Behavior of Gases Chapter 11

Gas Pressure Pressure is the force per unit area. Example: pounds per square inch Gas pressure is related to volume, temperature and the number of gas particles.

Pressure vs Number of Particles Increasing the number of gas particles in a container increases the pressure. More particles means more collisions with the walls and, therefore, greater pressure. Example: pumping air into a car or bike tire

Pressure vs Temperature Increasing the temperature of a gas increases the pressure. Increasing temperature gives particles more KE. They move faster and collide with walls more often and with greater force, therefore causing greater pressure.

Standard Temperature and Pressure (STP) 0ºC 273 K Standard Pressure 760 mmHg 101.3 kPa 1 atm 14.7 psi 29.92 inHg 760 torr

Temperature Conversions All temperatures in gas problems must be converted to Kelvin. K = Cº + 273 Cº = K – 273 (Remember: There are no negative Kelvin temperatures) Example #1: Convert -15ºC to Kelvin K = -15ºC + 273 = 258 K Example #2: Convert 355 K to ºC Cº = 355 – 273 = 82ºC

Pressure Conversions Use standard pressures for pressure conversions.

Manometers Manometer Barometer device used to measure the pressure of a gas. Barometer Device used to measure air pressure. Made by Torricelli in mid-1600s Air pushes Hg up a vacuum tube (see diagram). Measured in mmHg, inHg (weather forecasts), kPa. https://www.youtube.com/watch?v=EkDhlzA-lwI

Manometers- closed x and y points in the picture are at same level, thus pressures acting on these points are equal. Pressure at point x is the pressure of gas and pressure at point y is the pressure of mercury at h height. In this CLOSED system pressure of gas is equal to; Pgas=h

Manometers- open I: Since height is greater on AIR side, gas pressure is greater than atm pressure Pgas = Po +h II: Air and gas pressure are exactly equal Pgas = Po III: Since height is less on AIR side, gas pressure is less than atm pressure Pgas = Po -h

Sphygmomanometer Device used to measure blood pressure. How it works A cuff is wrapped around the arm and inflated until the cuff stops the blood flow in your artery. A valve lets some of the air out of the cuff which allows blood to start flowing. A stethoscope is used to listen to the blood flowing/rushing back through the artery. The first thumping sound is called the systolic blood pressure. When the sound is no longer heard, it is called the diastolic blood pressure. All pressures are measured in mmHg but are usually reported as numbers. Example – “120 over 80”

Ideal Gas Behavior An ideal gas has two properties: 1. It has mass but no volume (called a point mass). 2. It has no mutual attraction for other particles.

Real Gas Behavior All gases are “real” gases. However, most behave like ideal gases (no significant volume and little attraction for other molecules). The gases behave near ideal conditions until they are at: 1. extremely low temperatures 2. extremely high pressures For 1st year chemistry, we will assume that all gases behave like ideal gases.

Boyle’s Law Assuming temperature is held constant: The pressure and volume of a gas at constant temperature are inversely proportional. Assuming temperature is held constant: If pressure increases, volume decreases. If pressure decreases, volume increases. i.e., pressure and volume move in opposite directions.

Boyle’s Law Boyle’s law equation: PV = k If temperature, k, is held constant…we can manipulate the equation to: P1V1 = P2V2 P = pressure (kPa, mmHg, psi, atm, …) V = volume (L, mL, dm3, cm3)

Example #1: A gas is compressed at constant temperature from 25.0 L to 4.0 L. If the initial pressure was 0.50 atm, what is the new pressure?

Example #1: A gas is compressed at constant temperature from 25.0 L to 4.0 L. If the initial pressure was 0.50 atm, what is the new pressure? P1V1 = P2V2 (0.50 atm)(25.0 L) = (P2)(4.0 L) P2 = 3.1 atm

Example #2: A sample of argon gas at STP has a volume of 450. mL. If the temperature remains constant and the pressure changes to 685 mmHg, what is the new volume? *Remember, STP = standard temperature and pressure

Example #2: A sample of argon gas at STP has a volume of 450. mL. If the temperature remains constant and the pressure changes to 685 mmHg, what is the new volume? *Remember, STP = standard temperature and pressure P1V1 = P2V2 (760 mmHg)(450. mL) = (685 mmHg)(V2) V2 = 499 mL

Example #3: A 24.5 L sample of hydrogen gas exerts 125 kPa of pressure at 18ºC. What is the pressure of the hydrogen if expanded to 65.0 L at 18ºC? What is that pressure equal to in psi?

Example #3: A 24.5 L sample of hydrogen gas exerts 125 kPa of pressure at 18ºC. What is the pressure of the hydrogen if expanded to 65.0 L at 18ºC? What is that pressure equal to in psi? P1V1 = P2V2 (125 kPa)(24.5 L) = (P2)(65.0 L) P2 = 47.1 kPa

Dalton’s Law of Partial Pressures The total pressure of a gas mixture is the sum of the partial pressures of the component gases Ptotal = P1 + P2 + P3 +…

Example #1 A mixture of oxygen, hydrogen, and nitrogen gases exerts a total pressure of 278 kPa. If the partial pressures of oxygen and hydrogen are 112 kPa and 101 kPa respectively, what would be the partial pressure exerted by the nitrogen?

Example #1 A mixture of oxygen, hydrogen, and nitrogen gases exerts a total pressure of 278 kPa. If the partial pressures of oxygen and hydrogen are 112 kPa and 101 kPa respectively, what would be the partial pressure exerted by the nitrogen? 278 kPa = 112 kPa + 101 kPa + P3 P3 = 65 kPa

Example #2 752 cm3 of oxygen gas is collected over water with a temperature of 32C. The total pressure of the gases is 102.4 kPa. What is the partial pressure of the dry gas? Vapor pressure of water at 32C = 4.8 kPa (appendix B-8)

Example #2 752 cm3 of oxygen gas is collected over water with a temperature of 32C. The total pressure of the gases is 102.4 kPa. What is the partial pressure of the dry gas? Vapor pressure of water at 32C = 4.8 kPa (appendix B-8) 102.4kPa = 4.8 kPa + Poxygen Poxygen = 97.6 kPa

Charles’s Law At constant pressure, the volume of a gas is directly proportional to its Kelvin temperature. Assuming pressure is held constant: If temperature increases, volume increases. If temperature decreases, volume decreases. i.e., temperature and volume move in the same direction.

Charles Law-- Equation Charles’s Law equation: V/T = k   V1 = V2 T1 T2 V = volume (L, mL, dm3, cm3) T = temperature (K) The equation could also be written without fractions as: V1T2 = V2T1 Temperatures must be Kelvin temperatures. Convert from Celsius to Kelvin.

Example #1: A balloon is filled with a 3.0 L of helium at 310 K and 1 atm. The balloon is placed in an oven where the temperature reaches 450 K with constant pressure. What is the new volume of the balloon?

Example #1: A balloon is filled with a 3.0 L of helium at 310 K and 1 atm. The balloon is placed in an oven where the temperature reaches 450 K with constant pressure. What is the new volume of the balloon? V1 / T1 = V2 / T2 3.0 L / 310 K = V2 / 450 K V2 = 4.4 L

Example #2: A flask contains 90.0 mL of hydrogen gas at 695 torr and 22ºC. What is the new temperature if the volume changes to 70.0 mL at 695 torr?

Example #2: A flask contains 90.0 mL of hydrogen gas at 695 torr and 22ºC. What is the new temperature if the volume changes to 70.0 mL at 695 torr? V1 / T1 = V2 / T2 22ºC = 295 K 90.0 mL / 295 K = 70.0 mL / T2 T2 = 229 K = -44ºC

Gay – Lussac’s Law P/T = k As the temperature of an enclosed gas increases, the pressure increases if volume is constant Directly proportional, temp in K P1 = P2 T1 T2

Example #1: A cylinder of gas has a pressure of 4.40 atm at 25C. At what temperature, in C, will it reach a pressure of 6.50 atm?

Example #1: A cylinder of gas has a pressure of 4.40 atm at 25C. At what temperature, in C, will it reach a pressure of 6.50 atm?

Combined Gas Law PV = k T P1V1 =P2 V2 T1 T2

Avogadro’s Law Equal volumes of gases at the same temperature and pressure contain equal numbers of molecules. V = kn k = constant, n = amount of gas in moles Under STP conditions, one mole of any gas (6.02 x 1023) takes up a volume of 22.4L

Example #1 What volume does 0.0685 mol of gas occupy at STP?

Example #1 What volume does 0.0685 mol of gas occupy at STP? 0.0686mol x 22.4L = 1.53L 1mol Conversion Factor

Graham’s Law of Effusion The rates of effusion of gases at the same temperature and pressure are inversely proportional to the square roots of their molar masses. ie heavier gases travel slower **remember difference between effusion & diffusion?

Example #1 Compare the rates of effusion of hydrogen and oxygen at the same temperature and pressure.

Example #1 Compare the rates of effusion of hydrogen and oxygen at the same temperature and pressure. Rate of H2 = √MO2 = √32.00 = 3.98 Rate of O2 √MH2 √2.02 Hydrogen effuses 3.98 x’s faster than oxygen

Ideal Gas Law The mathematical relationship between pressure, volume, temperature and the number of moles of a gas PV = nRT R = the ideal gas constant = 0.0821L•atm/(mol•K) = 8.314L•kPa/(mol•K)

Example #1 What is the pressure in atmospheres exerted by a 0.500 mol sample of nitrogen gas in a 10.0L container at 298K?

Example #1 What is the pressure in atmospheres exerted by a 0.500 mol sample of nitrogen gas in a 10.0L container at 298K? P(10.0L) = (0.500mol)(0.0821L•atm)(298K) (mol•K) P = 1.22atm

Example #2 Determine the mass of 5.60L of O2 at 1.75atm and 250.0K.

Example #2 Determine the mass of 5.60L of O2 at 1.75atm and 250.0K. (1.75atm)(5.60L) = n(0.0821L•atm)(250.0K) (mol•K) n = .477mol .477mol O2 32.00g O2 1mol O2 = 15.3g O2

Gas Stoichiometry Under Non-standard Conditions We know all there is to know about the gas laws, now let’s add some stoichiometry…

Example #1 4.00mol C8H18 25molO2 ?LO2 2mol C8H18 1molO2 1320LO2 How do we know what to put in the last conversion factor? PV = nRT V = RT **this will give us L/mol for oxygen n P V = (0.0821Latm)(308K) = 26.5L/molO2 n (molK) (.953atm) 1320LO2

Example #2 PV = nRT (1.3atm)(20.0L) = n(0.0821Latm)(263K) = 1.2molO2 (molK) 1.2molO2 18molH2O 18.02gH2O = 16gH2O 25molO2 1molH2O