Presentation on theme: "Chapter 13 – Gases 13.1 The Gas Laws"— Presentation transcript:
1Chapter 13 – Gases 13.1 The Gas Laws 13.2 The Ideal Gas Law (Part I – Equation Only)13.3 Gas Stoichiometry
2Section 13.2 The Ideal Gas Law The ideal gas law relates the number of particles to pressure, temperature, and volume.Relate the amount of gas present to its pressure, temperature, and volume using the ideal gas law.
3Section 13.2 The Ideal Gas Law Key ConceptThe ideal gas law relates the amount of a gas present to its pressure, temperature, and volumePV = nRT
4Gas Laws – Mechanistic and Empirical “Named” laws described in section were empirically determined from experiments – stated relationships “fit the data”Example: Boyle experimentally determined that P & V are inversely related but there was no underlying understanding of why this occurs – this is an empirical model
5Gas Laws – Mechanistic and Empirical Ideal gas law is mechanistic – it can be derived by using the basic theory (kinetic- molecular) of how gases behave; the underlying mechanism that leads to the equation is known and understoodIdeal gas law was formulated after the empirical (named) laws were discoveredIt predicted the same behavior as the empirical laws but provided a way to understand why they were true
6Kinetic-Molecular Theory (KMT) of Gases (see section 12.1) Mostly empty spaceNo attractions/repulsions between particlesConstant straight-line motion until collisions with other particles or wallsCollisions perfectly elastic: KE constantT constant, KE constantAll gases have same KEavg at a given T
7R = Gas Constant – value depends upon units used for P Ideal Gas LawRelationship between pressure P, volume V, Kelvin temperature T, and number of moles nPV = nRTT must be expressed in units of kelvinEquation derived using KMT assumptionsR = Gas Constant – value depends upon units used for P
9Ideal Gas Law – Dependence on n PV = nRTFixed V, T thenP nFixed P, TV n
10Ideal Gas Law: Example Problem 13.6 Calculate number of moles of NH3 for V = 3.0 L, T = 3.00x102 K, P = 1.50 atmPV = nRT n = PV / RTn = (1.50 atm 3.0 L) / (8.21x10-2 L atm/mol K 3.00x102 K)n = 0.18 molNote that value of R was chosen that has atm as the pressure unit
12Ideal Gas Law and the Combined Gas Law PV = nRTIf the # of moles of gas (n) is fixed, thenPV/T = nR = constantIf P,V,T conditions change (from 1 to 2)P1V1/T1 = P2V2/T2(have derived the Combined Gas Law)
13Combined & Named Gas Laws (Subject of Section 13.1) For given mass (fixed # moles) of gasP1V1/T1 = P2V2/T23 laws can be derived from this:Fix T: P1V1 = P2V2 (Boyle’s)Fix P: V1/T1 = V2/T2 (Charles’)Fix V: P1/T1 = P2/T2 (Gay-Lussac’s)
14Chapter 13 – Gases 13.1 The Gas Laws 13.2 The Ideal Gas Law (Part I) 13.3 Gas Stoichiometry
15Section The Gas LawsFor a fixed amount of gas, a change in one variable—pressure, temperature, or volume—affects the other two.State the relationships among pressure, temperature, and volume of a constant amount of gas.Apply the gas laws to problems involving the pressure, temperature, and volume of a constant amount of gas.
16Section 13.1 The Gas Laws Key Concepts Boyle’s law states that the volume of a fixed amount of gas is inversely proportional to its pressure at constant temperature.P1V1 = P2V2Charles’s law states that the volume of a fixed amount of gas is directly proportional to its kelvin temperature at constant pressure.
17Section 13.1 The Gas Laws Key Concepts Gay-Lussac’s law states that the pressure of a fixed amount of gas is directly proportional to its kelvin temperature at constant volume.The combined gas law relates pressure, temperature, and volume in a single statement.
18Boyle’s LawVolume of given mass (fixed # of moles) of gas held at constant T varies inversely with pressureP1V1 = P2V2PV = constantV2 = V1 (P1/P2)
21Boyle’s Law – Example Problem 13.1 Diver blows bubble of volume = 0.75 L at 10 m depth & pressure = 2.25 atm. Volume of bubble at surface when pressure = 1.03 atm?P1V1 = P2V2V2 = V1 (P1/P2)V2 = 0.75 L (2.25 atm / 1.03 atm)V2 = 1.6 L
27Volume vs T (Kelvin) – Different Gases Gases follow Charles’s Law but liquefy at different T– dashed lines are extrapolations of gas phase behaviorIdeal gases occupy no volume at absolute zero
28Charles’s Law: Example Problem 13.2 Balloon has volume of 2.32 L at 40.0 C. Volume at 75.0 C if P remains constant?V1/T1 = V2/T2T1 = 40.0 C = KT2 = 75.0 C = KV2 = V1 (T2/T1)V2 = 2.32 L (348.2 K/ K)V2 = 2.58 L
30Joseph Louis Gay-Lussac (1778 – 1850) French chemist & physicist. Known mostly for 2 laws related to gases, & for his work on alcohol-water mixtures, which led to the degrees Gay-Lussac used to measure alcoholic beverages in many countries.Prof. of physics at Sorbonne (1808 to 1832), a post which he only resigned for the chair of chemistry at the Jardindes Plantes. In 1802, first formulated law stating that if mass and pressure of gas are held constant, then gas volume increases linearly as temperature rises.
31P change with T at fixed V Gay-Lussac’s LawP change with T at fixed V
32Gay-Lussac’s LawPressure of given mass (fixed # of moles) of gas in a constant volume is directly proportional to its kelvin temperatureP1/T1 = P2/T2P2 = P1(T2/T1)TK = TC
33At 0 K, ideal gas exerts zero pressure Gay-Lussac’s LawAt 0 K, ideal gas exerts zero pressure
34Gay-Lussac’s Law: Example Problem 13.3 Pressure of O2 in canister is 5.00 atm at 25.0 C. New pressure at T of C?P1/T1 = P2/T2T1 = 25.0 C = KT2 = C = KP2 = P1(T2/T1)P2 = 5.00 atm (263.2 K / K)P2 = 4.41 atm
36If P,V,T conditions change (from 1 to 2) Combined Gas LawIf P,V,T conditions change (from 1 to 2)P1V1/T1 = P2V2/T2
37Combined Gas Law: Practice Problem 13.4 Gas at 110 kPa and 30.0 C fills flexible container with initial V of 2.00 L. If T and P raised to 80.0 C and 440 kPa, new V?T1 = 30.0 C = KT2 = 80.0 C = KP1V1/T1 = P2V2/T2V2 = V1 (P2 /P1) (T1/T2)V2 = 2.00 L (440 kPa/ 110 kPa) (303.2 K/ K)V2 = 0.58 L
39Gas Laws Summary – Table 13.1 All require fixed n (moles of gas)
40Chapter 13 – Gases 13.1 The Gas Laws (includes Combined) 13.2 The Ideal Gas Law (Part II - Avogadro’s Principle, Gas Density, Real Gases)13.3 Gas Stoichiometry
41Section 13.2 The Ideal Gas Law The ideal gas law relates the number of particles to pressure, temperature, and volume.Relate number of particles and volume using Avogadro’s principle.Compare the properties of real and ideal gases.
42Section 13.2 The Ideal Gas Law Key ConceptsAvogadro’s principle states that equal volumes of gases at the same pressure and temperature contain equal numbers of particles.The ideal gas law can be used to find molar mass if the mass of the gas is known, or the density of the gas if its molar mass is known.At very high pressures and very low temperatures, real gases behave differently than ideal gases.
43Avogadro’s PrincipleEqual volumes of gases at same T and P contain equal numbers of particlesSize of gas particles negligible compared to volume occupied, so it doesn’t matter which gas you are using
44Molar Volume of GasVolume occupied by one mole of gas at 0.00 C and 1.00 atmosphere pressureT, P conditions known as STP – Standard Temperature and PressureAt STP, 1 mole of any gas occupies 22.4 LKey to lots of problems involving gases at STP
45Molar Volume of Gas: Example Problem 13.5 V of 2.00 kg of methane at STP?2.00 kg CH4 1.00x103 g/kg mol CH4/16.05 g CH4 = 125 mol CH4125 mol CH4 22.4 L CH4 / mol CH4 (at STP)V = 2.80x103 L
47Ideal Gas Law: Calculate M or Density PV = nRTn = # moles = mass /molar mass = m / MPV = mRT / M MP = mRT / V M = mRT / PVM = (m/V) RT / P (m/V) = density = DM = D RT / P or D = MP / RTFor a given P, T , if know:1. mass m & V, can compute molar mass M2. density D, can compute molar mass M3. molar mass M, can compute density D
48Practice (Molar Mass, Density from Ideal Gas Law) Problems 70, 72, 76 page 469Problems 17 – 18, page 985
49Ideal Gas Law - Deviations No gas truly ideal, but most behave as ideal gases over wide range of T & PDeviations happen at high P and low T where assumptions of KMT break downVolume of gas no longer negligibleAn ideal gas can’t be liquefiedIntermolecular forces more significantPolar molecules especially affected
50Ideal Gas Law - Deviations Polar molecules have larger attractive forces between particles and generally do not behave as ideal gases at low T where slower-moving particles can be affected by these forcesLarge gas particles occupy more space and deviate more from ideal gases at high density (high P, low T) where there is less empty space
51Real Gases - PV/RT vs P for 1 Mole of Gas Approach ideal gas at low P2.01.0PVRTP (atm)
52Chapter 13 – Gases 13.1 The Gas Laws 13.2 The Ideal Gas Law 13.3 Gas Stoichiometry
53Section 13.3 Gas Stoichiometry When gases react, the coefficients in the balanced chemical equation represent both molar amounts and relative volumes.Determine volume ratios for gaseous reactants and products by using coefficients from chemical equations.Apply gas laws to calculate amounts of gaseous reactants and products in a chemical reaction.
54Section 13.3 Gas Stoichiometry Key ConceptsThe coefficients in a balanced chemical equation specify volume ratios for gaseous reactants and products.The gas laws can be used along with balanced chemical equations to calculate the amount of a gaseous reactant or product in a reaction.
55Balanced Equation & Gas Volumes Gas laws can be applied to calculate stoichiometry of reactions in which gases are reactants or products2H2(g) + O2(g) → 2H2O(g)2 mol H2 reacts with 1 mol O2 to produce 2 mol water vaporCoefficients in balanced equation represent volume ratios for gases if P and T are fixed during reaction
56Balanced Equation & Gas Volumes CH4(g) O2(g) CO2(g) H2O(g)There is a one-to-one relationship between number of moles and gas volume
57Balanced Equation & Gas Volumes 2C4H10(g) + 13O2(g) 8CO2(g) + 10H2O(g)Interpreted as moles:2 mol butane + 13 mol oxygen 8 mol carbon dioxide + 10 mol waterAvogadro’s principle can also be volume2 L butane + 13 L oxygen 8 L carbon dioxide + 10 L water
59Practice – Gas Stoichiometry How many mol of hydrogen gas are required to react with 1.50 mol oxygen gas in the following reaction?2H2(g) + O2(g) → 2H2O(g)ABCD?
60V toV Stoichiometry: Practice Problem 13.7 Volume of oxygen gas needed for complete combustion of 4.00 L of propane (C3H8) assuming P and T remain constant?C3H8(g) + 5O2(g) 3CO2(g) + 4H2O(g)4.00 L C3H8 5 L O2 / L C3H8 = 20.0 L O2
61Practice – Gas Stoich. (all V to V) Problems 38 – 41, page 461Problems 84 – 86, page 469Problems 19 – 20, page 985
62Combo Mass/Volume Problems Need:Balanced equationAt least one mass or V for product or reactantConditions under which volumes measured
63Combo Mass/Volume – Prob 13.8 N2(g) + 3H2(g) 2NH3(g)5.00 L N2 reacts completelyP = 3.00 atm T = 298 KMass of ammonia produced?5.00 L N2 [2 L NH3/1 L N2] =volume ratio from equation10.0 L NH3
64Combo Mass/Volume – Prob 13.8 N2(g) + 3H2(g) 2NH3(g)10.0 L NH3 P = 3.00 atm T = 298 Kn = PV/RT = ( atm 10.0 L ) (298 K L atm/mol K)= 1.23 mol NH3Molar mass NH3 = g/mol1.23 mol NH3 g NH3/mol NH3= 21.0 g NH3