Ideal Gas Law (Equation): P = Pressure PV = nRT V = Volume n = number of moles R = ideal gas constant T = Temperature (K)

R = PV = (1 atm )(22.4 L) R = 0.0821 L * atm We can calculate R by looking at 1 mol of a gas at STP: STP = 273 K and 1 atm @ STP, 1 mol of a gas occupies 22.4 L R = PV = (1 atm )(22.4 L) nT (1 mol)(273 K) R = 0.0821 L * atm mol * K

Use ideal gas law, PV=nRT. Practice Problem A 23.8-L cylinder contains oxygen gas at 20.0 0C and 732 torr. How many moles of oxygen are in the cylinder? PV = n RT Use ideal gas law, PV=nRT. Pressure must be in atm, Temperature in K: 732 torr x 1 atm = 0.963 atm 760 torr (0.963 atm)(23.8 L) = n (0.0821 L atm/mole K)(293 K ) = 0.953 mol

Gas Quiz - Rd 1 PV = nRT R = 0.0821 L atm mol K P1V1 = P2V2 T1 T2 1 atm = 760 mm Hg A 5.8 L sample of gas at 5.0oC and 750 mm Hg is cooled so that the volume becomes 9.6 L and the pressure 380 mmHg. What is the temperature in oC? What is the volume of 0.750 mole of oxygen gas at 25.0 oC and 725 torr?

P1V1 = P2V2 T1 T2 (750 mm Hg)(5.8 L) = (9.6 L)(380 mm Hg) 278 K T2 A 5.8 L sample of gas at 5.0 oC and 750 mm Hg is cooled so that the volume becomes 9.6 L and the pressure 380 mm Hg. What is the temp in oC? P1V1 = P2V2 T1 T2 (750 mm Hg)(5.8 L) = (9.6 L)(380 mm Hg) 278 K T2 T2 = 233 (230) K - 273 = -40oC (-43oC)

PV = nRT V = nRT P P = 725 torr x 1 atm = 0.954 atm 760 torr What is the volume of 0.750 mole of oxygen gas at 25.0 oC and 725 torr? PV = nRT V = nRT P P = 725 torr x 1 atm = 0.954 atm 760 torr V = (0.750 mole)(0.0821 L* atm/mol*K)(298 K) 0.954 atm V = 19.2 L