Gases

What do we know? 1.Gases have mass. 2.Gases are easily compressed. 3.Gases uniformly and completely fill their containers. 4.Different gases move through each other quite rapidly. 5.Gases exert pressure. 6.The pressure of a gas depends on its temperature.

Measuring Gases Amount of gas (moles = n) n = mass (g) / molar mass (g/mol) Volume (V) V gas = V container Measured in liters

Measuring Gases Temperature (T) Always measured in Kelvin (K) T(K) = T( o C) Pressure (P) Gases exert pressure when particles collide with walls and the force is spread over the area of the container STP = 1 atm pressure, 0 o C (273 K)

Measuring Pressure Pressure of air is measured with a barometer (developed by Torricelli in 1643)Pressure of air is measured with a barometer (developed by Torricelli in 1643) Hg rises in tube until force of Hg (down) balances the force of atmosphere (pushing up).Hg rises in tube until force of Hg (down) balances the force of atmosphere (pushing up).

Measuring Pressure atm (standard atmosphere) = mm Hg (millimeters of mercury) = 760 torr (named after Torricelli) = psi (pounds per square inch) = 101,325 Pa (pascals) The Pascal is the SI unit for pressure, but it is used sparingly because it is so small. Just use the equalities as conversion factors to get from unit to another.

The Gas Laws Boyle’s Law (pressure-volume relationship) Charles’s Law (volume-temperature relationship) Gay-Lussac’s Law (pressure-temperature relationship) Combined Gas Law (relates pressure, volume, and temperature) Dalton’s Law of Partial Pressures The Ideal Gas Law

Boyle’s Law The pressure of a given sample of a gas is inversely proportional to the volume of the gas at constant temperature. If we know the volume of a gas at a given pressure, we can predict the new volume if the pressure is changed, as long as the temperature and the amount of the gas remain the same. P 1 V 1 = P 2 V 2

Boyle’s Law Freon-12 (the common name for the compound CCl 2 F 2 ) was once widely used in refrigeration systems, but has now been replaced by other compounds that do not lead to the breakdown of the protective ozone in the upper atmosphere. Consider a 1.5-L sample of gaseous CCl 2 F 2 at a pressure of 56 torr. If the pressure is changed to 150 torr at a constant temperature, will the volume of the gas increase or decrease? What will be the new volume of the gas?

Boyle’s Law Given: P 1 = 56 torr V 1 = 1.5 L P 2 = 150 torr Temperature is constant Pressure increases so volume decreases. Boyle’s Law: P 1 V 1 = P 2 V 2 (56 torr)(1.5 L) = (150 torr) V 2 V 2 = (56 torr)(1.5 L) / (150 torr) V 2 = 0.56 L Unknown: V 2 = ? L

Boyle’s Law In an automobile engine the gaseous fuel-air mixture enters the cylinder and is compressed by a moving piston before it is ignited. In a certain engine the initial cylinder volume is L. After the piston moves up, the volume is L. The fuel-air mixture initially has a pressure of 1.00 atm. Calculate the pressure of the compressed fuel-air mixture, assuming that both the temperature and the amount of gas remain constant.

Boyle’s Law Given: P 1 = 1.00 atm V 1 = L V 2 = L Temperature and amount of gas are constant Volume decreases so pressure increases. Boyle’s Law: P 1 V 1 = P 2 V 2 (1.00 atm)(0.725 L) = P 2 (0.075 L) P 2 = (1.00 atm)(0.725 L) / (0.075 L) P 2 = 9.7 atm Unknown: P 2 = ? atm

Charles’s Law The volume of a given sample of a gas is directly proportional to the temperature of the gas at constant pressure. Initial values can also be used to find final values V 1 /T 1 = V 2 /T 2

Charles’s Law A 2.0-L sample of air is collected at 298 K and then cooled to 278 K. The pressure is held constant at 1.0 atm. Does the volume increase or decrease? Calculate the volume of the air at 278 K.

Charles’s Law Given: T 1 = 298 K V 1 = 2.0 L T 2 = 278 K Pressure is constant Temperature decreases so volume decreases. Unknown: V 2 = ? L

298 K Charles’s Law 2.0 L 298 K 278 K V 2 = V 1 /T 1 = V 2 /T 2 V 2 = 2.0 L x 278 K V 2 = 1.9 L (2.0 L)(278 K) = V 2 (298K)

Charles’s Law A sample of gas at 15 o C (at 1 atm) has a volume of 2.58 L. The temperature is then raised to 38 o C (at 1 atm). Does the volume of the gas increase or decrease? Calculate the new volume.

Charles’s Law Given: T 1 = 15 o C V 1 = 2.58 L T 2 = 38 o C Pressure is constant Temperature increases so volume increases Unknown: V 2 = ? L = 288 K = 311 K

288 K Charles’s Law 2.58 L 288 K 311 K V 2 = V 1 /T 1 = V 2 /T 2 V 2 = 2.58 L x 311 K V 2 = 2.79 L (2.58 L)(311 K) = V 2 (288K)

Equal volumes of gases at the same temperature and pressure contain an equal number of particles. Initial values can also be used to find final values V 1 /n 1 = V 2 /n 2 Avogadro’s Law

The pressure of a given sample of a gas is directly proportional to the temperature of the gas at constant volume. Initial values can also be used to find final values P 1 /T 1 = P 2 /T 2 Gay-Lussac’s Law

Calculate the final pressure inside a scuba tank after it cools from 1000 °C to 25.0 °C. The initial pressure in the tank is atm.

Gay-Lussac’s Law Given: P 1 = 130 atm T 1 = 1000 o C T 2 = 25 o C Volume is constant Temperature decreases so pressure decreases. Unknown: P 2 = ? atm = 1273 K = 298 K

1273 K Gay-Lussac’s Law atm 1273 K 298 K P 2 = P 1 /T 1 = P 2 /T 2 P 2 = 130 atm x 298 K P 2 = 30.4 atm (130 atm)(298 K) = P 2 (1273 K)

Gay-Lussac’s Law If a gas in a closed container is pressurized from 15.0 atmospheres to 16.0 atmospheres and its original temperature was 25.0 °C, what would the final temperature of the gas be?

Gay-Lussac’s Law Given: P 1 = 15 atm P 2 = 16 atm T 1 = 25.0 o C Volume is constant Pressure increases so temperature increases. Unknown: T 2 = ? K = 298 K

15 atm Gay-Lussac’s Law 15.0 atm 298 K T2T atm = P 1 /T 1 = P 2 /T 2 T 2 = 16 atm x 298 K T 2 = 318 K (15 atm)T 2 = (16 atm)(298 K)

Summary of Laws LawStatementEquationConstant Boyle's P inversely proportional to V P 1 V 1 = P 2 V 2 T, n Charles'sV directly proportional to T V 1 /T 1 = V 2 /T 2 P, n Avogadro'sV directly proportional to n V 1 /n 1 = V 2 /n 2 P, T Gay-Lussac’sP directly proportional to T P 1 /T 1 = P 2 /T 2 V, n

Combined Gas Law

Combines Boyle’s, Charles’s, and Gay-Lussac’s laws to relate the pressure, volume, and temperature of constant amounts of gases. Initial values can also be used to find final values P 1 V 1 /T 1 = P 2 V 2 /T 2

Combined Gas Law A toy balloon has an internal pressure of 1.05 atm and a volume of 5.0 L. If the temperature above where the balloon is released is 20 o C, what will happen to the volume when the ballooon rises to an altitude where the pressure is 0.65 atm and the temperature is -15 o C?

Combined Gas Law Given: P 1 = 1.05 atm V 1 = 5.0 L T 1 = 20 o C P 2 = 0.65 atm T 2 = -15 o C Unknown: V 2 = ? L = 293 K = 258 K

(0.65 atm)(293 K) Combined Gas Law (1.05 atm)(5.0 L) 293 K 258 K (0.65 atm)V 2 = P 1 V 1 /T 1 = P 2 V 2 /T 2 V 2 = (1.05 atm)(5.0 L)(258 K) V 2 = 7.1 L (1.05 atm)(5.0 L)(258 K) = (0.65 atm)V 2 (293 K)

Combined Gas Law If I initially have 4.0 L of gas at a pressure of 1.1 atm and a temperature of 25 o C, what will happen to the pressure if I decrease the volume to 3.0 L and decrease the temperature to 20 o C?

Combined Gas Law Given: V 1 = 4.0 L P 1 = 1.1 atm T 1 = 25 o C V 2 = 3.0 L T 2 = 20 o C Unknown: P 2 = ? L = 298 K = 293 K

(3.0 L)(298 K) Combined Gas Law (1.1 atm)(4.0 L) 298 K 293 K P 2 (3.0 L) = P 1 V 1 /T 1 = P 2 V 2 /T 2 P 2 = (1.1 atm)(4.0 L)(293 K) P 2 = 1.4 atm (1.1 atm)(4.0 L)(293 K) = P 2 (3.0 L)(298 K)

The Ideal Gas Law

Combines Boyle’s, Charles’s, and Avogadro’s Laws to describe the behavior of gases as dependent upon volume, pressure, temperature, and the number of moles present. PV = nRT where R is the combined constant called the universal gas constant R = L atm / mol K

The Ideal Gas Law A gas that obeys this equation is said to behave ideally, thus the name “Ideal Gas Law” There is no such thing as an ideal gas, but many real gases behave ideally at pressures of approximately 1 atm and lower and temperatures of approximately 273 K or higher.

The Ideal Gas Law A sample of hydrogen gas has a volume of 8.56 L at a temperature of 0 o C and a pressure of 1.5 atm. Calculate the number of moles of H 2 present in this gas sample. Assume that the gas behaves ideally.

The Ideal Gas Law Given: P = 1.5 atm T = 0 o C = 273 K V = 8.56 L PV = nRT (1.5 atm)(8.56 L) = n( L-atm/mol-K)(273 K) Unknown: n = ? (1.5 atm)(8.56 L) ( L-atm/mol-K)(273 K) n = n = 0.57 mol

The Ideal Gas Law What volume is occupied by mol of carbon dioxide gas at 25 o C and 371 torr? Because the gas constant is measured in L-atm/mol-K, volume must be measured in liters, pressure in atmospheres, amount of gas in moles, and temperature in Kelvin. 25 o C = 298 K 371 torr x 760 torr 1.00 atm = atm

The Ideal Gas Law Given: P = atm T = 298 K n = mol CO 2 PV = nRT (0.488 atm)V = (0.250 mol)( L-atm/mol-K)(298K) Unknown: V = ? L (0.250 mol) (0.488 atm) ( L-atm/mol-K) (298 K) V = V = 12.5 L

Dalton’s Law of Partial Pressures

For a mixture of gases in a container, the total pressure exerted is the sum of the partial pressures of the gases present The partial pressure of a gas is the pressure that the gas would exert if it were alone in the container. P total = P 1 + P 2 + P 3 Each gas is responsible for only part of the total pressure.

Dalton’s Law of Partial Pressures What is the atmospheric pressure if the partial pressure of nitrogen, oxygen, and argon are mm Hg, mm Hg, and 0.5 mm Hg, respectively? P T = P N 2 + P O 2 + P Ar P T = mm Hg mm Hg mm Hg P T = mm Hg

Dalton’s Law of Partial Pressures A gas mixture contains hydrogen, helium, neon, and argon. The total pressure of the mixture is 93.6 kPa. The partial pressures of helium, neon, and argon are 15.4 kPa, 25.7 kPa, and 35.6 kPa, respectively. What is the pressure exerted by hydrogen? P T = P H 2 + P He + P Ne + P Ar 93.6 kPa = P H kPa kPa kPa P H 2 = 93.6 – ( ) = 16.9 kPa

Dalton’s Law of Partial Pressures Mixtures of helium and oxygen are used in the “air” tanks of underwater divers for deep dives. For a particular dive, 12 L of O 2 at 25 o C and 1.0 atm and 46 L of He at 25 o C and 1.0 atm were pumped into a 5.0-L tank. Calculate the partial pressure of each gas and the total pressure in the tank at 25 o C.

Dalton’s Law of Partial Pressures O2O2 V 1 = 12 L P 1 = 1.0 atm T 1 = 298 K He V 1 = 46 L P 1 = 1.0 atm T 1 = 298 K Mixture V 2 = 5.0 L T 2 = 298 K P 2 = ? Since only V and P are changing, use Boyle’s Law to solve for the pressure of each gas. (Do TWO Boyle’s Law problems to solve for the final pressure of each gas in the mixture!)

O2O2 P 1 V 1 = P 2 V 2 (1.0 atm)(12 L) = P 2 (5.0 L) P 2 = 2.4 atm (partial pressure of oxygen in the mixture) He P 1 V 1 = P 2 V 2 (1.0 atm)(46 L) = P 2 (5.0 L) P 2 = 9.2 atm (partial pressure of oxygen in the mixture) P T = 2.4 atm atm = 11.6 atm

Understanding the Laws

What do we want to know? What are the characteristics of the individual gas particles that cause a gas to behave like it does? Remember: Laws tell us what happens, not why. Scientists develop models or theories to explain why nature behaves the way it does.

The Kinetic Molecular Theory of Gases (KMT) A simple model that attempts to explain the behavior of an ideal gas Based on the behavior of the individual particles (atoms or molecules) in a gas

Assumptions/Postulates of the KMT of Ideal Gases 1.Gases consist of tiny particles (atoms or molecules). 2.These particles are so small, compared with the distances between them, that the volume (size) of the individual particles can be assumed to be negligible (zero). 3.The particles are in constant random motion, colliding with the walls of the container. These collisions cause the pressure exerted by the gas. 4.The particles are assumed not to attract or repel one another. In other words, the particles don’t interact. 5.The average kinetic energy of the gas particles is directly proportional to the Kelvin temperature of the gas.

Postulates True for Real Gases? 1.Yes, all gases are made up of atoms or molecules. 2.Fair assumption, but not perfect. The particles are in fact very small, but their volumes are not zero. 3.Yes, all gases have particles in constant, random motion that produce pressure. 4.Depends on the gas. Some gases have stronger intermolecular forces than others. 5.Yes, increased temperature does increase motion of the particles in a real gas.

The KMT of Gases Temperature reflects how rapidly the individual particles of a gas are moving. High temps = fast movement = lots of collisions with walls of container Low temps = slow movement = fewer collisions with walls of container Consider a container with an expandable volume. What will happen if the external pressure remains the same and the temperature of the gas is increased? Higher temp = faster movement = more collisions with the walls of the container Increased pressure on walls  increase in volume Proves Charles’s and Gay-Lussac’s Laws!