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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

- 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 describes the behavior of ideal gases.

An ideal gas is one that conforms to the KMT.

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).

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.

There are no attractive or repulsive forces between gas molecules.

Adjacent molecules do not attract or repel each other.

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.

- Gases in constant motion, speed depends on temperature.
- Molecules have negligible volume.
- No intermolecular forces.
- Elastic collisions. No change in energy.

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

… where the lowest possible temperature is zero degrees.

Use Kelvin degrees!

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.

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

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

Standard sea level pressure is…

1.00 atmospheres (atm)

760 mm Hg

760 torr (from Torricelli)

101.3 kilopascals (kPa)

14.7 lb/in2

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

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

Avogadro’s Law

Dalton’s Law

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

Pressure

Volume

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

1/Pressure

Volume

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.

-273 C

Pressure

Extrapolate to the x-axis

-300 -150 0 100 200 300

Temperature (C)

Volume

Temperature

At constant pressure, volume is directly proportional to temp.

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 gas studies published by Gay-Lussac in1802.

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

equation by adding a constant.

PV

= k

kT

T

P =

V

Rearranging the equation gives:

PV

= k

T

Combining the laws of Boyle, Amonton and Charles produces the 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

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.

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

Start with the equation for the combined gas law.

100 mL

65 mL

P1V1

P2V2

=

T1

T2

100 mL

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

65 mL

P1V1

=

P2V2

P1V1

P2

=

V2

This becomes Boyle’s Law

100 mL

Next, solve for P2.

65 mL

(0.950 atm)(100.0 mL)

P2

=

65.0 mL

P1V1

P2

=

P2

=

1.46 atm

V2

100 mL

65 mL

- 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

The answer is

- 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

The answer is

- 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

The answer is

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 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

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 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.

“Collecting a gas over water”

Inverted gas collecting bottle

Rubber tubing carrying H2 gas

Flask with metal and HC l

Water in trough

“Collecting a gas over water”

Inverted gas collecting bottle

Rubber tubing carrying H2 gas

Flask with metal and HC l

Water in trough

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

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

H2 gas

Flask with metal and HC l

Water in trough

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

Ptotal = PH2O + PH2

H2 gas

Flask with metal and HC l

Water in trough

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

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

PH2 = Ptotal - PH2O

PH2 = 712.7 torr – 16.5 torr = 696.2 torr

H2 gas

Flask with metal and HC l

Water in trough

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.

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

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

PHg = Ptotal - Pair

PHg = 693 torr– 684 torr

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

- Number of molecules
- Temperature
- Volume of the container

Think in terms of the number of collisions.

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

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

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

P =

R

T

n

V

… is usually written as …

=

R

T

P

V

n

L atm

mol K

P =

V

n

R

T

R is the “gas constant”

R =

0.0821

P =

V

n

R

T

Can R be in units

other than

L atm

?

mol K

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

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

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

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

PV = nRT

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

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 =

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

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

The answer is

2.46 g SO3

Start with the ideal gas equation:

PV = nRT

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

P1V1 = nRT1

P2V2 = nRT2

and

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

P1V1 = nRT1

P2V2 = nRT2

and

P1V1

= nR

P2V2

T1

= nR

T2

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

and

P1V1

= nR

P2V2

T1

= nR

T2

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

and

P1V1

P2V2

T1

T2

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

=

P1V1

P2V2

T1

T2

This is 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

m

n =

M

Start with the equation for density:

m

D =

V

And an equation for “moles”:

Where m = mass

and M = molar mass

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

DRT

P =

M

PM

D =

RT

Solving for density,

becomes:

PM

D =

RT

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

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.

The answers are

1.25 g/L,

and 0.996 g/L.

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

The answer is

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?

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”.

The temperature of the solution.

Why is the champagne chilled?

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

The solubility of gases decreases with increasing temperature.

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

Thermal pollution

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

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

High dissolved oxygen.

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.

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.

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

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

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

Answer:

CO2

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

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.

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.

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.

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.

Gas A

Gas B

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

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 …

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

Start with the equation for 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

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.

Gas A

Gas B

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

Gas A

Gas B

… it reacts with water …

NH3(g) + HOH(l)

NH4+ +

OH-

OH-

… which is a basic solution,

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