Section 4.1 pg. 150 – 160
Homework: Boyle’s Law: pg. 152 # 6-9 Charles’ Law: pg. 156 # 14-17 Combined Gas Law: pg. 159 # 20-23 They are based on the temperature, pressure and volume relationships that all gases have in common
1. Boyle’s Law P1V1 = P2V2
2. Charles’ Law V1 = V2 T1 = T2
3. Combined Gas Law P1V1 =P2V2 T1 = T2 Boyle’s Law
Anglo- Irish chemist Robert Boyle (1627-1691) was a founding member of the Royal Society of London. Considers the effects of pressure on the volume of a gas only while temperature is held constant
This is an inverse relationship:
P = V P = V
Boyle’s Law states: “as the pressure on a gas increases, the volume of the gas decreases proportionally if temperature and mass are constant”
Boyle’s Equation: P1V1 = P2V2 “As the pressure on a gas increases, the volume of the gas decreases proportionally if temperature and mass are constant” Pressure Volume PV (kPa) (L) (kPa•L) 100 3.00 300 200 1.52 304 300 1.01 303 400 0.74 296 500 0.60 300 Cartesian Diver Demo ◦ “The Cartesian diver is named after the French philosopher, Rene Descartes (1596-1650), and is a very old experiment. The volume of a gas decreases as the pressure on the gas increases. As you squeeze the bottle, the pressure is transferred from your hand to the water and from the water to the air trapped inside the diver. As the volume of air in the diver gets smaller, more water enters the diver, making it heavier and less buoyant, and the diver sinks to the bottom. As the pressure is released, the air inside the diver expands and increases the buoyancy so that the diver rises.”
Expanding Marshmallow Demo ◦ “The marshmallow expands as the volume in the syringe increases and the pressure decreases. It shrinks as the volume is reduced and the pressure is increased” 1. A sample of gas at 1.0 atm is in a 1.0 L container. What is the pressure when the volume is changed to 2.0L? (Assume T and m are constant)
P1V1 = P2V2 (1.0 atm) (1.0 L) = (x) (2.0 L) 1.0 atm•L = 2.0 L • (x) 1.0 atm•L = 2.0 L • (x) 2.0L 2.0L
0.5 atm = P2
According to Boyle’s Law, the gas would now have a pressure of 0.5 atm 2. A 2.0 L party balloon at 98 kPa is taken to the top of a mountain where the pressure is 75 kPa. Assume that the temperature and mass of the gas remain the same. What is the new volume of the balloon.
P1V1 = P2V2 (98 kPa) (2.0 L) = (75 kPa) (x) 196 kPa•L = 75 kPa • (x) 196 kPa•L = 75 kPa • (x) 75 kPa 75 kPa
2.6 L = V2
According to Boyle’s Law, the balloon would have a new volume of 2.6L Pg. 152 #6-9 Charles’ Law
Jacque Charles (1746- 1823)
He made the first flight of a hydrogen balloon on August 27, 1783. This balloon was destroyed by terrified peasants when it landed outside of Paris. Shows the relationship between temperature (must be in Kelvin) and volume of gas if pressure and mass are constant
This is a direct relationship: T = V T = V
Charles’ Law states: “as the temperature of a gas increases, the volume increases proportionally, provided that the pressure and mass remain constant”
Charles’ Equation: V1 = V2
T1 = T2 “As the temperature of a gas increases, the volume increases proportionally, provided that the pressure and mass remain constant” When the graphs of several careful volume-temperature experiments are extrapolated, all the lines meet at absolute zero, 0K or -273° C Ivory Soap Demo ◦ “It demonstrates Charles' Law, which states the volume of a gas increases with its temperature. The microwaves impart energy into the soap, water, and air molecules, causing them to move faster and further away from each other. The result is that the soap puffs up. Other brands of soap don't contain as much whipped air and simply melt in the microwave.” 1. A gas inside a cylinder with a movable piston is heated to 315°C. The initial volume of gas in the cylinder is 0.30 L at 25°C. What will be the final volume when the temperature is 315°C?
T1 = (25 + 273) = 298K V1 = V2
T2 = (315 + 273) = 588K T1 = T2 * Remember temperature
has to be in Kelvin V2 = V1 T2 T1 = (0.30 L)(588K) According to Charles’ Law, 298K the final volume will be 0.59 L = 0.59 L pg. 156 #14 - 17 The Combined Gas Law
You can get Boyle’s Law back by assuming temperature is constant.
You can get Charles’ Combined Gas Law Law back by assuming pressure is constant When Boyle’s and Charles’ laws are combined, the resulting combined gas law produces a relationship among the volume, temperature, and pressure of any fixed mass of gas.
The combined gas law is a useful starting point for all cases with gases, even if one of the variables is a constant. ◦ A variable that is constant can easily be eliminated from the combined gas law equation, reducing it back to Boyle’s or Charles’ Law 1. A gas cylinder with a fixed volume contains a gas at a pressure of 652 kPa and a temperature of 25°C. If the cylinder is heated to 150°C, use the combined gas law to calculate the new pressure.
P V = P V Because volume is constant we can 1 1 2 2 cancel V1 and V2 because V1 = V2 T1 = T2
T1 = (25 + 273) = 298 K P2 = P1T2 T2 = (150 + 273) = 423 K T1
P2 = (625kPa) (423K) The gas will have a new 298 K pressure of 925 kPa P2 = 925 kPa 2. A balloon containing helium gas at 20 °C and a pressure of 100 kPa has a volume of 7.50 L. Calculate the volume of the balloon after it rises 10 km into the upper atmosphere, where the temperature is –36 °C and the outside air pressure is 28 kPa. (Assume that no gas escapes and that the balloon is free to expand so that the gas pressure within it remains equal to the air pressure outside.)
P1V1 = P2V2 T1 = (20 + 273) = 293 K T1 = T2 T2 = (-36 + 273) = 237 K V2 = P1V1T2 T1P2
According to the Combined V2 = (100kPa)(7.50 L)(237 K) Gas law, the volume of the (293 K)(28 kPa) balloon in the upper atmosphere will be 22L V2 = 21.6 L = 22 L STP: 0 °C and 101.325 kPa (exact values) SATP: 25 °C and 100 kPa (exact values) 101.325 kPa = 1 atm = 760 mm Hg (exact values) absolute zero = 0 K or –273.15 °C K = (°C) + 273 (for calculation)
Boyle’s Law P1V1 = P2V2
Charles’ Law V1 = V2 T1 = T2
Combined Gas Law P1V1 = P2V2
T1 = T2 Pg. 159 # 20 - 23 Boyle’s Law: pg. 152 # 6-9 Charles’ Law: pg. 156 # 14-17 Combined Gas Law: pg. 159 # 20 – 23 Gas Laws Worksheet