The Chemistry of Gases

A Gas Uniformly fills any container. Mixes completely with any other gas Exerts pressure on its surroundings.

Pressure is equal to force/unit area SI units = Newton/meter2 = 1 Pascal (Pa) 1 standard atmosphere = 101,325 Pa 1 standard atmosphere = 1 atm = 760 mm Hg = 760 torr

Effect of Atmospheric Pressure

Standard Pressure Units 1 atmosphere = 760 mm Hg = 760 torr 1.013 x 105 Pa = 101.3 kPa 14.7 psi (pounds per square inch)

Pressure Conversions The pressure of a gas is measured as 49 torr. Represent this as atmospheres and as Pascals. 49 torr| 1 atm = 6.4 x 10-2 atm | 760 torr 49 torr | 1.013 x 105 Pa = 6.5 x 103 Pa | 760 torr

Boyle’s Law* Pressure  Volume = Constant (T = constant) P1V1 = P2V2 (T = constant) V  1/P (T = constant) (*Holds precisely only at very low pressures.)

A gas that strictly obeys Boyle’s Law is called an ideal gas.

1.53 L of sulfur dioxide gas at 5600 Pa is changed to 15000 Pa at constant temperature. What is its new volume? P1V1 =P2V2 5600 x 1.53 = 15000 x V2 V2 = 0.57 L 1.0 L of hydrogen gas at 345 torr is connected to an empty 2.5 L flask and allowed to combine. What is its new pressure? 345 x 1.0 = P2 x 3.5 P2 = 98.6 torr

Charles’s Law The volume of a gas is directly proportional to temperature, and extrapolates to zero at zero Kelvin. V = bT (P = constant) b = a proportionality constant

Balloon with Liquid N2

Charles’s Law

Always Change to KELVIN! K = C + 273

A sample of gas at 15o C and 1 atm has a volume of 2. 58 L A sample of gas at 15o C and 1 atm has a volume of 2.58 L. What is its new volume at 38o C and 1 atm pressure? Pressure remains constant at 1 atm T1 = 15 + 273 = 288 K T2 = 38 + 273 = 311 K V1 = V2 or V1T2 = V2T1 T1 T2 2.58 x 311 = V2 x 288 V2 = 2.79 L

Gay-Lussac’s Law Pressure is directly proportional to the Kelvin Temperature P = k T or P1 = P2 T1 T2

Standard Temperature and Pressure “STP” P = 1 atmosphere T = C The molar volume of an ideal gas is 22.42 liters at STP

Combined Gas Law When all three change use: P1 V1 = P2 V2 T1 T2 or P1V1 T2 = P2V2 T1

A cylinder contains 255 mL of CO2 at 27o C and 812 torr A cylinder contains 255 mL of CO2 at 27o C and 812 torr. What volume would this occupy at STP? P1= 812 torr T1= 27o C + 273= 300 K V1= 255 mL P2= 1 atm= 760 torr T2 = 0o C = 273 K P1V1T2=P2V2T1 (812)(255)(273)=V2(760)(300) V2= 247.9 mL

Avogadro’s Law For a gas at constant temperature and pressure, the volume is directly proportional to the number of moles of gas (at low pressures). V = an a = proportionality constant V = volume of the gas n = number of moles of gas

Ideal Gas Law An equation of state for a gas. “state” is the condition of the gas at a given time. PV = nRT

Ideal Gas Law PV = nRT Holds closely at P < 1 atm R = proportionality constant = 0.08206 L atm  mol P = pressure in atm V = volume in liters n = moles T = temperature in Kelvins Holds closely at P < 1 atm

Ideal Gas Laws Can be used to calculate molar masses by gas density density= mass = molar mass volume molar volume At STP, molar volume is 22.4 L

Suppose you have a sample of ammonia gas with a volume of 856 mL at a pressure of 932 torr and 28o C. What is the mass of the ammonia present? P= 932 torr / 760 torr = 1.23 atm V = 0.856 L T = 28 + 273= 301 K n = PV = (1.23)(.856) = 0.0426 moles RT (.0821)(301) mass = moles x MM= 0.0426 x 17= 0.724g

Dalton’s Law of Partial Pressures For a mixture of gases in a container, PTotal = P1 + P2 + P3 + . . . Used primarily when collecting gases over water.

Mixtures of helium and oxygen are used for deep scuba diving Mixtures of helium and oxygen are used for deep scuba diving. If 46 L oxygen and 12 L He at 1.0 atm are put into a 5.0 L tank at 25oC, calculate the partial pressures of each gas and the total pressure in the tank. n = PV = (1)(46) and (1)(12) RT (.0821)(298) (.0821)(298) nO2= 1.9 mol nHe = 0.49 mol

In the tank, V= 5.0 L PO2 =nRT / V = (1.9)(.0821)(298) / 5.0 = 9.3 atm Phe = (0.49)(.0821)(298)/ 5.0 = 2.4 atm Ptotal = 9.3 + 2.4 = 11.7 atm

Homework #1 p223ff 8, 12, 15, 17, 18, 30

Stoichiometry and Gases A sample of solid potassium chlorate decomposes by the equation 2 KClO3 (s) -----> 2 KCl (s) + 3 O2 (g) If 650 mL of oxygen at 22oC and 754 torr is collected over water, how much KClO3 decomposed? PH2O = 21 torr (from table of vapor pressure) PO2 = 754 - 21 = 733 torr / 760 = 0.964 atm nO2 = PV /RT = (.964)(.650) / (.0821)(295) = 0.0259 0.0259 mol O2 | 2 mol KClO3 | 122.6 g KClO3 = 2.11 g | 3 mol O2 | 1 mol KClO3 KClO3

Homework #2 p226ff 32, 35, 38, 39, 41, 42, 44

Kinetic Molecular Theory 1. Volume of individual particles is  zero. 2. Collisions of particles with container walls cause pressure exerted by gas. 3. Particles exert no forces on each other. 4. Average kinetic energy  Kelvin temperature of a gas.

The Meaning of Temperature Kelvin temperature is an index of the random motions of gas particles (higher T means greater motion.)

Diffusion: describes the mixing of gases Diffusion: describes the mixing of gases. The rate of diffusion is the rate of gas mixing. Effusion: describes the passage of gas into an evacuated chamber.

Diffusion Rates

Effusion: Diffusion:

Rate H2 = MMUF6 = 352 =13.2 Rate UF6 MMH2 2 Find the ratio of the effusion rates of UF6 versus H2. Rate H2 = MMUF6 = 352 =13.2 Rate UF6 MMH2 2

Homework #3 p228ff 53, 54, 68, 70, 72, 76 #4 p229ff 77, 89, 91, 92 Super XCR #97

Real Gases Must correct ideal gas behavior when at high pressure (smaller volume) and low temperature (attractive forces become important).

Real Gases   corrected pressure corrected volume Pideal Videal