John Bookstaver St. Charles Community College Cottleville, MO Chemistry, The Central Science, 11th edition Theodore L. Brown; H. Eugene LeMay, Jr.; and Bruce E. Bursten Chapter 10 Gases John Bookstaver St. Charles Community College Cottleville, MO  2009, Prentice-Hall, Inc.

Characteristics of Gases Unlike liquids and solids, gases expand to fill their containers; are highly compressible; have extremely low densities.  2009, Prentice-Hall, Inc.

Gas Pressure gas molecules are constantly in motion as they move and strike a surface, they push on that surface push = force if we could measure the total amount of force exerted by gas molecules hitting the entire surface at any one instant, we would know the pressure the gas is exerting pressure = force per unit area

Pressure: force per unit area newton (N): Unit of force pascal (Pa): Unit of pressure

Units of Pressure mm Hg or torr These units are literally the difference in the heights measured in mm (h) of two connected columns of mercury. Atmosphere 1.00 atm = 760 mm Hg Barometer  2009, Prentice-Hall, Inc.

Standard Pressure Normal atmospheric pressure at sea level is referred to as standard pressure. It is equal to 1.00 atm 760 torr (760 mm Hg) 101.325 kPa  2009, Prentice-Hall, Inc.

Manometer This device is used to measure the difference in pressure between atmospheric pressure and that of a gas in a vessel.  2009, Prentice-Hall, Inc.

The Gas Laws Gas laws – empirical relationships among gas parameters Volume (V) Pressure (P) Temperature (T) Amount of a gas (n) Copyright McGraw-Hill 2009

Boyle’s Law The volume of a fixed quantity of gas at constant temperature is inversely proportional to the pressure.  2009, Prentice-Hall, Inc.

As P and V are inversely proportional A plot of V versus P results in a curve. Since V = k (1/P) This means a plot of V versus 1/P will be a straight line. PV = k  2009, Prentice-Hall, Inc.

Example When you double the pressure on a gas, the volume is cut in half (as long as the temperature and amount of gas do not change)

Charles’s Law The volume of a fixed amount of gas at constant pressure is directly proportional to its absolute temperature. i.e., V = kT A plot of V versus T will be a straight line.  2009, Prentice-Hall, Inc.

Avogadro’s Law V a number of moles (n) V = constant x n Constant temperature - Constant pressure V a number of moles (n) V = constant x n

Avogadro’s Law The volume of a gas at constant temperature and pressure is directly proportional to the number of moles of the gas. Mathematically, this means V = kn  2009, Prentice-Hall, Inc.

Ideal-Gas Equation V  nT P So far we’ve seen that V  1/P (Boyle’s law) V  T (Charles’s law) V  n (Avogadro’s law) Combining these, we get V  nT P  2009, Prentice-Hall, Inc.

Ideal-Gas Equation PV = nRT nT P V  nT P V = R or The relationship then becomes or PV = nRT  2009, Prentice-Hall, Inc.

PV = nRT Ideal Gas Law P = pressure in atm V = volume in liters n = moles R = Gas constant = 0.08206 L. atm. Mol-1. K-1 T = temperature in Kelvin

Ideal-Gas Equation The constant of proportionality is known as R, the gas constant.  2009, Prentice-Hall, Inc.

For an ideal gas, calculate the pressure of the gas if 0.215 mol occupies 338 mL at 32.0ºC. n = 0.215 mol V = 338 mL = 0.338 L T = 32.0 + 273.15 = 305.15 K P = ? PV = nRT  = 15.9 atm

Example Air occupies 24.8 cm3 at 1.12 atm. The pressure has been increased to 2.64 atm. Assuming constant temperature, what is the new volume of air (in L)? P2 = 2.64 atm P1 = 1.12 atm V1 = 24.8 cm3 V2 = unknown PV = nRT → PV = Constant P1V1 = P2V2 P1V1 P2 V2 = 1.12 atm 2.46 atm = 0.0105 L = 0.0248 L

Example 2 NO2(g) N2O4(g) Consider the above chemical reaction. If 25 ml of NO2 gas completely converted to N2O4 gas under the same conditions, what volume will N2O4 will occupy? V RT T = Constant P = Constant PV = nRT = = constant n P VNO2 VN2O4 VN2O4 0.025 = = constant = n NO2 n N2O4 n N2O4 2 x n N2O4 2 x VN2O4 = 0.025 VN2O4 = 0.0125 L = 12.5 ml

A steel cylinder with a volume of 68.0 L contains O2 at a pressure of 15,900 kPa at 23ºC. What is the volume of this gas at Pressure = 1 atm and Temperature = 0 ºC ? P1 P2 = 1 atm T1 = 23 + 273 = 296 K T2 = 273 K V1 = 68.0 L V2 = ? PV = nRT 

Standard Condition STP The condition 0 °C and 1 atm called standard condition or standard temperature and pressure (STP) STP - Standard pressure = 1 atm - Standard temperature = 273 K = 0 °C

Molar Volume Molar Volume is volume of 1 mole of gas at STP (1 atm, 273.15 K) Calculate the volume of 1 mole of gas at STP condition? n =1 mole R = 0.08206 L. atm. Mol-1. K-1 at STP T= 273.15 K P = 1 atm nRT PV = nRT V= P 1 mole x 0.082 L.atm. Mol-1.K-1 x 273.15 K V = = 22.4 L 1 atm

Molar Volume

Example How many grams of O2 are present in 8.0 L of gas at STP? n = PV RT PV = nRT n = 1 atm x 8.0 L 0.082 L.atm. Mol-1.K-1 x 273.15 K = 0.36 mole Mass of O2 = 0.36 mole x 32 g/mole = 11.4 g of O2

Example 30.2 mL of 1.00 M HCl are reacted with excess FeS. What is the volume of H2S gas generated at STP? 2 HCl + FeS  FeCl2 + H2S(g) Moles of H2S produced = ½ moles of HCl = ½ x 1.00 mole/L x 0.0302 L = 0.0151 mole V = nRT P PV = nRT = 0.0151 mole x 0.082 L.atm. Mol-1.K-1 x 273.15 K 1 atm = 0.339 L

Densities of Gases n P V = RT If we divide both sides of the ideal-gas equation by V and by RT, we get n V P RT =  2009, Prentice-Hall, Inc.

Densities of Gases n   = m P RT m V = We know that moles  molecular mass = mass n   = m So multiplying both sides by the molecular mass ( ) gives P RT m V =  2009, Prentice-Hall, Inc.

Densities of Gases P RT m V = d = Mass  volume = density So, P RT m V = d = Note: One only needs to know the molecular mass, the pressure, and the temperature to calculate the density of a gas.  2009, Prentice-Hall, Inc.

Molecular Mass P d = RT dRT P  = We can manipulate the density equation to enable us to find the molecular mass of a gas: P RT d = Becomes dRT P  =  2009, Prentice-Hall, Inc.

Example Find the density of CO2 (in g/L) at STP (00C and 1 atm) at room conditions (20.0C and 1.00 atm). d = RT M x P 44.01 g/mol x 1atm atm*L mol*K 0.0821 x 273.15K (a) = 1.96 g/L  2009, Prentice-Hall, Inc.

Continue d = 44.01 g/mol x 1 atm x 293K atm*L mol*K 0.0821 (b) = 1.83 g/L  2009, Prentice-Hall, Inc.

Dalton’s Law of Partial Pressures The total pressure of a mixture of gases equals the sum of the pressures that each would exert if it were present alone. In other words, Ptotal = P1 + P2 + P3 + …  2009, Prentice-Hall, Inc.

Schematic of Dalton’s Law PN2 Ptotal = PN2 + PO2 Add O2

Partial Pressures When one collects a gas over water, there is water vapor mixed in with the gas. To find only the pressure of the desired gas, one must subtract the vapor pressure of water from the total pressure.  2009, Prentice-Hall, Inc.

Oxygen was produced and collected over water at 22ºC and a pressure of 754 torr. 2 KClO3(s)  2 KCl(s) + 3 O2(g) 325 mL of gas were collected and the vapor pressure of water at 22ºC is 21 torr. Calculate the number of moles of O2 and the mass of KClO3 decomposed.

Ptotal = PO2 + PH2O = PO2 + 21 torr = 754 torr PO2 = 754 torr – 21 torr = 733 torr = 733 / 760 atm V = 325 mL = 0.325 L T = 22ºC + 273 = 295 K 2 KClO3(s)  2 KCl(s) + 3 O2(g)

Effusion Effusion is the escape of gas molecules through a tiny hole into an evacuated space.  2009, Prentice-Hall, Inc.

Effusion The difference in the rates of effusion for helium and nitrogen, for example, explains a helium balloon would deflate faster.  2009, Prentice-Hall, Inc.

Diffusion Diffusion is the spread of one substance throughout a space or throughout a second substance.  2009, Prentice-Hall, Inc.

Graham's Law KE1 KE2 = 1/2 m1v12 1/2 m2v22 = = m1 m2 v22 v12 m1 m2  2009, Prentice-Hall, Inc.