Gases Chapter 5 Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

Elements that exist as gases at 25 0 C and 1 atmosphere 5.1

Gases assume the volume and shape of their containers. Gases are the most compressible state of matter. Gases will mix evenly and completely when confined to the same container. Gases have much lower densities than liquids and solids. 5.1 Physical Characteristics of Gases

Units of Pressure 1 pascal (Pa) = 1 N/m 2 1 atm = 760 mmHg = 760 torr 1 atm = 101,325 Pa 5.2 Barometer Pressure = Force Area (force = mass x acceleration)

Sea level1 atm 4 miles0.5 atm 10 miles0.2 atm 5.2

Manometers Used to Measure Gas Pressures

5.3 As P (h) increases V decreases Apparatus for Studying the Relationship Between Pressure and Volume of a Gas

The volume of a gas depends on its Temperature, its pressure, & the amount of gas (moles of gas). The gas laws define the dependence of the volume on each of the factors individually and in combinations. Boyle’s Law: Dependence of volume (V) on pressure (P) Charles’ Law: Dependence of volume (V) on temperature (T) Avogadro’s Law: Dependence of Volume (V) on amount of gas (n) Combined Gas Law: Dependence of Volume (V) on P & T Ideal Gas Law: Dependence of Volume on P, T and n Gas Laws

V  1/P P x V = constant P 1 x V 1 = P 2 x V Boyle’s Law Constant temperature Constant amount of gas

Boyle’s Law: The volume of a fixed amount of gas at constant T varies inversely as its pressure. When pressure increases volume decreases and vice versa. Prediction of Boyle’s Law: volume = 0 when pressure = infinity. Reality: When the pressure is increased, the volume decreases and then the gas condenses to become a liquid ( as long as the temperature is below critical temperature). Boyle’s law is not valid for real gases at very high pressures.

5.3

A sample of chlorine gas occupies a volume of 946 mL at a pressure of 726 mmHg. What is the pressure of the gas (in mmHg) if the volume is reduced at constant temperature to 154 mL? P 1 x V 1 = P 2 x V 2 P 1 = 726 mmHg V 1 = 946 mL P 2 = ? V 2 = 154 mL P 2 = P 1 x V 1 V2V2 726 mmHg x 946 mL 154 mL = = 4460 mmHg 5.3 P x V = constant

As T increasesV increases 5.3

Variation of gas volume with temperature at constant pressure. 5.3 V  TV  T V = constant x T V 1 /T 1 = V 2 /T 2 T (K) = t ( 0 C) Charles’ & Gay-Lussac’s Law Temperature must be in Kelvin

Charles’s law predicts that the volume of a gas increases with increasing temperature and decreases with decreasing temperature. Charles’ law predicts that the volume of a gas = 0 when its temperature = 0K =  C Reality: As the gas cooled at constant pressure, its volume decreases up to the point when it condenses. Gas laws only apply to gases. Charles’ law fails at low temperatures.

5.3

A sample of carbon monoxide gas occupies 3.20 L at C. At what temperature will the gas occupy a volume of 1.54 L if the pressure remains constant? V 1 = 3.20 L T 1 = K V 2 = 1.54 L T 2 = ? T 2 = V 2 x T 1 V1V L x K 3.20 L = = 192 K 5.3 V 1 /T 1 = V 2 /T 2 T 1 = 125 ( 0 C) (K) = K

Avogadro’s Law V  number of moles (n) V = constant x n V 1 / n 1 = V 2 / n Constant temperature Constant pressure

Avogadro’s law states that for a gas at constant Temperature & pressure, the volume occupied is directly proportional to the amount of gas (mol). In chemical reactions carried out at fixed T & P involving gases as reactants and products, the Avogadro’s law predicts that the ratio of volumes of gaseous reactants used to the volumes of gaseous products formed = ratio of moles of gaseous reactants used to the moles of gaseous products formed. That is for a reaction such as: aA(g) + bB(g)  cC(g) + dD(g) Moles of A used : moles of B used : moles of C formed : moles of D formed = a : b: c : d Volume of A used : volume of B used : volume of C formed : volume of D formed = a : b : c : d

Ammonia burns in oxygen to form nitric oxide (NO) and water vapor. How many volumes of NO are obtained from one volume of ammonia at the same temperature and pressure? 4NH 3 + 5O 2 4NO + 6H 2 O 1 mole NH 3 1 mole NO At constant T and P 1 volume NH 3 1 volume NO 5.3 How many volumes of NO would be obtained from 2.0 L of O2? 1.6 L NO

5.3

Combined Gas Law This law is a combination of Charles’ law & Boyle’s law. It describes the dependence of the volume of a fixed amount of gas when pressure and temperature are varied It is only valid and high temperatures AND low pressures.

Ideal Gas Equation 5.4 Charles’ law: V  T  (at constant n and P) Avogadro’s law: V  n  (at constant P and T) Boyle’s law: V  (at constant n and T) 1 P V V  nT P V = constant x = R nT P P R is the gas constant PV = nRT

The conditions 0 0 C and 1 atm are called standard temperature and pressure (STP). PV = nRT R = PV nT = (1 atm)(22.414L) (1 mol)( K) R = L atm / (mol K) 5.4 Experiments show that at STP, 1 mole of an ideal gas occupies L.

Boyles’ law: P 1 V 1 = P 2 V 2 (constant T, n) Charles’ Law: V 1 /T 1 = V 2 /T 2 (constant (P, n); T in K Avogadro’s law: V 1 /n 1 = V 2 /n 2 (constant T, P); units of n = mole Combined gas law: P 1 V 1 /T 1 = P 2 V 2 /T 2 ; T in K Ideal Gas Law: PV = nRT d,Density of a vapor (gas) = P x Molar Mass/(RT); T in K, P in atm, units of d = g/L Units of PUnits of VValue of R atmL L atm/ (K-mol) Pa (SI)m3m J/(K-mol)

What is the volume (in liters) occupied by 49.8 g of HCl(g) at STP? PV = nRT V = nRT P T = 0 0 C = K P = 1 atm n = 49.8 g x 1 mol HCl g HCl = 1.37 mol V = 1 atm 1.37 mol x x K Latm molK V = 30.6 L 5.4

Argon is an inert gas used in lightbulbs to retard the vaporization of the filament. A certain lightbulb containing argon at 1.20 atm and 18 0 C is heated to 85 0 C at constant volume. What is the final pressure of argon in the lightbulb (in atm)? PV = nRT n, V and R are constant nR V = P T = constant P1P1 T1T1 P2P2 T2T2 = P 1 = 1.20 atm T 1 = 291 K P 2 = ? T 2 = 358 K P 2 = P 1 x T2T2 T1T1 = 1.20 atm x 358 K 291 K = 1.48 atm 5.4

Density (d) Calculations d = m V = PMPM RT m is the mass of the gas in g M is the molar mass of the gas Molar Mass ( M ) of a Gaseous Substance dRT P M = d is the density of the gas in g/L 5.4

A 2.10-L vessel contains 4.65 g of a gas at 1.00 atm and C. What is the molar mass of the gas? 5.4 dRT P M = d = m V 4.65 g 2.10 L = = 2.21 g L M = 2.21 g L 1 atm x x K Latm molK M = 54.6 g/mol

1.A fixed quantity of gas at constant temperature occupies 20.5 L at 737 torr. What is the pressure if the volume is 16.0L? 1.24 atm 2.A fixed quantity of gas at a constant pressure occupies a volume of 8.50 L at 29 C. What is the volume at 125C? 11.2 L 3.At 46 C and 0.88atm, a gas occupies 600 mL. What will be the volume at 0C and 156 torr? 2.20 L 4.1 mole of O2(g) occupies 22.4 L at STP. What will be the volume occupied by 4.00g of O2 at STP? 2.8 L 5. Calculate the density of SO2 gas at 730 torr and 35C g/L 6.If 1.84 g of a gas occupies 500 mL at 15 C and 825 torr, what is the molar mass? 80.1 g/mol 7.What is the volume occupied by mol of an ideal gas at 145C, 59.0 torr? 29.6 L

Gas Stoichiometry What is the volume of CO 2 produced at 37 0 C and 1.00 atm when 5.60 g of glucose are used up in the reaction: C 6 H 12 O 6 (s) + 6O 2 (g) 6CO 2 (g) + 6H 2 O (l) g C 6 H 12 O 6 mol C 6 H 12 O 6 mol CO 2 V CO g C 6 H 12 O 6 1 mol C 6 H 12 O g C 6 H 12 O 6 x 6 mol CO 2 1 mol C 6 H 12 O 6 x = mol CO 2 V = nRT P mol x x K Latm molK 1.00 atm = = 4.76 L 5.5

1. Calculate the volume of NH 3 (g) needed at 20.0  C and 25.0 atm to react with 150 kg of H 2 SO 4 2NH 3 (g) + H 2 SO 4 (aq)  (NH 4 ) 2 SO 4 (aq) 2. I 2 (s) + 5F 2 (g)  2IF 5 (g) A 5.00L flask contains excess I 2 and 10.0g F 2 (g). a.What is mass of I 2 used if the reaction is carried out at 125  C? 13.4 g (you do not need gas laws to solve this part of the problem) b.What is pressure due to IF 5 (g) formed at this temperature? 10.0 g F2 used = mol F2 used = mol IF5 formed = atm (ideal gas equation)

1. The equation for the combustion of propanol, C 3 H 7 OH, is 2C 3 H 7 OH + 9O 2  6CO 2 + 8H 2 O What is the volume of CO 2 (g) formed at 26  C and 767torr when 40.0g of propanol, C 3 H 7 OH, is completely combusted? 2. The balanced equation for the reaction between Al and O 2 forming Aluminum oxide, Al 2 O 3 is 4 Al(s) + 3O 2 (g)  2Al 2 O 3 (s) Calculate the volume of O 2 (g) at a pressure of 782 torr and a temperature of 25°C required to completely react with 54.0 g of Al.

Dalton’s Law of Partial Pressures V and T are constant P1P1 P2P2 P total = P 1 + P The total pressure of a mixture of gases at constant V & T = sum of their individual pressures or partial pressures. A mixture of ideal gases also obeys ideal gas law; P total =n total RT/V; where n total = n1 + n2 + … = the sum of moles of all gases in mixture.

1.What is the total pressure of a gas mixture containing mol He(g), mol Ne(g) and mol Ar(g) occupying a 7.00L flask at 25.0  C? 2.A mixture containing 4.00g each of CH4(g), C2H2(g), C2H4(g) is contained in a 1.50 L flask at 273K. What is the pressure due to this mixture? 1.P total = n total RT/V n total = ( ) MOL = mol T = 298K; V = 7.00L P = mol (0.0821L atm/K-mol) 298K/7.00L = 3.34 atm 2. P total = n total RT/V n total = (4.00/16.05) mol + (4.00/26.04) mol + (4.00/28.06) mol = mol T = 273 K; V = 1.50 L P = mol (0.0821L atm/K-mol) 273K/1.50L = 8.14 atm

1. A mixture of 4.05g N 2, 6.05g Kr are placed in a 6.10L vessel at 25.0  C. Assume that the gases behave ideally a. Calculate the partial pressures of each of the gases in the mixture. b. Calculate the total pressure due to the mixture of gases. 2. A gas mixture contains 2.80 g N 2 (g) and g O 2 (g) in a 2.00 L container at 23.0°C. Calculate the pressure due to this mixture.

Pressure of a component gas A in a container of Volume V & temperature T, P A = n A RT/V Total pressure of the mixture of gases in the same container and at the same temperature of which A is a component, P total = n total RT/V P A /P total =(n A RT/V)/(n total RT/V) = n A /n Total = mole fraction of A in mixture = X A. P A /P total = X A = n A /n total Partial pressure of a component/Total pressure of gas mixture = mole fraction of the component gas = moles of component gas/moles of all gases in the mixture.

Consider a case in which two gases, A and B, are in a container of volume V. P A = n A RT V P B = n B RT V n A is the number of moles of A n B is the number of moles of B P T = P A + P B X A = nAnA n A + n B X B = nBnB n A + n B P A = X A P T P B = X B P T P i = X i P T 5.6 mole fraction (X i ) = nini nTnT

1.The total pressure of a gas mixture consisting of mol N 2, mol O 2, mol CO 2 is 1.32 atm. Calculate the partial pressures of the three component gases. 2.The total pressure of a gas mixture consisting of 3.50g N 2, 2.15g H 2, 5.27g NH 3 is 2.50 atm. Calculate the partial pressures of the three component gases. 1. P A /P total = X A = n A /n total P total = 1.32 atm; n total = ( )mol = mol X N2 = n N2 /n total = mol/0.850 mol = P N2 = X N2 P total = x 1.32 atm = atm X O2 = n O2 /n total = mol/0.850 mol = P O2 = X O2 P total = x 1.32 atm = atm X CO2 = n CO2 /n total = mol/0.850 mol = P CO2 = X CO2 P total = x 1.32 atm = atm

1.A gas mixture contains g N 2, 8.15 g NH 3 and 1.98 g H 2. If the total pressure due to this mixture is 2.35 atm, calculate the partial pressure of NH 3 in this mixture. 2. What are the partial pressures of each component in a mixture of g N 2, 12.45g H 2, 2.38g NH 3. if the total pressure of the mixture is 25.4 atm?

g H 2 = (2.15g x 1mol/2.02g) = 1.06 mol H g N 2 = (3.50 g x 1mol/28.02g) = mol N g NH 3 = (5.27g x 1mol/17.04g) = mol NH 3 P total = 2.50 atm n total = ( ) mol = 1.49 mol Calculate partial pressures of N 2, H 2, NH 3 at home Answers: Partial pressure of N2 = atm Partial pressure of H2: 1.78 Partial pressure of NH3 = 0.518

A sample of natural gas contains 8.24 moles of CH 4, moles of C 2 H 6, and moles of C 3 H 8. If the total pressure of the gases is 1.37 atm, what is the partial pressure of propane (C 3 H 8 )? P i = X i P T X propane = P T = 1.37 atm = P propane = x 1.37 atm= atm 5.6

2KClO 3 (s) 2KCl (s) + 3O 2 (g) Bottle full of oxygen gas and water vapor P T = P O + P H O

Chemistry in Action: Scuba Diving and the Gas Laws PV Depth (ft)Pressure (atm)

Kinetic Molecular Theory of Gases 1.A gas is composed of molecules that are separated from each other by distances far greater than their own sizes. The molecules can be considered to be points; that is, they possess mass but have negligible volume (volume of an individual gas molecule is insignificant compared to the container volume). 2.Gas molecules are in constant motion in random directions, and they frequently collide with one another. Collisions among molecules are perfectly elastic. 3.Gas molecules exert neither attractive nor repulsive forces on one another. 4.The average kinetic energy of the molecules is proportional to the temperature of the gas in kelvins. Any two gases at the same temperature will have the same average kinetic energy 5.7 KE = (3/2) RT; R = J/K-mol

Kinetic theory of gases and … Compressibility of Gases Boyle’s Law P  collision rate with wall Collision rate  number density Number density  1/V P  1/V Charles’ Law P  collision rate with wall Collision rate  average kinetic energy of gas molecules Average kinetic energy  T P  TP  T 5.7

Kinetic theory of gases and … Avogadro’s Law P  collision rate with wall Collision rate  number density Number density  n P  nP  n Dalton’s Law of Partial Pressures Molecules do not attract or repel one another P exerted by one type of molecule is unaffected by the presence of another gas P total =  P i 5.7

Apparatus for studying molecular speed distribution 5.7

The distribution of speeds for nitrogen gas molecules at three different temperatures The distribution of speeds of three different gases at the same temperature 5.7 u rms = 3RT M  Units of molar mass in the equation below: kg/mol R = J/K-mol T in K u rms = root mean square velocity

Chemistry in Action: Super Cold Atoms Gaseous Rb Atoms 1.7 x K Bose-Einstein Condensate

Gas diffusion is the gradual mixing of molecules of one gas with molecules of another by virtue of their kinetic properties. 5.7 NH 3 17 g/mol HCl 36 g/mol NH 4 Cl r1r1 r2r2 M2M2 M1M1  = The rate or speed at which a gas diffuses is inversely proportional to the square root of its molar mass. Larger the molar mass, smaller the rate of diffusion, r. For two different gases, 1 and 2, the rates are related by the equation on the left.

1.Compare the rates of diffusion of He (molar mass = 4g/mol) and methane, CH 4 (molar mass = 16 g/mol): calculate r He /rCH 4. Which gas diffuses faster 2.He diffuses 4 times faster than a certain gas X. Calculate the molar mass of that gas & identify the gas from the following list: (a) CH 4, (b) CO 2 (c) SO 2 (d) Ar 1.He diffuses faster; r He /rCH 4 = √ (M CH4 /M He ) = √ (16/4) = 2 2.r He /r X = √ (M X /M He ) = 4; (M X /M He ) = 4 2 = 16; (M X /4) = 16 M X = 16 x 4 = 64; gas = SO 2

Gas effusion is the is the process by which gas under pressure escapes from one compartment of a container to another by passing through a small opening. 5.7 r1r1 r2r2 t2t2 t1t1 M2M2 M1M1  == Nickel forms a gaseous compound of the formula Ni(CO) x What is the value of x given that under the same conditions methane (CH 4 ) effuses 3.3 times faster than the compound? r 1 = 3.3 x r 2 M 1 = 16 g/mol M 2 = r1r1 r2r2 ( ) 2 x M 1 = (3.3) 2 x 16 = x 28 = x = 4.1 ~ 4

Deviations from Ideal Behavior 1 mole of ideal gas PV = nRT n = PV RT = Repulsive Forces Attractive Forces

Effect of intermolecular forces on the pressure exerted by a gas. 5.8

Van der Waals equation nonideal gas P + (V – nb) = nRT an 2 V2V2 () } corrected pressure } corrected volume