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# Gases - PowerPoint PPT Presentation

Gases. Revision history: 10/5/01 2/20/03 5/21/03 6/24/04 12/27/06 12/29/06 01/05/10 01/09/10 050212. Pisgah High School M. Jones. The Properties of Gases. Part I. Properties of Gases. Gases expand to fill the container. Gases take on the shape of the container.

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Gases

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## Gases

Revision history:

10/5/01

2/20/03

5/21/03

6/24/04

12/27/06

12/29/06

01/05/10

01/09/10

050212

Pisgah High School

M. Jones

## The Properties of Gases

Part I

### Properties of Gases

• Gases expand to fill the container.

• Gases take on the shape of the container.

• Gases are highly compressible. (Can be liquefied at high pressures).

• Gases have low densities.

• Gases mix uniformly.

### The Kinetic Molecular Theory

The kinetic molecular theory describes the behavior of ideal gases.

An ideal gas is one that conforms to the KMT.

### 1. Molecules are in constant random motion

Temperature is proportional to the average kinetic energy of the molecules. KE = ½ mv2

KE = ½ mass times speed squared

The speed is proportional to the absolute temperature (Kelvin).

### 2. A gas is mostly empty space

Molecules are far apart from each other. This accounts for the low density and high compressibility.

The volume of the individual molecules is negligible compared to the volume of the gas.

### 3. No intermolecular forces

There are no attractive or repulsive forces between gas molecules.

Adjacent molecules do not attract or repel each other.

### 4. Collisions are elastic

When gas molecules collide with each other they may speed up or slow down, BUT …

The net (total) energy of the gas molecules does not change.

### Kinetic Molecular Theory

• Gases in constant motion, speed depends on temperature.

• Molecules have negligible volume.

• No intermolecular forces.

• Elastic collisions. No change in energy.

### Temperature reminder

When doing calculations, temperature must always be in an absolute temperature scale …

… where the lowest possible temperature is zero degrees.

Use Kelvin degrees!

### Temperature conversion

K =

C +

273

Pressure

1. Pressure is the force per unit area exerted by the gas molecules.

2. Pressure is proportional to the number of collisions between the gas molecules and the walls of the container.

### Pressure

P =

1. Pressure is a measure of the force per unit area.

force

Pressure can be in pounds per square inch (PSI), or …

area

… newtons per square meter (N/m2)

pascal (Pa) = N/m2

### Pressure

P =

At sea level, air pressure holds up a column of mercury 760 mm high.

1. Pressure is a measure of the force per unit area.

force

area

Glass tube with Hg

Torricelli

Bowl of Hg

### Pressure Measurements

Standard sea level pressure is…

1.00 atmospheres (atm)

760 mm Hg

760 torr (from Torricelli)

101.3 kilopascals (kPa)

14.7 lb/in2

### Pressure Measurements

Standard sea level pressure is…

1.00 atmospheres (atm)

760 mm Hg

760 torr (from Torricelli)

101.3 kilopascals (kPa)

14.7 lb/in2

Exact

### Pressure

2. Pressure is proportional to the number of collisions between the gas molecules and the walls of the container.

If you change the number of collisions, you change the pressure.

## The Gas Laws

Part II

Boyle’s Law

Amonton’s Law

Charles’s Law

Combined Gas Law

Gay-Lussac’s Law

Dalton’s Law

### Boyle’s Law

At a constant temperature, pressure is inversely proportional to volume.

Pressure

Volume

### Boyle’s Law

At a constant temperature, pressure is inversely proportional to volume.

1/Pressure

Volume

### Boyle’s Law

1

P µ

V

At a constant temperature, pressure is inversely proportional to volume.

PV = k

P1V1 = P2V2

Year: 1662

Amonton’s Law

P µ

T

P1

P2

=

T1

T2

“Air thermometer”, 1695.

This is not Gay-Lussac’s law.

Diagram from http://chemed.chem.purdue.edu/genchem/topicreview/bp/ch4/gaslaws3.html

Amonton’s Law

P µ

T

P1

P2

=

T1

T2

This is why you measure your tire pressure when the tire is cold. Tire pressures vary

with temperature.

Amonton’s Law

Amonton’s air thermometer was used to find the value of absolute zero.

Measure the pressures of a gas at various temperatures at a constant volume.

### Finding Absolute Zero

-273 C

Pressure

Extrapolate to the x-axis

-300 -150 0 100 200 300

Temperature (C)

### Charles’s Law

Volume

Temperature

At constant pressure, volume is directly proportional to temp.

### Charles’s Law

V µ

T

V

V1

V2

= k

=

T

T1

T2

At constant pressure, volume is directly proportional to temperature.

Charles’s Law

• Studied gases during 1780’s.

• Hydrogen balloon assents, 3000 m, in 1783.

• Collaborated with the Montgolfier brothers on hot air balloons, 1783.

Charles’s Law

Hydrogen balloon assent, 3000 m, 1783.

T

1

P µ

T

P µ

P µ

V

V

gives

T

P µ

V

kT

P =

V

to an

Next we convert

### Combined Gas Law

PV

= k

kT

T

P =

V

Rearranging the equation gives:

### Combined Gas Law

PV

= k

T

Combining the laws of Boyle, Amonton and Charles produces the combined gas law.

### Combined Gas Law

P2V2

P1V1

k

k

=

=

T2

T1

Consider a confined gas at two sets of conditions. Since the number of molecules is the same and the values of k are the same, then we can combine the two equations.

P2V2

P1V1

k

k

=

=

T2

T1

P1V1

P2V2

=

T1

T2

### Combined Gas Law

P1V1

P2V2

=

T1

T2

Use the combined gas law whenever you are asked to find a new P, V or T after changes to a confined gas.

### Sample Combined Gas Law Problem

Consider a confined gas in a cylinder with a movable piston. The pressure is 0.950 atm. Find the new pressure when the volume is reduced from 100.0 mL to 65.0 mL, while the temperature remains constant?

100 mL

65 mL

100 mL

65 mL

### Sample Combined Gas Law Problem

P1V1

P2V2

=

T1

T2

100 mL

Since the temperature is constant, we can cancel out T1 and T2.

65 mL

### Sample Combined Gas Law Problem

P1V1

=

P2V2

P1V1

P2

=

V2

This becomes Boyle’s Law

100 mL

Next, solve for P2.

65 mL

### Sample Combined Gas Law Problem

(0.950 atm)(100.0 mL)

P2

=

65.0 mL

P1V1

P2

=

P2

=

1.46 atm

V2

100 mL

65 mL

### Combined Gas Law Problems

• 1. A sample of neon gas has a volume of 2.00 L at 20.0 C and 0.900 atm. What is the new pressure when the volume is reduced to 0.750 L and the temperature increases to 24.0 C?

2.43 atm

### Combined Gas Law Problems

• 2. Some “left over” propane gas in a rigid steel cylinder has a pressure of 24.6 atm at a temperature of 20.C. When thrown into a campfire the temperature in the cylinder rises to 313C. What will be the pressure of the propane?

49.2 atm

### Combined Gas Law Problems

• 3. Consider the fuel mixture in the cylinder of a diesel engine. At its maximum, the volume is 816 cc. The mixture comes in at 0.988 atm and 31 C. What will be the temperature (in C) when the gas is compressed to 132 cc and 42.4 atm?

1837 C

### Gay-Lussac’s Law

Also known as the

law of combining volumes.

At a given temperature and pressure, the volumes of reacting gases are in a ratio of small, whole numbers.

Year: 1802

### Gay-Lussac’s Law

Gay-Lussac found that the volumes of gases in a reaction were in ratios of small, whole numbers.

2 H2(g) + O2(g)  2 H2O(g)

200 mL of hydrogen reacts with 100 mL of oxygen.

2:1

### Gay-Lussac’s Law

2 H2(g) + O2(g)  2 H2O(g)

The ratio of volumes of gases come from the ratios of the coefficients in the balanced equation.

V µ

n

V

V1

V2

= k

=

n

n1

n2

Avogadro’s law followed Dalton’s atomic theory and Gay-Lussac’s law.

Year: 1811.

Equal volumes of gases at the same temperature and pressure have equal numbers of molecules.

### Dalton’s Law

Dalton’s law of partial pressures deals with mixtures of gases.

The total pressure is the sum of the partial pressures:

Ptotal = P1 + P2 + P3 …

Use when dealing with the pressure of H2O(g) when collecting a gas over water.

### Dalton’s Law

“Collecting a gas over water”

Inverted gas collecting bottle

Rubber tubing carrying H2 gas

Flask with metal and HC l

Water in trough

### Dalton’s Law

“Collecting a gas over water”

Inverted gas collecting bottle

Rubber tubing carrying H2 gas

Flask with metal and HC l

Water in trough

### Dalton’s Law

Water is displaced through the mouth of the bottle as H2 gas bubbles in.

H2 gas

Flask with metal and HC l

Water in trough

### Dalton’s Law

A mixture of hydrogen gas and water vapor is in the collection bottle.

H2 gas

Flask with metal and HC l

Water in trough

### Dalton’s Law

The pressure of the mixture is the sum of the pressures of H2O gas and H2 gas.

H2 gas

Flask with metal and HC l

Water in trough

### Dalton’s Law

Ptotal = PH2O + PH2

H2 gas

Flask with metal and HC l

Water in trough

### Sample Problem

The total pressure in the bottle is 712.7 torr. The temperature is 19 C. What is the H2 pressure?

H2 gas

Flask with metal and HC l

Water in trough

### Sample Problem

At 19C, the vapor pressure of water is 16.5 torr. The H2 pressure is …

H2 gas

Flask with metal and HC l

Water in trough

### Sample Problem

PH2 = Ptotal - PH2O

PH2 = 712.7 torr – 16.5 torr = 696.2 torr

H2 gas

Flask with metal and HC l

Water in trough

### Another Problem

A pressure sensor is attached to a sealed flask containing 80.0 mL of mercury at 50.0 C. The pressure sensor indicates 693 torr. The pressure of the air inside the empty flask at 50.0 C is 684 torr. Determine the vapor pressure of mercury at 50.0 C.

### Another Problem

A pressure sensor is attached to a sealed flask containing 80.0 mL of mercury at 50.0 C. The pressure sensor indicates 693 torr. The pressure of the air inside the empty flask at 50.0 C is 684 torr. Determine the vapor pressure of mercury at 50.0 C.

Ptotal = P1 + P2

### Another Problem

A pressure sensor is attached to a sealed flask containing 80.0 mL of mercury at 50.0 C. The pressure sensor indicates 693 torr. The pressure of the air inside the empty flask at 50.0 C is 684 torr. Determine the vapor pressure of mercury at 50.0 C.

Ptotal = Pair+ PHg

### Another Problem

A pressure sensor is attached to a sealed flask containing 80.0 mL of mercury at 50.0 C. The pressure sensor indicates 693 torr. The pressure of the air inside the empty flask at 50.0 C is 684 torr. Determine the vapor pressure of mercury at 50.0 C.

PHg = Ptotal - Pair

### Another Problem

A pressure sensor is attached to a sealed flask containing 80.0 mL of mercury at 50.0 C. The pressure sensor indicates 693 torr. The pressure of the air inside the empty flask at 50.0 C is 684 torr. Determine the vapor pressure of mercury at 50.0 C.

PHg = 693 torr– 684 torr

### Another Problem

A pressure sensor is attached to a sealed flask containing 80.0 mL of mercury at 50.0 C. The pressure sensor indicates 693 torr. The pressure of the air inside the empty flask at 50.0 C is 684 torr. Determine the vapor pressure of mercury at 50.0 C.

PHg = 9torr

More on the gas laws:

http://chemed.chem.purdue.edu/genchem/topicreview/bp/ch4/gaslaws3.html

Part III

## Gases and Moles

The Ideal Gas Equation

### What factors affect the pressure of a confined gas?

• Number of molecules

• Temperature

• Volume of the container

Think in terms of the number of collisions.

### Number of molecules

Increasing the number of molecules increases the number of collisions …

… which increases the pressure.

Where n is the number of moles of molecules

P µ n

### Temperature

Increasing the temperature makes the molecules move faster, increasing the number of collisions …

… which increases the pressure.

P µ T

Where T is the absolute temperature

### Volume

1

V

Increasing the volume of the container decreases the number of collisions …

… which decreases the pressure.

P µ

Where V is the volume

P µ

1

P µ

V

P µ n

P µ T

n

T

V

P =

P µ

n

T

V

n

R

T

V

### The Ideal Gas Equation

P =

R

T

n

V

… is usually written as …

=

R

T

P

V

n

### The Ideal Gas Equation

L atm

mol K

P =

V

n

R

T

R is the “gas constant”

R =

0.0821

### The Ideal Gas Equation

P =

V

n

R

T

Can R be in units

other than

L atm

?

mol K

### The Ideal Gas Equation

P =

V

n

R

T

R = 0.0821 L atm/mol K

R = 8.314 L kPa/mol K

R = 62.4 L torr/mol K

### The Ideal Gas Equation

P =

V

n

R

T

R = 0.0821 L atm/mol K

R = 8.314 L kPa/mol K

R = 62.4 L torr/mol K

### The Ideal Gas Equation

P =

V

n

R

T

R = 0.0821 L atm/mol K

R = 8.314 L kPa/mol K

R = 62.4 L torr/mol K

### The Ideal Gas Equation

P =

V

n

R

T

R = 0.0821 L atm/mol K

R = 8.314 L kPa/mol K

R = 62.4 L torr/mol K

### Ideal Gas Equation

The ideal gas equation relates pressure, volume, temperature and the number of moles of a quantity of gas.

PV = nRT

### Ideal Gas Equation

Use the ideal gas equation whenever the problem gives you mass or moles, or asks for a mass or a number of moles.

PV = nRT

### Ideal gas equation problem:

Some ammonia gas (NH3) is contained in a 2.50 L flask at a temperature of 20.0 C. If there are 0.0931 moles of the gas, what is its pressure?

L atm

(0.0821

mol K

PV = nRT

P = (nRT)/V

)(293 K)

= (0.0931 mol)

2.50 L

0.896 atm

P =

### Here’s another one

Find the volume of 1.00 mole of nitrogen gas (N2) at 0.0 C and 1.00 atm of pressure.

L atm

(0.0821

mol K

PV = nRT

V = (nRT)/P

)(273 K)

V = (1.00 mol)

1.00 atm

V =

22.4 L

### Ideal gas equation problem:

How many grams of sulfur trioxide are in an 855 mL container at a pressure of 1585 torr and a temperature of 434 C?

2.46 g SO3

PV = nRT

### The Combined Gas Law

Suppose the volume, pressure and temperature change to give a new pressure, volume and temperature.

P1V1 = nRT1

P2V2 = nRT2

and

### The Combined Gas Law

Now, solve for what doesn’t change, the constants n and R:

P1V1 = nRT1

P2V2 = nRT2

and

### The Combined Gas Law

P1V1

= nR

P2V2

T1

= nR

T2

Now, solve for what doesn’t change, the constants n and R:

and

### The Combined Gas Law

P1V1

= nR

P2V2

T1

= nR

T2

Since both are equal to nR, we can make a new equation.

and

### The Combined Gas Law

P1V1

P2V2

T1

T2

Since both are equal to nR, we can make a new equation.

=

### The Combined Gas Law

P1V1

P2V2

T1

T2

This is the Combined Gas Law

=

### The Combined Gas Law

P1V1

P2V2

T1

T2

=

It can be derived from the laws of Boyle, Amonton and Charles, or

the Ideal Gas Equation

### Density calculations

m

n =

M

m

D =

V

And an equation for “moles”:

Where m = mass

and M = molar mass

### Density calculations

m

n =

M

mRT

PV =

M

Now substitute

into the ideal gas equation …

PV = nRT

and get

mRT

PV =

M

mRT

P =

VM

Now rearrange

to get

m

D =

V

mRT

DRT

P =

P =

VM

M

Recall that

### Density calculations

DRT

P =

M

PM

D =

RT

Solving for density,

becomes:

### Density calculations

PM

D =

RT

The density of a gas depends on the molar mass and the pressure and temperature.

### Density Problem

1. Determine the density of nitrogen, N2, gas

(a) at STP

(b) at a pressure of 695 torr and a temperature of 40.0 C.

1.25 g/L,

and 0.996 g/L.

### Another Problem

2. Determine the molar mass of a gas which has a density of 8.53 g/L at a pressure of 2.50 atm and a temperature of 500.0 K?

140. g/mol

### Solutions of Gases

Many gases are soluble in water. Were it not for the solubility of oxygen, fish would have wear scuba tanks.

Besides the characteristics of the gas itself, what are the two external factors which affect the solubility of a gas?

### Solutions of Gases

Two factors affect the solubility of a gas:

The temperature of the solution.

The pressure of the gas above the solution. See “Henry’s Law”.

### Solutions of Gases

The temperature of the solution.

Why is the champagne chilled?

### Solutions of Gases

More CO2 is released from the champagne as its temperature rises.

### Solutions of Gases

The solubility of gases decreases with increasing temperature.

### Solutions of Gases

Thermal pollution

This 1988 thermal image of the Hudson River highlights temperature changes caused by discharge of 2.5 billion gallons of water each day from the Indian Point power plant. The plant sits in the upper right of the photo — hot water in the discharge canal is visible in yellow and red, spreading and cooling across the entire width of the river. Two additional outflows from the Lovett coal-fired power plant are also clearly visible against the natural temperature of the water, in green and blue.

http://www.switchstudio.com/waterkeeper/issues/Winter%2007/fight_power_plants.html

### Solutions of Gases

Thermal pollution

The result is a decrease in dissolved oxygen which is needed by the plants and animals living in the heated water.

### Solutions of Gases

Why do we find trout in cold mountain streams, often near riffles and waterfalls?

High dissolved oxygen.

### Solutions of Gases

The second factor is the pressure of the dissolved gas above the solution.

CO2 is dissolved in a can of Pepsi. Above the liquid in the can is CO2 gas. The high pressure of the CO2 in the can keeps the can rigid and the CO2 in solution.

### Solutions of Gases

Henry’s Law:

At a constant temperature, the amount of a given gas dissolved in a solution is directly proportional to the partial pressure of that gas in equilibrium with the solution.

### Solutions of Gases

p = kHc

Henry’s Law:

Where p is the partial pressure of the gas, kH is the Henry’s law constant, and c is the concentration of the gas in moles per liter.

• Some Henry’s law constants at 298K

• O2 769.2 Latm/mol

• CO2 29.4 Latm/mol

• H21282.1 Latm/mol

### Solutions of Gases

p = kHc

Henry’s Law:

The pressure of the CO2 in the headspace of can of Pepsi is 4.50 atm at 298K. What is the concentration of the CO2?

• Some Henry’s law constants at 298K

• O2 769.2 Latm/mol

• CO2 29.4 Latm/mol

• H21282.1 Latm/mol

Ans:

0.153 M

### Solutions of Gases

p = kHc

Henry’s Law:

How do we predict the relative solubilities of gases based on the values of kH? Which gas below is the most soluble in water?

• Some Henry’s law constants at 298K

• O2 769.2 Latm/mol

• CO2 29.4 Latm/mol

• H21282.1 Latm/mol

CO2

Graham’s Law

### Graham’s Law

Molecules at the same temperature have the same average kinetic energy. (T µ KE)

KE is proportional to the speed of the molecules and the mass of the molecules. KE = ½ mv2

### Graham’s Law

Diffusion – “a gas spreading out”

Effusion – a gas moving through a small hole

Graham’s Law deals with the effusion of two gases into each other.

### Graham’s Law

Gas A

Gas B

Consider two gases at the same temperature and pressure in a box with two partitions, separated by a wall with a hole that can be opened.

### Graham’s Law

Gas A effuses into Gas B, and …

Gas B effuses into Gas A, so that …

Gas A

Gas B

…eventually both gases will be evenly distributed in the box.

### Graham’s Law

Gas A

Gas B

The ratio of the rates at which Gas A and Gas B effuse into each other is …

… inversely proportional to the square roots of their molar masses.

### Graham’s Law

Gas A

Gas B

The bottom line: the lighter the gas, the faster it moves!

### Graham’s Law

Gas A

Gas B

Suppose nitrogen and an unknown gas, both at the same temperature and pressure, are in the box. The rates of effusion are …

### Graham’s Law

Gas A

Gas B

… 0.0160 moles/L/min for the nitrogen and 0.0205 mol/L/min for the unknown. What is the molecular weight of the unknown?

Gas A

Gas B

### Graham’s Law

Gas A

Gas B

Square both sides, then solve for the molecular mass of compound x

Gas A

Gas B

Gas A

Gas B

Mx = 17.0g/mol

### Graham’s Law

Gas A

Gas B

What is the unknown gas?

The gas is “smelly”, dissolves in water to make a basic solution, and has a molar mass of 17 g/mol.

### Graham’s Law

Gas A

Gas B

Ammonia gas, NH3, has a strong smell, and ...

### Graham’s Law

Gas A

Gas B

… it reacts with water …

NH3(g) + HOH(l) 

NH4+ +

OH-

OH-

… which is a basic solution,

### Graham’s Law

Gas A

Gas B

… and ammonia, NH3, has a molar mass of 17.0 g/mol.

1 x 14.0 + 3 x 1.0 = 17.0 g/mol