Gases, Liquids, and Solids

Gases Chapters 13.1 and 14 2

Kinetic Molecular Theory “Particles of matter are always in motion and this motion has consequences.”

Kinetic Molecular Theory (for gases) Size Gases consist of large numbers of tiny particles, which have mass. The distance between particles is great.  Gas particles are neither attracted to nor repelled by each other.

Kinetic Molecular Theory (for gases) Motion a) Gas particles are in constant, rapid, straight-line, random motion. They possess KE. b) Gas particles have elastic collisions (with each other and container walls) (no net loss of KE) e.g. elastic : pool balls (do not lose KE) inelastic: car crash (lose lots of KE)

Kinetic Molecular Theory Energy 3. The average KE of the gas particles is directly proportional to the Kelvin temperature of the gas. (Reminder: KE = ½ mv2)

Properties of Gases very low density (1/1000 that of solids or liquids) indefinite volume, expand and contract fluid diffuse through each other have mass exert pressure 1 mol at STP = 22.4 L (STP = 273 K, 1 atm)

Effusion vs. Diffusion Effusion: escape rate of gas through a small opening Diffusion: one material moving through another – gases diffuse through each other

Graham’s Law The kinetic energies of any two gases at the same temperature are equal: recall KE = ½ mv2

Graham’s Law Rate of gas effusion is related to MM of gas KE= ½ mv2 (m= molar mass, v=velocity (m/s) ) ½ mava2= ½ mbvb2

Practice Problems The molar mass of gas "b" is 16.04 g/mol and gas "a" is 44.04 g/mol. If gas "b" is travelling at 5.25 x 109 m/s, how fast is gas "a" travelling? Both gases have the same KE. (3.17 x 109 m/s)

Practice Problems 2.An unknown gas effuses through an opening at a rate 3.53 times slower than nitrogen gas. What is the molecular mass of the unknown gas? (349 g/mol)

Gas Pressure Pressure = Force/Area Units Force in N Pressure in: Pascals (N/m2) Torr mm Hg atmospheres Standard Pressure 101.3 kPa 760 torr 760 mm Hg 1 atm

Units: Pascals: 1 Pa = 1 N/m2, 1 kPa = 1000 Pa mm Hg (or Torr) psi = lbs/in2 atm = atmospheres Standard Pressure @ sea level 1atm = 101.325 kPa = 760. mm Hg = 760. Torr = 14.7 psi

Barometer instrument used to measure atmospheric P, using a column of Hg invented by Evangelista Torricelli in 1643

Manometer measures P of an enclosed gas relative to atmospheric P (open end) Gas P = atmospheric P ± P of liquid in U-tube Ask: Is the gas P higher or lower than atmospheric P? - If higher, add the pressure of the liquid to atm P. - If lower, subtract the pressure of the liquid from atm P.

Manometer practice problems

Manometer practice problems

Dalton’s Law of Partial Pressures Ptotal= Pa + Pb + …

Dalton’s Law Practice Problem A 1 L sample contains 78% N2, 21% O2 and 1.0% Ar. The sample is at a pressure of 1 atm (760. mm Hg). a) What is the partial pressure of each gas in mm Hg? b) What is the partial volume of each gas in mL?

Application of Dalton’s Law Collecting gas by displacement of water Ptotal = total pressure (given) PH2O varies at different temperatures (see table…)

Collecting Gas by Water Displacement The gas bubbles through the water in the jar and collects at the top due to its lower density. The gas has water vapor mixed with it. Ptotal = Pgas + PH2O Pdry gas = Ptotal – PH2O Ptotal is what is measured (= atmospheric P) PH2O can be found in standard tables of vapor pressure of water at different temperatures

Vapor pressure of H2O at various temperatures Practice Problem: Hydrogen gas is collected over water at a total pressure of 95.0 kPa and temperature of 25oC. What is the partial pressure of the dry hydrogen gas? (A: 91.8 kPa)

Mole Fraction The partial pressure of oxygen was observed to be 156 torr in air with a total atmospheric pressure of 743 torr. Calculate the mole fraction of O2 present.  The partial pressure of nitrogen was observed to be 590 mm Hg in air with a total atmospheric pressure of 760. mm Hg. Calculate the mole fraction of N2 present.

Partial Pressure problems 1. Determine the partial pressure of oxygen (O2) collected over water if the temperature is 20.0oC and the total (atmospheric) gas pressure is 98.0 kPa. (95.7 kPa) 2. The barometer at an indoor pool reads 105.00 kPa. If the temperature in the room is 30.0oC, what is the partial pressure of the “dry” air? (100.76 kPa) 3. What is the mole fraction of hydrogen (H2) in a gas mixture that has a PH2 of 5.26 kPa? The other gases in the mixture are oxygen (O2), with a PO2 of 35.2 kPa and carbon dioxide with a PCO2 of 16.1 kPa. (0.0929)

Describing Gases  “Gas Laws” To describe a gas, you need: Volume Pressure Temperature (K) # particles (moles)  “Gas Laws”

Constant Temperature What happens to P when V decreases Constant Temperature What happens to P when V decreases? Constant Pressure What happens to V when T increases? Constant Volume What happens to P when T increases?

Constant Volume and Temperature What happens to P when the # of particles is increased? Constant Temperature and Pressure What happens to V when the # of particles is increased?

The Combined Gas Law What happens to a gas when various conditions are changed? The combined gas law includes Boyle’s, Charles’s, and Gay-Lussac’s Laws….

Boyle’s Law (constant T) Demonstrates an inverse relationship between pressure and volume:

Charles’s Law (constant P) Demonstrates a direct relationship between temperature and volume:

Gay Lussac’s Law (constant V) Demonstrates a direct relationship between temperature and pressure

Boyle's Law P1V1 = P2V2 The pressure on 2.50 L of anaesthetic gas is changed from 760. mm Hg to 304 mm Hg. What will be the new volume if the temperature remains constant? (6.25 L)

Charles's Law V1 = V2 T1 T2 If a sample of gas occupies 6.8 L at 327oC, what will be its volume at 27oC if the pressure does not change? (3.4 L)

Gay-Lussac's Law P1 = P2 T1 T2 A gas has a pressure of 50.0 mm Hg at 540. K. What will be the temperature, in oC, if the pressure is 70.0 mm Hg and the volume does not change? (483oC)

Combined Gas Law P1V1 = P2V2 T1 T2 If a gas has a pressure of 2.35 atm at 25oC, and fills a container of 543 mL, what is the new pressure if the container is increased to 750. mL at 50.1oC? (1.84 atm)

Combined Gas Law A sample of methane that initially occupies 250. mL at 500. Pa and 500. K is expanded to a volume of 700. mL. To what temperature will the gas need to be heated to lower the pressure of the gas to 200. Pa? (560. K)

Ideal Gases Based on kinetic molecular theory Follows gas laws at all T and P Assumes particles: - have no V  impossible - have no attraction to each other  if true, would be impossible to liquefy gases (e.g. CO2 is liquid at ³ 5.1 atm, < 56.6oC Real Gases Because particles of real gases occupy space: Follow gas laws at most T and P At high P, individual volumes count At low T, attractions count The more polar the molecule, the more attraction counts  P decreases

The Ideal Gas Law: PV = nRT To describe a gas completely you need to identify: V – Volume P – Pressure T – Temperature in K n - # of moles The Ideal Gas Law: Can be used to derive the combined gas law Is usually used to determine a missing piece of information about a gas (requires the ideal gas constant R) 39

Most gases act like ideal gases most of the time. Ideal Gas Law Most gases act like ideal gases most of the time. PV = nRT P, V inversely related BOYLE PT directly related GAY-LUSSAC V, T directly related CHARLES V, n directly related AVOGADRO R = universal gas constant

R: The Ideal Gas Constant Derived from ideal gas law using STP conditions: standard temperature: 273K standard pressure: 1 atm volume of 1 mole of gas: 22.4 L The value of R depends on units of pressure used:

Ideal Gas Law Solve for R at STP: T = 0oC + 273 = 273 K; P = 1 atm R = PV = (1 atm)(22.4 L) = 0.0821 LŸatm nT (1 mol)(273 K) molŸK Note that the value of R depends on the units of pressure

Ideal Gas Law Calculate the pressure, in atmospheres, of 1.65 g of helium gas at 16.0oC and occupying a volume of 3.25 L. P = ? PV = nRT V = 3.25 L n = 1.65 g (1 mol He) = 0.412 mol He 4.00 g He R = 0.0821 LŸatm/molŸK T = 16.0oC + 273 = 289 K P = nRT = (0.412 mol)(0.0821 LŸatm)(289 K) V 3.25 L molŸK = 3.01 atm

Combined Gas Law Warmup A sample of neon gas has a pressure of 7.43 atm in a container with a volume of 45.1 L. This sample is transferred to a container with a volume of 18.4 L. What is the new pressure of the neon gas? Assume constant temperature. (A: 18.2 atm) A 2.45 L sample of nitrogen gas is collected at 0oC and heated to 52oC. Calculate the volume of the nitrogen gas at 52oC. Assume constant pressure. (A: 2.92 L)

Combined Gas Law Warmup 3. Consider a gas with a volume of 5.65 L at 27oC and 1 atm. At what temperature, in oC, will this gas have a volume of 6.69 L? Pressure stays at 1 atm. (A: 82oC, 355 K)

Combined Gas Law Warmup The volume of a gas-filled balloon is 30.0 L at 40oC and 153 kPa. What volume will the balloon have at STP? (A: 39.5 L) A 3.50-L gas sample at 20oC and a pressure of 86.7 kPa expands to a volume of 8.00 L. The final pressure of the gas is 56.7 kPa. What is the final temperature of the gas, in oC? (A: 165oC, 438 K)

Ideal Gas Law Practice A sample of carbon dioxide with a mass of 0.250 g was placed in a 350. mL container at 127oC. What is the pressure, in kPa, exerted by the gas? (53.9 kPa)

Ideal Gas Law Practice A 500. g block of dry ice (solid CO2) vaporizes to a gas at room temperature. Calculate the volume of gas produced at 25oC and 975 kPa. (29.0 L CO2)

Ideal Gas Law Practice At what temperature will 7.0 mol of helium gas exert a pressure of 1.2 atm in a 25.0 kL tank? (5.2 x 104 K)

Ideal Gas Law Practice What mass of chlorine (Cl2) is contained in a 10.0 L tank at 27oC and 3.50 atm? Hint: begin by solving for n. (101 g)

Density of Gases Measured in g/L (liquids and solids: g/mL) 1 mole of any gas = 22.4 L at STP (273 K and 1 atmosphere) You can use 22.4 L = 1 mol as a conversion factor at STP For non-standard conditions, use the ideal gas law.

Finding Molar Mass of a Gas Using Its Density, at STP What is the molar mass of a gas that has a density of 1.28 g/L at STP? (28.7 g/mol) A 0.519 g gas sample is found to have a volume of 200. mL at STP. What is the molar mass of this gas? (58.1 g/mol) 3. A chemical reaction produced 98.0 mL of sulfur dioxide gas (SO2) at STP. What was the mass (in grams) of the gas produced? (0.280 g SO2)

Finding Molar Mass of a Gas Using the Ideal Gas Law under non-standard conditions 4. A 1.25 g sample of the gaseous product of a chemical reaction was found to have a volume of 350. mL at 20.0oC and 750. mm Hg. What is the molar mass of this gas? 2 steps – 1. Find the number of moles (n), using the Ideal Gas Law 2. Divide the mass of the gas given in the problem by n  g/mol (86.8 g/mol) (More practice – Gases WS #5)

Density of Gases (Honors) Practice: Derive density from the ideal gas law (RQ 14.3): PV = nRT n = m/MM Rearrange to isolate M (molar mass) Apply these variations to HW on p. 438, #46-50

Warmup a) What is the density of a gas that has a mass of 0.0256 g and a volume of 178 mL? (A: 1.44 x 10-1 g/L or 1.44 x 10-4 g/mL) b) If this is the density under STP, what is the molar mass of this gas? (A: 3.22 or 3.23 g/mol) What is the molar mass of a gas that has a mass of 1.23 g and a volume of 580. mL at STP? (A: 47.5 g/mol) What is the molar mass of a gas that has a mass of 1.53 g and volume of 825 mL at 55oC and 95 kPa? (A: 53 g/mol)

Stoichiometry and Gases at STP (review) Calcium carbonate reacts with phosphoric acid to produce calcium phosphate, carbon dioxide, and water. 3 CaCO3(s) + 2 H3PO4(aq)  Ca3(PO4)2(aq) + 3 CO2(g) + 3 H2O(l) How many grams of phosphoric acid, H3PO4, react with excess calcium carbonate, CaCO3, to produce 3.74 g of Ca3(PO4)2? (2.36 g H3PO4) Assuming STP, how many liters of carbon dioxide are produced when 5.74 g of H3PO4 reacts with an excess of CaCO3? (1.97 L)

Stoichiometry of Gases non-STP conditions Context is stoichiometry To get to moles of your known asap, you either perform a mass  conversion, or use PV=nRT – solve for # moles The last step of stoich is to convert moles of your unknown into required units. That means either moles  mass or volume, for which you use PV=nRT – solves for L

Stoichiometry and Gases under non-STP Conditions Either stoichiometry first (known = solid or liquid) to find moles of unknown, followed by the Ideal Gas Law to find the volume of a gaseous product (unknown), OR use the Ideal Gas Law to find the # moles of a gaseous reactant (known), followed by stoichiometry to find the amount of a solid or liquid product (unknown).

Stoichiometry and Gases under non-STP Conditions If water is added to magnesium nitride, ammonia gas is produced when the mixture is heated. Mg3N2(s) + 3 H2O(l)  3 MgO(s) + 2 NH3(g) If 10.3 g of magnesium nitride is treated with water, what volume of ammonia gas would be collected at 24oC and 752 mm Hg? (A: 5.03 L) 2. When you produce 16.2 L of ammonia gas at 100.oC and 802 mm Hg, how many grams of magnesium oxide are also produced? (A: 33.7 g MgO)

Warm up – Stoichiometry and Gases at STP The formation of aluminum oxide from its constituent elements is represented by this equation. 4 Al(s) + 3 O2(g)  2 Al2O3(s) How many grams of aluminum are required to react with excess oxygen, to produce 3.74 g of Al2O3? (1.98 g Al) Assuming STP, how many liters of oxygen are produced when 5.74 g of Al2O3 reacts with an excess of aluminum? (1.89 L O2)

Warmup - Stoichiometry and Gases at STP If 650. mL of hydrogen gas is produced through a replacement reaction involving solid iron and sulfuric acid (H2SO4) at STP, how many grams of iron (II) sulfate are also produced?

Stoichiometry and Gases non-STP conditions How many liters of oxygen at 27oC and 188 mm Hg are needed to burn 65.5 g of carbon according to the equation 2 C(s) + O2(g)  2 CO(g) (A: 272 L)

Stoichiometry and Gas Laws non-STP conditions From Gases WS #6: WO3 (s) + 3 H2 (g)  W (s) + 3 H2O (l) How many liters of hydrogen at 35oC and 745 mm Hg are needed to react completely with 875 g of tungsten oxide? Begin with stoich or PV = nRT? (A:292 L)

Stoichiometry and Gas Laws non-standard conditions Magnesium will “burn” in carbon dioxide to produce elemental carbon and magnesium oxide. What mass of magnesium will react with a 250 mL container of CO2 at 77oC and 65 kPa? Begin with stoich or PV=nRT? (A: 0.27 g Mg)

Summary of Gas Laws (check your reference sheet – 1st page of notes) Grahams Law Dalton’s Law Ideal Gas Law Combined Gas Law Boyle’s Law (on top) Charles’s Law (likes T.V.) Gay Lussac’s Law (what’s left…)

Forces of Attraction Liquids and Solids Phase Changes 13.2-13.4

Forces of Attraction Bonding forces: Ionic, Metallic and Covalent Ionic and metallic bonding forces hold atoms of compounds together: Intramolecular forces (covalent bonds) hold atoms of individual molecules together: Intermolecular forces exist between molecules of covalently-bonded compounds - Relatively weak compared to intramolecular forces 3 types: Dispersion (London) forces Dipole-dipole forces Hydrogen bonds

“London” Forces (aka van der Waals or Dispersion forces) Weakest intermolecular force occurs between all molecules Results from an induced “temporary dipole” which induces a dipole in a nearby molecule Acts on all molecules all the time The only intermolecular force acting among noble gas atoms and non-polar molecules larger # of electrons  larger temporary dipole  stronger attractions between molecules  higher m.p. and b.p. F2 and Cl2 are gases at room T Br2 is liquid at room T (more electrons than F2 and Cl2) I2 is solid at room T (largest number of electrons)

Dipole Forces Attractions between polar molecules, stronger than London dispersion forces: (-) end of one polar molecule attracts the (+) end of another polar molecule more polar  stronger dipole force closer together  stronger dipole force

Hydrogen Bonding Always involves H attached to an O, F or N (small, high electronegativity) Strongest intermolecular force: 5% of the strength of a covalent bond! Increased b.p. and viscosity Accounts for high b.p. of H2O

Liquids and Solids Property Gases Liquids Solids Spacing Movement How are liquids and solids similar to and different from gases? (in terms of the KMT) Property Gases Liquids Solids Spacing Movement Avg. KE Attraction between particles Disorder/Order Volume Shape Fluidity Density Compressibility Diffusing Ability

Liquids and Solids Property Gases Liquids Solids Spacing Far apart much closer tog. than gases most closely packed Movement Very fast slower, slip by each other vibrate in position Avg. KE high lower lowest Attraction between particles Very low more effective, e.g. H-bonding most effective, e.g. ionic bond Disorder Highly disordered less disordered (due to IM forces  less mobility most ordered Volume Indefinite definite (like solids) definite

Liquids and Solids Property Gases Liquids Solids Shape Indefinite Fluidity Yes Yes (like gases) No (but some amorphous solids flow at a very slow rate, e.g. glasses) Density Very low Relatively high (closer to solids) – 10% less than in their solid state Highest – more closely packed (osmium is the densest) Compressibility Very compressible (1000 X) Relatively incompressible (~ 4%), like solids) less compressible than liquids

Liquids and Solids Property Gases Liquids Solids Diffusing Ability Very yes, but much slower than gases (closer tog., attractive forces get in the way; increased T  increased diffusion) millions of times slower in solids than in liquids

Liquids Viscosity: a measure of resistance to flow. In liquids, viscosity is determined by Intermolecular forces – more attractions  greater viscosity Molecular shape – longer chains  greater viscosity Temperature – colder temp  greater viscosity  Surface tension: E required to increase the SA of a liquid by a given amount Stronger intermolecular forces  greater surface tension Surfactants: compounds that lower the surface tension of water (e.g. detergent or soap)  

Liquids Capillary action – the result of cohesion and adhesion Cohesion = force of attraction between identical molecules Adhesion = force of attraction between different types of molecules e.g. Water in capillary tube – adhesion between water molecules and glass > cohesion between water molecules

Solids Crystalline solids Density of solids is higher than density of most liquids, and may be Crystalline or Amorphous Crystalline solids Made of crystals: particles arranged in orderly, geometric, repeating patterns Have definite geometric shape Crystal lattice = total 3-D array of points that describe the arrangements of particles, smallest unit is the unit cell Crystal has same symmetry as its unit cell Abrupt melting point - all bonds break at once

Crystalline Solids 7 shapes, based on arrangement of atoms in unit cell, cell lengths and cell angles:

Categories of Crystalline Solids Atomic (e.g. noble gases) Molecular (e.g. table sugar, proteins) Covalent network (e.g. diamond, quartz) Ionic Metallic Xenon and hydrogen -potential hydrogen fuel storage 79

Binding Forces in Crystals (see Bonding notes) Atomic and Covalent molecular crystals Weak intermolecular forces, low m.p., easily vaporized, relatively soft, good insulators Covalent network (e.g. diamond, graphite) 3-D covalent bonds (giant covalent molecules)  very hard, brittle, high m.p., nonconductors or semiconductors; some are planar, e.g. graphite – in sheets Ionic (e.g. NaCl) strong positive and negative ions, electrostatically attracted to one another  hard, brittle, high m.p., good insulators Metallic (metals) positive ions surrounded by a cloud of electrons (electrons can move freely through lattice)  high electrical conductivity, malleability, ductility

Amorphous solids Glasses and plastics – particles arranged randomly  nearly any shape, depending on molding No definite melting point, gradually soften to  thick, sticky liquids Cool too fast for crystals to form

Phase Changes process Phases Involved Endo/Exothermic? melting solid  liquid endothermic

Phase Changes process Phases Involved Endo/Exothermic? Melting Solid to liquid Endothermic Freezing Liquid to solid Exothermic Vaporization Liquid to gas Condensation Gas to liquid Sublimation Solid to gas Deposition Gas to solid

Phase Change Terminology Melting point (m.p.): T at which forces holding the crystal lattice of a crystalline solid together are broken and it becomes a liquid. Note: Amorphous solids act somewhat like liquids even when solid Vapor Pressure (VP): P exerted by a vapor over a liquid Boiling point (b.p.): T at which the VP of a liquid = atmospheric P Evaporation: Vaporization only at surface of liquid, below b.p. Freezing point: T at which liquid is converted into crystalline solid

Phase Diagrams Graph showing the relationships between solid, liquid and gaseous phases over a range of conditions, e.g. P vs. T Triple point = T and P conditions at which the solid, liquid and vapor of a substance can coexist at equilibrium Critical T = the highest T at which a gas can be liquified by P alone Critical P = P exerted by a substance at the critical T

Phase Diagrams

Phase Diagrams 1. What variables are plotted on a phase diagram? 2. How many phases of water are represented in its phase diagram? What are they? 3. What phases of water coexist at each point along the red curve? Along the yellow curve? Look at the phase diagram for carbon dioxide. Above which pressure and temperature is carbon dioxide unable to exist as a liquid? At which pressure and temperature do the solid, liquid, and gaseous phases of carbon dioxide coexist?

Phase Diagrams 1. What variables are plotted on a phase diagram? Pressure and temperature 2. How many phases of water are represented in its phase diagram? What are they? Three: solid, liquid, and vapor (gas) What phases of water coexist at each point along the red curve? Solid and liquid along the yellow curve? Solid and gas Look at the phase diagram for carbon dioxide. Above which pressure and temperature is carbon dioxide unable to exist as a liquid? 73 atm, 31oC At which pressure and temperature do the solid, liquid, and gaseous phases of carbon dioxide coexist? 5.1 atm, -57oC

Phase Diagrams Temperature (oC) Pressure (atm) Phase 200 1 -2 150 100 0.001 30 0.8 Liquid 100.00 vapor

Phase Diagrams Temperature (oC) Pressure (atm) Phase 200 1 vapor -2 solid 150 100 Liquid 0.001 30 0.8 liquid 0.00 < T < 100.00 100.00 < 1.00 atm

Warmup – Stoichiometry and Gases 1. The formation of aluminum oxide from its constituent elements is represented by this equation. 4 Al(s) + 3 O2(g)  2 Al2O3(s) Assuming STP, how many liters of oxygen are produced from 5.74 g of Al2O3? (1.89 L O2) 2. Consider the following chemical equation: 2 Cu2S(s) + 3 O2(g)  2 Cu2O(s) + 2 SO2(g) What volume of oxygen gas, measured at 27oC and 0.998 atm, is required to react completely with 25.0 g of copper(I) sulfide? (A: 5.82 L O2) What mass of NaCl can be produced by the reaction of Na(s) with 3.65 L Cl2(g) at 25oC and 105 kPa? (Hint: Write the chemical equation first.) (A: 18.1 g NaCl)

Gas vs. Vapor Gas State of particles at room temperature Vapor Gas formed from a substance that normally exists as a solid or liquid at room T and P Vaporization Conversion of a liquid to gas or vapor Boiling point T at which vapor pressure of liquid = atmospheric pressure

Boiling Point Boiling point at 1 atm = “normal boiling point” H2O enters vapor state within liquid. Vapor is less dense, rises to surface Needs constant energy (heat) to keep it boiling – cooling process Liquid never rises above its boiling point! (at constant P) If atm P ¯, less E is required for particles to escape atm P Mountains vs. pressure cooker

Evaporation Evaporation occurs when particles have enough KE to overcome their I.M. forces In a contained vessel (closed): “dynamic equilibrium” occurs when rate of vaporization = rate of condensation VP (which = partial pressure of a vapor above a liquid) depends on: 1. # gas particles: ­­ ­­­­Ó­ # particles  Ó­ vapor pressure 2. Temperature: as T Ó­  v.p. ­Ó­ (particles have more energy to escape) 3. Intermolecular forces: Stronger I.M. forces  ¯ v.p. (fewer particles have enough energy to break the I.M. “bonds” and escape)

Evaporation In an uncontained vessel (open): evaporation is a cooling process: the most energetic particles leave so average KE of remaining particles is lower Rate of evaporation ­ Ó­ with Ó­ ­ in: T (more particles have energy to escape the liquid) air currents surface area

Evaporation vs. Boiling takes place at liquid surface below boiling T Boiling occurs throughout liquid at boiling T

Vapor Pressure What is the VP of ethanol at 60oC? of water at the same T (60oC)? Which compound boils at a lower T? How can you tell? Which exhibits stronger IMFs? ethanol or water? Explain your answer.

Vapor Pressure What is the VP of ethanol at 60oC? about 402 torr of water at the same T (60oC)? about 180 torr Which compound boils at a lower T? How can you tell? Which exhibits stronger IMFs? ethanol or water?

Gas Stoichiometry Volume – volume: Since all gases take up the same volume at the same temperature, the mole ratio can be used as a volume ratio of two gases in the equation. Volume – mass: Requires conditions under which reaction takes place - use ideal gas law.