Ch. 13.4 - Liquids & Solids III. Changes of State C. Johannesson

A. Phase Changes C. Johannesson

A. Phase Changes Evaporation molecules at the surface gain enough energy to overcome intermolecular forces Volatility measure of evaporation rate depends on temp & intermolecular forces C. Johannesson

A. Phase Changes temp volatility IMF volatility Boltzmann Distribution # of Particles volatility IMF volatility C. Johannesson Kinetic Energy

A. Phase Changes Equilibrium trapped molecules reach a balance between evaporation & condensation C. Johannesson

A. Phase Changes temp v.p. IMF v.p. Vapor Pressure pressure of vapor above a liquid at equilibrium p.478 v.p. depends on temp & IMF directly related to volatility temp temp v.p. IMF v.p. C. Johannesson

A. Phase Changes Patm b.p. IMF b.p. Boiling Point temp at which v.p. of liquid equals external pressure depends on Patm & IMF Normal B.P. - b.p. at 1 atm Patm b.p. IMF b.p. C. Johannesson

A. Phase Changes IMF m.p. Melting Point equal to freezing point Which has a higher m.p.? polar or nonpolar? covalent or ionic? polar ionic C. Johannesson

A. Phase Changes Sublimation solid  gas v.p. of solid equals external pressure EX: dry ice, mothballs, solid air fresheners C. Johannesson

B. Heating Curves Gas - KE  Boiling - PE  Liquid - KE  Melting - PE  Solid - KE  C. Johannesson

B. Heating Curves Temperature Change change in KE (molecular motion) depends on heat capacity Heat Capacity energy required to raise the temp of 1 gram of a substance by 1°C Water has a high heat capacity C. Johannesson

B. Heating Curves Phase Change change in PE (molecular arrangement) temp remains constant Heat of Fusion (Hfus) energy required to melt 1 gram of a substance at its m.p. C. Johannesson

B. Heating Curves Heat of Vaporization (Hvap) energy required to boil 1 gram of a substance at its b.p. usually larger than Hfus…why? EX: sweating, steam burns C. Johannesson

C. Phase Diagrams Show the phases of a substance at different temps and pressures. C. Johannesson

Phase diagram C. Johannesson

I. Physical Properties (p. 450-455) Ch. 14.1 - Gases I. Physical Properties (p. 450-455) C. Johannesson

A. Kinetic Molecular Theory Particles in an ideal gas… have no volume. have elastic collisions. are in constant, random, straight-line motion. don’t attract or repel each other. have an avg. KE directly related to Kelvin temperature. C. Johannesson

B. Real Gases Particles in a REAL gas… have their own volume attract each other Gas behavior is most ideal… at low pressures at high temperatures in nonpolar atoms/molecules C. Johannesson

C. Characteristics of Gases Gases expand to fill any container. random motion, no attraction Gases are fluids (like liquids). no attraction Gases have very low densities. no volume = lots of empty space C. Johannesson

C. Characteristics of Gases Gases can be compressed. no volume = lots of empty space Gases undergo diffusion & effusion. random motion C. Johannesson

D. Temperature Always use absolute temperature (Kelvin) when working with gases. ºF ºC K -459 32 212 -273 100 273 373 K = ºC + 273 C. Johannesson

E. Pressure Which shoes create the most pressure? C. Johannesson

E. Pressure Barometer measures atmospheric pressure Aneroid Barometer Mercury Barometer Aneroid Barometer C. Johannesson

E. Pressure Manometer measures contained gas pressure U-tube Manometer Bourdon-tube gauge C. Johannesson

E. Pressure KEY UNITS AT SEA LEVEL 101.325 kPa (kilopascal) 1 atm 760 mm Hg 760 torr 14.7 psi C. Johannesson

Standard Temperature & Pressure F. STP STP Standard Temperature & Pressure 0°C 273 K 1 atm 101.325 kPa -OR- C. Johannesson

Ch. 14.2 - Gases II. The Gas Laws P V T C. Johannesson

A. Boyle’s Law P V P1V1=P2V2 C. Johannesson

Boyles Law C. Johannesson

A. Boyle’s Law The pressure and volume of a gas are inversely related at constant mass & temp P V P1V1=P2V2 C. Johannesson

Applying Boyles Law  

Practice Problems What pressure will be needed to reduce the volume of 77.4 L of helium at 98.0 kPa to a volume of 60.0 L? A 250.0 mL sample of chlorine gas is collected when the barometric pressure is 105.2 kPa. What is the volume of the sample after the barometer drops to 100.3 kPa A weather balloon contains 59.5 L of helium at sea level, where the atmospheric pressure is 101.3 kPa. The balloon is released from a 4000m mountaintop where the pressure is 61.7 kPa. What is the volume of the balloon when it is released?

Answers 126 kPa 262.2 mL 97.7 L C. Johannesson

B. Charles’ Law V T C. Johannesson

B. Charles’ Law The volume and absolute temperature (K) of a gas are directly related at constant mass & pressure V T C. Johannesson

Example  

Practice Problems A sample of SO2 gas has a volume of 1.16 L at a temperature of 23°C. At what temperature will the gas have a volume of 1.25 L? A balloon is inflated with 6.22 L of helium at a temperature of 36°C. What is the volume of the balloon when the temperature is 22°C? A student collects a 125.0 mL sample of hydrogen. Later the sample is found to have a volume of 128.6 mL at a temperature of 26°C. At what temperature was the sample collected?

Answers 46°C (319K) 5.94 L 18°C (291K)

C. Gay-Lussac’s Law P T C. Johannesson

C. Gay-Lussac’s Law The pressure and absolute temperature (K) of a gas are directly related at constant mass & volume P T C. Johannesson

Combined Gas Law All of the gas laws can be combined into a single law You can find the value of any one of the variables if you know the other five C. Johannesson

D. Combined Gas Law P1V1 T1 = P2V2 T2 P1V1T2 = P2V2T1 C. Johannesson

Applying the combined gas law   C. Johannesson

A sample of ammonia gas occupies a volume of 1 A sample of ammonia gas occupies a volume of 1.58 L at 22° C and a pressure of 0.983 atm. What volume will the sample occupy at 1.00 atm and 0° C? A student collects 285 mL of O2 gas at a temperature of 15° C and a pressure of 99.3 kPa. The next day, the same sample occupies 292 mL at a temperature of 11° C. What is the new pressure of the gas? C. Johannesson

Answers 1.44 L 95.6 kPa C. Johannesson

Avogadro’s principal One mole of any gas occupies a volume of 22.4 L Given that the mass of a mole of gas is the molecular mass expressed in grams Avogadro’s principal allows you to interrelate mass, moles, pressure, volume and temperature for any sample of gas C. Johannesson

Applying Avogadro’s principal   C. Johannesson

Applying Avogadro’s principal   C. Johannesson

Practice Problems How many moles of acetylene (C2H2) gas occupy a volume of 3.25 moles at STP? Determine the volume of 12.3 g of formaldehyde gas (CH2O) at STP What is the volume of 1.000 kg of helium gas at 36°C and a pressure of 98.7 kPa C. Johannesson

Answers 0.145 mol 9.18 L C. Johannesson

The Ideal Gas Law   C. Johannesson

Applying the gas law 2   C. Johannesson

Practice Problems What is the pressure in atm of 10.5 mol of acetylene in a 55.0 L cylinder at 37°C? What volume does 0.056 mol of H2 gas occupy at 25°C and 1.11atm of pressure A sample of carbon monoxide has a volume of 344 mL at 85°C and a pressure of 88.4 kPa. Determine the amount in moles of CO present. C. Johannesson

Answers 4.86 atm 5.45 x 10-4 mol C. Johannesson

Using Mass with the Ideal Gas Law   C. Johannesson

Determining the Molar Mass   C. Johannesson

Practice Problems A 250.0 mL sample of a noble gas collected at 88.1 kPa and 7°C has a mass of 0.378 g. What is the molar mass of the gas. Identify the sample A sample of gas is known to be either H2S or SO2. A 2.00 g sample of the gas occupies a volume of 725 mL at a temperature of 13°C and a pressure of 102.4 kPa. What are the molar mass and the identity of the gas? C. Johannesson

Answers 40.0 g Argon 64.1 g/mol Sulfur dioxide C. Johannesson

Gas Stoichiometry   C. Johannesson

  C. Johannesson