b The good news is that you don’t have to remember all three gas laws! Since they are all related to each other, we can combine them into a single equation. P 1 V 1 P 2 V 2 = T 1 T 2 No, it’s not related to R2D2 Combined Gas Law

If you should only need one of the other gas laws, you can cover up the item that is constant and you will get that gas law! P1P1 V1V1 T1T1 P2P2 V2V2 T2T2 Boyle’s Law Charles’ Law Gay-Lussac’s Law Combined Gas Law

A sample of helium gas has a volume of L, a pressure of atm and a temperature of 29°C. What is the new temperature of the gas at a volume of 90.0 mL and a pressure of 3.20 atm? Set up Data Table P 1 = atm V 1 =.180 L T 1 = 302 K P 2 = 3.20 atm V 2 =.090 L T 2 = ?? Combined Gas Law Problem

P 1 = atm V 1 = 180 mL T 1 = 302 K P 2 = 3.20 atm V 2 = 90 mL T 2 = ?? P 1 V 1 P 2 V 2 *Cross Multiply - P 1 V 1 T 2 = P 2 V 2 T 1 = T 1 T 2 T 2 = P 2 V 2 T 1 P 1 V 1 T 2 = 3.20 atm x 90.0 mL x 302 K atm x mL = 604 K Calculation

Combined Problem #2 b 2.00 L of a gas is collected at 25.0 °C and mmHg. What is the volume of the gas at STP? Set up Data Table P 1 = 745 mmHg V 1 = 2.00 L T 1 = 298 K P 2 = 760 mmHg V 2 = ?? T 2 = 273 K

P 1 = 745 mmHgV 1 = 2.00 L T 1 = 298 K P 2 = 760 mmHg V 2 = ? L T 2 = 273 K P 1 V 1 P 2 V 2 *Cross Multiply - P 1 V 1 T 2 = P 2 V 2 T 1 = T 1 T 2 V 2 = P 1 V 1 T 2 P 2 T 1 V 2 = 745 x 2.00 x 273 K 760 x 298 = 1.80 L Calculation

Ideal Gas Law b The Ideal Gas law is “Ideal” because it accounts for all four variables that can affect gases. PV = nRT b P = Pressure (In atm) b V = Volume (In Liters) b n = Moles (In mol) b R = Gas Constant = b T = Temperature (in Kelvin)

Ideal Gas Law Constant b R is called the gas constant. b Using a known constant that doesn’t change allows us to calculate other parts of the equation. b The value of R depends on the units of Pressure and Volume. b For us, R = 0.082, but Pressure must be in atmospheres and Volume must be in Liters.

Ideal Gas Law Problem 1 A 7.50 liter sealed jar at 18 °C contains moles of oxygen and moles of nitrogen gas. What is the pressure in the container? Set up Data Table: P = ? atmV = 7.50 L n = mol R = T = 291 K

Ideal Gas Law Problem 1 P = ? atmV = 7.50 L n = mol R = T = 291 K PV = nRT (P)(7.50) = (0.250)(0.082)(291) P = atm

Ideal Gas Law Problem 2 b A 4.0 liter container is filled with Argon gas. At 26 °C, the total pressure of the gas is 98.0 kPa. How many grams of Argon did it take to fill the container? Set up Data Table: * First solve for moles, then convert to grams. P = atm V = 4.0 L n = ? mol R = T = 299 K

Ideal Gas Law Problem 2 P = atm V = 4.0 L n = ? mol R = T = 299 K PV = nRT (0.967)(4.0) = (n)(0.082)(299) n = moles Ar - But we want Grams! mol Ar g Ar = 6.30 g Ar 1 mol Ar