Section 2 – The Gas Laws 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.

The gas laws Since gases are highly compressible and will expand when heated, these properties have been studied extensively. 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.

Units we will be using Volume liters, although other units could be used. Temperature Must use an absolute scale. K - Kelvin is most often used. Pressure Atm, torr, mmHg, lb/in2. - use what is appropriate. Moles We specify the amounts in molar quantities.

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 The Ideal Gas Equation combines several of these laws into a single relationship.

Temperature and number of moles must be held constant! Boyle’s law The volume of a gas is inversely proportional to its pressure. PV = k or P1 V1 = P2 V2 Temperature and number of moles must be held constant!

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

Practice Problem A sample of oxygen gas has a volume of 150. mL when its pressure is 0.947 atm. What will the volume of the gas be at a pressure of 0.987 atm if the temperature remains constant? Given: V1 of O2 = 150. mL; P1 of O2 = 0.947 atm; P2 of O2 = 0.987 atm Unknown: V2 of O2 in mL

Rearrange the equation for Boyle’s law (P1V1 = P2V2) to obtain V2 Given: V1 of O2 = 150. mL; P1 of O2 = 0.947 atm; P2 of O2 = 0.987 atm Unknown: V2 of O2 in mL Rearrange the equation for Boyle’s law (P1V1 = P2V2) to obtain V2

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

A gas has a pressure of 1. 26 atm and occupies a volume of 7. 40 L 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

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

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

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

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.

The temperature −273.15°C is referred to as absolute zero and is given a value of zero in the Kelvin scale K = 273 + °C V = kT or V/T = k

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 + 273 = 298 K; T2 of Ne = 50°C + 273 = 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

Given: V1 of Ne = 752 mL; T1 of Ne = 25°C + 273 = 298 K; T2 of Ne = 50°C + 273 = 323 K Unknown: V2 of Ne in mL

A helium-filled balloon has a volume of 2. 75 L at 20. °C 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

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

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

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.

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 H2 (g) + O2 (g) 2 H2O (g) + H2 O2 H2O

The gas in an aerosol can is at a pressure of 3. 00 atm at 25°C 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 + 273 = 298 K; T2 of gas = 52°C + 273 = 325 K Unknown: P2 of gas in atm

Given: P1 of gas = 3.00 atm; T1 of gas = 25°C + 273 = 298 K; T2 of gas = 52°C + 273 = 325 K Unknown: P2 of gas in atm

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

At 120. °C, the pressure of a sample of nitrogen is 1. 07 atm 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

A sample of helium gas has a pressure of 1. 20 atm at 22°C 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

Avogadro’s law Equal volumes of gas at the same temperature and pressure contain equal numbers of molecules. V = k n V1 V2 n1 n2 =

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

Standard conditions (STP) Remember the following standard conditions. Standard temperature = 273.15 K Standard pressure = 1 atm At these conditions: One mole of a gas has a volume of 22.4 liters.

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 = 0.082 06 L atm/K mol (gas law constant)

Example What is the volume of 2.00 moles of gas at 3.50 atm and 310.0 K? PV = nRT V = nRT / P = (2.00 mol)(0.08206 L atm K-1mol-1)(310.0 K) (3.50 atm) = 14.5 L

Ideal gas law PV R = nT R = = 0.08206 atm L mol-1 K-1 R can be determined from standard conditions. PV nT R = ( 1 atm ) ( 22.4 L ) ( 1 mol ) ( 273.15 K) R = = 0.08206 atm L mol-1 K-1

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

Ideal gas law P1V1 P2V2 n1T1 n2T2 = R = This one equation says it all. Anything held constant will “cancels out” of the equation

Ideal gas law P1V1 P2V2 n1T1 n2T2 = P1V1 = P2V2 Example - if n and T are held constant P1V1 n1T1 P2V2 n2T2 = This leaves us P1V1 = P2V2 Boyle’s Law

Example If a gas has a volume of 3.0 liters at 250 K, what volume will it have at 450 K ? Cancel P and n They don’t change P1V1 n1T1 P2V2 n2T2 = We end up with Charles’ Law V1 T1 V2 T2 =

Example If a gas has a volume of 3.0 liters at 250 K, what volume will it have at 450 K ? P1V1 n1T1 P2V2 n2T2 V1 T1 = V2 T2 = V2 = (3.0 l) (450 K) (250 K) = 5.4 L

Dalton’s law of partial pressures The total pressure of a gaseous mixture is the sum of the partial pressure of all the gases. PT = P1 + P2 + P3 + ..... Air is a mixture of gases - each adds it own pressure to the total. Pair = PN2 + PO2 + PAr + PCO2 + PH2O

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 O2 and 12 liters of He were pumped in to a 5 liter tank. Both gases were added at 1.0 atm pressure at 25oC. Determine the partial pressure for both gases in the scuba tank at 25oC.

Partial pressure example First calculate the number of moles of each gas using PV = nRT. nO2 = = 1.9 mol nHe = = 0.49 mol (1.0 atm) (46 l) (0.08206 l atm K-1 mol-1)(298.15K) (1.0 atm) (12 l) (0.08206 l atm K-1 mol-1)(298.15K)

Partial pressure example Now calculate the partial pressures of each. PO2 = = 9.3 atm PHe = = 2.4 atm Total pressure in the tank is 11.7 atm. (1.9 mol) (298.15 K) (0.08206 l atm K-1 mol-1) (5.0 l) (0.49 mol) (298.15 K) (0.08206 l atm K-1 mol-1) (5.0 l)

Diffusion

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

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 0.855 atm and 10.°C? Given: V1 of He = 50.0 L; T1 of He = 25°C + 273 = 298 K; T2 of He = 10°C + 273 = 283 K; P1 of He = 1.08 atm; P2 of He = 0.855 atm Unknown: V2 of He in L

Given: V1 of He = 50.0 L; T1 of He = 25°C + 273 = 298 K; T2 of He = 10°C + 273 = 283 K; P1 of He = 1.08 atm; P2 of He = 0.855 atm Unknown: V2 of He in L

The volume of a gas is 27. 5 mL at 22. 0°C and 0. 974 atm The volume of a gas is 27.5 mL at 22.0°C and 0.974 atm. What will the volume be at 15.0°C and 0.993 atm? Answer 26.3 mL

A 700. mL gas sample at STP is compressed to a volume of 200 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 × 105 Pa, or 394 kPa

Given: PT = Patm = 731. 0 torr; PH2O = 17 Given: PT = Patm = 731.0 torr; PH2O = 17.5 torr (vapor pressure of water at 20.0°C); Patm = PO2 + PH2O Unknown: PO2 in torr PO2 = Patm − PH2O PO2 = 731.0 torr − 17.5 torr = 713.5 torr

Some hydrogen gas is collected over water at 20. 0°C 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 742.5 torr. What is the barometric pressure at the time the gas is collected? Answer 760.0 torr

Helium gas is collected over water at 25°C Helium gas is collected over water at 25°C.What is the partial pressure of the helium, given that the barometric pressure is 750.0 mm Hg? Answer 726.2 mm Hg