Chemistry 1411 Joanna Sabey Chapter 5: Gases Chemistry 1411 Joanna Sabey

Substances as Gases

Properties of Gases Variable shape and volume: A gas has the same shape and volume as the container it is in. Expand uniformly: When a container increases its volume, a gas expands and the gas molecules distribute uniformly throughout the container. Compress uniformly: When a container decreases in volume, a gas expands and the molecules are closer to each other. Have low density: The density of air is about 0.001g/mL, the density of water is 1 g/mL. Mix uniformly with other gases in the same container.

Atmospheric Pressure Gas Pressure: the result of constantly moving molecules striking the inside surface of their container. If the molecules collide more often, the gas pressure increases. If the molecules collide with more energy the pressure increases. Atmospheric pressure: the result of air molecules in the environment. Standard atmospheric pressure is 1 atmosphere, atm. Unit Standard pressure Atmosphere 1 atm (exactly) Inches of mercury 29.9 in Hg Centimeters of mercury 76 cm Hg(exactly) Millimeters of mercury 760 mm Hg(exactly) Torr 760 torr(exactly) Pounds per square inch 14.7 psi Kilopascal 101 kPa

Gas Pressure Conversions Given the pressure inside a car tire is 34.0 psi, express the pressure in each of the following: Atm: For every 14.7 psi there is 1 atm 34.0 psi X (1 atm/14.7 psi) = 2.31 atm Cm Hg: For every 76 cm Hg, there is 14.7 psi 34.0 psi X (76 cm Hg/14.7 psi) = 176 cm Hg Torr: For every 14.7 psi, there are 760 torr. 34.0 psi X(760 torr / 14.7 psi) = 1758 torr kPa: For every 14.7 psi there are 101 kPa. 34.0 psi X (101 kPa / 14.7 psi) = 234 kPa

Variable Affecting Gas Pressure Increase or decrease the volume of the container: increase in volume, will decrease pressure and a decrease in volume will increase the pressure on the gas. Increase or decrease the temperature of the gas: Increase in the temperature, will increase the pressure and a decrease in temperature will decrease pressure. Increase or decrease in the number of molecules: Increase in number of molecules will increase the pressure and a decrease in the number of molecules will decrease pressure.

Variable Affecting Gas Pressure

Boyle’s Law Boyle’s Law: The volume of a gas is inversely proportional to the pressure when the temperature remains constant. P a 1/V When used to consider a sample of gas under changing conditions, with the temperature remains constant. P1 x V1 = P2 x V2

Boyle’s Law A 1.50 L sample of methane gas exerts a pressure of 1650 mmHg, Calculate the final pressure if the volume changes to 7.00 L. Assume no temperature change. Step One: what equation will be used: P1 X V1 = P2 X V2 Step Two: What is being solved for? P2 Step Three: Rearrange the equation: P2= P 1 X V 1 V 2 Step Four: Plug in known values and solve: P2 = 1650 𝑚𝑚𝐻𝑔 1.50 𝐿 7.00 𝐿 = 354 mmHg

Charles's Law Charles’s Law: The volume of a gas is directly proportional to the Kelvin temperature if the pressure remains constant. V = T This can be used to consider a sample of gas under changing conditions with the pressure remaining constant: V 1 T 1 = V 2 T 2 The equation can be rearranged to solve for the unknown value.

Charles’s Law A 275 L Helium balloon is heated from 20 oC to 40oC. Calculate the final volume assuming the pressure remains constant. Step One: Convert to Kelvin: 20oC + 273 = 293 K 40oC + 273 = 313 K Step Two: Decide which equation to use: V 1 T 1 = V 2 T 2 Step Three: Rearrange the equation to solve for unknown: V2 = 𝑉 1 𝑋 𝑇 2 𝑇 1 Step Four: Plug-in given values: V1 = 275 L T1 = 293 K T2 =313 K V2 = 275 𝐿 𝑋 313 𝐾 293 𝐾 = 294 L

Gay-Lussac’s Law Gay-Lussac’s Law: The pressure of a gas is directly proportional to the Kelvin Temperature if the volume remains constant. P = T This can be used to consider a sample of gas under changing conditions with the volume remaining constant: 𝑃 1 𝑇 1 = 𝑃 2 𝑇 2

Gay-Lussac’s Law A steel container filled with nitrous oxide at 15.0 atm is cooled from 25 oC to -40 oC. Calculate the final pressure assuming the volume remains constant. Step One: Convert to Kelvin: 25 oC +273 = 293 K -40 oC +273 = 233 K Step Two: Decide which equation to use: 𝑃 1 𝑇 1 = 𝑃 2 𝑇 2 Step Three: Rearrange the equation to solve for unknown: P2 = P 1 X 𝑇 2 T 1 Step Four: Plug-in given values: P2 = 15.0 atm X 233 K 293 K = 11.9 atm

Avogadro’s Law Avogadro’s Law: At constant temperature and pressure, the volume of gas is directly proportional to the number of moles of the gas present. V a number of moles (n) V1 / n1 = V2 / n2

Ideal Gas Law Kinetic Theory of an ideal gas: Gases are made up of tiny molecules: gas molecules are relatively small and the distance between them is quite large. Gas molecules demonstrate rapid motion, move in straight lines, and travel in random directions. Gas molecules have no attraction for one another: when they collide they simply bounce off of one another. Gas molecules collide without losing energy: they are said to have elastic collisions. The average kinetic energy of gas molecules is proportional to the Kelvin temperature, KE = T at the same temperature, all gas molecules have equal kinetic energy.

Ideal Gas Law Ideal Gas Law: The relationship of Pressure, Volume, Temperature and number of moles using the ideal gas constant: PV = nRT R = 0.0821 𝐿 𝑋 𝑎𝑡𝑚 𝑚𝑜𝑙 𝑋 𝐾 P must be in atm V must be in L n must be in moles T must be in K

Ideal Gas Law How many moles of hydrogen gas occupy a volume of 0.500L at STP? Step One: Convert to necessary units: at STP P = 1 atm and T = 0 oC 0 oC + 273 = 273 K Step Two: Decide which equation to use: PV=nRT Step Three: Rearrange the equation: n = 𝑃𝑉 𝑅𝑇 Step Four: Plug-in known values: P = 1 atm V = 0.500L T = 273 K R = 0.0821 L X atm mol X K n = 1 atm X 0.500 L 0.0821 L X atm mol X K X 273 K =0.0223 mol

Combined Gas Law Combined Gas Law: the combination of all three variables P, V, and T. The combined gas law is used when considering variable changes for all three conditions. 𝑃 1 𝑉 1 𝑇 1 = 𝑃 2 𝑉 2 𝑇 2 Standard conditions, STP, for a gas are 0o C and 1 atm.

Combined Gas Law A nitrogen gas sample occupies 50.5 mL at -80 oC and 1250 torr. What is the volume at STP? Step One: Convert to Kelvin: -80 oC + 273 = 193 K 0 oC + 273 = 273 K Step Two: Decide which equation to use: 𝑃 1 𝑉 1 𝑇 1 = 𝑃 2 𝑉 2 𝑇 2 Step Three: Rearrange the equation: V2 = 𝑃 1 𝑉 1 𝑇 2 𝑃 2 𝑇 1 Step Four: Plug-in given values: P1= 1250 torr V1 = 50.5 mL T1 = 193 K P2 = 760 torr T2 = 273 K V2 = 1250 𝑡𝑜𝑟𝑟 𝑋 50.5 𝑚𝐿 𝑋 273 𝐾 760 𝑡𝑜𝑟𝑟 𝑋 193 𝐾 = 117 mL

Gaseous Substances Density (d) Calculations m is the mass of the gas in g d = m V = PM RT M is the molar mass of the gas Molar Mass (M ) of a Gaseous Substance dRT P M = d is the density of the gas in g/L

Density Calculate the density of carbon dioxide (CO2) in grams per liter (g/L) at 0.990 atm and 55°C. D = P X MM RT MM CO2 = 44.01 g/mol P = 0.990 atm R = 0.0821 L X atm mol X K T = 328 K Plug into Equation: D = (0.990 atm) X (44.01 g mol ) 0.0821 L X atm mol X K X 328 K = 1.62 g/L

Gas Stoichiometry Calculate the volume of O2 (in liters) required for the complete combustion of 7.64 L of acetylene (C2H2) measured at the same temperature and pressure. 2 C2H2(g) + 5 O2(g)  4 CO2(g) + 2 H2O(l) For every 2 moles of C2H2 there are 5 moles of O2 7.64 L C2H2 X 5 L O 2 2 L C 2 H 2 = 19.1 L O2

Dalton’s Law of Partial Pressure Dalton’s Law of Partial Pressure: The total pressure of a gaseous mixture is equal to the sum of the individual pressures of each gas. P1 + P2 + P3 +… = Ptotal Pi = Xi PT ni mole fraction (Xi ) = nT

Dalton’s Law of Partial Pressure If the partial pressure of hydrogen, ammonia, and methane are 275 torr, 125 torr, and 340 torr, respectively, what was the partial pressure of the water vapor? Assuming the pressure in the container was STP, 760 torr. 275 torr + 125 torr+ 340 torr + Pwater vapor = 760 torr Pwater vapor = 760 torr- (275 torr+ 125 torr +340 torr) Pwater vapor = 760 torr – 740 torr = 20 torr

Dalton’s Law of Partial Pressure A mixture of gases contains 4.46 moles of neon (Ne), 0.74 mole of argon (Ar), and 2.15 moles of xenon (Xe). Calculate the partial pressures of the gases if the total pressure is 2.00 atm at a certain temperature. Calculate the mole fraction: Ne= 4.46 mol Ne 4.46 mol Ne+0.74 mol Ar+2.15 mol Xe =0.607 Ne Ar= 0.74 mol Ne 4.46 mol Ne+0.74 mol Ar+2.15 mol Xe = 0.10 Ar Xe= 2.15 mol Ne 4.46 mol Ne+0.74 mol Ar+2.15 mol Xe = 0.293 Xe Use the mole fraction equation for each gas: Ne: PNe= (0.607 moles Ne) (2.00 atm) = 1.21 atm Ar: PAr=(0.10 moles Ar)(2.00atm)= 0.20 atm Xe: PXe=(20.293 moles Xe)(2.00atm)= 0.586 atm

Dalton’s Law of Partial Pressure Oxygen gas generated by the decomposition of potassium chlorate is collected. The volume of oxygen collected at 24°C and atmospheric pressure of 762 mmHg is 128 mL. Calculate the mass (in grams) of oxygen gas obtained. The pressure of the water vapor at 24°C is 22.4 mmHg. Calculate the Partial Pressure of O2: PT = PO2+PH2O PO2 = 762 mmHg – 22.4 mmHg = 739.6 mmHg Use the Ideal Gas Law to determine mass obtained: PV=nRT= m MM RT m = 𝑃𝑉 𝑀𝑀 𝑅𝑇 = 739.6 760 mm Hg 0.128 L (32.0 g mol ) 0.0821 L X atm mol X K X 297 K = 0.164 g

Gas Diffusion and Effusion Gas diffusion: the gradual mixing of molecules of one gas with molecules of another by virtue of their kinetic properties. Gas effusion: the process by which gas under pressure escapes from one compartment of a container to another by passing through a small opening. r 1 r 2 = M M 1 MM 2 r1 and r2 are the diffusion rates

Gas Diffusion and Effusion A flammable gas made up only of carbon and hydrogen is found to effuse through a porous barrier in 1.50 min. Under the same conditions of temperature and pressure, it takes an equal volume of bromine vapor 4.73 min to effuse through the same barrier. r 1 r 2 = M M 1 MM 2 Bromine Gas= 159.80 g/mol 1.50 𝑚𝑖𝑛 4.73 𝑚𝑖𝑛 = M M 1 159.80 𝑔/𝑚𝑜𝑙 MM1 = ( 1.50 4.73 )2 X 159.80 g/mol = 16.1 g/mol