Chapter 10 States of Matter & KMT Kinetic Molecular Theory (KMT) Is based on the idea that particles of matter are always in motion This theory is used to explain properties of a s, l, and g. We will focus on gases.

Chapter 10 States of Matter and KMT Ideal Gas: Conditions are low pressure and high temp. They have no volume. (particles are very far apart) Collisions are elastic. (particle to particle & particle to container wall) Gas particles are in continuous, rapid, random straight motion. There are no forces of attraction between gas particles. Avg. temp. depends on the avg. KE of the particles.

Chapter 10 States of Matter and KMT KE = ½ m v2 m = mass and v = velocity For any gas, as the temp. increases, the avg. speed and KE will increase. If the temp. decreases, the avg. speed and KE will decrease. All gases at the same temp. and pressure have the same avg. KE.

Chapter 10 States of Matter and KMT Characteristics of Gases Particles expand to fill any container (assumption#3) Gases are fluids, like liquids (assumption#4) Have very low densities & can be compressed (assumption#1)

Chapter 10 States of Matter and KMT Diffusion is the random and continuous mixing of the particles of two substances. Effusion is the process by which gas particles pass through a tiny opening.

Chapter 10 States of Matter and KMT Real gas: Conditions high pressure, low temp. and nonpolar atoms/molecules Does not behave completely according to the assumptions of the KMT. Have their own volume Attract each other

Chapter 10 States of Matter and KMT -459 32 212 -273 100 273 373 Temperature Always use absolute temperature (Kelvin) when working with gases. 0C = 5/9(0F-32) K = 0C + 273

Chapter 11 Gases Pressure (P) The force per unit area on a surface. P = Force/Area In which picture does the ballet dancer apply more pressure? What changes? The force applied by the dancer or the area over which the force was applied.

Chapter 11 Gases Measuring Pressure Barometer measures atmospheric pressure (force applies by a gas) Torricelli: Water pumps could only raise water 34 ft. He concluded that this was due to the weight of water compare to weight of air. Used mercury (14 times as dense as water and would only rise 1/14 the height of water) to make the first barometer.

Chapter 11 Gases Pressure Standard Temperature and Pressue (STP) is O0C at 1 atm or 273K at 1 atm 101.325 kPa (kilopascal – pressure of 1 newton acting on one square meter) 101.325 kPa = 1 atm = 760 mm Hg = 760 torr = 14.7 psi Pressure Problem: The average atmospheric pressure in Denver is .830 atm. Express this pressure in (a) mm Hg and (b) kPa. (a) ? mm Hg (b) ? kPa

Chapter 11 Gases Partial Pressure Dalton’s Law of partial pressure: total pressure of a mixture of gases is equal to the sum of partial pressure. PT=P1 + P2 + P3 + …. Often collected by water displacement.

Chapter 11 Gases Mixture A, B, and C have a total pressure of 6.11 atm. A has a pressure of 1.68 atm and B has a pressure of 3.89 atm. What is the pressure of C? C = ? Air is composed of carbon dioxide, oxygen, and nitrogen. At standard pressure, what is the partial pressure of oxygen? CO2 = .285 torr N2 = 593.525 torr O2 = ?

Chapter 11 Gases Oxygen was collected by water displacement in an experiment. The barometric pressure and the temperature were 731.0 torr and 200C. What is the partial pressure of the oxygen collected? (Use Table B-8 in App. B) It is 17.5 mm Hg = 17.5 torr Patm = Pgas + Pwater O2 = ?

Chapter 11 Gases Gas laws: Boyle’s law Charles’s law Gay-Lussac’s law Combined gas law.

Chapter 11 Gas Laws Boyle’s law states that the volume of a fixed mass of gas varies inversely with the pressure at a constant temperature. PV = K The inverse relationships between changes of P and V can be expressed as: P1V1=P2V2

Chapter 11 Gas Laws A sample of oxygen gas has a volume of 150. mL when its pressure is 0.947 atm. What will the volume of the gas be at a pressure of 0.987 atm if the temperature remains constant? ? of O2 A helium balloon contains 125 mL of gas at a pressure of 0.974 atm. What volume will the gas occupy at STP (standard temperature and pressure)? STP = 1 atm at 00C ? of He

Chapter 11 Gas Laws Charles’s law states that the volume of a fixed mass of gas at constant pressure varies directly with the temperature in kelvins. V=kT or k=V/T The relationships between the changes of V and T can be expressed as: V1/T1=V2/T2

Side Note He found that when he plotted the V of a gas against its T in degrees C0, it produced a straight line that extrapolated to zero V at -2730C This became know as Absolute zero, the lowest possible achievable temperature. It is assigned a value of zero on the K temperature scale. The K temperature scale had advantages. When the K temperature of a gas doubles, the volume doubles. The same occurs when the K temperature decrease.

Chapter 11 Gas Laws A sample of neon gas occupies a volume of 752 mL at 250C. What volume will the gas occupy at 50.0C if the pressure remains constant? 815 mL of Ne A sample of N gas is contained in a piston with a freely moving cylinder. At 0.00C, the volume of the gas is 375 mL. To what temperature in 0C must the gas be heated to occupy a volume of 500.0 mL? 910C

Chapter 11 Gas Laws Gay-Lussac’s law states that the pressure of a fixed mass of gas at constant volume varies directly with the temperature in kelvins. P=kT or k=P/T The relationships between the changes of V and T can be expressed as: P1/T1=P2/T2

Chapter 11 Gas Laws The gas in a container is at a pressure of 3.00 atm at 250C. Directions on the container warn the user not to keep it in a place where the temperature exceeds 520C. What would be the gas pressure in the container at 520C? 3.27 atm At 120.00C, the pressure of a sample of N is 1.07 atm. What will the pressure be at 2050C, assuming constant volume? 1.30 atm

Chapter 11 Gas Laws Combined gas law expresses the relationship between pressure, volume, and temperature of a fixed amount of gas. PV/T = k The relationships between the changes of P, V, and T can be expressed as: P1V1/T1=P2V2/T2 A gas often undergoes changes in temp, pressure, and volume all at the same time. When this happens you deal with all three variables at once. In these situations we combine all three gas laws together to in a single expression, the combined gas law.

Chapter 11 Gas Laws A helium-filled balloon has a volume of 50.0 L at 250C and 1.08 atm. What volume will it have at 0.855 atm and 10.00C? 60.0 L He The volume of a gas is 27.5 mL at 22.00C and 0.974 atm. What will the volume be at 15.00C and 0.933 atm? 28.0 mL

Chapter 11 Gas Laws In this section, you will study the reaction between the volume of gases that react with each other. We will learn about the relationship between molar amount of gas and volume. And the law that unifies all the basic gas laws. It will be important to remember the diatomic molecules: H2, N2, O2, F2, Cl2, Br2, and I2 I mole of any gas = 22.4 L at STP

Chapter 11 Gas Laws 2H2 + O2  2H2O Avogadro’s law states that equal volumes of gases at the same temperature and pressure contain equal numbers of molecules. V=kn n is the amount of gas in moles The coefficients in chem reactions involving gases can indicate # of molecules, # of moles, an volumes. 2H2 + O2  2H2O

Chapter 11 Gas Laws All gases have a volume of 22.4 L under standard conditions (STP) regardless of their mass. This is known as the standard molar volume of a gas So, 1 mole of a gas = 22.4 L We will use this as conversion factor between moles and liters. Sample Problems: What volume does 0.0658 mol of gas occupy at STP? 1.53L What quantity of gas, in moles, is contained in 2.2L at STP? 0.0987 mol

Chapter 11 Gas Laws Ideal gas law is a mathematical relationship among, pressure, volume, temperature, and the number of moles of a gas. PV=nRT Merge this with the combined gas law and you get  R=PV/nT - R is the Ideal gas constant Its value is 0.0821 L*atm/mol*K or 8.315 dm3*kPa/mol*K

Chapter 11 Gas Laws What is the pressure in atmosphere exerted by a 0.500 mol sample of nitrogen gas in a 10.0 L container at 298K? 1.22 atm What pressure, in atmosphere, is exerted by a 0.325 mol of hydrogen gas in a 4.08 L container at 350C? 2.01 atm

Chapter 11 Gas Laws Converting Moles of a gas to Liters of a gas at STP use 22.4 L/mol Not at STP: Given liters of gas? Start with ideal gas law Looking for liters of gas? Start with Stoichiometry conversion

Chapter 11 Gas Laws What volume of CO2 forms from 5.25 g of CaCO3 at 103 kPa at 250C? CaCO3  CaO + CO2 1.26 dm3 CO2 Start by calculating moles of CO2 Then plug the moles of CO2 into the ideal gas law to find liters PV=nRT

Chapter 11 Gas Laws How many grams of Al2O3 are formed from 15.0 L of O2 at 97.3 kPa at 210C? 4Al + 3O2  2Al2O3 Not at STP, so start with ideal gas law and calculate moles of O2. Now convert moles of O2 to moles of Al2O3.