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Scientists have been studying physical properties of gases for hundreds of years. In 1662, Robert Boyle discovered that gas pressure and volume are related mathematically. The observations of Boyle and others led to the development of the gas laws. The gas laws are simple mathematical relationships between the volume, temperature, pressure, and amount of a gas. Section 2 – The Gas Laws

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5 - 2 The gas laws Since gases are highly compressible and will expand when heated, these properties have been studied extensively. gas laws The relationships between volume, pressure, temperature and moles are referred to as the gas laws. To understand the relationships, we must look a few concepts.

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5 - 3 Units we will be using Volume Volumeliters, although other units could be used. Temperature TemperatureMust use an absolute scale. K - Kelvin is most often used. Pressure PressureAtm, torr, mm Hg, lb/in 2. - use what is appropriate. Moles MolesWe specify the amounts in molar quantities.

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5 - 4 Gas laws Laws that show the relationship between volume and various properties of gases Boyle’s Law Charles’ Law Gay-Lussac’s Law Avogadro’s Law Ideal Gas Equation The Ideal Gas Equation combines several of these laws into a single relationship.

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5 - 5 Boyle’s law The volume of a gas is inversely proportional to its pressure. PV = k or P 1 V 1 = P 2 V 2 Temperature and number of moles must be held constant!

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Boyle’s law can be used to compare changing conditions for a gas Using P1 and V1 to stand for initial conditions and P2 and V2 to stand for new conditions results in the following equations P1V1 =k P2V2 = k P1V1 = P2V2

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Practice Problem A sample of oxygen gas has a volume of 150. mL when its pressure is atm. What will the volume of the gas be at a pressure of atm if the temperature remains constant? Given: V1 of O2 = 150. mL; P1 of O2 = atm; P2 of O2 = atm Unknown: V2 of O2 in mL

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Given: V1 of O2 = 150. mL; P1 of O2 = atm; P2 of O2 = atm Unknown: V2 of O2 in mL Rearrange the equation for Boyle’s law (P1V1 = P2V2) to obtain V2

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Practice Problems A balloon filled with helium gas has a volume of 500 mL at a pressure of 1 atm.The balloon is released and reaches an altitude of 6.5 km, where the pressure is 0.5 atm. Assuming that the temperature has remained the same, what volume does the gas occupy at this height? Answer 1000 mL He

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A gas has a pressure of 1.26 atm and occupies a volume of 7.40 L. If the gas is compressed to a volume of 2.93 L, what will its pressure be, assuming constant temperature? Answer 3.18 atm

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Divers know that the pressure exerted by the water increases about 100 kPa with every 10.2 m of depth. This means that at 10.2 m below the surface, the pressure is 201 kPa; at 20.4 m, the pressure is 301 kPa; and so forth. Given that the volume of a balloon is 3.5 L at STP and that the temperature of the water remains the same, what is the volume 51 m below the water’s surface? Answer 0.59 L

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Charles’s Law: Volume-Temperature Charles’s law the volume of a fixed mass of gas at constant pressure varies directly with the Kelvin temperature

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Charles’ law When you heat a sample of a gas, its volume increases. The pressure and number of moles must be held constant.

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Charles’ Law Placing an air filled balloon near liquid nitrogen (77 K) will cause the volume to be reduced. Pressure and the number of moles are constant.

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The temperature − °C is referred to as absolute zero and is given a value of zero in the Kelvin scale K = °C V = kT or V/T = k

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Practice Problems A sample of neon gas occupies a volume of 752 mL at 25°C.What volume will the gas occupy at 50°C if the pressure remains constant? Given: V1 of Ne = 752 mL; T1 of Ne = 25°C = 298 K; T2 of Ne = 50°C = 323 K Note that Celsius temperatures have been converted to kelvins.This is a very important step for working the problems in this chapter Unknown: V2 of Ne in mL

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Given: V1 of Ne = 752 mL; T1 of Ne = 25°C = 298 K; T2 of Ne = 50°C = 323 K Unknown: V2 of Ne in mL

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A helium-filled balloon has a volume of 2.75 L at 20.°C.The volume of the balloon decreases to 2.46 L after it is placed outside on a cold day. What is the outside temperature in K? in °C? Answer 262 K, or − 11°C

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A gas at 65°C occupies 4.22 L. At what Celsius temperature will the volume be 3.87 L, assuming the same pressure? Answer 37°C

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Gay-Lussac’s Law: Pressure-Temperature Gay-Lussac’s law: The pressure of a fixed mass of gas at constant volume varies directly with the Kelvin temperature P = kT or P/T = k

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Gay-Lussac’s Law Law of of Combining Volumes. At constant temperature and pressure, the volumes of gases involved in a chemical reaction are in the ratios of small whole numbers. Studies by Joseph Gay-Lussac led to a better understanding of molecules and their reactions.

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Gay-Lussac’s Law Example. Reaction of hydrogen and oxygen gases. Two ‘volumes’ of hydrogen will combine with one ‘volume’ of oxygen to produce two volumes of water. We now know that the equation is: 2 H 2 (g) + O 2 (g) 2 H 2 O (g) +H2H2 H2H2 O2O2 H2OH2OH2OH2O

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The gas in an aerosol can is at a pressure of 3.00 atm at 25°C. Directions on the can warn the user not to keep the can in a place where the temperature exceeds 52°C. What would the gas pressure in the can be at 52°C? Given: P1 of gas = 3.00 atm; T1 of gas = 25°C = 298 K; T2 of gas = 52°C = 325 K Unknown: P2 of gas in atm

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Given: P1 of gas = 3.00 atm; T1 of gas = 25°C = 298 K; T2 of gas = 52°C = 325 K Unknown: P2 of gas in atm

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Before a trip from New York to Boston, the pressure in an automobile tire is 1.8 atm at 20.°C. At the end of the trip, the pressure gauge reads 1.9 atm. What is the new Celsius temperature of the air inside the tire? (Assume tires with constant volume.) Answer 36°C

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At 120.°C, the pressure of a sample of nitrogen is 1.07 atm. What will the pressure be at 205°C, assuming constant volume? Answer 1.30 atm

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A sample of helium gas has a pressure of 1.20 atm at 22°C. At what Celsius temperature will the helium reach a pressure of 2.00 atm? Answer 219°C

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Avogadro’s law Equal volumes of gas at the same temperature and pressure contain equal numbers of molecules. V = k n V 1 V 2 n 1 n 2 =

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Avogadro’s law If you have more moles of a gas, it takes up more space at the same temperature and pressure.

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Standard conditions (STP) Remember the following standard conditions. Standard temperature = K Standard pressure= 1 atm At these conditions: One mole of a gas has 22.4 a volume of 22.4 liters.

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The ideal gas law A combination of Boyle’s, Charles’ and Avogadro’s Laws PV = nRT P = pressure, atm V = volume, L n = moles T = temperature, K R = L atm/K mol (gas law constant)

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Example What is the volume of 2.00 moles of gas at 3.50 atm and K? PV = nRT V = nRT / P = (2.00 mol)( L atm K -1 mol -1 )(310.0 K) (3.50 atm) = 14.5 L

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Ideal gas law PV nT ( 1 atm ) ( 22.4 L ) ( 1 mol ) ( K) R = = atm L mol -1 K -1 R = R can be determined from standard conditions.

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Ideal gas law When you only allow volume and one other factor to vary, you end up with one of the other gas laws. Just remember Boyle Pressure Charles Temperature Avogadro Moles

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Ideal gas law P1V1n1T1P1V1n1T1 = R = P2V2n2T2P2V2n2T2 This one equation says it all. Anything held constant will “cancels out” of the equation

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Ideal gas law Example - if n and T are held constant P1V1n1T1P1V1n1T1 = P2V2n2T2P2V2n2T2 This leaves us P 1 V 1 = P 2 V 2 Boyle’s Law

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Example If a gas has a volume of 3.0 liters at 250 K, what volume will it have at 450 K ? P1V1n1T1P1V1n1T1 = P2V2n2T2P2V2n2T2 Cancel P and n They don’t change V1T1V1T1 = V2T2V2T2 We end up with Charles’ Law

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Example If a gas has a volume of 3.0 liters at 250 K, what volume will it have at 450 K ? V1T1V1T1 = V2T2V2T2 V 2 =(3.0 l) (450 K) (250 K) = 5.4 L P1V1n1T1P1V1n1T1 = P2V2n2T2P2V2n2T2

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Dalton’s law of partial pressures The total pressure of a gaseous mixture is the sum of the partial pressure of all the gases. P T =P 1 + P 2 + P Air is a mixture of gases - each adds it own pressure to the total. P air = P N 2 + P O 2 + P Ar + P CO 2 + P H 2 O

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Partial pressure example Mixtures of helium and oxygen are used in scuba diving tanks to help prevent “the bends.” For a particular dive, 46 liters of O 2 and 12 liters of He were pumped in to a 5 liter tank. Both gases were added at 1.0 atm pressure at 25 o C. Determine the partial pressure for both gases in the scuba tank at 25 o C.

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Partial pressure example First calculate the number of moles of each gas using PV = nRT. n O 2 = = 1.9 mol n He = = 0.49 mol (1.0 atm) (46 l) ( l atm K -1 mol -1 )(298.15K) (1.0 atm) (12 l) ( l atm K -1 mol -1 )(298.15K)

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Partial pressure example Now calculate the partial pressures of each. P O 2 = = 9.3 atm P He = = 2.4 atm Total pressure in the tank is 11.7 atm. (1.9 mol) ( K) ( l atm K -1 mol -1 ) (5.0 l) (0.49 mol) ( K) ( l atm K -1 mol -1 ) (5.0 l)

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Diffusion

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Diffusion and effusion Diffusion The random and spontaneous mixing of molecules.Effusion The escape of molecules through small holes in a barrier.

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Practice Problems A helium-filled balloon has a volume of 50.0 L at 25°C and 1.08 atm.What volume will it have at atm and 10.°C? Given: V1 of He = 50.0 L; T1 of He = 25°C = 298 K; T2 of He = 10°C = 283 K; P1 of He = 1.08 atm; P2 of He = atm Unknown: V2 of He in L

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Given: V1 of He = 50.0 L; T1 of He = 25°C = 298 K; T2 of He = 10°C = 283 K; P1 of He = 1.08 atm; P2 of He = atm Unknown: V2 of He in L

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The volume of a gas is 27.5 mL at 22.0°C and atm. What will the volume be at 15.0°C and atm? Answer 26.3 mL

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A 700. mL gas sample at STP is compressed to a volume of 200. mL, and the temperature is increased to 30.0°C. What is the new pressure of the gas in Pa? Answer 3.94 × 10 5 Pa, or 394 kPa

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Given: P T = P atm = torr; P H2 O = 17.5 torr (vapor pressure of water at 20.0°C); P at m = P O2 + P H2O Unknown: P O2 in torr P O2 = P atm − P H2O P O2 = torr − 17.5 torr = torr

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Some hydrogen gas is collected over water at 20.0°C.The levels of water inside and outside the gas-collection bottle are the same. The partial pressure of hydrogen is torr. What is the barometric pressure at the time the gas is collected? Answer torr

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Helium gas is collected over water at 25°C.What is the partial pressure of the helium, given that the barometric pressure is mm Hg? Answer mm Hg

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