TOPICS 1. Intermolecular Forces 2. Properties of Gases 3. Pressure 4. Gas Laws Boyle, Charles, Lussac 5. Ideal Gas Law 6. Gas Stoichiometry 7. Partial Pressure 8. Kinetic Molecular Theory 9. Effusion & Diffusion 10. Real Gases
5.1 Pressure Units for pressure STP 5.2 Boyle Charles Lussac Avogadro IMPORTANT CONCEPTS 5.3 Ideal Gas Law 5.4 Gas Stoichiometry Gas density Molar mass 5.5 Partial Pressure Collecting gas 5.6 Kinetic Molecular Theory 5.7 Effusion Diffusion 5.8 & 5.9 Real gases
2. PROPERTIES OF GASES H, N, O, F and Cl are diatomic
What causes matter to exist in different states? Intermolecular Forces Heat/Temperature
1. INTERMOLECULAR FORCES & TEMPERATURE OBJECTIVE: Forces that determine states of matter
1. IM FORCES & TEMPERATURE What are intermolecular forces? Attractive forces BETWEEN MOLECULES IMPORTANT: THESE ARE NOT BONDS
1. IM FORCES & TEMPERATURE 3 Types 1. Dipole Dipole 2. Hydrogen Bonds 3. London Dispersion Force
1. IM FORCES & TEMPERATURE ATTRACTION CAUSED BY LONDON DISPERSION DIPOLE-DIPOLE HYDROGEN BOND KEYWORDS/ SYMBOLS Your own notes
1. IM FORCES & TEMPERATURE 1. Dipole Dipole Dipole means partial charge Dipoles result from uneven sharing for electron
Symbol for a dipole is d + d - 1. IM FORCES & TEMPERATURE
1. IM FORCES & TEMPERATURE Attractive forces caused by electrostatic attraction. Basically, opposite charges attract, like charges repel
1. IM FORCES & TEMPERATURE
1. IM FORCES & TEMPERATURE 2: Hydrogen Bond - special type of Dipole Dipole - so cause of attraction is same as D-D - STRONGER than normal D-D - involves Hydrogen bonded to O, N, or F (elements with HIGH electronegativity)
2. Hydrogen Bond 1. INTERMOLECULAR FORCES Hydrogen becomes partially positive when bonded with an atom with high electronegativity
1. IM FORCES & TEMPERATURE 3: London Dispersion Forces Weakest of the three Attraction caused by TEMPORARY dipole How are temporary dipoles created? Temporary dipoles are created when electrons are temporarily clustered to one side
1. IM FORCES & TEMPERATURE
1. IM FORCES & TEMPERATURE 3: London Dispersion Force The larger the molar mass of the molecular, the stronger the LD Force. WHY? more mass = more electrons more electrons = more chances of...
1. IM FORCES & TEMPERATURE ATTRACTION CAUSED BY LONDON DISPERSION DIPOLE-DIPOLE HYDROGEN BOND KEYWORDS/ SYMBOLS Your own notes
1. IM FORCES & TEMPERATURE The IMF s shape the state of matter of an object. Example: When boiling water, we are adding energy in the form of heat to break the DD (Hydrogen Bond) between the water molecules so they become separated into a gaseous state.
What causes matter to exist in different states? Intermolecular Forces Heat/Temperature
What is temperature? How hot or cold it is. What does it mean for something to be hot?
1. IM FORCES & TEMPERATURE HEAT is a form of energy is kinetic energy What does this mean? Heat = how fast atoms/molecules are vibrating/moving
1. IM FORCES & TEMPERATURE What if we could measure how fast the atom/molecules are vibrating/moving? We can! We call this temperature!
Temperature measures heat 1. IM FORCES & TEMPERATURE measures average kinetic energy of particles
1. IM FORCES & TEMPERATURE A high temperature means atoms/molecules And a low temperature means
1. IM FORCES & TEMPERATURE Summary & Review 1. Understand WHAT IMFs are; their causes, strengths, and features 2. Understand the difference between heat and temperature 3. Understand how #1 and #2 effects the state of matter of an object
5.1 & 5.2 PROPERTIES OF GASES OBJECTIVE: Properties of Gases
Play file 75185 and 75187 REVIEW
5.1 PROPERTIES OF GASES 1. Has mass 2. Has volume 3. Low density 4. Creates Pressure
5.1 & 5.2 PROPERTIES OF GASES What is Pressure? We live at the bottom of an ocean of air each collision = PRESSURE
5.1 & 5.2 PROPERTIES OF GASES UNITS for pressure What is Pressure? We live at the bottom of an ocean of air 1 atm 760 torr 760 mm Hg each collision = PRESSURE 101.325 kpa
5.1 & 5.2 PROPERTIES OF GASES Summary & Review Properties of a gases: volume, shape, compressibility, mixing, density What is pressure? Pressure has various units
5.3 GAS LAWS OBJECTIVE: Properties of gases with numbers
5.3 GAS LAWS Boyle s Law Relationship between PRESSURE and VOLUME when Temperature is held constant. P 1 V 1 = P 2 V 2
5.3 GAS LAWS P 1 = initial pressure V 1 = initial volume P 2 = pressure after V 2 = volume after P 1 V 1 = P 2 V 2
5.3 GAS LAWS UNITS P = kpa, atm, mm Hg, torr Volume = ml or L
5.3 GAS LAWS
5.3 GAS LAWS Volume Pressure, pressure, volume Volume Pressure, pressure, volume
BOYLE S LAW Practice Problem A 2.5 L balloon has a pressure of 110.0 kpa. You shrunk the balloon s volume to 1.25 L. What is the new pressure? P 1 V 1 = P 2 V 2 (110.0 kpa)(2.5 L) = P 2 (1.25 L) (110.0 kpa)(2.5 L) = P 2 (1.25 L)
BOYLE S LAW Practice Problem A car airbag inflated to a volume of 4.55 L with a pressure of 5.5 atm. The original pressure was 0.5 atm. What was the original volume? P 1 V 1 = P 2 V 2 (0.5 atm)v 1 = (5.5 atm)(4.55 L) V 1 = (5.5 atm)(4.55 L) (0.5 atm)
CHARLES S LAW OBJECTIVE: VOLUME and TEMPERATURE
5.3 GAS LAWS CHARLES S LAW Relationship between TEMPERATURE and VOLUME at constant PRESSURE
CHARLES S LAW V 1 = Initial Volume T 1 = Initial Temperature V 2 = Volume After T 2 = Temperature After
CHARLES S LAW UNITS Volume = cm 3, ml L Temperature = K Celsius + 273 = K
CHARLES S LAW
6. CHARLES S LAW Volume, Temperature Temperature, Volume Volume, Temperature Temperature, Volume
6. CHARLES S LAW A balloon has a volume of 15.5 L and is in a room at 20.0 C. The temperature drops to 7.00 C, what is the new volume? V 1 = V 2 T 1 T 2 15.5 L = V 2 (20.0 C + 273) (7.00 C + 273)
6. CHARLES S LAW Soda container has volume of 62.2L at 150 C. New volume is 24.4 L, What is the new temperature?
LUSSACS S LAW OBJECTIVE: PRESSURE and TEMPERATURE
LUSSAC S LAW Relationship between TEMPERATURE and PRESSURE when Volume is constant
LUSSAC S LAW P 1 = Initial Pressure T 1 = Initial Temperature P 2 = Pressure After T 2 = Temperature After
LUSSAC S LAW Units P = kpa, atm, mm Hg, torr, psi T = C, K
LUSSAC S LAW Pressure, Temperature Temperature, Pressure Pressure, Temperature Temperature, Pressure
Which law does this graph represent?
LUSSAC S LAW The pressure in a tire is 101 kpa at 10.0 C. What will be the pressure at 45.0 C? (T must be in K)
LUSSAC S LAW The pressure in a soda bottle is 505 kpa at 20.0 C. What is the pressure if bottle is warmed to 65.0 C? (T must be in K)
AVOGADRO S LAW OBJECTIVE: VOLUME and MOLE
AVOGADRO S LAW Relationship between VOLUME and MOLE When Pressure AND Temperature is held constant
AVOGADRO S LAW V 1 = initial volume n 1 = initial moles V 2 = volume after n 2 = moles after V = V 1 2 n n 1 2
AVOGADRO S LAW Gases at the SAME temperature and pressure contain the same # of moles
AVOGADRO S LAW Gases at the SAME temperature and pressure contain the same # of moles Basically, when temperature and pressure doesn t change, the number of moles determines the volume of a gas
8. AVOGADRO S LAW
AVOGADRO S LAW Volume, # moles # moles, Volume Volume, # moles # moles, Volume
AVOGADRO S LAW during STP STP = Standard Temperature & Pressure 0 C and 1 atm
GAS LAW REVIEW V1 n1 = V2 n2
WHICH LAW DOES EACH GRAPH REPRESENT?
Summary & Review 5.3 GAS LAW 1. Understand the relationship between pressure, temperature, and volume for each of the laws 2. Recognize these relationships when graphed 3. Be able to use each of the laws to answer calculation questions
5.4 IDEAL GAS LAW OBJECTIVE: Combining the Gas Laws
Combined Gas Law Use when 2 things change 2 more laws ugh 5.4 IDEAL GAS LAW Ideal Gas Law P 1 V 1 = P 2 V 2 T 1 T 2
Combined Gas Law Use when 2 things change 2 more laws 5.4 IDEAL GAS LAW A gas is at pressure 12 atm, volume of 23 L, and temperature of 200K P 1 V 1 = P 2 V 2 T 1 T 2 What is the new volume if the pressure increases to 14 atm and temperature rises to 300k?
Combined Gas Law Use when 2 things change P 1 V 1 = P 2 V 2 T 1 T 2 2 more laws ugh 5.4 IDEAL GAS LAW A gas at volume of 23 L, temperature of 318 K and unknown pressure has increased the volume to 34, and decreased temperature to 308 K. The changes resulted in a final pressure of 2.0 atm. What was the original pressure?
Combined Gas Law 2 more laws ugh 5.4 IDEAL GAS LAW Ideal Gas Law Use when 2 things change Used to measure density and molar mass of a gas. P 1 V 1 = P 2 V 2 T 1 T 2 PV = nrt
5.4 IDEAL GAS LAW Combing all 4 laws Formula: PV = nrt
5.4 IDEAL GAS LAW PV = nrt P = pressure V = volume n = # of mol T = temperature in KELVIN
5.4 IDEAL GAS LAW PV = nrt R = 8.31 L kpa R = 0.0821 L atm mol K mol K
5.4 IDEAL GAS LAW 2.5 mol of a gas has a volume of 5.00 L, and a temperature of 25 C. What is the pressure in BOTH kpa and atm? PV = nrt P (5.00 L) = (2.5 mol)(8.314) (298 K)
5.4 IDEAL GAS LAW 4.5 mol of a gas has a temperature of 25 C and a pressure of 100 kpa. What is the volume in L? PV = nrt (100 kpa) V = (4.5 mol) (298 K) (8.314)
5.4 IDEAL GAS LAW A gas has a pressure of 27 atm, volume of 44.8 L, and a temperature of 25 C. How many mols? PV = nrt (27 atm) (44.8 L) = n (0.0821) (8.314) (298 K)
5.4 IDEAL GAS LAW 5.00 mol of a gas has a pressure of 125 kpa, and a volume of 75 L. What is the temperature in Kelvin? PV = nrt (125 kpa) (75 L) = (5.00 mol) T (8.314)
5.4 IDEAL GAS LAW Ideal Gas Law can be used to 1. Calculate density of a gas 2. Find molar mass & formula of an unknown gas
5.4 IDEAL GAS LAW Ideal Gas Law can be used to 1. Calculate density of a gas PV = nrt n = moles PV = mass RT molar mass moles = mass molar mass substitute above for n
5.4 IDEAL GAS LAW Ideal Gas Law can be used to 1. Calculate density of a gas PV = mass RT molar mass density = mass volume P (molar mass) = mass RT volume rearrange IGL to have mass volume
5.4 IDEAL GAS LAW Ideal Gas Law can be used to 1. Calculate density of a gas P (molar mass) = mass RT volume units for density = g/l Calculate the density of carbon dioxide in grams/l at 0.990 atm and 55 C
5.4 IDEAL GAS LAW Ideal Gas Law can be used to 1. Calculate density of a gas P (molar mass) = mass RT volume units for density = g/l What is the density in g/l of UF 6 at 779 mm Hg and 62 C R = 62.4
5.4 IDEAL GAS LAW Ideal Gas Law can be used to 1. Calculate density of a gas 2. Find molar mass & formula of an unknown gas
PV = mass RT molar mass molar mass = mass RT PV molar mass = density RT P units = g/mol 5.4 IDEAL GAS LAW Ideal Gas Law can be used to 2. Find molar mass & formula of an unknown gas A chemist synthesized a greenish-yellow gas of chlorine and oxygen, and found that its density is 7.71 g/l at 36 C and 2.88 atm What is the molar mass and molecular formula?
5.4 IDEAL GAS LAW Ideal Gas Law can be used to 2. Find molar mass & formula of an unknown gas
5.4 IDEAL GAS LAW Ideal Gas Law can be used to 2. Find molar mass & formula of an unknown gas
GASES 1. How many moles of gases are contained in a can with a volume of 555mL and a pressure of 600.0 kpa at 25C? 2. Calculate the pressure exerted by 43 mol of nitrogen in a 65L cylinder at 5C. 3. What will be the volume of 111 mole of nitrogen in the stratosphere where the temperature is -57C and pressure is 7.30 kpa.
5.5 GAS STOICHIOMETRY OBJECTIVE: Ideal Gas Law + Stoichiometry
5.4 IDEAL GAS LAW 2 TYPES First Type Ideal Gas Law THEN Stoichiometry Second Type Stoichiometry THEN Ideal Gas Law
10. GAS STOICHIOMETRY First Type Ideal Gas Law THEN Stoichiometry 2H 2 + O 2 + 2H 2 O A. How many moles of O 2 when temperature is 555 C, volume is 8.50 L and pressure is 3.27 atm? B. How many grams of H 2 O can be made from that many moles of O 2?
10. GAS STOICHIOMETRY First Type Ideal Gas Law THEN Stoichiometry 2H 2 + O 2 + 2H 2 O A. How many moles of H 2 when temperature is 327 C, volume is 8.50 L and pressure is 5.55 atm? B. How many grams of H 2 O can be made from that many moles of H 2?
10. GAS STOICHIOMETRY Second Type Stoichiometry THEN Ideal Gas Law 2KClO 3 2KCl +3O 2 A. 366 grams of KClO 3. How many moles of O 2 gas produced? (stoic) B. Part A took place in conditions with 25 C and 101 kpa. What is the volume in L of O 2 gas? (IG Law)
10. GAS STOICHIOMETRY Second Type Stoichiometry THEN Ideal Gas Law 2Na + 2H 2 O 2NaOH + H 2 A. 115 g Na produces how many moles of H 2? B. What is the temperature in K of H2 if volume is 7.23 L and pressure is 92.5 kpa?
1. What will be the volume of 111 mole of nitrogen in the stratosphere where the temperature is -57C and pressure is 7.30 kpa. 2.73 x 10 4 L (ID) 2. What is the volume of 4.35 moles of a gas at a pressure of 85.6kPa and 26.0 degrees C? 126L (ID) 3. A deep underground cavern contains 2.24 x 10 6 L of methane gas at a pressure of 1500kPa and a temperature of 42 degrees C. How many kilograms of methane does this gas deposit contain? 2.05 x 10 4 kg of methane gas. (ID + STOIC) 4. What volume of oxygen, collected at 25 degrees C and 101kPa, can be prepared by decomposition of 37.9g of potassium chlorate? 11.4L (STOIC + ID) 2KClO 3 2KCl +3O 2 5. Liquid hydrogen and oxygen are burned in a rocked. What volume of water vapor, at 555 degrees C and 76.4kPa, can be produced from 4.67kg of H2? 2.08 x 10^5 L H 2 (ID + STOIC) 2H 2 + O 2 + 2H 2 O 6. How many grams of sodium are needed to produce 2.24Lof hydrogen collected at 23 degrees C and 92.5kPa? 3.87g Na (ID + STOIC) 2Na + 2H2O 2NaOH + H2
Ideal Gas Law Can be used to calculate molar volume GASES Can be use to determine molar mass if P, V, g and T are known. Likewise, can be used to calculate density of a gas if conditions of P, T, and molar mass is known.
Calculate the density of C 4 H 10 at 25 degrees C and 104.5 kpa GASES A copper penny + nitric acid (HNO 3 ) results in a red-brown gaseous compound containing nitrogen and oxygen. Calculate the molar mass of the gas and suggest a reasonable chemical formula for the compound. Pressure =.957atm temperature of 18 C weighs 0.289 g in a flask with a volume of 157.0 ml.
5.5 PARTIAL PRESSURE & MOLE FRACTION OBJECTIVE: Details on Pressure
5.5 PARTIAL PRESSURE & MOLE FRACTION Air = N 2, O 2, Ar, CO 2, Air pressure right now = 1atm Partial Pressure N 2 = 0.70 atm O 2 = 0.20 atm Ar, CO 2 = 0.10 atm
5.5 PARTIAL PRESSURE & MOLE FRACTION Partial Pressure P A + P B + P C = P total
5.5 PARTIAL PRESSURE & MOLE FRACTION A scuba diving tank contains 96 g of O 2 and 48 g of He. The volume of the tank is 10.0 L. The temperature of the gases is 20.0 C. 1. What is the partial pressure in atm of each gas? 2. What is the total pressure?
5.5 PARTIAL PRESSURE & MOLE FRACTION Partial Pressure Mole Fraction = X P a = n a P total = n total n a n total = X
5.5 PARTIAL PRESSURE & MOLE FRACTION The total pressure on a small mountain is 743 torr. The partial pressure of O 2 is 156 torr. What is the mole fraction of O 2? Mole Fraction = X P a P total = X
5.5 PARTIAL PRESSURE & MOLE FRACTION A scuba tank has a O 2 has a mole fraction of 0.050 and a partial pressure of 0.21 atm. What is the total pressure? Mole Fraction = X P a P total = X
5.5 PARTIAL PRESSURE & MOLE FRACTION Application of Partial Pressure Collecting Gas over Water
5.5 PARTIAL PRESSURE & MOLE FRACTION Application of Partial Pressure
5.5 PARTIAL PRESSURE & MOLE FRACTION Collecting Gas over Water 2KClO 3(s) 2KCl (s) + 3O 2(g) A sample of KClO 3 was heated in a test tube. The O 2 produced was collected by displacement of water at 22 C at a total pressure of 754 torr. The volume of O2 collected is 0.650 L. The vapor pressure at this temperature is 21 torr. Calculate the partial pressure of O2 and the mass of KClO3 decomposed
5.5 PARTIAL PRESSURE & MOLE FRACTION Summary & Review Concept of Partial Pressure Calculating Partial Pressure Using Mole Fraction to find PP Using Mole Fraction, PP, and the gas laws Demo of application of PP on Monday
5.6 KINETIC MOLECULAR THEORY OBJECTIVE: The theory behind the gas laws
5.6 KMT Kinetic Molecular Theory Explains properties of gases Ideal Gas Law MEASURES properties of gases
5.6 KMT Kinetic Molecular Theory Makes 3 important assumptions 1. Gas molecules have no volume 2. The AVERAGE kinetic energy of gas is related to temperature
Kinetic Molecular Theory 5.6 KMT Makes 3 important assumptions 1. Gas molecules have no volume 2. The AVERAGE kinetic energy of gas is related to temperature 3. When gases collide, they do not lose energy
5.6 KMT 1.When the temperature increases, what happens to the motion of the particles? 2.Why is it possible to compress a gas? 3.How does a gas create pressure?
5.6 KMT 1. When the temperature increases, what happens to the motion of the particles? Ans. Many of the particles move faster. 2. Why is it possible to compress a gas? Ans. Because the particles of a gas are far apart, they can be pushed closer together. 3. How does a gas exert pressure? Ans. A gas exerts pressure through the collision of its particles with the walls of its container.
5.6 KMT Graham s Law of Effusion How fast gas travels
5.6 KMT UNITS Graham s Law of Effusion v A = speed of Gas A in meters/sec v B = speed of Gas B in meters/sec M A = Molar Mass of Gas A in g M B = Molar Mass of Gas B in g
5.6 KMT Graham s Law of Effusion O 2 travels at 480 m/s. What is the speed of He when compared to O 2?
5.6 KMT Root Mean Square Velocity measuring speed of one gas u rms = 3RT M U rms = velocity R = 8.314 J/K*mol T = temperature in K M = molar mass in KG
5.6 KMT Calculate the root mean square velocity Helium at 25 C u rms = 3RT M U rms = velocity R = 8.314 J/K*mol T = temperature in K M = molar mass in KG
Smaller the molar mass, the faster the gas molecules travel
Summary & Review 5.6 KMT How the KMT relates to each gas law - Why increase in T results in increase of P - Why decrease in V results in increase of P - Temperature = AVERAGE kinetic energy (graph) - How fast gas molecules move inverse of MM of gas - 2 formulas for calculating speed of gas molecules
5.8 REAL GASES OBJECTIVE: When gases deviate from ideal behavior
Ideal Gases vs Real Gases KMT assumes gases have NO IM forces gases have NO volume 5.8 REAL GASES Gases deviate from ideal behavior when Temperature is low because Pressure is high because
GASES Ideal Gases vs Real Gases So...actual volume needs to be taken into consideration How large a factor the actual volume is dependent on how many atoms/molecules are present, or the number of moles (V nb)
GASES Ideal Gases vs Real Gases IM forces also need to be taken into consideration. Generally, IM forces can be ignored for gases until they are compressed by increasing pressure or a reduction in volume. How would IM forces make RG different from IG?
GASES IM forces effect behavior of RG by reducing pressure. Frequency and strength of attraction of IM forces need to be considered Frequency is based on concentration: number of stuff in a given area (n/v) 2 Constant, a, is also added to consider the relative strength of the IM forces.
GASES Putting it all together (P + an 2 /v 2 ) (V-bn) = nrt
GASES So what is an IDEAL gas? An IDEAL gas = a gas that PERFECTLY follows the gas laws. It does not condense into a liquid a low temperatures Does not have forces of attraction and repulsion between particles. This law describes the behavior of real gases quite well at room temperature and pressure.
GASES Although we have these laws, no gas perfectly follows them under all conditions, but the laws are helpful for most gases under most conditions. So another way to understand gases is to assume that they behave as an IDEAL gas.