Chapter 5 Gases

Elements that exist as gases at 250C and 1 atmosphere 5.1

5.1

Physical Characteristics of Gases 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

Force Pressure = Area Units of Pressure 1 pascal (Pa) = 1 N/m2 Barometer Pressure = Units of Pressure 1 pascal (Pa) = 1 N/m2 1 atm = 760 mmHg = 760 torr 1 atm = 101,325 Pa 5.2

10 miles 0.2 atm 4 miles 0.5 atm Sea level 1 atm 5.2

h = pressure difference 5.2

As P (h) increases V decreases 5.3

Boyle’s Law P a 1/V Constant temperature P x V = constant Constant amount of gas P x V = constant P1 x V1 = P2 x V2 5.3

Boyle’s Law : P - V Relationship Pressure is inversely proportional to volume Or pressure is directly proportional to 1/V P = or V = or PV=k If n (number of moles) and T are constant P1V1 = k P2V2 = k Then : P1V1 = P2V2 k V k P

P1 x V1 = P2 x V2 P1 = 726 mmHg P2 = ? V1 = 946 mL V2 = 154 mL P1 x V1 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? P1 x V1 = P2 x V2 P1 = 726 mmHg P2 = ? V1 = 946 mL V2 = 154 mL P1 x V1 V2 726 mmHg x 946 mL 154 mL = P2 = = 4460 mmHg 5.3

Charles Law - V - T- Relationship Volume proportional to T (in K) : V = kT n and P must be constant V/T = k or T1 V1 T2 = V2 V1 T1 V2 = T2 or Temperatures must be expressed in Kelvin .

Charles Law Problem A sample of carbon monoxide, a poisonous gas, occupies 3.20 L at 125 oC. Calculate the temperature (oC) at which the gas will occupy 1.54 L if the pressure remains constant. V1 = 3.20 L T1 = 125oC = 398 K V2 = 1.54 L T2 = ? T2 = 192 K oC = K - 273.15 = 192 - 273 oC = -81oC

Constant same for all gases Avogadro’s Law Constant temperature Constant pressure V a number of moles (n) V = constant x n Constant same for all gases V1/n1 = V2/n2 5.3

4NH3 + 5O2 4NO + 6H2O 1 mole NH3 1 mole NO At constant T and P Ammonia burns in oxygen to form nitric oxide (NO) and water vapor. How many liters of NO are obtained from 10 liters of ammonia at the same temperature and pressure? How many liters of oxygen are required? 4NH3 + 5O2 4NO + 6H2O 1 mole NH3 1 mole NO At constant T and P 1 volume NH3 1 volume NO 5.3

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

PV = nRT PV (1 atm)(22.414L) R = = nT (1 mol)(273.15 K) The conditions 0 0C and 1 atm are called standard temperature and pressure (STP). Experiments show that at STP, 1 mole of an ideal gas occupies 22.414 L. PV = nRT R = PV nT = (1 atm)(22.414L) (1 mol)(273.15 K) R = 0.082057 L • atm / (mol • K) 5.4

Problems for Chapter 5 Pages 163-156 1-5, 8, 13-16, 18, 20, 22, 24-30, 32, 34, 36, 38, 40, 42, 44, 48, 50, 52, 60, 62, 65, 66, 72, 76, 80

PV = nRT nRT V = P 1.37 mol x 0.0821 x 273.15 K V = 1 atm V = 30.6 L What is the volume (in liters) occupied by 49.8 g of HCl at STP? T = 0 0C = 273.15 K P = 1 atm PV = nRT n = 49.8 g x 1 mol HCl 36.45 g HCl = 1.37 mol V = nRT P V = 1 atm 1.37 mol x 0.0821 x 273.15 K L•atm mol•K V = 30.6 L 5.4

What is the pressure exerted by 168 g of argon in a 38 What is the pressure exerted by 168 g of argon in a 38.0 L container at 27oC?

PV = nRT n, V and R are constant nR V = P T = constant P1 T1 P2 T2 = 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 0C is heated to 85 0C 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 P1 = 1.20 atm T1 = 291 K P2 = ? T2 = 358 K P1 T1 P2 T2 = P2 = P1 x T2 T1 = 1.20 atm x 358 K 291 K = 1.48 atm 5.4

Practice Exercise 5.4 A gas initially at 4.0 L, 1.2 atm, and 66oC undergoes a change so that its temperature and pressure become 42oC and 1.7 atm. What is its final volume?

d is the density of the gas in g/L Density (d) Calculations m is the mass of the gas in g m V = PM RT d = 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

Example 5.5 A chemist has synthesized a greenish-yellow compound of Cl and O and finds its density to be 7.71g/L at 36oC and 2.88 atm. Calculate the molar mass and determine its molecular formula. A 2.10 L vessel contains 4.65 g of gas at 1.00 atm and 27.0oC. What is its density and molar mass?

C6H12O6 (s) + 6O2 (g) 6CO2 (g) + 6H2O (l) Gas Stoichiometry 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 L•atm mol•K 1.00 atm = nRT P V = = 4.76 L 5.5

Ideal Gas Law and Reaction Stoichiometry Sodium azide (NaN3) is used in some air bags in automobiles. Calculate the volume of nitrogen gas generated at 21 oC and 823 mm Hg by the decomposition of 60.0 g of NaN3 . 2 NaN3 (s) 2 Na (s) + 3 N2 (g)

EH Assignment Due March ? Unit 7 Minimum score Sec 1 80 Sec 2 90 Sec 3 Sec 4 85 Sec 5

Dalton’s Law of Partial Pressures V and T are constant P1 P2 Ptotal = P1 + P2 5.6

Dalton’s Law of Partial Pressures Definition: In a mixture of gases, each gas contributes to the total pressure the amount it would exert if the gas were present in the container by itself. This is called the partial pressure of the gas. To obtain a total pressure, add all of the partial pressures: Ptotal = p1+p2+p3+...

Dalton’s Law of Partial Pressures A 2.00 L flask contains 3.00 g of CO2 and 0.10 g of helium at a temperature of 17.0 oC. What are the partial pressures of each gas, and the total pressure? T = 17 oC + 273 = 290 K nCO2 = 3.00 g CO2 x 1 mole CO2 / 44.01 g CO2 = 0.0682 mole CO2 pCO2 = nCO2RT/V pCO2 = pCO2 = 0.812 atm ( 0.0682 mol CO2) ( 0.08206 L atm/mol K) ( 290 K) (2.00 L)

Dalton’s Law Problem - cont. nHe = 0.10 g He / 1 mole He / 4.003 g He = 0.025 mole He pHe = nHeRT/V pHe = pHe = 0.30 atm PTotal = pCO2 + pHe = 0.812 atm + 0.30 atm PTotal = 1.11 atm (0.025 mol) ( 0.08206 L atm / mol K) ( 290 K ) ( 2.00 L )

PA = nART V PB = nBRT V XA = nA nA + nB XB = nB nA + nB PT = PA + PB Consider a case in which two gases, A and B, are in a container of volume V. PA = nART V nA is the number of moles of A PB = nBRT V nB is the number of moles of B XA = nA nA + nB XB = nB nA + nB PT = PA + PB PA = XA PT PB = XB PT Pi = Xi PT 5.6

Pi = Xi PT PT = 1.37 atm 0.116 8.24 + 0.421 + 0.116 Xpropane = A sample of natural gas contains 8.24 moles of CH4, 0.421 moles of C2H6, and 0.116 moles of C3H8. If the total pressure of the gases is 1.37 atm, what is the partial pressure of propane (C3H8)? Pi = Xi PT PT = 1.37 atm 0.116 8.24 + 0.421 + 0.116 Xpropane = = 0.0132 Ppropane = 0.0132 x 1.37 atm = 0.0181 atm 5.6

Dalton’s Law is used for calculations when a gas is collected over water. Whenever water (or any liquid) is placed in a closed container some water will evaporate and reach a state of equilibrium where the partial pressure of the water remains constant. This pressure is called the equilibrium vapor pressure of the liquid. It depends only on the temperature.

PT = PO + PH O 2KClO3 (s) 2KCl (s) + 3O2 (g) 5.6 Bottle full of oxygen gas and water vapor 2KClO3 (s) 2KCl (s) + 3O2 (g) PT = PO + PH O 2 5.6

Relative Humidity pressure of water in air Rel Hum = x 100% Example : the partial pressure of water at 15oC is 6.54 mm Hg, what is the relative humidity? Rel Hum =(6.54 mm Hg/ 12.79 mm Hg )x100% = 51.1 % What if the temperature is increased to 25oC? pressure of water in air maximum vapor pressure of water

Collection of Hydrogen Gas over Water - Vapor Pressure 2 HCl(aq) + Zn(s) ZnCl2 (aq) + H2 (g) Calculate the mass of hydrogen gas collected over water if 156 ml of gas is collected at 20oC and 769 mm Hg. PTotal = p H2 + pH2O pH2 = PTotal - pH2O pH2 = 769 mm Hg - 17.4 mm Hg = 752 mm Hg = 0.987 atm T = 20oC + 273 = 293 K V = 0.156 L

Kinetic Molecular Theory of Gases A gas is composed of molecules that are separated from each other by distances far greater than their own dimensions. The molecules can be considered to be points; that is, they possess mass but have negligible volume. Gas molecules are in constant motion in random directions. Collisions are perfectly elastic. Pressure is caused by collisions of molecules with the walls of the container. Gas molecules exert neither attractive nor repulsive forces on one another. 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

Kinetic theory of gases explains the gas laws and properties of gases Compressibility of Gases Boyle’s Law - Why does a decrease in volume cause an increase in pressure? Charles’ Law – Why does an increase in temperature cause an increase in pressure? 5.7

Velocity and Energy Kinetic Energy = 1/2mu2 All gas molecules have the same average kinetic energy at the same temperature. Therefore heavier molecules must have slower average speeds. Consider samples of N2 and He gas at the same temperature. Which has the higher average kinetic energy? Which has the higher average speed?

Next EH assignment due Mar 12 Molecular Mass and Molecular Speeds Problem: Calculate the molecular speeds of the molecules of hydrogen, methane, and carbon dioxide at 300K! Compare speeds and kinetic energies. Next EH assignment due Mar 12 Section 7 – Properties of Gases

Molecular Mass and Molecular Speeds - III Molecule H2 CH4 CO2 Molecular Mass (g/mol) 2.016 16.04 44.01 Kinetic Energy (J/molecule) 6.213 x 10 - 21 6.0213 x 10 - 21 6.213 x 10 - 21 Velocity (m/s) 1,926 683.8 412.4

Diffusion vs. Effusion Diffusion - One gas mixing into another gas, or gases, of which the molecules are colliding with each other, and exchanging energy between molecules. Effusion - A gas escaping from a container into a vacuum. Graham’s Law

NH3 (g) + HCl(g) = NH4Cl (s) HCl = 36.46 g/mol NH3 = 17.03 g/mol RateNH3/ RateHCl = 1.463

Gas diffusion is the gradual mixing of molecules of one gas with molecules of another by virtue of their kinetic properties. NH4Cl NH3 17 g/mol HCl 36 g/mol 5.7

Apparatus for studying molecular speed distribution 5.7

 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  5.7

Deviations from Ideal Behavior 1 mole of ideal gas Repulsive Forces PV = nRT n = PV RT = 1.0 Attractive Forces

For a gas to be an ideal gas its molecules must have zero volume and there must be no forces between the molecules.

Gases show the greatest deviation from ideal behavior at 1 Gases show the greatest deviation from ideal behavior at 1. High pressures 2. Low temperatures Why?

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

The van der Waals Equation n2a V2 P + (V-nb) = nRT Gas a b atm L2 mol2 L mol He 0.034 0.0237 Ne 0.211 0.0171 Ar 1.35 0.0322 Kr 2.32 0.0398 Xe 4.19 0.0511 H2 0.244 0.0266 N2 1.39 0.0391 O2 6.49 0.0318 Cl2 3.59 0.0562 CO2 2.25 0.0428 NH3 4.17 0.0371 H2O 5.46 0.0305

Determination of the Gas Constant R THIS WEEK EXPERIMENT #9 Determination of the Gas Constant R