1. Gases Unit 2

2. Combined Gas Law Combines the laws of Boyle (P,V), Charles (V, T), and Gay-Lussac (P, T) Relates: P, V, Twhen n is constant Equation:P 1 V 1 = P 2 V 2 T 1 T 2

3. Combined Gas Law Has 3 variables and can also be used as any of the above minor gas laws Suggestion: Organize your info before you start!

4. Combined Gas Law Examples: A gas has a volume of 800.0 mL at 23.00 o C and 300.0 torr. What would the volume of the gas be at 227.0 o C and 600.0 torr of pressure?

5. Combined Gas Law V 1 = 800.0 mL T 1 = 23.00 o C = 296 K P 1 = 300.0 torr V 2 = ? mL T 2 = 227.0 o C = 500 K P 2 = 600.0 torr

6. Combined Gas Law P 1 V 1 = P 2 V 2 T 1 T 2 (300.0 torr)(800.0 mL) = (600.0 torr)(V 2 ) 296 K 500 K 810.811 = 1.20(V 2 ) 1.20 V 2 = 675.7 mL

7. Combined Gas Law A gas sample occupies 3.25 L at 24.5 o C and 350 kPa. Determine the temperature (in Celsius) at which the gas will occupy 4.25 L at 150 kPa. V 1 = 3.25 L T 1 = 24.5 o C = 297.5 K P 1 = 350 kPa V 2 = 4.25 L T 2 = ? o C P 2 = 150 kPa

8. Combined Gas Law P 1 V 1 = P 2 V 2 T 1 T 2 (350 kPa)(3.25 L) = (150 kPa)(4.25 L) 297.5 KT 2 3.82 = 637.5 1 T 2 Cross multiply…

9. Combined Gas Law (3.82)(T 2 ) = 637.5 3.82 3.82 T 2 = 167 K -273 T 2 = -106 o C

10. Combined Gas Law 0.73 L of nitrogen at STP is heated to 80.0 o C and the volume is increased to 4.53 L. What is the new pressure in mm Hg? V 1 = 0.73 L T 1 = 273 K P 1 = 760 mm Hg V 2 = 4.53 L T 2 = 80.0 o C = 353 K P 2 = ? mm Hg

11. Combined Gas Law P 1 V 1 = P 2 V 2 T 1 T 2 (760 mm Hg)(0.73 L) = P 2 (4.53 L) 273 K353 K 2.03 = 0.0128(P 2 ).0128 P 2 = 160 mm Hg

12. Combined Gas Law If a gas in a rigid container has a pressure of 1.75 atm at standard temperature, what is the new pressure when it is heated to 50.0 o C? P 1 = 1.75 atm T 1 = 273 K P 2 = ? atm T 2 = 50.0 o C = 323 K V 1 = V 2

13. Combined Gas Law P 1 V 1 = P 2 V 2 T 1 T 2 (1.75 atm)(V) = (P 2 )(V) 273 K 323 K 565.25(V) = (273)(P 2 )(V)273(V) P 2 = 2.07 atm