Unformatted text preview: Gases and Gas Laws
Introduction to Chemistry T. R. Dickson 8th Edition Pressure Force per unit area P = F/A Bike tire example psi = 1 lb object on 1 in2 mm Hg = torr Pressure Torr Evangelista Torricelli Early 1600s Barometer 760 mm or 760 torr 1 atm Pressure SI system International System Pascal (Pa) = newtons/square meter = N/m2 1 atm = 101,325 Pa Boyle's Law V 1/P Boyle's Law Boyle's Law V 1/P PV = k P1V1 = k and P2V2 = k P1 V1 = P2 V2
PV1 PV1 1 1 P2 = and V2 = V2 P2 Boyle's Law Example A gas occupying a volume of 725 mL at a pressure of 0.970 atm is allowed to expand at constant temperature until its pressure becomes 0.541 atm. What is its final volume? Boyle's Law Example A gas occupying a volume of 725 mL at a pressure of 0.970 atm is allowed to expand at constant temperature until its pressure becomes 0.541 atm. What is its final volume? Use P1V1 = P2V2 Boyle's Law Example P 1 V 1 = P2 V 2 Solve for V2: V2 = P1V1 P2 Boyle's Law Example
V2 = P1V1 P2 V2 = (725 mL)(0.940 atm) 0.541 atm Boyle's Law Example
V2 = (725 mL)(0.940 atm) 0.541 atm V2 = 1300 mL Boyle's Law Example
V2 = (725 mL)(0.940 atm) 0.541 atm V2 = 1300 mL 3 significant digits V2 = 1300 mL or 1.30 x 103 mL Charles's Law V T Liquid Nitrogen and a Balloon Charles's Law Charles's Law
V T V =k T V1 V2 = T1 T2 Charles's Law Example A 36.4 L volume of methane gas is heated from 25C to 88C at constant pressure. What is the final volume of the gas? Use V1 = V2 T1 = T2 Charles's Law Example A 36.4 L volume of methane gas is heated from 25C to 88C at constant pressure. What is the final volume of the gas? Tk = Tc + 273.15 Tk1 = 25 + 273.15 = 298.15 K Charles's Law Example A 36.4 L volume of methane gas is heated from 25C to 88C at constant pressure. What is the final volume of the gas? Tk = Tc + 273.15 Tk1 = 25 + 273.15 = 298.15 K Tk = Tc + 273.15 Tk2 = 88 + 273.15 = 361.15 K Charles's Law Example Use V1 = V2 T1 = T2 V 2 = V1 T 2 T1 Charles's Law Example Use V1 = V2 T1 = T2 V2 = V1T2 = (36.4 L)(361.15 K) T1 298.15 K Charles's Law Example Use V1 = V2 T1 = T2 V2 = V1T2 = (36.4 L)(361.15 K) = 44.0914 T1 298.15 K Charles's Law Example
V2 = V1T2 = (36.4 L)(361.15 K) = 44.0914 T1 298.15 K Rounds to 44 temperature is only 2 sig. digits GayLussac's Law Pressure and Temperature Law PT GayLussac's Law GayLussac's Law
P T
P =k T P P2 1 = T1 T2 GayLussac's Law Example A certain amount of gas at 25C and at a pressure of 0.800 atm is contained in a glass vessel. Suppose that the vessel can withstand a pressure of 2.00 atm. How high can you increase the temperature of the gas without bursting the vessel? GayLussac's Law Example A certain amount of gas at 25C and at a pressure of 0.800 atm is contained in a glass vessel. Suppose that the vessel can withstand a pressure of 2.00 atm. How high can you increase the temperature of the gas without bursting the vessel? Convert temp. from C to K GayLussac's Law Example Convert temp. from C to K Tk = Tc + 273.15 Tk = 25C + 273.15 = 298.15 K GayLussac's Law Example
P P2 1 = T1 T2 P 1 T2 = P2 T1 T1 T2 = P2 P 1 GayLussac's Law Example
T1 T2 = P2 P 1 298.15 K T2 = (2.00atm) 0.800atm GayLussac's Law Example
T1 T2 = P2 P 1 298.15 K T2 = (2.00atm) = 745 K 0.800atm GayLussac's Law Example
T1 T2 = P2 P 1 298.15K T2 = (2.00atm) = 745.375 K 0.800atm Tc = Tk  273.15 = 745.375 273.15 = 472.225C GayLussac's Law Example
298.15K T2 = (2.00atm) = 745.375 K 0.800atm Tc = Tk  273.15 = 745.375 273.15 = 472.225C Rounds to 470C Avogadro's Law Volume (V) number of moles (n) V = kn V =k n V1 V2 = n1 n2 Avogadro's Law V =k n
STP: 0C 273.15 K 1 atm 760 mm Hg Avogadro's Law
STP: 0C 273.15 K 1 atm 760 mm Hg 1 mol gas = 22.4 L Example Use the standard molar volume of a gas at STP (22.4 L) to find how many moles of air at STP are in a room measuring 4.11 m wide by 5.36 m long by 2.58 m high. Example Use the standard molar volume of a gas at STP (22.4 L) to find how many moles of air at STP are in a room measuring 4.11 m wide by 5.36 m long by 2.58 m high. Calculate volume of room: Example Use the standard molar volume of a gas at STP (22.4 L) to find how many moles of air at STP are in a room measuring 4.11 m wide by 5.36 m long by 2.58 m high. Calculate volume of room: Volume = (4.11 m)(5.36 m)(2.58 m) = 56.8 m3 = (56.8 m3)(1000 L/1 m3) = 5.68 x
4 Example Use the standard molar volume of a gas at STP (22.4 L) to find how many moles of air at STP are in a room measuring 4.11 m wide by 5.36 m long by 2.58 m high. Calculate number of moles: (5.68 x 104L)(1 mol/22.4 L) = 2.54 x 103 mol Ideal Gas Law
Boyle's law Charle's law k V= P V = bT (at constant T and n) (at constant P and n) (at constant T and P) Avogadro's law V = an Tn V = R P PV = nRT Ideal Gas Law PV = nRT
LV atm P in atmospheres: R = 0.0821 mol K L mmHg P in millimeters of Hg: R = 62.4 mol K Ideal Gas Law Example How many moles of air are in the lungs of an average person with a total lung capacity of 3.8 L? Assume that the person is at 1.0 atm pressure and has a normal body temperature of 37C. Ideal Gas Law Example How many moles of air are in the lungs of an average person with a total lung capacity of 3.8 L? Assume that the person is at 1.0 atm pressure and has a normal body temperature of 37C. Identify known information: P = 1.0 atm, V = 3.8 L T = 37C = 310 K Ideal Gas Law Example Identify known information: P = 1.0 atm, V = 3.8 L T = 37C = 310 K Identify the equation: PV= nRT LV atm R = 0.0821 mol K PV n= RT Ideal Gas Law Example Identify known information: P = 1.0 atm, V = 3.8 L T = 37C = 310 K Solve: PV (1.0atm)(3.8 L) n= = = 0.15 mol L atm RT 0.0821 (310 K) mol g K Molecular View of The Ideal Gas Law STP 1 mole ideal gas 0C (273.15K) 1 atm STP 1 mole ideal gas 0C (273.15K) 1 atm nRT (1.000mol )(0.08206 Lg / K g )(273.15 K ) atm mol V= = = 22.4 L P 1.000atm Molar Mass of a Gas PV = nRT n P = V RT Molar Mass of a Gas n P = V RT # of moles (n) = m n= M Molar Mass of a Gas n P = V RT m P = MV RT Molar Mass of a Gas m P = MV RT m PM d= = V RT Molar Mass of Gas m PM d= = V RT
dRT M = P Molar Mass from Gas Density A chemist has synthesized a greenishyellow gaseous compound of chlorine and oxygen and finds that its density is 7.81 g/L at 36C and 2.88 atm. Calculate the molar mass of the compound and determine its molecular formula. Molar Mass from Gas Density
dRT M = P (7.81g / L)(0.0821Lg / K g )(36 + 273) K atm mol M = 2.88atm Molar Mass from Gas Density
dRT M = P (7.81g / L)(0.0821Lg / K g )(36 + 273) K atm mol M = 2.88atm M = 67.9 g / mol Summary of Gas Laws
Gas Law Boyle's Law Charles's Law GayLussac's Law Combined Gas Law Avogadro's Law Ideal Gas Law P1V1 = P2V2 V1/T1 = V2/T2 P1/T1 = P2/T2 P1V1/T1 = P2V2/T2 V1/n1 = V2/n2 PV= nRT Variables P, V V, T P, T P, V, T V, n P, V, T, n Constant n, T n, P n, V n P, T R Dalton's law of partial pressures Total pressure of a mixture of gases is just the sum of the pressures that each gas would exert if it were present alone. Ptotal = P1 + P2 + P3 + ... Dalton's law of partial pressures
n1 RT n2 RT P= P2 = 1 V V n3 RT P3 = V Dalton's law of partial pressures
n1 RT n2 RT P= P2 = 1 V V Ptotal n3 RT P3 = V n3 RT n1 RT n2 RT = P + P2 + P3 + ... = + + + ... 1 V V V Dalton's law of partial pressures
Ptotal n3 RT n1 RT n2 RT = P + P2 + P3 + ... = + + + ... 1 V V V RT = (n1 + n2 +n3 +...) V RT = ntotal V Dalton's law of partial pressures PT = PA + PB Ptotal Ptotal nA RT nB RT = + V V RT = (nA + nB ) V Dalton's law of partial pressures
PA = PT (n + n ) RT A B V nA = nA + nB = XA nA RT V Dalton's Law Dry Air: 21% oxygen 78% nitrogen 1% argon Dalton's Law Ptotal = Pgas1 + Pgas2 + Pgas3 + ... Ptotal = PO2 + PN2 + PAr O2 N2 Ar 0.21 x 760 mm Hg = 160 mm Hg 0.78 x 760 mm Hg = 593 mm Hg 0.01 x 760 mm Hg = 7 mm Hg Dalton's Law Example Humid air on a warm summer day is approximately 20% oxygen, 75% nitrogen, 4% water vapor, and 1% argon. What is the partial pressure of each component is the atmospheric pressure is 750 mm Hg? Dalton's Law Example Humid air on a warm summer day is approximately 20% oxygen, 75% nitrogen, 4% water vapor, and 1% argon. What is the partial pressure of each component is the atmospheric pressure is 750 mm Hg? Ptotal = PO2 + PN2 + PH2O + PAr Dalton's Law Example Humid air on a warm summer day is approximately 20% oxygen, 75% nitrogen, 4% water vapor, and 1% argon. What is the partial pressure of each component is the atmospheric pressure is 750 mm Hg? PO2 = 0.20 x 750 mm Hg = 150 mm Hg Dalton's Law Example Humid air on a warm summer day is approximately 20% oxygen, 75% nitrogen, 4% water vapor, and 1% argon. atmospheric pressure is 750 mm Hg PO2 = 0.20 x 750 mm Hg = 150 mm Hg PN2 = 0.75 x 750 mm Hg = 560 mm Hg Dalton's Law Example Humid air on a warm summer day is approximately 20% oxygen, 75% nitrogen, 4% water vapor, and 1% argon. atmospheric pressure is 750 mm Hg PO2 = 0.20 x 750 mm Hg = 150 mm Hg PN2 = 0.75 x 750 mm Hg = 560 mm Hg PH2O = 0.20 x 750 mm Hg = 150 mm Hg Dalton's Law Example Humid air on a warm summer day is approximately 20% oxygen, 75% nitrogen, 4% water vapor, and 1% argon. atmospheric pressure is 750 mm Hg PO2 = 0.20 x 750 mm Hg = 150 mm Hg PN2 = 0.75 x 750 mm Hg = 560 mm Hg PH2O = 0.04 x 750 mm Hg = 30 mm Hg PAr = 0.01 x 750 mm Hg = 8 mm Hg KineticMolecular Theory Ludwig Boltzmann James Clerk Maxwell Verbal description of an ideal gas KineticMolecular Theory Large distance between molecules KineticMolecular Theory Large distance between molecules Random motion Collisions elastic KineticMolecular Theory Large distance between molecules Random motion Collisions elastic No attractive or repulsive forces KineticMolecular Theory Large distance between molecules Random motion Collisions elastic No attractive or repulsive forces Average kinetic energy temperature 1 2 KE = mu 2 Root Mean Square Velocity
urms 3RT = M urms = rootmeansquare speed (m/s) T = temperature (K) M = molar mass (in kg) R = gas constant (8.314 J/K mol) Kinetic Molecular Theory Graham's Law of Effusion
Effusion Diffusion Graham's Law of Effusion
Rate of effusion for gas 1 = Rate of effusion for gas 2 M2 M1 Graham's Law of Effusion
Rate of effusion for gas 1 = Rate of effusion for gas 2 M2 M1 32.0 g Rate of effusion for gas N 2 = = 1.069 Rate of effusion for gas O 2 28.0 g
Rounds to 1.07 Graham's Law of Effusion
Rate of effusion for gas A = Rate of effusion for gas B = MB MA 17 g Ag = 0.68 = 0.68 Graham's Law of Effusion
MB MA = 0.68 =
2 17 g Ag
2 17 g ( 0.68) = Ag 17 g 0.4624 = Ag Graham's Law of Effusion
17 g 0.4624 = Ag 17 g Ag= = 36.7647 0.4624 Rounds to 37 g Real Gases Real gases agree within 5% Low temp or high pressure Real gases deviate significantly J. D. van der Waals 1873 Van der Waals Equation Volume of real gas too large Subtract term from volume (b) P(Vnb) = nRT Pressure small and volume large nb term makes no difference Van der Waals Equation Small force of attraction between molecules Gases condense Pressure of real gas smaller than expected Added term an2/V2 an 2 P (V + 2  nb ) = nRT V Summary Pressure Boyle's Law Charles's Law GayLussac's Law Avogadro's Law Ideal Gas Law Dalton's Law of Partial Pressure Kinetic Molecular Theory ...
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 Spring '09
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 Chemistry, Law Example

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