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Unit 7
Gas Laws and Thermodynamics
Chapters 11 & 16
Gases
CHAPTER 11
Chapter 11 – Section 1: Gases and Pressure
Pressure and Force
• Pressure is the force per unit area on a surface.
Pressure =
Force
Area
Chapter 11 – Section 1: Gases and Pressure
Gases in the Atmosphere
• The atmosphere of Earth is a layer of gases
surrounding the planet that is retained by
Earth's gravity.
• By volume, dry air
is 78% nitrogen,
21% oxygen,
0.9% argon,
0.04% CO2, and
small amounts of
other gases.
Chapter 11 – Section 1: Gases and Pressure
Atmospheric Pressure
• Atmospheric pressure is the force per unit
area exerted on a surface by the weight of the
gases that make up the atmosphere above it.
Chapter 11 – Section 1: Gases and Pressure
Measuring Pressure
• A common unit of pressure is millimeters of
mercury (mm Hg).
• 1 mm Hg is also called 1 torr in
honor of Evangelista Torricelli who
invented the barometer (used to
measure atmospheric pressure).
• The average atmospheric pressure
at sea level at 0°C is 760 mm Hg,
so one atmosphere (atm) of
pressure is 760 mm Hg.
Chapter 11 – Section 1: Gases and Pressure
Measuring Pressure (continued)
•Pressure can also be measured in pascals (Pa):
1 Pa = 1 N/m2.
•One pascal is very
small, so usually
kilopascals (kPa)
are used instead.
•One atm is equal
to 101.3 kPa.
1 atm = 760 mm Hg (Torr) = 101.3 kPa
Chapter 11 – Section 1: Gases and Pressure
Units of Pressure
Chapter 11 – Section 1: Gases and Pressure
Converting Pressure
Sample Problem
The average atmospheric pressure in Denver, CO is
0.830 atm. Express this pressure in:
a. millimeters of mercury (mm Hg)
0.830 atm x 760 mm Hg = 631 mm Hg
1 atm
b. kilopascals (kPa)
0.830 atm x 101.3 kPa
1 atm
= 84.1 kPa
Chapter 11 – Section 1: Gases and Pressure
Dalton’s Law of Partial Pressures
• Dalton’s law of partial pressures - the total
pressure of a gas mixture is the sum of the
partial pressures of the component gases.
PT = P1 + P2 + P3 …
Chapter 11 – Section 1: Gases and Pressure
Dalton’s Law of Partial Pressures
Sample Problem
A container holds a mixture of gases A, B & C. Gas
A has a pressure of 0.5 atm, Gas B has a pressure of
0.7 atm, and Gas C has a pressure of 1.2 atm.
a. What is the total pressure of this system?
PT = P1 + P2 + P3 …
PT = 0.5 atm + 0.7 atm + 1.2 atm = 2.4 atm
b. What is the total pressure in mm Hg?
2.4 atm x 760 mm Hg = 1800 mm Hg
1 atm
Chapter 11 – Section 2: The Gas Laws
Gases and Pressure
• Gas pressure is caused by collisions of the gas
molecules with each other and with the walls of
their container.
• The greater the number of collisions, the higher
the pressure will be.
Chapter 11 – Section 2: The Gas Laws
Pressure – Volume Relationship
• When the volume of a gas is decreased,
more collisions will occur.
• Pressure is caused
by collisions.
• Therefore, pressure
will increase.
• This relationship
between pressure and
volume is inversely proportional.
Chapter 11 – Section 2: The Gas Laws
Boyle’s Law
• Boyle’s Law – The volume of a fixed mass of
gas varies inversely with the pressure at a
constant temperature.
P1V1 = P2V2
• P1 and V1 represent
initial conditions, and
P2 and V2 represent
another set of
conditions.
Chapter 11 – Section 2: The Gas Laws
Boyle’s Law
Sample Problem
A sample of oxygen gas has a volume of 150.0 mL
when its pressure is 0.947 atm.
What will the volume of the gas be at a pressure of
0.987 atm if the temperature remains constant?
Solution:
P1V1 = P2V2
(0.947 atm)(150.0 mL) = (0.987 atm)V2
(0.947 atm)(150.0 mL)
= 144 mL
V2 =
(0.987 atm)
Chapter 11 – Section 2: The Gas Laws
Volume – Temperature Relationship
• the pressure of gas inside
and outside the balloon
are the same.
• at low temperatures, the
gas molecules don’t move
as much – therefore the
volume is small.
• at high temperatures, the
gas molecules move more
– causing the volume to
become larger.
Chapter 11 – Section 2: The Gas Laws
Charles’s Law
• Charles’s Law – The volume of a fixed mass of
gas at constant pressure varies directly with
the Kelvin temperature.
V1
V2
=
T1
T2
• V1 and T1 represent
initial conditions,
and V2 and T2
represent another
set of conditions.
Chapter 11 – Section 2: The Gas Laws
The Kelvin Temperature Scale
• Absolute zero – The theoretical lowest possible
temperature where all molecular motion stops.
• The Kelvin temperature
scale starts at absolute
zero (-273oC.)
• This gives the following
relationship between the
two temperature scales:
K = oC + 273
Chapter 11 – Section 2: The Gas Laws
Charles’s Law
Sample Problem
A sample of neon gas occupies a volume of 752 mL
at 25°C. What volume will the gas occupy at 50°C if
the pressure remains constant?
Solution:
K = oC + 273
V1
V2
T1 = 25 + 273 = 298
=
T1
T2
T2 = 50 + 273 = 323
752 mL
V2
=
298 K
323 K
752 mL
V2 =
x 323 K = 815 mL
298 K
Chapter 11 – Section 2: The Gas Laws
Pressure – Temperature Relationship
• Increasing temperature means increasing
kinetic energy of the particles.
• The energy and frequency of collisions depend
on the average kinetic energy of the molecules.
• Therefore, if
volume is kept
constant, the
pressure of a gas
increases with
increasing
temperature.
Chapter 11 – Section 2: The Gas Laws
Gay-Lussac’s Law
• Gay-Lussac’s Law – The pressure of a fixed
mass of gas varies directly with the Kelvin
temperature.
P1
P2
=
T1
T2
• P1 and T1 represent
initial conditions.
P2 and T2 represent
another set of conditions.
Chapter 11 – Section 2: The Gas Laws
The Combined Gas Law
• The combined gas law is written as follows:
P1 V1 P2 V2
=
T1
T2
• Each of the other gas laws can be obtained
from the combined gas law when the proper
variable is kept constant.
Chapter 11 – Section 2: The Gas Laws
The Combined Gas Law
Sample Problem
A helium-filled balloon has a volume of 50.0 L at
25°C and 1.08 atm. What volume will it have at
0.855 atm and 10.0°C?
K = oC + 273
Solution:
T1 = 25 + 273 = 298
P1V1 P2V2
T2 = 10 + 273 = 283
=
T1
T2
(1.08 atm)(50.0 L) (0.855 atm) V2
=
298 K
283 K
(1.08 atm)(50.0 L) (283 K)
V2 =
= 60.0 L
(298 K) (0.855 atm)
Chapter 11 – Section 3: Gas Volumes and the Ideal Gas Law
Avogadro’s Law
• In 1811, Amedeo Avogadro discovered that the
volume of a gas is proportional to the number of
molecules (or number of moles.)
• Avogadro’s Law - equal volumes of gases at the
same temperature and pressure contain equal
numbers of molecules, or:
V1
V2
=
n1
n2
Chapter 11 – Section 3: Gas Volumes and the Ideal Gas Law
Standard Molar Volume
• Standard Temperature and Pressure (STP) is
0oC and 1 atm.
• The Standard Molar Volume of a gas is the
volume occupied by one
mole of a gas at STP.
It has been found to
be 22.4 L.
Chapter 11 – Section 3: Gas Volumes and the Ideal Gas Law
Molar Volume Conversion Factor
• Standard Molar Volume can be used as a
conversion factor to convert from the
number of moles of a gas at STP to volume
(L), or vice versa.
Chapter 11 – Section 3: Gas Volumes and the Ideal Gas Law
Molar Volume Conversion
Sample Problem
a. What quantity of gas, in moles, is contained
in 5.00 L at STP?
5.00 L x
1 mol
22.4 L
= 0.223 mol
b. What volume does 0.768 moles of a gas
occupy at STP?
0.768 mol x
22.4 L
1 mol
= 17.2 L
Chapter 11 – Section 3: Gas Volumes and the Ideal Gas Law
Volume Ratios
• You can use the volume ratios as conversion
factors just like you would use mole ratios.
2CO(g)
+ O2(g)
→
2 molecules 1 molecule
2 mole
1 mole
2 volumes
1 volume
2CO2(g)
2 molecules
2 mol
2 volumes
• Example: What volume of O2 is needed to react
completely with 0.626 L of CO to form CO2?
0.626 L CO x
1 L O2
2 L CO
= 0.313 L O2
Chapter 11 – Section 3: Gas Volumes and the Ideal Gas Law
The Mole Map
• You can now convert between number of particles,
mass (g), and volume (L) by going through moles.
Chapter 11 – Section 3: Gas Volumes and the Ideal Gas Law
Gas Stoichiometry
Sample Problem
Assume that 5.61 L H2 at STP reacts with excess CuO
according to the following equation:
CuO(s) + H2(g) → Cu(s) + H2O(g)
a. How many moles of H2 react?
5.61 L H2 x 1 mol H2 = 0.250 mol H2
22.4 L H2
b. How many grams of Cu are produced?
5.61 L H2 x 1 mol H2 x 1 mol Cu x 63.5 g Cu = 15.9 g
22.4 L H2 1 mol H2 1 mol Cu
Cu
Chapter 11 – Section 3: Gas Volumes and the Ideal Gas Law
The Ideal Gas Law
• All of the gas laws you have learned so far can
be combined into a single equation,
the ideal gas law:
PV = nRT
• R represents the ideal gas constant which has
a value of 0.0821 (L•atm)/(mol•K).
Chapter 11 – Section 3: Gas Volumes and the Ideal Gas Law
The Ideal Gas Law
Sample Problem
What is the pressure in atmospheres exerted by
a 0.500 mol sample of nitrogen gas in a 10.0 L
container at 298 K?
Solution:
PV = nRT
P (10.0 L) = (0.500 mol)(0.0821 L•atm/mol•K) (298 K)
(0.500 mol) (0.0821 L•atm/mol•K) (298 K)
P=
(10.0 L)
1.22
=
atm
Chapter 11 – Section 4: Diffusion and Effusion
Diffusion and Effusion
• Diffusion is the gradual mixing of two or more
gases due to their spontaneous, random
motion.
• Effusion is the process
whereby the molecules of
a gas confined in a
container randomly pass
through a tiny opening in
the container.
Chapter 11 – Section 4: Diffusion and Effusion
Graham’s Law of Effusion
• Light molecules move faster than heavy ones.
• Graham’s
law of
effusion says
the greater
the molar
mass of a gas,
the slower it
will effuse.
Thermochemistry
CHAPTER 16
Chapter 16 – Section 1: Thermochemistry
Heat and Temperature
• Temperature – a measure of the average kinetic
energy of the particles in a sample of matter.
• The greater the kinetic energy of the particles in a
sample, the hotter it feels.
• Heat – energy transferred between samples of
matter due to a difference in their temperatures.
• Heat always moves
spontaneously from
matter at a higher
temperature to matter
at a lower temperature.
Chapter 16 – Section 1: Thermochemistry
Measuring Heat
• Heat energy is measured in
joules (or calories – food only)
• Chemical reactions usually
either absorb or release energy
as heat.
• The energy absorbed or
released as heat in a chemical
or physical change is measured
in a calorimeter.
Chapter 16 – Section 1: Thermochemistry
Specific Heat
• A quantity called specific
heat can be used to
compare heat absorption
capacities for different
materials.
• Specific heat – the amount of energy required to
raise the temperature of one gram of a substance
by 1°C or 1 K.
• Specific heat can be measured in units of J/(g•°C),
J/(g•K), cal/(g•°C), or cal/(g•K).
Chapter 16 – Section 1: Thermochemistry
Heat Transfer Equation
• Specific heat can be used to find the quantity of
heat energy gained or lost with a change in
temperature according to the following equation:
Q = m x cp x ∆T
• Where the variables stand for the following:
Q = heat transferred (joules or calories)
m = mass (g)
cp = specific heat
∆T = change in temperature (oC or K)
Chapter 16 – Section 1: Thermochemistry
Heat Transfer Equation
Sample Problem
A 4.0 g sample of glass was heated from 274 K to 314 K,
a temperature increase of 40. K, and was found to have
absorbed 32 J of energy as heat.
a. What is the specific heat of this type of glass?
Q = m x cp x ∆T
32 J = (4.0 g)(cp) (40. K)
32 J
= 0.20
cp =
(4.0 g)(40. K) J/(g•K)
b. How much energy will the same glass sample
gain when it is heated from 314 K to 344 K?
Q = m x cp x ∆T
Q = (4.0 g)(0.20 J/(g•K))(30 K) = 24 J
Chapter 16 – Section 1: Thermochemistry
Enthalpy of Reaction
• The enthalpy of reaction (H) is the quantity of
energy transferred as heat during a chemical
reaction.
• The change in enthalpy (∆H) of a
reaction is always the difference
between the enthalpies of the products
and the reactants.
∆H = Hproducts - Hreactants
Chapter 16 – Section 1: Thermochemistry
Exothermic Reactions
• In an exothermic reaction,
energy is released. Therefore,
the energy of the products
must be less than the energy
of the reactants, and ∆H is negative.
• The great majority of
chemical reactions in
nature are exothermic.
Chapter 16 – Section 1: Thermochemistry
Endothermic Reactions
• In an endothermic reaction, energy is
absorbed. Therefore, the energy of the
products must be greater than the energy of
the reactants, and ∆H is positive.
Chapter 16 – Section 1: Thermochemistry
Entropy and Reaction Tendency
• Entropy – a measure of the degree of
randomness in a system.
• Processes in nature are
driven in two directions:
towards decreasing
enthalpy and towards
increasing entropy.
• The combined enthalpy-entropy function is
called the Gibbs free energy (G).
Chapter 16 – Section 1: Thermochemistry
Hess’s Law
• Hess’s Law – the overall enthalpy change in a
reaction is equal to the sum of enthalpy changes
for the individual steps in the process.
• Possible steps in Hess’s Law:
1. Reverse equation/
change sign on ΔH.
2. Multiply or Divide
coefficients/multiply
or divide ΔH.
Chapter 16 – Section 1: Thermochemistry
Hess’s Law
Sample Problem
Calculate the enthalpy of formation for CH4:
C(s) + 2H2(g) → CH4(g)
∆Hf = ?-74.3 kJ
The component reactions are:
C(s) + O2(g) → CO2(g)
∆Hc = -393.5 kJ
-571.6 kJ
kJ
2H2(g) + ½O2(g) →2 H2O(l)
∆Hc = -285.8
CH4(g) + 2O2(g) → CO2(g) + 2H2O(l) ∆Hc = -890.8 kJ
CO2(g) + 2H2O(l) → CH4(g) + 2O2(g) ∆Hc = +890.8 kJ
-74.3 kJ
- 2 moles of H2 are used to make CH4,
so multiply the 2nd equation by 2 (including ∆H.)
- CH4 is on the products side, not the reactants side, so
reverse the 3rd reaction and change the sign on ∆H.
- Cancel unwanted terms and add the ∆H’s.