2. 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

3. 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

4. A sample of helium gas has a volume of 0.180 L, a pressure of 0.800 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 = 0.800 atm V 1 =.180 L T 1 = 302 K P 2 = 3.20 atm V 2 =.090 L T 2 = ?? Combined Gas Law Problem

5. P 1 = 0.800 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 0.800 atm x 180.0 mL = 604 K Calculation

6. Combined Problem #2 b 2.00 L of a gas is collected at 25.0 °C and 745.0 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

7. 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

8. 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 = 0.082 b T = Temperature (in Kelvin)

9. 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.

10. Ideal Gas Law Problem 1 A 7.50 liter sealed jar at 18 °C contains 0.125 moles of oxygen and 0.125 moles of nitrogen gas. What is the pressure in the container? Set up Data Table: P = ? atmV = 7.50 L n = 0.250 mol R = 0.082 T = 291 K

11. Ideal Gas Law Problem 1 P = ? atmV = 7.50 L n = 0.250 mol R = 0.082 T = 291 K PV = nRT (P)(7.50) = (0.250)(0.082)(291) P = 0.795 atm

12. 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 = 0.967 atm V = 4.0 L n = ? mol R = 0.082 T = 299 K

13. Ideal Gas Law Problem 2 P = 0.967 atm V = 4.0 L n = ? mol R = 0.082 T = 299 K PV = nRT (0.967)(4.0) = (n)(0.082)(299) n = 0.1577 moles Ar - But we want Grams! 0.1577 mol Ar 39.95 g Ar = 6.30 g Ar 1 mol Ar