GASES

Chapter 14

From last chapter… Kinetic Molecular Theory

• Particles in an ideal gas…– have no volume.– have elastic collisions. – are in constant, random, straight-line

motion.– don’t attract or repel each other.– have an avg. KE directly related to Kelvin

temperature.

Real Gases

• Particles in a REAL gas…– have their own volume– attract each other

• Gas behavior is most ideal…– at low pressures– at high temperatures– in nonpolar atoms/molecules

Properties of Gases

• Compressibility – gases are easily compressed because of the space between the particles in a gas

• Gases expand to take the shape and volume of their container

Factors Affecting Gas Pressure• Amount of gas – more particles have more collisions

with the container walls and thus create more pressure

• Volume – if you reduce the volume of the container, the particles are more compressed and exert a greater pressure on the walls of the container

• Temperature – increasing temperature increases the kinetic energy of the particles, which then strike the walls of the container with more energy

Remember?Units of Pressure

• KEY UNITS AT SEA LEVEL

101.325 kPa (kilopascal)

1 atm

760 mm Hg

760 torr

14.7 psi

*These are all equivalent amounts of pressure

Standard Temperature & PressureStandard Temperature & Pressure

0°C 273 K

1 atm 101.325 kPa-OR-

STP

The Gas Laws14.2

The Gas Laws14.2

Boyle’s Law• The pressure and volume of a gas

are inversely related – at constant mass & temp

P

V

P1V1 = P2V2

GIVEN:

V1 = 100. mL

P1 = 150. kPa

V2 = ?

P2 = 200. kPa

WORK:

Gas Law Problem• A gas occupies 100. mL at 150. kPa. Find its

volume at 200. kPa.

BOYLE’S LAW

P V

V

T

Charles’ Law• The volume and absolute

temperature (K) of a gas are directly related – at constant mass & pressure

2

2

1

1

T

V

T

V

GIVEN:

V1 = 473 cm3

T1 = 36°C = 309K

V2 = ?

T2 = 94°C = 367K

WORK:

Gas Law Problem• A gas occupies 473 cm3 at 36°C. Find its

volume at 94°C.

CHARLES’ LAW

T V

P

T

Gay-Lussac’s Law• The pressure and absolute

temperature (K) of a gas are directly related – at constant mass & volume

2

2

1

1

T

P

T

P

GIVEN:

P1 = 765 torr

T1 = 23°C = 296K

P2 = 560. torr

T2 = ?

WORK:

Gas Law Problem• A gas’ pressure is 765 torr at 23°C. At what

temperature will the pressure be 560. torr?

GAY-LUSSAC’S LAW

P T

Combined Gas Law

P1V1

T1

=P2V2

T2

GIVEN:

V1 = 7.84 cm3

P1 = 71.8 kPa

T1 = 25°C = 298 K

V2 = ?

P2 = 101.325 kPa

T2 = 273 K

WORK:

Gas Law Problem• A gas occupies 7.84 cm3 at 71.8 kPa & 25°C.

Find its volume at STP.

P T VCOMBINED GAS LAW

V

n

Avogadro’s Law

• The volume and number of moles of a gas are directly related – at constant temperature & pressure

2

2

1

1

n

V

n

V

GIVEN:

V1 = 36.7 L

n1 = 1.5 mol

V2 = 16.5 L

n2 = ?

WORK:

Gas Law Problem• Consider two sample of N2 gas. Sample 1 contains 1.5 mol

of N2 and has a volume of 36.7 L at 25°C and 1 atm. Sample 2 has a volume of 16.5 L at 25°C and 1 atm. Calculate the number of moles of N2 in Sample 2.

AVOGADRO’S LAW

n V

14.3 - Ideal Gas Law

UNIVERSAL GAS CONSTANTR=0.0821

Latm/molKR=8.315

dm3kPa/molK

PV=nRT

GIVEN:

P = ? atm

n = 0.412 mol

T = 16°C = 289 K

V = 3.25 LR = 0.0821Latm/molK

WORK:

Ideal Gas Law Problem• Calculate the pressure in atmospheres of 0.412

mol of He at 16°C & occupying 3.25 L.

14.4 - Dalton’s Law of Partial Pressures

• The partial pressure of a gas is the pressure that the gas would exert if it were alone in the container.

• Dalton’s Law of Partial Pressures says that the total pressure of a mixture of gas is equal to the sum of the partial pressures of all gases in the mixture. Or,

Ptotal = P1 + P2 + P3 +…Note: you can calculate the partial pressures of the gases if they

behave ideally using the ideal gas law (P = nRT/V)

Dalton’s Law Example

• A 2.0 L flask contains a mixture of nitrogen gas and oxygen gas at 25°C. The total pressure of the gaseous mixture is 0.91 atm, and the mixture is known to contain 0.050 mol of N2. Calculate the partial pressure of oxygen and the moles of oxygen present.

Graham’s Law of Effusion

• Diffusion is the tendency of molecules to move toward areas of lower concentration until the concentration is uniform throughout

• Effusion is when a gas escapes through a tiny hole in its container

• Gases of lower molar mass diffuse and effuse faster than gases of higher molar mass

Graham’s Law of Effusion

• The rate of effusion of a gas is inversely proportional to the square root of the gas’s molar mass.

• This equation compares effusion rates for two gases

Graham’s Law Problem

• Calculate the ratio of the velocity of hydrogen molecules (H2) to the velocity of carbon dioxide (CO2) molecules at the same temperature.

Gas Stoichiometry

• Molar volume of a gas is the volume that is occupied by 1 mol of an ideal gas at STP.–1 mol of gas occupies 22.4 L

• Yes, we are going back to those 3 step problems…

Gas Stoichiometry Problem

Quicklime, CaO, is produced by heating calcium carbonate. Calculate the volume of carbon dioxide produced at STP from the decomposition of 152 g of calcium carbonate according to the reaction

CaCO3 (s) CaO (s) + CO2 (g)