Answer the following problem in complete solution. (Points to be considered: Formula used. given values. solution. and anal-war,l A. Boyle's Law 1. If a gas at 28C occupies 3.42 liters at a pressure of 2.02 atm, what will be its volume at a pressure of 2.81 atm? B. Charle's Lawr 1. Calculate the decrease in temperature (in Celsius) when 4.05 L at 21 .0 C is compressed to 1.04 L. C. Gay-Lussac's Law 1. Determine the pressure change when a constant volume of gas at 1.08 atm is heated from 222C to 31C. Avogadro's Law 1. 8.12 L of a gas is known to contain 0.985 mol. If the amount of gas is increased to 1.80 mol, what new volume will result (at an unchanged temperature and pressure)? Combined Gas Law 1. A gas has a volume of 821 mL at 23.0C and 300 torr. What would the volume of the gas be at 227C and 800 torr of pressure\"? Ideal Gas Law 1. At what temperature will 0.854 moles of neon gas occupv 12.30 liters at 1.05 atmospheres\"? BOYLE'S LAW Example: A gas occupies 12.3 liters at a pressure of 40.0 mmHg. What is the volume when the pressure is increased to 60.0 mmHg? Formula: Solution: P1V1 = P2V2 (40.0 mmHg) (12.3L) = (60.0 mmHg) (V2) Given: 492 mmHg*L= (60.0 mmHg)(V2) P1 = 40.0 mmHg 492 mm}/g*L 60. mittg V 2 V1 = 12.3 L 60.0 mmHg P2 = 60.0 mmHg 8.2 L = V2 V2 = ?CHARLES' LAW Example: A car tire filled with air has a volume of 100 L at 10C. What will the expanded volume of the tire be after driving the car has raised the temperature of the tire to 40C? in. | -'.. in. : I 1: 100L _ v 283K _ 313K ('100L)(313K) = (283K) v2 31300 K 23L 2 (283K) v1 6', 1' 313002; =+= L (283};wa 283K 283K: Answer; GAY-LUSSAC'S LAW Example: A 30.0 L sample of nitrogen inside a rigid, metal container at 20.0 C s placed inside an oven whose temperature is 50.0 C. The pressure nside the container at 20.0 'C was at 3.0 atm. What is the pressure of the nitrogen after its temperature is increased to 50.0 'C? Formula: Solution: P2 3.0 atm P2 T1 T2 cross multiply 293K 323K Given: P1 = 3.0 atm (3.0 atm) (323K) = (293K) P2 T1 = 20 C (293K) 969 K * atm = (293K) P2 Answer: P2 = ? 969K * atm (293K) P2 3.31 atm = P2 T2 = 50 C (323K) 293K 293KAVOGADRO'S LAW Example: If 2.4 moles of gas occupies a volume of 60L at a certain temperature, what volume will 3.7 moles of gas occupy? Formula: V2 = nz Solution: V2 60L V2 n2 3.7 mol cross multiply 2.4 mol Given: (60L) (3.7 mol) = (2.4 mol) V2 n1 = 2.4 moles V1 = 60L 222 L * mol = (2.4 mol) V2 Answer: n2 = 3.7 moles 92.5 = V2 222 L * mol (2.4 mol) V2 V2 = ? E 2.4 mol 24 molCOMBINED GAS LAW Example: A sample of N2 gas was placed in a 9L container at 300K at a pressure of 1.5 atm. The container was compressed to a volume of 3L and the temperature increased to 600K. What will be the new pressure inside the container? 9L Formula: Given: P1 = 1.5 atm V1 T = 300K P1V1 P2 V2 P2 = ? V2 = 3L T2 = 600K T1 T2 (1.5atm) (9L) (600K) = P2 (3L) (300K) Solution: (1.5atm) (9L) P2 (3L) 8100 atm*L*K =P2 (900 L * K) Answer: 300K 600K 8100 atm*L*K P2 (900 ]* K) 9 atn = P2 = cross multiply 900 Z *X 900 L * KIDEAL GAS LAW A 2.5L container holds 0.45mol of N2 gas at 315K. What is the pressure inside the container in atm? Values of R Given: 0.082057 L atm mol-1 K-1 Formula: P =? V = 2.5L 62.364 L Torr mol-1 K-1 PV = nRT 8.3145 m3 Pa mol-1 K-1 T = 315K n = 0.45mol 8.3145 J moll K-1" Solution: P (2.5L) = (0.45mol) (0.082057) (315K) P (2.5L) = (0.45mol) (0.082057 L atm moll K1) (315K) P (2.5L) = 11.63 L*atm Answer: P (2.5L) 11.63 L * atm = P = 4.65 atm 2/5 L 2.5 L