Chapter 11 Gas Laws

Gases Chapter #11 Kinetic Molecular theory Pressure Boyle’s Law Charles’s Law Combined Gas Law Avogtadro”s Law The Ideal Gas Law Dalton’s Law of Partial Pressures Molar Volume STP

11.1 The Gas Phase Gases have no distinct volume or shape. Gases expand to fill the volume of their container. Gas particles are miscible with each other. Evidence for gas particles being far apart : We can see through gases We can walk through gases Gases are compressible Gases have low densities

11.1 The Air We Breathe Composition of Earth’s Atmosphere Compound %(Volume) Mole Fractiona Nitrogen 78.08 0.7808 Oxygen 20.95 0.2095 Argon 0.934 0.00934 Carbon dioxide 0.033 0.00033 Methane 2 x 10-4 2 x 10-6 Hydrogen 5 x 10-5 5 x 10-7 a. mole fraction = mol component/total mol in mixture.

11.2 Kinetic Theory of Gases Kinetic Theory Postulates: Gas particles are sizeless relative to the volume of the gas Gas particles are in constant rapid motion Gas particles have elastic collisions; means no kinetic energy is lost on impact. The absolute temperature is directly proportional to the kinetic energy of a gas. Gas particles have no attraction to each other; i.e. no inter particle froces.

Parameters Affecting Gases Pressure (P); atm, mmHg, torr, lbs/in2 Volume (V); L, mL Temperature (T); K (only) Number of Moles (n)

11.3 Pressure Pressure is equal to force/unit area (P =F/A) lbs/in2 Force is a push which comes from gas particles striking a container wall

11.3 Pressure Units SI units = Newton/meter2 = 1 Pascal (Pa) 1 standard atmosphere (atm) = 101,325 Pa 1 atm =760 mm Hg 1 atm = 760 torr (torr is abbreviation of mmHg) 1 atm = 14.7 lbs/in2 1 atm = 1.013 barr Barr = 100 kPa

11.3 Measurement of Pressure What is above mercury?

11.3 Measurement of Pressure

11.3 Atmospheric Pressure

Units for Expressing Pressure Value Atmosphere 1 atm Pascal (Pa) 1 atm = 1.01325 x 105 Pa Kilopascal (kPa) 1 atm = 101.325 kPa mmHg 1 atm = 760 mmHg Torr 1 atm = 760 torr Bar 1 atm = 1.01325 bar mbar 1 atm = 1013.25 mbar psi 1 atm = 14.7 psi

11.3 Pressure Measurement Is the atmosphere or the gas in the canister pushing harder? Open Tube Manometer = 15 mm

11.3 Pressure Measurement Is the atmosphere or the gas in the canister pushing harder? Gas in the canister If the atmospheric pressure is 766 mm, then what is the pressure of the canister? Open Tube Manometer = 15 mm

11.3 Pressure Measurement Is the atmosphere or the gas in the canister pushing harder? Gas in the canister If the atmospheric pressure is 766 mm, then what is the pressure of the canister? P = 766 + 15 = 781 mm (torr) Open Tube Manometer = 15 mm gas

11.3 Pressure Measurement Open Tube Manometer Is the atmosphere or the gas in the canister pushing harder? gas

11.3 Pressure Measurement Open Tube Manometer Is the atmosphere or the gas in the canister pushing harder? The atmosphere What is the pressure of the gas if the atmosphere is 766 mm? = 13 mm gas

11.3 Pressure Measurement Open Tube Manometer Is the atmosphere or the gas in the canister pushing harder? The atmosphere What is the pressure of the gas if the atmosphere is 766 mm? 753 mm = 13 mm gas

11.3 Pressure Measurement Open Tube Manometer Now what is pushing harder, the gas or the atomosphere? gas

11.3 Pressure Measurement Open Tube Manometer Now what is pushing harder, the gas or the atmosphere? Neither, both the same. gas

11.3 Pressure Measurement Open Tube Manometer Now what is pushing harder, the gas or the atmosphere? Neither, both the same. Is the gas canister empty? gas

11.3 Pressure Measurement Open Tube Manometer Now what is pushing harder, the gas or the atmosphere? Neither, both the same. Is the gas canister empty? No, completely full of gas! gas

11.9 Dalton’s Law of Partial Pressures For a mixture of gases in a container PTotal = P1 + P2 + P3 + . . .

11.4 Boyles Law Consider a gas in a closed system containing a movable plunger. If the plunger is not moving up or down, what can be said about the pressure of the gas relative to the atmospheric pressure? Atm ● ● ● ●

11.4 Boyles Law Suppose we add some red gas to the container, what would happen to the collisions of gas particles with container walls. Would they increase, decrease or stay the same? Atm ● ● ● ● ● ● ●

11.4 Boyles Law Suppose we add some red gas to the container, what would happen to the collisions of gas particles with container walls. Would they increase, decrease or stay the same? More particles, more collisions, and more pressure. What happens to the plunger? Atm ● ● ● ● ● ● ●

11.4 Boyles Law Suppose we add some red gas to the container, what would happen to the collisions of gas particles with container walls. Would they increase, decrease or stay the same? More particles, more collisions, and more pressure. What happens to the plunger? Atm ● ● ● ● ● ● ●

11.4 Boyles Law The number of particles remain the same, but the surface area they have to strike increases, thus the number of collisions per square inch decrease as the plunger goes up exposing more surface area causing a decrease in pressure. ● ● ● ● ● ● ● ● ● ●

11.4 Boyle’s Law P  1/V (T and n fixed) P  V = Constant P1V1 = P2V2 Pressure and volume are inversely proportional. P  1/V (T and n fixed) P  V = Constant P1V1 = P2V2 29

11.5 Charles’s Law The volume of a gas is directly proportional to Kelvin temperature, and extrapolates to zero at zero Kelvin. V  T (P & n are constant) V1 = V2 T1 T2

11.6 Combined Gas Law Combining the gas laws the relationship P T(n/V) can be obtained. If n (number of moles) is held constant, then PV/T = constant. Temperature, K (only) Pressure: Atm, mmHg, Torr, PSI, KPa Volume: L, mL, cm3, …

11.6 Example A balloon is filled with hydrogen to a pressure of 1.35 atm and has a volume of 2.54 L. If the temperature remains constant, what will the volume be when the pressure is increased to 2.50 atm? (1.35 atm) (2.54 L) (2.50atm)V2 = T1 T1 (1.35 atm) (2.54 L) V2 = Constant Temp. means T1=T2 (2.50atm) V2 = 1.37 L

11.6 Example A sample of oxygen gas is at 0.500 atm and occupies a volume of 11.2 L at 00C, what volume will the gas occupy at 6.00 atm at room temperature (250C)?

11.8 Ideal Gas Law PV = nRT R = universal gas constant = 0.08206 L atm K-1 mol-1 P = pressure in atm V = volume in liters n = moles T = temperature in Kelvin

11.9 STP “STP” means standard temperature and standard pressure P = 1 atmosphere T = 0C The molar volume of an ideal gas is 22.42 liters at STP (put 1 mole, 1 atm, R, and 273 K in the ideal gas law and calculate V)

11.9 Example Calculate the pressure of a 1.2 mol sample of methane gas in a 3.3 L container at 25°C.

11.9 Example Calculate the pressure of a 1.2 mol sample of methane gas in a 3.3 L container at 25°C. 0.0821 L-atm Mole-K

11.9 Example Calculate the pressure of a 1.2 mol sample of methane gas in a 3.3 L container at 25°C. 0.0821 L-atm Mole-K 3.3 L

11.9 Example Calculate the pressure of a 1.2 mol sample of methane gas in a 3.3 L container at 25°C. 0.0821 L-atm 298 K Mole-K 3.3 L

11.9 Example Calculate the pressure of a 1.2 mol sample of methane gas in a 3.3 L container at 25°C. 0.0821 L-atm 298 K 1.2 mole Mole-K 3.3 L

11.9 Example Calculate the pressure of a 1.2 mol sample of methane gas in a 3.3 L container at 25°C. 0.0821 L-atm 298 K 1.2 mole = 8.9 atm Mole-K 3.3 L

11.9 Example An experiment shows that a 0.495 g sample of an unknown gas occupies 127 mL at 98°C and 754 torr pressure. Calculate the molar mass of the gas.

11.9 Example An experiment shows that a 0.495 g sample of an unknown gas occupies 127 mL at 98°C and 754 torr pressure. Calculate the molar mass of the gas. 0.0821 L-atm mole-K

11.9 Example An experiment shows that a 0.495 g sample of an unknown gas occupies 127 mL at 98°C and 754 torr pressure. Calculate the molar mass of the gas. 0.0821 L-atm 0.495 g mole-K

11.9 Example An experiment shows that a 0.495 g sample of an unknown gas occupies 127 mL at 98°C and 754 torr pressure. Calculate the molar mass of the gas. 0.0821 L-atm 0.495 g mL mole-K 10-3 L

11.9 Example An experiment shows that a 0.495 g sample of an unknown gas occupies 127 mL at 98°C and 754 torr pressure. Calculate the molar mass of the gas. 0.0821 L-atm 0.495 g mL mole-K 10-3 L 127 mL

11.9 Example An experiment shows that a 0.495 g sample of an unknown gas occupies 127 mL at 98°C and 754 torr pressure. Calculate the molar mass of the gas. 0.0821 L-atm 0.495 g mL 760 torr mole-K 10-3 L 127 mL atm

11.9 Example An experiment shows that a 0.495 g sample of an unknown gas occupies 127 mL at 98°C and 754 torr pressure. Calculate the molar mass of the gas. 0.0821 L-atm 0.495 g mL 760 torr mole-K 10-3 L 127 mL atm 754 torr

11.9 Example An experiment shows that a 0.495 g sample of an unknown gas occupies 127 mL at 98°C and 754 torr pressure. Calculate the molar mass of the gas. 0.0821 L-atm 0.495 g mL 760 torr 371 K mole-K 10-3 L 127 mL atm 754 torr

11.9 Example An experiment shows that a 0.495 g sample of an unknown gas occupies 127 mL at 98°C and 754 torr pressure. Calculate the molar mass of the gas. 0.0821 L-atm 0.495 g mL 760 torr 371 K mole-K 10-3 L 127 mL atm 754 torr = 120 g/mole

11.9 Collecting a Gas Over Water 51

11.9 Practice A sample of KClO3 is heated and decomposes to produce O2 gas. The gas is collected by water displacement at 25°C. The total volume of the collected gas is 229 mL at a pressure of 755 torr. How many moles of oxygen formed? Hint: The gas collected is a mixture so use Dalton’s Law to calculate the pressure of oxygen then the ideal gas law to find the number of moles oxygen. PT = P + P O2 H2O 52

11.9 Vapor Pressure of Water

11.9 Practice A sample of KClO3 is heated and decomposes to produce O2 gas. The gas is collected by water displacement at 25°C. The total volume of the collected gas is 229 mL at a pressure of 755 torr. How many moles of oxygen formed? Hint: The gas collected is a mixture so use Dalton’s Law to calculate the pressure of oxygen then the ideal gas law to find the number of moles oxygen. PT = P + P O2 H2O 755 torr = P + 23.8 torr O2 P = 755 – 23.8 = 731 torr O2 54

11.9 Practice A sample of KClO3 is heated and decomposes to produce O2 gas. The gas is collected by water displacement at 25°C. The total volume of the collected gas is 229 mL at a pressure of 755 torr. How many moles of oxygen formed? mole-K 0.0821 L-atm

11.9 Practice A sample of KClO3 is heated and decomposes to produce O2 gas. The gas is collected by water displacement at 25°C. The total volume of the collected gas is 229 mL at a pressure of 755 torr. How many moles of oxygen formed? mole-K atm 0.0821 L-atm 760 torr

11.9 Practice A sample of KClO3 is heated and decomposes to produce O2 gas. The gas is collected by water displacement at 25°C. The total volume of the collected gas is 229 mL at a pressure of 755 torr. How many moles of oxygen formed? mole-K atm 731 torr 0.0821 L-atm 760 torr

11.9 Practice A sample of KClO3 is heated and decomposes to produce O2 gas. The gas is collected by water displacement at 25°C. The total volume of the collected gas is 229 mL at a pressure of 755 torr. How many moles of oxygen formed? mole-K atm 731 torr 0.0821 L-atm 760 torr 298 K

11.9 Practice A sample of KClO3 is heated and decomposes to produce O2 gas. The gas is collected by water displacement at 25°C. The total volume of the collected gas is 229 mL at a pressure of 755 torr. How many moles of oxygen formed? mole-K atm 731 torr 10-3 L 0.0821 L-atm 760 torr 298 K mL

11.9 Practice A sample of KClO3 is heated and decomposes to produce O2 gas. The gas is collected by water displacement at 25°C. The total volume of the collected gas is 229 mL at a pressure of 755 torr. How many moles of oxygen formed? mole-K atm 731 torr 10-3 L 229 mL 0.0821 L-atm 760 torr 298 K mL = 9.00 X 10-3 mole

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