Transcript Chapter 6 - Gases - Montville Township School District

Gases and Gas
Laws
Introduction
The properties of gases will be
introduced along with five ways of
predicting the behavior of gases: Boyle’s
Law, Charles’ Law, Gay-Lussac’s Law,
Avogadro’s Law and the Ideal Gas Law.
Properties of Gases
You can predict the behavior of gases
based on the following properties:
Pressure
Volume
Amount (moles)
Temperature
Lets review each of these briefly…
Physical Characteristics of Gases

Although gases have different chemical properties, gases have
remarkably similar physical properties.

Gases always fill their containers (recall solids and liquids). No definite
shape and volume

Gases are highly compressible: Volume decreases as pressure
increases. Volume increases as pressure decreases.

Gases diffuse (move spontaneously throughout any available space).

Temperature affects either the volume or the pressure of a gas, or both.
Definition of a Gas

Therefore a definition for gas is: a
substance that fills and assumes the
shape of its container, diffuses rapidly, and
mixes readily with other gases.
Pressure

Automobile tires, basketballs, balloons,
and soda bottles.
Pressure
• Evangelista Torricelli
invented the
– BAROMETER: device
to measure pressure
– At sea level the height
of the column of
mercury is 760 mm.
Pressure exerted by the atmospheric gases
on the surface of the mercury in the dish
keeps the mercury in the tube.
Measuring pressure of a confined gas
Atmospheric pressure > gas
Manometer
Atmospheric pressure < gas
Pressure
Units of Pressure
-1mm Hg = 1 torr
Standard atmosphere (atm)
1 atm. = 760.0mm Hg = 760.0 torr
Problem
The height of mercury in a mercury barometer is
measured to be 732 mm Hg. Represent this
pressure in atm and torr.
1 atm. = 760.0mm Hg = 760.0 torr
Problem
The height of mercury in a mercury barometer is
measured to be 732 mm Hg. Represent this
pressure in atm and torr.
1 atm. = 760.0mm Hg = 760.0 torr
Answer: 0.963 atm; 732 torr
Kelvin Temperature Scale
Kelvin Temp Scale: based on absolute zero — all kinetic
motion stops

Formulas
°C = K - 273
K= °C+273
0°C = 273K
30°C =303 K
-20°C = 253 K
Various gas laws describe the relationships among
four of the important physical properties of gases:
1. Volume (liters)
2. Pressure (usually in atms)
3. Temperature (Kelvin)
4. Amount (moles)
Five Gas Laws

Boyle’s Law

Charles Law

Gay-Lussac’s Law

Avogadro’s Law

Ideal Gas Law
Boyle’s Law

Describes the pressure-volume
relationship of gases if the temperature
and amount are kept constant.

P1V1 = P2V2

P1V1 = initial pressure and volume
P2V2 = final pressure and volume

Boyle’s Law
Volume and
pressure are
inversely
proportional
Illustration of Boyle’s law.
P1V1= P2V2
Pressure and Volume: Boyle’s Law

A sample of neon gas has a pressure of
7.43 atm in a container with a volume of
45.1 L. This sample is transferred to a
container with a volume of 18.4 L. What
is the new pressure of the neon gas?
P1V1= P2V2
18.2 atm
Pressure and Volume: Boyle’s Law

A steel tank of oxygen gas has a volume
of 2.00L. If all of the oxygen is
transferred to a new tank with a volume
of 5.50 L, the pressure is measured to be
6.75 atm. What was the original
pressure of the oxygen gas?
18.6 atm
Charles’s Law

Describes the temperature-volume
relationship of gases if the pressure and
amount are kept constant.

the volume of a gas increases
proportionally as the temperature of the
gas increases.
V1 = V2
T1 T2
Charles’s Law shows
that the volume of a
gas (at constant pressure)
increases with the
temperature.
Charles’s Law is used to
explain what happens to a
balloon when placed in a
freezer.
Volume and Temperature:
Charles’s Law

A sample of methane gas is collected at
285 K and cooled to 245K. At 245 K the
volume of the gas is 75.0 L. Calculate
the volume of the methane gas at 285K.
V1 = V2
T1 T2
V1 = 87.2 L
Volume and Temperature:
Charles’s Law

A 2.45 L sample of nitrogen gas is
collected at 273 K and heated to 325K.
Calculate the volume of the nitrogen gas
at 325 K.
V1 = V2
T1 T2
V2 = 2.92 L
Volume and Temperature:
Charles’s Law

Consider a gas with a volume of 5.65 L
at 27 C and 1 atm pressure. At what
temperature will this gas have a volume
of 6.69 L and 1 atm pressure.
V1 = V2
T1 T2
T2 = 82oC (355K)
Volume and Temperature:
Charles’s Law

Consider a gas with a volume of 9.25L at
47oC and 1 atm pressure. At what
temperature does this gas have a
volume of 3.50 L and 1 atm pressure.
T2 = -152oC (121K)
Gay-Lussac’s Law

Describes the temperature-pressure
relationship of gases if the volume and
amount are kept constant.

the pressure of a gas increases
proportionally as the temperature of the
gas increases.
P1 = P2
T1 T2
Pressure and Temperature: GayLussac’s Law

If you have a tank of gas at 800 torr
pressure and a temperature of 250
Kelvin, and it is heated to 400 Kelvin,
what is the new pressure?
P1 = P2
T1 T2
P2 = 1,280 torr
Sample Problem 13.3
Page 448
Classwork
Page 448, Problems 8 and 9
Combined Gas Law

This is when all variables (T,P, and V) are
changing
P1 V1 = P2 V2
T1
T2
The Combined Gas Law
Consider a sample of helium gas at 23oC
with a volume of 5.60 L at a pressure of
2.45 atm. The pressure is changed to
8.75 atm and the gas is cooled to 15oC.
Calculate the new volume of the gas using
the combined gas law equation.
P1 V1 = P2 V2
T1
T2
V2 = 1.53 L
The Combined Gas Law
Consider a sample of helium gas at 28oC
with a volume of 3.80 L at a pressure of
3.15 atm. The gas expands to a volume of
9.50 L and the gas is heated to 43oC.
Calculate the new pressure of the gas
using the combined gas law equation.
P2 = 1.32 atm
Avogadro’s Law

Describes the amount-volume relationship of
gases if the pressure and temperature are
kept constant.

Equal volume of gases at the same
temperature and pressure contain equal
number of moles of gas.

the volume of a gas is directly proportional to
the number of moles of gas
V1 = V2
n1 n2
Avogadro’s Law
V1 = V2
n1 n2
Volume and Moles: Avogadro’s
Law
If 2.55 mol of helium gas occupies a volume
of 59.5 L at a particular temperature and
pressure, what volume does 7.83 mol of
helium occupy under the same conditions?
V1 = V2
n1 n2
V2 = 183 L
Volume and Moles: Avogadro’s
Law
If 4.35 g of neon gas occupies a volume of
15.0 L at a particular temperature and
pressure, what volume does 2.00 g of
neon gas occupy under the same
conditions?
V2 = 6.90 L
Avogadro’s Law
A very useful consequence of Avogadro’s law is that the volume
of a mole gas can be calculated at any temperature and pressure.
An extremely useful form to know when calculating the volume
of a mole of gas is 1 mole of any gas at STP occupies 22.4 liters.
STP stands for standard temperature and pressure.
Standard Pressure: 1.00 atm (760 torr or 760 mm Hg)
Standard Temperature: 273K
Molar Volume

When gases are at STP:
1
mole of any gas = 22.4 L/mol
 We
will return to this concept with gas
stoichiometry problems.
Ideal Gas Equation
PV=nRT
The Ideal Gas Law
PV=nRT
R=Universal gas constant (proportionality
constant)
R= 0.08206 L atm/ K
Pressure is in atm.
Volume is in liters
Temperature is in Kelvin (K)
The Ideal Gas Law
A sample of neon gas has a volume of 3.45 L
at 25oC and a pressure of 565 torr.
Calculate the number of moles of neon
present in the gas sample.
PV=nRT
R=0.0821 L atm/K
Convert pressure to atm and temperature to K
n = 0.105 mol
The Ideal Gas Law
A 0.250 mol sample of argon gas has a
volume of 9.00 L at a pressure of 875 torr.
What is the temperature (in K) of the gas?
PV=nRT
T = 505K
At 28oC and 0.974 atm, what is the volume of 5.15 gra
of O2 gas ?
Given: P of gas = 0.974 atm
V of gas = ?
T of gas = 28oC + 273 = 301K
m of gas = 5.16g, convert to moles
PV = nRT
Dalton’s Law of Partial Pressures
Objectives: To understand the relationship
between the partial and total pressures of a
gas mixture, and to use this relationship in
calculations.
Dalton’s Law of Partial Pressures
John Dalton: For a mixture of gases in a container,
the total pressure exerted is the sum of the
partial pressures of the gases present.
Partial pressure: pressure that each gas would
exert if it were alone in the container.
The pressure that each gas exerts in the mixture is
independent of that exerted by other gases
present
Dalton’s law of partial pressures:
Ptotal = P1 + P2 + P3
P1, P2 and P3 are the partial pressure of component
gases 1, 2 and 3.
Ptotal=ntotal (RT/V)
The volume of the individual gas particles
is not important.
Gases Collected by Water Displacement
Gases produced in the laboratory are often
collected over water.
The gas produced by the reaction displaces
the water, which is more dense.
You can apply Dalton’s law of partial pressures
in calculating the pressure of gases collected
in this way.
The production of oxygen by
thermal decomposition.
A gas collected by water displacement is not pure
but is always mixed with water vapor.
Water molecules at the liquid surface evaporate and
mix with the gas molecules.
Water vapor, like other gases, exerts a pressure,
known as water-vapor pressure.
Patm = Pgas + PH20
Patm is read from a barometer in the lab and
is equal to Ptotal
PH20 varies with temperature and is in a reference table
Dalton’s Law of Partial Pressures
A sample of solid potassium chlorate KClO3, was
heated in a test tube and decomposed
according to the reaction:
2KClO3(s)
2KCl(s) + 3O2
The oxygen produced was collected by
displacement of water. The barometric
pressure and the temperature during the
experiment were 731.0 torr and 20oC,
respectively. The vapor pressure of water at
20oC is 17.5 torr. What is the partial pressure
of oxygen collected?
Dalton’s Law of Partial Pressures
Ptotal = Patm = PO2 + PH2O
PH2O = 17.5 torr (vapor pressure of water at 20oC)
731.0 torr = PO2 + 17.5 torr
PO2= 713.5 torr
Dalton’s Law of Partial Pressures
A sample of solid potassium chlorate KClO3, was
heated in a test tube and decomposed
according to the reaction:
2KClO3(s)
2KCl(s) + 3O2
The oxygen produced was collected by
displacement of water at 22oC. The resulting
mixture of O2 and H2O vapor had a total
pressure of 754 torr and a volume of 0.650L.
Calculate the partial pressure of O2 in the gas
collected and the number of moles of O2
present. The vapor pressure of water at 22oC
is 21 torr.
Dalton’s Law of Partial Pressures
Ptotal=PO2 + PH2O
754=PO2 + 21
PO2= 733 torr
nO2 =PO2V
733/760=0.964 atm
RT
nO2= (0.964 atm)(0.650L)
(0.08206) (295 K)
n = .026 moles
Hydrogen gas is collected over water at 22oC. The
resulting mixture of H2 and H2O vapor had a
total pressure of 740 torr and a volume of 1.2L.
Calculate the partial pressure of H2 in the gas
collected and the number of moles of H2
present.
The vapor pressure of water at is 21 torr.
Dalton’s Law of Partial Pressures
Ptotal=PH2 + PH2O
740=PH2 + 21
PH2= 719 torr
nH2 =PH2V
719/760=0.946 atm
RT
nH2= (0.946 atm)(1.20L)
(0.08206) (295 K)
n = 0.047 moles
Dalton’s Law of Partial Pressures
A sample of oxygen gas is saturated with
water vapor at 30.0oC. The total
pressure is 753 torr and the vapor
pressure of water at 30.0 oC is 31.824
torr. What is the partial pressure of the
oxygen gas in atm?
0.949 atm
Gas Stoichiometry
Objectives:
1) To understand the molar volume of an
ideal gas.
2) To review the definition of STP
3) To use these concepts and the ideal gas
equation.
Gas Stoichiometry
For 1 mole of a gas at 0oC (273K) and 1 atm, the
volume will be.
V= nRT/P = (1.0 mol)(0.08206)(273K) = 22.4L
1 atm.
22.4 L is called the molar volume at standard
temperature and pressure (abbreviated STP).
Contains 1 mole of an ideal gas at STP.
Gas Stoichiometry
When magnesium reacts with hydrochloric
acid, hydrogen gas is produced:
Mg(s) + 2HCl
MgCl2(aq) + H2(g)
Calculate the volume of hydrogen gas
produced at STP by reacting 5.0g Mg
and an excess of HCl (aq)
5.0 g Mg
1 molMg
24.0 g Mg
1mol H 2
1mol Mg
= 0.21 moles H2
Units match
PV=nRT
(1.0 atm)(V) = 0.21(0.082)(273K)
Volume = 4.66 L
5.0 g Mg
1 molMg
24.0 g Mg
1mol H 2
1mol Mg
Volume = 4.66 L
22.4 L
1 mol H2
Gas Stoichiometry
When subjected to an electric current, water
decomposes to hydrogen and oxygen
gas: 2H2O(l)
2H2(g) + O2(g)
If 25.0g of water is decomposed, what
volume of oxygen gas is produced at
STP?
25.0 g H20
1 molH2O
18.0g H2O
1molO2
2molH20
= 0.69 moles O2
Units match
PV=nRT
(1.0 atm)(V) = 0.69(0.082)(273K)
Volume = 15.5 L
Method 2
1 mole of a gas occupies 22.4 liters of volume
25.0 g H2O
1 mol H2O
1 mol O2
22.4 L O2
18.0 g H2O
2 mol H2O
1 mol O2
= 15.5 L O2
How many liters of oxygen gas, at STP, can be
collected from the complete decomposition of
50.0 grams of potassium chlorate?
2 KClO3(s)  2 KCl(s) + 3 O2(g)
50.0 g KClO3
1 mol KClO3
122.55 g KClO3
3 mol O2
22.4 L O2
2 mol KClO3
1 mol O2
= 13.7 L O2