Apparatus for Studying the Relationship Between Pressure and Volume of a Gas As P (h) increases V decreases

Boyle’s Law P a 1/V Constant temperature P x V = constant Constant amount of gas P x V = constant P1 x V1 = P2 x V2

P x V = constant P1 x V1 = P2 x V2 P1 = 726 mmHg P2 = ? V1 = 946 mL A sample of chlorine gas occupies a volume of 946 mL at a pressure of 726 mmHg. What is the pressure of the gas (in mmHg) if the volume is reduced at constant temperature to 154 mL? P x V = constant P1 x V1 = P2 x V2 P1 = 726 mmHg P2 = ? V1 = 946 mL V2 = 154 mL P1 x V1 V2 726 mmHg x 946 mL 154 mL = P2 = = 4460 mmHg

As T increases V increases Variation in Gas Volume with Temperature at Constant Pressure As T increases V increases

Variation of Gas Volume with Temperature at Constant Pressure Charles’ & Gay-Lussac’s Law V a T Temperature must be in Kelvin V = constant x T V1/T1 = V2 /T2 T (K) = t (0C) + 273.15

A sample of carbon monoxide gas occupies 3. 20 L at 125 0C A sample of carbon monoxide gas occupies 3.20 L at 125 0C. At what temperature will the gas occupy a volume of 1.54 L if the pressure remains constant? V1 /T1 = V2 /T2 V1 = 3.20 L V2 = 1.54 L T1 = 398.15 K T2 = ? T1 = 125 (0C) + 273.15 (K) = 398.15 K V2 x T1 V1 1.54 L x 398.15 K 3.20 L = T2 = = 192 K

Avogadro’s Law V a number of moles (n) V = constant x n Constant temperature Constant pressure V a number of moles (n) V = constant x n V1 / n1 = V2 / n2

4NH3 + 5O2 4NO + 6H2O 1 mole NH3 1 mole NO At constant T and P Ammonia burns in oxygen to form nitric oxide (NO) and water vapor. How many volumes of NO are obtained from one volume of ammonia at the same temperature and pressure? 4NH3 + 5O2 4NO + 6H2O 1 mole NH3 1 mole NO At constant T and P 1 volume NH3 1 volume NO

Summary of Gas Laws Boyle’s Law

Charles Law

Avogadro’s Law

Ideal Gas Equation 1 Boyle’s law: P a (at constant n and T) V Charles’ law: V a T (at constant n and P) Avogadro’s law: V a n (at constant P and T) V a nT P V = constant x = R nT P R is the gas constant PV = nRT

PV = nRT PV (1 atm)(22.414L) R = = nT (1 mol)(273.15 K) The conditions 0 0C and 1 atm are called standard temperature and pressure (STP). Experiments show that at STP, 1 mole of an ideal gas occupies 22.414 L. PV = nRT R = PV nT = (1 atm)(22.414L) (1 mol)(273.15 K) R = 0.082057 L • atm / (mol • K)

PV = nRT nRT V = P 1.37 mol x 0.0821 x 273.15 K V = 1 atm V = 30.7 L What is the volume (in liters) occupied by 49.8 g of HCl at STP? T = 0 0C = 273.15 K P = 1 atm PV = nRT n = 49.8 g x 1 mol HCl 36.45 g HCl = 1.37 mol V = nRT P V = 1 atm 1.37 mol x 0.0821 x 273.15 K L•atm mol•K V = 30.7 L

PV = nRT n, V and R are constant nR V = P T = constant P1 T1 P2 T2 = Argon is an inert gas used in lightbulbs to retard the vaporization of the filament. A certain lightbulb containing argon at 1.20 atm and 18 0C is heated to 85 0C at constant volume. What is the final pressure of argon in the lightbulb (in atm)? PV = nRT n, V and R are constant nR V = P T = constant P1 = 1.20 atm T1 = 291 K P2 = ? T2 = 358 K P1 T1 P2 T2 = P2 = P1 x T2 T1 = 1.20 atm x 358 K 291 K = 1.48 atm

Sample Problem 5.2 Applying the Volume-Pressure Relationship PROBLEM: Boyle’s apprentice finds that the air trapped in a J tube occupies 24.8 cm3 at 1.12 atm. By adding mercury to the tube, he increases the pressure on the trapped air to 2.64 atm. Assuming constant temperature, what is the new volume of air (in L)? PLAN: SOLUTION: P and T are constant V1 in cm3 P1 = 1.12 atm P2 = 2.64 atm unit conversion 1cm3=1mL V1 = 24.8 cm3 V2 = unknown V1 in mL 1 mL 1 cm3 L 103 mL 103 mL=1L 24.8 cm3 = 0.0248 L V1 in L gas law calculation xP1/P2 P1V1 n1T1 P2V2 n2T2 P1V1 = P2V2 = V2 in L P1V1 P2 V2 = 1.12 atm 2.46 atm = 0.0105 L = 0.0248 L

Sample Problem 5.3 Applying the Pressure-Temperature Relationship PROBLEM: A steel tank used for fuel delivery is fitted with a safety valve that opens when the internal pressure exceeds 1.00x103 torr. It is filled with methane at 230C and 0.991 atm and placed in boiling water at exactly 1000C. Will the safety valve open? PLAN: SOLUTION: P1(atm) T1 and T2(0C) P1 = 0.991atm P2 = unknown 1atm=760torr K=0C+273.15 T1 = 230C T2 = 1000C P1(torr) T1 and T2(K) P1V1 n1T1 P2V2 n2T2 = P1 T1 P2 T2 = x T2/T1 P2(torr) 0.991 atm 1 atm 760 torr = 753 torr P2 = P1 T2 T1 = 753 torr 373K 296K = 949 torr

Sample Problem 5.4 Applying the Volume-Amount Relationship PROBLEM: A scale model of a blimp rises when it is filled with helium to a volume of 55 dm3. When 1.10 mol of He is added to the blimp, the volume is 26.2 dm3. How many more grams of He must be added to make it rise? Assume constant T and P. PLAN: We are given initial n1 and V1 as well as the final V2. We have to find n2 and convert it from moles to grams. n1(mol) of He SOLUTION: P and T are constant x V2/V1 n1 = 1.10 mol n2 = unknown P1V1 n1T1 P2V2 n2T2 = n2(mol) of He V1 = 26.2 dm3 V2 = 55.0 dm3 subtract n1 V1 n1 V2 n2 = n2 = n1 V2 V1 mol to be added x M n2 = 1.10 mol 55.0 dm3 26.2 dm3 4.003 g He mol He g to be added = 9.24 g He = 2.31 mol

Sample Problem 5.5 Solving for an Unknown Gas Variable at Fixed Conditions PROBLEM: A steel tank has a volume of 438 L and is filled with 0.885 kg of O2. Calculate the pressure of O2 at 210C. PLAN: V, T and mass, which can be converted to moles (n), are given. We use the ideal gas law to find P. SOLUTION: V = 438 L T = 210C (convert to K) n = 0.885 kg (convert to mol) P = unknown 0.885kg 103 g kg mol O2 32.00 g O2 = 27.7 mol O2 210C + 273.15 = 294.15K 24.7 mol 294.15K atm*L mol*K 0.0821 x 438 L P = nRT V = 1.53 atm =

Sample Problem 5.6 Using Gas Laws to Determine a Balanced Equation PROBLEM: The piston-cylinders below depict a gaseous reaction carried out at constant pressure. Before the reaction, the temperature is 150K; when it is complete, the temperature is 300K. New figures go here. Which of the following balanced equations describes the reaction? (1) A2 + B2 2AB (2) 2AB + B2 2AB2 (3) A + B2 AB2 (4) 2AB2 A2 + 2B2 PLAN: We know P, T, and V, initial and final, from the pictures. Note that the volume doesn’t change even though the temperature is doubled. With a doubling of T then, the number of moles of gas must have been halved in order to maintain the volume. SOLUTION: Looking at the relationships, the equation that shows a decrease in the number of moles of gas from 2 to 1 is equation (3).