Gas Laws Scheffler 1

Gases Variable volume and shape Expand to occupy volume available Volume, Pressure, Temperature, and the number of moles present are interrelated Can be easily compressed Exert pressure on whatever surrounds them Easily diffuse into one another Scheffler 2

Mercury Barometer Used to define and measure atmospheric pressure On the average at sea level the column of mercury rises to a height of about 760 mm. This quantity is equal to 1 atmosphere It is also known as standard atmospheric pressure Scheffler 3

Pressure Units & Conversions The above represent some of the more common units for measuring pressure. The standard SI unit is the Pascal or kilopascal. The US Weather Bureaus commonly report atmospheric pressures in inches of mercury. Pounds per square inch or PSI is widely used in the United States. Most other countries use only the metric system. Scheffler 4

Boyle’s Law According to Boyle’s Law the pressure and volume of a gas are inversely proportional at constant pressure. PV = constant. P1V1 = P2V2 Scheffler 5

Boyle’s Law A graph of pressure and volume gives an inverse function A graph of pressure and the reciprocal of volume gives a straight line Scheffler 6

Sample Problem 1: If the pressure of helium gas in a balloon has a volume of 4.00 dm3 at 210 kPa, what will the pressure be at 2.50 dm3? P1 V1 = P2 V2 (210 kPa) (4.00 dm3) = P2(2.50 dm3) P2 = (210 kPa) (4.00 dm3) (2.50 dm3) = 340 kPa Scheffler 7

Charles’ Law According to Charles’ Law the volume of a gas is proportional to the Kelvin temperature as long as the pressure is constant V = kT Note: The temperature for gas laws must always be expressed in Kelvin where Kelvin = oC +273.15 (or 273 to 3 significant digits) V1 = T1 V2 T2 Scheffler 8

Charles’ Law A graph of temperature and volume yields a straight line. Where this line crosses the x axis (x intercept) is defined as absolute zero Scheffler 9

Sample Problem 2 A gas sample at 40 oC occupies a volume of 2.32 dm3. If the temperature is increased to 75 oC, what will be the final volume? V1 = V2 T1 T2 Convert temperatures to Kelvin. 40oC = 313K 75oC = 348K 2.32 dm3 = V2 313 K 348K ( V2) = (2.32 dm3)(348K) (313K) V2 = 2.58 dm3 Scheffler 10

Gay-Lussac’s Law P1 = P2 T2 T1 Gay-Lussac’s Law defines the relationship between pressure and temperature of a gas. The pressure and temperature of a gas are directly proportional P1 = P2 T2 T1 Scheffler 11

Sample Problem 3: The pressure of a gas in a tank is 3.20 atm at 22 oC. If the temperature rises to 60oC, what will be the pressure in the tank? P1 = P2 T1 T2 Convert temperatures to Kelvin. 22oC = 295K 60oC = 333K 3.20 atm = P2 295 K 333K ( P2) = (3.20 atm)(333K) (295K) P2 = 3.6 atm Scheffler 12

The Combined Gas Law 1. If the amount of the gas is constant, then Boyle’s Charles’ and Gay-Lussac’s Laws can be combined into one relationship 2. P1 V1 = P2 V2 T1 T2 Scheffler 13

Sample Problem 4: A gas at 110 kPa and 30 oC fills a container at 2.0 dm3. If the temperature rises to 80oC and the pressure increases to 440 kPa, what is the new volume? P1V1 = P2V2 T1 T2 Convert temperatures to Kelvin. 30oC = 303K 80oC = 353K V2 = V1 P1 T2 P2 T1 = (2.0 dm3) (110 kPa ) (353K) (440 kPa ) (303 K) V2 = 0.58 dm3 Scheffler 14

Advogadro’s Law Equal volumes of a gas under the same temperature and pressure contain the same number of particles. If the temperature and pressure are constant the volume of a gas is proportional to the number of moles of gas present V = constant * n where n is the number of moles of gas V/n = constant V1/n1 = constant = V2 /n2 V1/n1 = V2 /n2 Scheffler 15

Universal Gas Equation Based on the previous laws there are four factors that define the quantity of gas: Volume, Pressure, Kevin Temperature, and the number of moles of gas present (n). Putting these all together: PV nT = Constant = R The proportionality constant R is known as the universal gas constant Scheffler 16

Universal Gas Equation The Universal gas equation is usually written as PV = nRT Where P = pressure V = volume T = Kelvin Temperature n = number of moles The numerical value of R depends on the pressure unit (and perhaps the energy unit) Some common values of R include: R = 62.36 dm3 torr mol-1 K-1 = 0.0821 dm3 atm mol-1 K-1 = 8.314 dm3kPa mol-1 K-1 Scheffler 17

Standard Temperature and Pressure (STP) The volume of a gas varies with temperature and pressure. Therefore it is helpful to have a convenient reference point at which to compare gases. For this purpose standard temperature and pressure are defined as: Temperature = 0oC 273 K Pressure = 1 atmosphere = 760 torr = 101.3 kPa This point is often called STP Scheffler 18

Sample Problem 5 Example: What volume will 25.0 g O2 occupy at 20oC and a pressure of 0.880 atmospheres? : (25.0 g) n = ----------------- = 0.781 mol (32.0 g mol-1) Data Formula Calculation Answer V =? P = 0.880 atm; T = (20 + 273)K = 293K R = 0.08205 dm-3 atm mol-1 K-1 PV = nRT so V = nRT/P V = (0.781 mol)(0.08205 dm-3 atm mol-1 K-1)(293K) 0.880 atm V = 21.3 dm3 Scheffler 19

d is the density of the gas in g/L Universal Gas Equation –Alternate Forms Density (d) Calculations = PM RT m is the mass of the gas in g m V d = M is the molar mass of the gas Molar Mass (M ) of a Gaseous Substance dRT P d is the density of the gas in g/L M = Scheffler 20

Sample Problem 6 A 2.10 dm3 vessel contains 4.65 g of a gas at 1.00 atmospheres and 27.0oC. What is the molar mass of the gas? Scheffler 21

Sample Problem 6 Solution A 2.10 dm3 vessel contains 4.65 g of a gas at 1.00 atmospheres and 27.0oC. What is the molar mass of the gas? dRT P M = d = m V 4.65 g 2.10 dm3 = = 2.21 g dm3 2.21 g dm3 1 atm x 0.0821 x 300.15 K dm3•atm mol•K M = M = 54.6 g/mol Scheffler 22

Dalton’s Law of Partial Pressures The total pressure of a mixture of gases is equal to the sum of the pressures of the individual gases (partial pressures). PT = P1 + P2 + P3 + P4 + . . . . where PT = total pressure P1 = partial pressure of gas 1 P2 = partial pressure of gas 2 P3 = partial pressure of gas 3 P4 = partial pressure of gas 4 Scheffler 23

Dalton’s Law of Partial Pressures Applies to a mixture of gases Very useful correction when collecting gases over water since they inevitably contain some water vapor. Scheffler 24

Sample Problem 7 Henrietta Minkelspurg generates Hydrogen gas and collected it over water. If the volume of the gas is 250 cm3 and the barometric pressure is 765.0 torr at 25oC, what is the pressure of the “dry” hydrogen gas at STP? (PH2O = 23.8 torr at 25oC) Scheffler 25

Sample Problem 8 -- Solution Henrietta Minkelspurg generates Hydrogen gas and collected it over water. If the volume of the gas is 250 cm3 and the barometric pressure is 765.0 torr at 25oC, what is the pressure of the “dry” hydrogen gas at STP? (PH2O = 23.8 torr at 25oC) Scheffler 26

Sample Problem 9 Henrietta Minkelspurg generated Hydrogen gas and collects it over water. If the volume of the gas is 250 cm3 and the barometric pressure is 765 torr at 25oC, what is the volume of the “dry” oxygen gas at STP? Scheffler 27

Sample Problem 9 -- Solution Henrietta Minkelspurg generated Hydrogen gas and collects it over water. If the volume of the gas is 250 cm3 and the barometric pressure is 765 torr at 25oC, what is the volume of the “dry” oxygen gas at STP? From the previous calculation the adjusted pressure is 742.2 torr P1= PH2 = 742.2 torr; P2= Std Pressure = 760 torr V1= 250 cm3; T1= 298K; T2= 273K; V2= ? (V1P1/T1) = (V2P2/T2) therefore V2= (V1P1T2)/(T1P2) V2 = (250 cm3)(742.2 torr)(273K) (298K)(760.torr) V2 = 223.7 cm3 Scheffler 28

Kinetic Molecular Theory Matter consists of particles (atoms or molecules) that are in continuous, random, rapid motion The Volume occupied by the particles has a negligibly small effect on their behavior Collisions between particles are elastic Attractive forces between particles have a negligible effect on their behavior Gases have no fixed volume or shape, but take the volume and shape of the container The average kinetic energy of the particles is proportional to their Kelvin temperature Scheffler 29

Maxwell-Boltzman Distribution Molecules are in constant motion Not all particles have the same energy The average kinetic energy is related to the temperature An increase in temperature spreads out the distribution and the mean speed is shifted upward Scheffler 30

 Velocity of a Gas 3RT urms = M The distribution of speeds of three different gases at the same temperature The distribution of speeds for nitrogen gas molecules at three different temperatures urms = 3RT M  Scheffler 31

Diffusion Gas diffusion is the gradual mixing of molecules of one gas with molecules of another by virtue of their kinetic properties. NH4Cl NH3 17.0 g/mol HCl 36.5 g/mol Scheffler 32

DIFFUSION AND EFFUSION Diffusion is the gradual mixing of molecules of different gases. Effusion is the movement of molecules through a small hole into an empty container. Scheffler 33

Graham’s Law Graham’s law governs effusion and diffusion of gas molecules. KE=1/2 mv2 The rate of effusion is inversely proportional to its molar mass. Thomas Graham, 1805-1869. Professor in Glasgow and London. Scheffler 34

4 NH3(g) + 5 O2(g)  4 NO(g) + 6 H2O(g) Sample Problem 10 1 mole of oxygen gas and 2 moles of ammonia are placed in a container and allowed to react at 850oC according to the equation: 4 NH3(g) + 5 O2(g)  4 NO(g) + 6 H2O(g) Using Graham's Law, what is the ratio of the effusion rates of NH3(g) to O2(g)? Scheffler 35

Sample Problem 10 Solution 1 mole of oxygen gas and 2 moles of ammonia are placed in a container and allowed to react at 850oC according to the equation: 4 NH3(g) + 5 O2(g)  4 NO(g) + 6 H2O(g) Using Graham's Law, what is the ratio of the effusion rates of NH3(g) to O2(g)? Scheffler 36

Sample Problem 11 What is the rate of effusion for H2 if 15.00 cm3 of CO2 takes 4.55 sec to effuse out of a container? Scheffler 37

Sample Problem 11 Solution What is the rate of effusion for H2 if 15.00 cm3 of CO2 takes 4.55 sec to effuse out of a container? Rate for CO2 = 15.00 cm3/4.55 s = 3.30 cm3/s Scheffler 38

Sample Problem 12 What is the molar mass of gas X if it effuses 0.876 times as rapidly as N2(g)? Scheffler 39

Sample Problem 12 Solution What is the molar mass of gas X if it effuses 0.876 times as rapidly as N2(g)? Scheffler 40

Ideal Gases v Real Gases Ideal gases are gases that obey the Kinetic Molecular Theory perfectly. The gas laws apply to ideal gases, but in reality there is no perfectly ideal gas. Under normal conditions of temperature and pressure many real gases approximate ideal gases. Under more extreme conditions more polar gases show deviations from ideal behavior. Scheffler 41

In an Ideal Gas --- The particles (atoms or molecules) in continuous, random, rapid motion. The particles collide with no loss of momentum The volume occupied by the particles is essentially zero when compared to the volume of the container The particles are neither attracted to each other nor repelled The average kinetic energy of the particles is proportional to their Kelvin temperature At normal temperatures and pressures gases closely approximate idea behavior Scheffler 42

Real Gases For ideal gases the product of pressure and volume is constant. Real gases deviate somewhat as shown by the graph pressure vs. the ratio of observed volume to ideal volume below. These deviations occur because Real gases do not actually have zero volume Polar gas particles do attract if compressed Scheffler 43

van der Waals Equation (P + n2a/V2)(V - nb) = nRT The van der Waals equation shown below includes corrections added to the universal gas law to account for these deviations from ideal behavior (P + n2a/V2)(V - nb) = nRT where a => attractive forces between molecules b => residual volume or molecules The van der Waals constants for some elements are shown below Substance a (dm6atm mol-2) b (dm3 mol-1) He 0.0341 0.02370 CH4 2.25 0.0428 H2O 5.46 0.0305 CO2 3.59 0.0437 Scheffler 44

C6H12O6 (s) + 6O2 (g) 6CO2 (g) + 6H2O (l) Sample Problem 13 What is the volume of CO2 produced at 370 C and 1.00 atm when 5.60 g of glucose are used up in the reaction: C6H12O6 (s) + 6O2 (g) 6CO2 (g) + 6H2O (l) Scheffler 45

C6H12O6 (s) + 6O2 (g) 6CO2 (g) + 6H2O (l) Sample Problem 13 Solution What is the volume of CO2 produced at 370 C and 1.00 atm when 5.60 g of glucose are used up in the reaction: C6H12O6 (s) + 6O2 (g) 6CO2 (g) + 6H2O (l) g C6H12O6 mol C6H12O6 mol CO2 V CO2 1 mol C6H12O6 180 g C6H12O6 x 6 mol CO2 1 mol C6H12O6 x 5.60 g C6H12O6 = 0.187 mol CO2 0.187 mol x 0.0821 x 310.15 K dm3•atm mol•K 1.00 atm = nRT P V = = 4.76 dm3 Scheffler 46