Gas Laws Ideal Gas Law: PV = nRT where P = pressure V = volume n = moles T = temperature (K) R = gas constant R = 0.08206 atm.L mol.K R = 62.36 L.torr

Gas Laws The ideal gas law is used to describe the behavior of an ideal gas. Ideal gas: hypothetical gas that obeys kinetic molecular theory and the ideal gas law

Gas Laws The ideal gas law is used in calculations for a specific sample of gas that has a constant T, P, V, and n. i.e. no changes are being made to the sample of gas If you know 3 of the 4 variables, you can calculate the other using the ideal gas law.

Gas Laws Example: Calculate the volume of 1.00 mol of an ideal gas at 1.00 atm and 0.00oC. Given: P = 1.00 atm n = 1.00 mol T = 0.00oC + 273 = 273 K Find: V PV = nRT

Gas Laws PV = nRT Solve for V V = nRT P V = 1.00 mol x 0.08206 atm.L x 273 K 1.00 atm mol.K V = 22.4 L

Gas Laws The temperature and pressure used in the previous problem are commonly used to report the properties of gases. Standard Temperature and Pressure (STP): 0oC and 1 atm. KNOW THIS!!

Gas Laws Molar volume: the volume one mol of a gas occupies (L/mole) At STP, one mole of an ideal gas has a molar volume of 22.4 L: 22.4 L 1 mol

Gas Laws The ideal gas law applies only to ideal gases. Does not always accurately describe real gases The molar volumes for many real gases at STP differ slightly from 22.4 L/mol. In most cases, the differences between ideal gas behavior and real gas behavior is so small that we can ignore it

Gas Laws Example: A weather balloon contains 4.75 moles of He gas. What volume does the gas occupy at an altitude of 4300 m if the temperature is 0oC and the pressure is 0.595 atm? Given: P = 0.595 atm T = 0oC = 273K n = 4.75 mol Find: V PV = nRT

Gas Laws PV = nRT solve for V V = nRT P V = (4.75 mol) x (273K) x (0.08206 atm.L) (0.595 atm) mol.K V = 179 L

Gas Laws Example: A used aerosol can contains 0.0173 mol of gas and has a volume of 425 mL. Calculate the pressure in the can if it is accidentally heated to 395oC. (Warning: Don’t do this!!) Given: V = 425 mL n = 0.0173 mol T = 395oC + 273 = 668K Find: P PV = nRT

Gas Laws PV = nRT solve for P P = nRT V P = (0.0173 mol) (0.08206 atm.L) (668K) 425 mL mol K x 1000 mL 1 L P = 2.23 atm

Gas Laws Example: A tire with an interior volume of 3.50 L contains 0.357 mol of air at a pressure of 2.49 atm. What is the temperature of the air in the tire in K? In oC? Given: P = 2.49 atm V = 3.50 L n = 0.357 mol Find: T (K) PV = nRT

Gas Laws PV = nRT solve for T T = PV nR T = 2.49 atm x 3.50 L x mol.K 0.357 mol 0.08206 atm.L T = 297 K T = 297 – 273 = 24oC

Gas Laws The ideal gas law was useful in determining the properties of a specific sample of gas at constant T, P, V, and n. We often need to know how a change in one (or more) properties impacts the other properties for a sample of a gas.

Gas Laws Combined Gas Laws: P1V1 = P2V2 T1 T2 This equation is true when the number of moles of a gas is constant.

Gas Laws Special cases for the combined gas law: At constant temperature (T1 = T2), P1V1 = P2V2 At constant volume (V1 = V2), P1 = P2 T1 T2 At constant Pressure (P1 = P2), V1 = V2

Gas Laws Example: A helium-filled balloon occupies 6.00 L at 19.5oC and 0.989 atm. What volume will the balloon occupy on top of Pike’s Peak if the pressure is 0.605 atm and the temperature is constant? Given: V1 = 6.00L P1 = 0.989 atm T1 = T2 = 19.5oC P2 = 0.605 atm Find: V2 Since T is constant, P1V1 = P2V2

Gas Laws P1 V1 = P2 V2 (0.989 atm) x (6.00 L) = (0.605 atm) x V2 V2 = 9.81 L

Gas Laws Example: Suppose a used aerosol can contains a gas at 0.989 atm at 23oC. If this can is heated to 425oC, what is the pressure inside the can? Given: P1 = 0.989 atm T1 = 23oC + 273 = 296K T2 = 425 + 273 = 698K V1 = V2 Find: P2 P1V1 = P2V2 T1 T2

Gas Laws Since V1= V2: P1 = P2 T1 T2 P2 = 0.989 atm (698K) 296K P2 = (0.989 atm) x 698K = 2.33 atm 296K

Gas Laws--More Applications The ideal gas equation can be used to determine either the density or the molar mass of a gas. d = P M RT Where d = density P = pressure T = temperature in K R = gas constant M = molar mass

Gas Laws--More Applications The density of a gas depends on: pressure temperature molar mass At constant temperature and pressure the densities of gas samples are directly proportional to their molar masses: d = P M RT

Gas Laws--More Applications If molar mass and pressure are held constant, then the density of the gas will decrease with increasing temperature Hot air rises because density is inversely proportional to temperature d = P M RT

Gas Laws--More Applications Example: What is the density of helium gas at 1.00 atm and 25oC? Given: P = 1.00 atm T = 25 + 273 = 298K Find: d d = P M RT

Gas Laws--More Applications d = P M RT M = molar mass of He = 4.00 g/mol d = (1.00 atm) 4.00 g mol.K (298K) 1 mol 0.08206 atm.L d = 0.164 g/L Notice: density of gases is usually in g/L instead of g/mL

Gas Laws--More Applications Example: What is the average molar mass of dry air if it has a density of 1.17 g/L at 21oC and 740.0 torr? Given: P = 740.0 torr T = 21 + 273 = 294K d = 1.17 g/L Find: M d = P M RT

Gas Laws--More Applications d = P M RT M = dRT P M = 1.17 g 62.36 torr.L (294 K) mol.K 740.0 torr M = 29.0 g/mol

Gas Laws--More Applications Understanding the properties of gases is important because gases are often the reactants or products in a chemical reaction. Often need to calculate the volume of gas produced or consumed during a reaction

Gas Laws--More Applications Moles A B Molar ratio grams Molar mass Gas Data A PA, VA, TA PV = nRT Gas Data B PB, VB, TB PV = nRT

Gas Laws--More Applications Example: The air bag in a car is inflated by nitrogen gas formed by the decomposition of NaN3: 2 NaN3(s)  2 Na (s) + 3 N2 (g) If an inflated air bag has a volume of 36 L and is to be filled with N2 gas at a pressure of 1.15 atm at a temperature of 26oC, how many grams of NaN3 must be decomposed?

Gas Laws--More Applications Moles N2 Moles A Gas Data N2 PV = nRT Molar ratio Molar ratio grams NaN3 grams B Moles B Moles NaN3 Molar mass Molar mass

Gas Laws--More Applications 36L N2, 1.15 atm 26oC Moles N2 Moles A PV = nRT Find Molar ratio Molar ratio Given grams NaN3 grams B Moles NaN3 Moles B Molar mass Molar mass

Gas Laws--More Applications n = PV = (1.15 atm) ( 36 L) mol.K RT (299 K) 0.08206 atm.L n = 1.7 mol N2 g NaN3 = 1.7 mol N2 x 2 mol NaN3 x 65.0 g NaN3 3 mol N2 1 mol NaN3 g NaN3 = 74 g NaN3

Gas Laws--More Applications Example: How many mL of oxygen gas can be collected at STP when 1.0 g of KClO3 decomposes: 2 KClO3 (s)  2 KCl (s) + 3 O2 (g)

Gas Laws--More Applications mL O2 at STP (1.0atm, 0oC) Moles O2 Moles A PV = nRT Given Molar ratio Molar ratio Find grams KClO3 grams B Moles KClO3 Moles B Molar mass Molar mass

Gas Laws--More Applications Mol O2 = 1.0 g KClO3 x 1 mole KClO3 x 3 mol O2 122.5 g 2 mol KClO3 mol O2 = 0.012 mol O2 V = nRT = (0.012 mol) 0.0821 atm.L (273K) x 103 mL P 1.0 atm mol.K 1 L V = 270 mL

Gas Laws – More Applications Example: What volume of CO2 at 125oC and 1.15 atm will be produced by the combustion of 1.00 L of C2H6 at 25oC and 1.00 atm? 2 C2H6 (g) + 7 O2 (g)  4 CO2 (g) + 6 H2O (g) Given: VC2H6 = 1.00 L PC2H6 = 1.00 atm TC2H6 = 25oC = 298 K PCO2 = 1.15 atm TCO2 = 125oC = 398 K Find: VCO2