Apparatus for Studying the Relationship Between Pressure and Volume of a Gas As P (h) increases V decreases 1
Boyle s Law P α 1/V P x V = constant P 1 x V 1 = P 2 x V 2 Constant temperature Constant amount of gas 2
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 P 1 x V 1 = P 2 x V 2 P 1 = 726 mmhg P 2 =? V 1 = 946 ml V 2 = 154 ml P 2 = P 1 x V 1 V 2 726 mmhg x 946 ml = = 4460 mmhg 154 ml 3
Variation in Gas Volume with Temperature at Constant Pressure As T increases V increases 4
Variation of Gas Volume with Temperature at Constant Pressure Charles & Gay-Lussac s Law V α T V = constant x T Temperature must be in Kelvin V 1 /T 1 = V 2 /T 2 T (K) = t ( 0 C) + 273.15 5
A sample of carbon monoxide gas occupies 3.20 L at 125 0 C. At what temperature will the gas occupy a volume of 1.54 L if the pressure remains constant? V 1 /T 1 = V 2 /T 2 V 1 = 3.20 L T 1 = 398.15 K V 2 = 1.54 L T 2 =? T 1 = 125 ( 0 C) + 273.15 (K) = 398.15 K T 2 = V 2 x T 1 V 1 1.54 L x 398.15 K = = 192 K 3.20 L 6
Avogadro s Law V α number of moles (n) V = constant x n Constant temperature Constant pressure V 1 / n 1 = V 2 / n 2 7
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? 4NH 3 + 5O 2 4NO + 6H 2 O 1 mole NH 3 1 mole NO At constant T and P 1 volume NH 3 1 volume NO 8
Summary of Gas Laws Boyle s Law 9
Charles Law 10
Avogadro s Law 11
Ideal Gas Equation Boyle s law: P α 1 (at constant n and T) V Charles law: V α T (at constant n and P) Avogadro s law: V α n (at constant P and T) V α nt P V = constant x nt P = R nt P R is the gas constant PV = nrt 12
The conditions 0 0 C 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) 13
What is the volume (in liters) occupied by 49.8 g of HCl at STP? T = 0 0 C = 273.15 K PV = nrt V = nrt P P = 1 atm n = 49.8 g x 1 mol HCl 36.45 g HCl = 1.37 mol V = L atm 1.37 mol x 0.0821 x 273.15 K mol K 1 atm V = 30.7 L 14
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 0 C is heated to 85 0 C at constant volume. What is the final pressure of argon in the lightbulb (in atm)? PV = nrt nr V = P T = constant n, V and R are constant P 1 = 1.20 atm T 1 = 291 K P 2 =? T 2 = 358 K P 1 T 1 = P 2 T 2 P 2 = P 1 x T 2 T 1 = 1.20 atm x 358 K 291 K = 1.48 atm 15
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 cm 3 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: V 1 in cm 3 1cm 3 =1mL V 1 in ml 10 3 ml=1l V 1 in L xp 1 /P 2 V 2 in L unit conversion gas law calculation V 2 = SOLUTION: P 1 V 1 P 2 P 1 = 1.12 atm V 1 = 24.8 cm 3 24.8 cm 3 1 ml 1 cm 3 P 1 V 1 n 1 T 1 = P 2 V 2 n 2 T 2 = 0.0248 L P and T are constant P 2 = 2.64 atm V 2 = unknown L = 0.0248 L 10 3 ml P 1 V 1 = P 2 V 2 1.12 atm = 0.0105 L 2.46 atm
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 1.00x10 3 torr. It is filled with methane at 23 0 C and 0.991 atm and placed in boiling water at exactly 100 0 C. Will the safety valve open? PLAN: P 1 (atm) T 1 and T 2 ( 0 C) 1atm=760torr K= 0 C+273.15 P 1 (torr) T 1 and T 2 (K) x T 2 /T 1 P 2 (torr) 0.991 atm SOLUTION: P 1 = 0.991atm T 1 = 23 0 C P 1 V 1 P 2 V 2 = n 1 T 1 n 2 T 2 760 torr = 753 torr 1 atm P 2 = unknown T 2 = 100 0 C P 1 T 1 = P 2 T 2 P 2 = P 1 T 2 T 1 = 753 torr 373K 296K = 949 torr
Sample Problem 5.4 Applying the Volume-Amount Relationship PROBLEM: PLAN: A scale model of a blimp rises when it is filled with helium to a volume of 55 dm 3. When 1.10 mol of He is added to the blimp, the volume is 26.2 dm 3. How many more grams of He must be added to make it rise? Assume constant T and P. We are given initial n 1 and V 1 as well as the final V 2. We have to find n 2 and convert it from moles to grams. n 1 (mol) of He x V 2 /V 1 n 2 (mol) of He subtract n 1 mol to be added x M g to be added SOLUTION: V 1 n 1 n 1 = 1.10 mol n 2 = unknown V 1 = 26.2 dm 3 V 2 = 55.0 dm 3 V 2 = n 2 = n 1 n 2 n 2 = 1.10 mol 55.0 dm 3 26.2 dm 3 V 2 V 1 P and T are constant P 1 V 1 n 1 T 1 4.003 g He = 2.31 mol mol He = P 2 V 2 n 2 T 2 = 9.24 g He
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 O 2. Calculate the pressure of O 2 at 21 0 C. 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 = 21 0 C (convert to K) n = 0.885 kg (convert to mol) P = unknown 10 3 g 0.885kg kg nrt P = = V mol O 2 = 27.7 mol O 21 0 2 C + 273.15 = 294.15K 32.00 g O 2 atm*l 24.7 mol x 0.0821 x 294.15K mol*k = 1.53 atm 438 L
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) A 2 + B 2 2AB (2) 2AB + B 2 2AB 2 (3) A + B 2 AB 2 (4) 2AB 2 A 2 + 2B 2 PLAN: SOLUTION: 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. Looking at the relationships, the equation that shows a decrease in the number of moles of gas from 2 to 1 is equation (3).