2 Gas Law FoldableFold the left and right to the middle.
3 Dalton’s Law of Partial Pressure The pressure of a mixture of gases is equal to the sum of the partial pressures of the component gases.PTotal = P1 + P2 + P3….
4 A balloon is filled with air (O2, CO2, & N2) at a pressure of 1.3 atm. ExampleA balloon is filled with air (O2, CO2, & N2) at a pressure of 1.3 atm.If PO2 = 0.4 atm and PCO2 = 0.3 atm, what is the partial pressure of the nitrogen gas?
8 Pressure and Volume with constant Temperature Changing the volume of the container changes the amount of space between the particles. The less space, the more the particles collide with each other and the walls.
9 Uses of Boyle’s LawTesting materials stability and ability to maintain their shape under force.Compressing gases for use in cooking cylinders, SCUBA tanks, and shaving cream.Used to describe density relationships between gases.
10 ExampleIf a gas expands from a volume of 5 L to 25 L at an initial pressure of 3.5 atm, what will be its new pressure?P1 = 3.5 atmV1= 5 LP2 = ?V2 = 25 L
11 Charles’ Law: Volume vs Temperature At constant pressure, the volume of a fixed mass of gas varies directly with the Kelvin temperatureV1 = V2T1 T2*use if pressure is constant
23 The Combined Gas LawExpresses a relationship between pressure, volume, and temperature (and moles) of a fixed amount of gas.It takes all three gas laws: (Boyle’s, Charles’s, & Avogadro’s) and combines them form one usable equation.
24 Example 1A gas is cooled from 45˚C to 20˚C. The pressure changes from 103 kPa to kPa as the volumes settles to 16.0 L. What was the initial volume?P1 = 103 kPaV1 = ?T1 = 45oC = 318 KP2 = kPaV2 = 16.0 LT2 = 20oC = 293 K
26 Example 2 (on back of foldable) A 1.5 mole sample of methane was originally L at 25˚C at 1.1 atm. If we decreased the volume of the container to 0.25 L, increased the pressure to 2.0 atm and added 2.5 moles, what would be the new temperature in ˚C?P1 = 1.1 atmV1 = 0.5 LT1 = = 298 Kn1 = 1.5 molesP2 = 2.0 atmV2 = 0.25 LT2 = ?n2 = = 4 moles
27 P1 V1 = P2 V2 n1T1 n2T2 P1 V1n2T2 = P2 V2n1T1 P1 V1n2 P1 V1n2 *** easier if you solve for T2 first THEN plug in the #’s ***P1 V1n2T2 = P2 V2n1T1P1 V1n P1 V1n2= (2.0 atm)(0.25 L)(1.5 mol)(298K)(1.1 atm)(0.5 L)(4.0 mol)T2 = K – 273= ˚C