1. 12.4 The Ideal Gas Law

2. The Ideal Gas Law Considers the amount of the gas in the system – in addition to pressure, volume, and temperature Deriving the ideal gas law P 1 V 1 = P 2 V 2 T 1 T 2 T 1 n 1 T 2 n 2 PV/Tn is constant. Evaluating for this constant you can solve for n at any P, V, or T!!!

3. The value of R is calculated based on the following: – 1 mol of a gas – 22.4 L kPa – STP 101.3 kPa 273 K

4. The value of R is calculated based on the following: – 1 mol of a gas – 22.4 L kPa – STP 101.3 kPa 273 K R = PV Tn

5. The value of R is calculated based on the following: – 1 mol of a gas – 22.4 L kPa – STP 101.3 kPa 273 K R = PV = (101.3 kPa ) (22.4L) = 8.31 L· kPa Tn (273 K) (1 mol) mol · K

6. Ideal Gas Law PV = nRT In your Journal, create a 3 column Chart VariableDefinitionUnits

7. Example : A rigid cylinder has a volume of 10 L and is filled with O 2 gas to a final pressure of 900 kPa at 28°C. How many moles of oxygen gas does the cylinder contain? PV = nRT Fill in all of the given values: Rearrange to solve for n:

8. Example : A rigid cylinder has a volume of 10 L and is filled with O 2 gas to a final pressure of 900 kPa at 28°C. How many moles of oxygen gas does the cylinder contain? PV = nRT Fill in all of the given values: (900 kPa)(10L) = n (8.31 L·kPA/mol· K)(28° +273) Rearrange to solve for n:

9. Example : A rigid cylinder has a volume of 10 L and is filled with O 2 gas to a final pressure of 900 kPa at 28°C. How many moles of oxygen gas does the cylinder contain? PV = nRT Fill in all of the given values: (900 kPa)(10L) = n (8.31 L·kPA/mol· K)(28° +273) Rearrange to solve for n: (900 kPa)(10L) = n n = 3.599 (8.31 L·kPA/mol· K)(28° +273)

10. Your Turn: When the temperature of a rigid chamber containing 700 L of helium gas is held at 500 K, the pressure of the gas is 1.65 x 10 3 kPa. How many moles of helium does the chamber contain? P V= n R T

11. Your Turn: When the temperature of a rigid chamber containing 700 L of helium gas is held at 500 K, the pressure of the gas is 1.65 x 10 3 kPa. How many moles of helium does the chamber contain? P V= n R T (1.65 x 10 3 kPa)(700 L) = n (8.31) (500 K) (1.65 x 10 3 kPa)(700 L) = n n = 277.98 (8.31) (500 K)

12. What pressure will be exerted by 0.65 mol of a gas at 30° if it is contained in a 0.75 L cylinder? PV = nRT

13. What pressure will be exerted by 0.65 mol of a gas at 30° if it is contained in a 0.75 L cylinder? PV = nRT P (0.75 L) = (0.65 mol)(8.31)(30+273) P = (0.65 mol)(8.31)(30+273) (0.75 L) P = 2182.2