)
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Ideal Gas Laws Work
Real gas particles occupy negligible volume compared to the total volume occupied by the gas…
(unless under very high pressures)
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Real Gases
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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)
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Real Gas LawsIdeal 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 VReal gas law: [(P + n2a)/V2] [V – nb] = KT
Solve for V
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The Real Gas Law Solution
V = -(P + n2a) (P + n2a)2 – 4(KT)(P +n2a)(nb) 2KT
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Manometer Problems
A Closed Arm Manometer
165 torr x 101.3 kPa 760 torr= 22.0 kPa
= ? kPa
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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 torr116 kPa