Behavior of Gases

Compressibility Compressibility- a measure of how much the volume of matter decreases under pressure

Compressibility Gases are easily compressed because of the space between the particles in a gas. The distance between particles in a gas is much greater than the distance between particles in a liquid or solid. Under pressure, the particles in a gas are forced closer together.

Compressibility At room temperature, the distance between particles in an enclosed gas is about 10 times the diameter of a particle.

Factors Affecting Gas Pressure The amount of gas, volume, and temperature are factors that affect gas pressure. Four variables are generally used to describe a gas: –pressure (P) in kilopascals –volume (V) in liters –temperature (T) in kelvins –the number of moles (n)

Factors Affecting Gas Pressure Amount of Gas –You can use kinetic theory to predict and explain how gases will respond to a change of conditions. If you inflate an air raft, for example, the pressure inside the raft will increase.

Factors Affecting Gas Pressure Collisions of particles with the inside walls of the raft result in the pressure that is exerted by the enclosed gas. Increasing the number of particles increases the number of collisions, which is why the gas pressure increases.

Factors Affecting Gas Pressure If the gas pressure increases until it exceeds the strength of an enclosed, rigid container, the container will burst.

Factors Affecting Gas Pressure Aerosol Spray Paint

Factors Affecting Gas Pressure Volume –You can raise the pressure exerted by a contained gas by reducing its volume. The more a gas is compressed, the greater is the pressure that the gas exerts inside the container.

Factors Affecting Gas Pressure When the volume of the container is halved, the pressure the gas exerts is doubled.

Factors Affecting Gas Pressure Temperature –An increase in the temperature of an enclosed gas causes an increase in its pressure. –As a gas is heated, the average kinetic energy of the particles in the gas increases. Faster-moving particles strike the walls of their container with more energy.

Factors Affecting Gas Pressure When the Kelvin temperature of the enclosed gas doubles, the pressure of the enclosed gas doubles.

Boyle’s Law: Pressure and Volume -How are the pressure, volume, and temperature of a gas related? –If the temperature is constant, as the pressure of a gas increases, the volume decreases.

Boyle’s Law: Pressure and Volume Boyle’s law- states that for a given mass of gas at constant temperature, the volume of the gas varies inversely with pressure

Boyle’s Law: Pressure and Volume

Using Boyle’s Law Example: A balloon contains 30.0 L of helium gas at 103 kPa. What is the volume of the helium when the balloon rises to an altitude where the pressure is only 25.0 kPa? (Assume the temperature remains constant.) Knowns: P 1 = 103 kPa, V 1 = 30.0 L, P 2 = 25.0 kPa Unknowns: V 2 = ? L

Using Boyle’s Law Remember: Rearrange Boyle’s law: V 2 = (V 1 X P 1 )/P 2 V 2 = (30.0 L X 103 kPa)/25.0 kPa = 1.24 X 10 2 L

Charles’s Law: Temperature and Volume As the temperature of an enclosed gas increases, the volume increases, if the pressure is constant.

Charles’s Law: Temperature and Volume As the temperature of the water increases, the volume of the balloon increases.

Charles’s Law: Temperature and Volume Charles’s law- states that the volume of a fixed mass of gas is directly proportional to its Kelvin temperature if the pressure is kept constant.

Charles’s Law: Temperature and Volume

Using Charles’s Law Example: A balloon inflated in a room at 24°C has a volume of 4.00 L. The balloon is then heated to a temperature of 58˚C. What is the new volume if the pressure remains constant? Known: V 1 = 4.00L, T 1 = 24°C, T 2 = 58˚C Unknown: V 2 = ? L

Using Charles’s Law Remember: Convert to Kelvin T 1 = 24˚C = 297 K T 2 = 58˚C = 331 K

Using Charles’s Law Rearrange Charles’s Law V 2 = (V 1 X T 2 )/T 1 Solve: V 2 = (4.00 L X 331 K)/297 K = 4.46 L

Gay-Lussac’s Law: Pressure and Temperature As the temperature of an enclosed gas increases, the pressure increases, if the volume is constant.

Gay-Lussac’s Law: Pressure and Temperature When a gas is heated at constant volume, the pressure increases.

Gay-Lussac’s Law: Pressure and Temperature Gay-Lussac’s law- states that the pressure of a gas is directly proportional to the Kelvin temperature if the volume remains constant.

Gay-Lussac’s Law: Pressure and Temperature A pressure cooker demonstrates Gay-Lussac’s Law.

Using Gay-Lussac’s Law The gas in a used aerosol can is at a pressure of 103 kPa at 25°C. If the can is thrown onto a fire, what will the pressure be when the temperature reaches 928°C? Known: P 1 = 103 kPa, T 1 = 25°C, T 2 = 928°C Unknown: P 2 = ? kPa

Using Gay-Lussac’s Law Remember: Convert Temperatures to Kelvin: T 1 = 25°C = 298 K T 2 = 928°C = 1,201 K

Using Gay-Lussac’s Law Rearrange Gay-Lussac’s equation: P 2 = (P 1 X T 2 )/T 1 Solve: P 2 = (103 kPa X 1,201 K)/298 K = 415 kPa = 4.15 X 10 2 kPa

The Combined Gas Law Combined gas law- describes the relationship among the pressure, temperature, and volume of an enclosed gas. The combined gas law allows you to do calculations for situations in which only the amount of gas is constant.

Using the Combined Gas Law Example: The volume of a gas-filled balloon is 30.0 L at 313 K and 153 kPa pressure. What would the volume be at standard temperature and pressure (STP)? Known: V 1 = 30.0 L, T 1 = 313 K, P 1 = 153 kPa, T 2 = 273 K (standard temperature), P 2 = kPa (standard pressure)

Using the Combined Gas Law Unknown: V 2 = ? L Remember: Rearrange: V 2 = (V 1 X P 1 X T 2 )/(P 2 X T 1 )

Using the Combined Gas Law Solve: V 2 = (30.0 L X 153 kPa X 273 K) (101.3 kPa X 313 K) = 39.5 L

Ideal Gas Law –What is needed to calculate the amount of gas in a sample at given conditions of volume, temperature, and pressure? –To calculate the number of moles of a contained gas requires an expression that contains the variable n.

Ideal Gas Law The gas law that includes all four variables— P, V, T, and n—is called the ideal gas law. The ideal gas constant (R) has the value 8.31 (L·kPa)/(K·mol).

Using the Ideal Gas Law to Find the Amount of Gas Example: A deep underground cavern contains 2.24 X 10 6 L of methane (CH 4 ) gas at a pressure of 1.50 X 10 3 kPa and a temperature of 315 K. How many kilograms of CH 4 does the cavern contain? Known: P = 1.50 X 10 3 kPa, V = 2.24 X 10 6 L, T = 315 K, R = 8.31 (L· kPa)/(K · mol), molar mass CH 4 = 16.0 g

Using the Ideal Gas Law to Find the Amount of Gas Unknown = ? kg CH 4 Remember: Rearrange: n = (P X V)/(RXT)

Using the Ideal Gas Law to Find the Amount of Gas Solve: n = ((1.50 X 10 3 kPa)(2.24 X 10 6 L)) ((8.31 (L· kPa)/(K · mol)(315 K)) = 1.28 X 10 6 mol CH 4

Using the Ideal Gas Law to Find the Amount of Gas Convert to grams: 1.28 X 10 6 mol CH 4 X (16.0 g CH 4 ) (1 mol CH 4 ) = 20.5 X 10 6 g CH 4 = 2.05 X 10 7 g CH 4

Using the Ideal Gas Law to Find the Amount of Gas Convert to kilograms: 2.05 X 10 7 g CH 4 X 1kg = 2.05 X 10 4 kg CH g

Ideal Gases and Real Gases –Under what conditions are real gases most likely to differ from ideal gases? Real gases differ most from an ideal gas at low temperatures and high pressures.

Ideal Gases and Real Gases There are attractions between the particles in an ideal gas. Because of these attractions, a gas can condense,or even solidify, when it is compressed or cooled.

Ideal Gases and Real Gases

Dalton’s Law –How is the total pressure of a mixture of gases related to the partial pressures of the component gases?

Dalton’s Law Partial pressure- contribution each gas in a mixture makes to the total pressure

Dalton’s Law In a mixture of gases, the total pressure is the sum of the partial pressures of the gases.

Dalton’s Law Dalton’s law of partial pressures- states that, at constant volume and temperature, the total pressure exerted by a mixture of gases is equal to the sum of the partial pressures of the component gases.

Dalton’s Law Three gases are combined in container T.

Dalton’s Law The partial pressure of oxygen must be kPa or higher to support respiration in humans. The climber below needs an oxygen mask and a cylinder of compressed oxygen to survive.

Using Dalton’s Law of Partial Pressures Example: Air contains oxygen, nitrogen, carbon dioxide, and trace amounts of other gases. What is the partial pressure of oxygen (P o ) at kPa of total pressure if the partial pressures of nitrogen, carbon dioxide, and other gases are kPa, kPa, and 0.94 kPa, respectively?

Using Dalton’s Law of Partial Pressures Known: P N 2 = kPa P CO 2 = kPa P others = 0.94 kPa P total = kPa Unknown = P O 2 = ?kPa

Using Dalton’s Law of Partial Pressures Remember: P O 2 = P total – (P N 2 + P CO 2 + P others ) = kPa – (79.10 kPa kPa kPa) = kPa

Graham’s Law –How does the molar mass of a gas affect the rate at which the gas effuses or diffuses? –Diffusion- is the tendency of molecules to move toward areas of lower concentration until the concentration is uniform throughout.

Graham’s Law Bromine vapor is diffusing upward through the air in a graduated cylinder.

Graham’s Law After several hours, the bromine has diffused almost to the top of the cylinder.

Graham’s Law During effusion, 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 Thomas Graham’s Contribution –Graham’s law of effusion- states that the rate of effusion of a gas is inversely proportional to the square root of the gas’s molar mass. – can also be applied to the diffusion of gases

Graham’s Law Comparing Effusion Rates –A helium filled balloon will deflate sooner than an air-filled balloon. –Helium atoms are less massive than oxygen or nitrogen molecules. So the molecules in air move more slowly than helium atoms with the same kinetic energy.

Graham’s Law Because the rate of effusion is related only to a particle’s speed, Graham’s law can be written as follows for two gases, A and B.

Graham’s Law Helium effuses (and diffuses) nearly three times faster than nitrogen at the same temperature.