Gas Laws and Relationships between P, V, and T Boyle’s Law Charles’s Law Gay-Lusaac’s Law How to use each

Gases and their Variables Four Variables when exploring gases: P, V, T, and n P: Pressure in atm’s V: Volume in L T: Temperature in K n: quantity of matter in moles To solicit information about student prior knowledge about gases and the variables that affect gas behavior, I will ask these questions. 1.Does anyone know the relationships between P,V, and T? 2.Why do tire manufacturers tell you to check the tire pressure before traveling long distances? 3.How does a hot air balloon operate? How do they relate to one another? Let’s put on our PVT Cards and find out!

Let’s get out our PVT cards as we go through each gas law to visualize the relationships between the variables. We will explore three gas laws 1.Boyle’s 2. Charles’s 3. Gay-Lussac’s Each law relates two variables and holds one constant and can best be expressed using mathmatical expressions. Gas Properties can be modeled using MATH! And again, Scientists have done all of the work for us!

Boyle’s Law States: The volume of a sample of gas is inversely proportional to its pressure, if temperature remains constant. Translation: At constant temperature and n, 1 α P V Robert Boyle EX. Volume Decreases, Pressure Increases Volume Increases, Pressure Decreases You Probably started experimenting with Boyle's Law when you were a small child. When you squeeze a balloon, you might notice that the harder you push, the harder it seems to push back. When you lie back on an inflatable mattress, or pool float, it compresses up to a point and then seems to stop. This is because as you decrease the volume of a confined gas, the pressure that it exerts increases. So- in terms of our PVT cards, describe Boyle’s Law in terms of the relationship between pressure and volume. Which variable is held constant? Temperature- so cover up temperature as it is not being manipulated in this relationship. Then, you only have left P and V to deal with. Therefore, as pressure increases then the volume decreases. * Think of a ruler with flags. As one end increases, the other end decreases.* In an inverse relationship, the product of the two quantities is a constant. P1 x V1 = P2 x V2 P1 V1 = P2 V2

Boyle’s Law How do we graph inverse relationships on x and y axis? Inverse relationships are plotted as hyperbolas or curves

Then PLUG AND CHUG! Let’s work it together. Example 1 - A sample of gas collected in a 350 cm3 container exerts a pressure of 103 kPa. What would be the volume of this gas at 150 kPa of pressure? (Assume that the temperature remains constant.) Solving: If temperature remains constant - use Boyle’s Law. Write the original formula: P1V1 = P2V2 Then list what is given and what is unknown. P1 = 103 kPa V1= 350 cm3 P2 = 150 kPa V2 = ? Then PLUG AND CHUG! Let’s work it together. 103 kPa (350 cm3) = 150 kPa V2 150 kPa 150 kPa 240 cm3= V2

Charles’s Law States: The volume of a sample of gas is directly proportional to its Kelvin temperature, if pressure remains constant. Translation: At constant pressure and n, V α T **Temperature ALWAYS in Kelvin** Jacques Charles EX. Volume Increases, Temperature Increases Volume Decreases, Temperature Decreases In an direct relationship, the quotient of the two quantities is a constant. V1 / T1 = V2 / T2 V1 = V2 T1 T2 Gases expand as they are heated and they contract when they are cooled. So- in terms of our PVT cards, describe Charles’s Law in terms of the relationship between pressure and volume. Which variable is held constant? Pressure- so cover up pressure as it is not being manipulated in this relationship. Then, you only have left V and T to deal with. Therefore, as volume in/decreases then the temperature in/decreases. What would you expect to observe from each balloon in the following scenario: Two balloons of equal size filled with the same volume of gas. Balloon A is put into an open container, covered with room temperature water, and immersed in an ice bath. Balloon B is put into an open container, covered with room temperature water, and warmed on a hot plate. Assume constant pressure as the containers are the same size open to the atmosphere. When Balloon A is cooled, volume decreases (temperature slows moving particles, decreases volume) thus balloon shrinks. When Balloon B is warmed, volume increases (temperature excites moving particles, increases volume) thus balloon expands even further.

Charles’s Law What is a directly proportional relationship? Graphing directly proportional relationships gives us a straight line graph To see this in action, lets check out our demonstration. What’s going on inside the balloons that causes this behavior. As Temp decreases, volume decreases as the kinetic energy of the particles decreases. Opposite is true.

Write the original formula:V1 = V2 T1 T2 Ex. If a gas occupies 733 cm3 at 10.0 oC, at what temperature will it occupy 950 cm3? Assume that pressure remains constant. Solving: If pressure remains constant - use Charles’s Law. Write the original formula:V1 = V2 T1 T2 Then list what is given and what is unknown. V1 = 733 cm3 T1= 10.0 oC V2 = 950 cm3 T2 = ? Then PLUG AND CHUG! Let’s work it together. First convert oC to Kelvin: K = 10.0 oC + 273 = 283 K 733 cm3 = 950 cm3 283 K T2 366.7 or 370 K= T2

How does a hot air balloon work? Q: Predict how a hot air balloon operates. Warm air is introduced to the balloon at a constant pressure thus Charles's Law holds true. The temperature change or warm air causes the density of the air inside the balloon to become less dense than the air outside the balloon. Warm air is less dense than cold air therefore the balloon rises upward. The density of air inside the top of the balloon decreases and the balloon rises upward. To descend, the pilot allows the less dense air to escape, and then balloon descends. Warmer air rises in cooler air. Essentially, hot air is lighter than cool air. Modern hot air balloons heat the air by burning propane Hot air balloons also have a cord to open the parachute valve at the top of the envelope. When the pilot pulls the attached cord, some hot air can escape from the envelope, decreasing the inner air temperature. This causes the balloon to slow its ascent Two controls… heat to make the balloon rise and venting to make it sink. TIP: Think about Charles’s Law.

Gay-Lussac’s Law States: The Kelvin temperature of a sample of gas is directly proportional to pressure, if volume remains constant. Translation: At constant volume and n, T α P **Temperature ALWAYS in Kelvin** Joseph Gay-Lussac EX. Temperature Increases, Pressure Increases Temperature Decreases, Pressure Decreases So- in terms of our PVT cards, describe Gay-Lussac’s Law in terms of the relationship between pressure and temperature. Which variable is held constant? Volume- so cover up volume as it is not being manipulated in this relationship. Then, you only have left P and T to deal with. Therefore, as pressure in/decreases then the temperature in/decreases. Q.Why do car manufacturers suggest checking tire pressure before driving distances? Increase in temperature increases the pressure inside the tires – could rupture while driving. How does a pressure cooker pot work? Why would cooks use this tool? Pressure cookers trap steam and increase the temperature of that steam. Increasing the temperature also increases the pressure (according to Gay-Lussac’s Law) and cooks food faster due to the higher temperatures. In an direct relationship, the quotient of the two quantities is a constant. P1 / T1 = P2 / T2 P1 = P2 T1 T2

Gay-Lussac’s Law Q: Aerosol cans have warnings on the cans that say not to store above certain temperatures or incinerate. Why? Because, as Gay-Lussac discovered, when the temperature of a fixed volume of gas (aerosol can) increases or is heated (in a fire, through a window), the pressure inside the container also increases proportionally. Therefore, if you put an aerosol can on a fire, the increased pressure causes the can to explode. (a) A pressurized can of hair spray or deodorant has a constant volume. Pressure increases as temperature increases. Heating the can could cause it to explode or rupture. (b) A can of food must be punctured or opened before (c) heating directly over a campfire

Then PLUG AND CHUG! Let’s work it together. EX. If a gas is cooled from 323.0 K to 273.15 K and the volume is kept constant what final pressure would result if the original pressure was 750.0 mm Hg? Solving: If volume remains constant - use Gay-Lussac’s Law. Write the original formula:P1 = P2 T1 T2 Then list what is given and what is unknown. P1 = 750.0 mm Hg T1= 323.0 K P2 = ? T2 = 273.15 K Then PLUG AND CHUG! Let’s work it together. 750.0 mm Hg = P2 323.0 K 273.15 K 634.2 mm Hg= P2

ONE MORE FUN FACT! Standard Temperature and Pressure (STP) At STP: Temperature = 273 K or 0 oC Pressure = 1 atm = 760 mm Hg Because the volume of a gas varies with the temperature and pressure of the gas, a standard set of conditions is necessary to use as a reference point to compare gases.

Let’s Practice! 1. A sample of neon has a volume of 239 cm3 at 2.00 atm of pressure. What would the pressure have to be in order for the gas to have a volume of 5.00 x 102 cm3? 2. A 30.0 L sample of nitrogen inside a rigid, metal container at 20.0 °C is placed inside an oven whose temperature is 50.0 °C. The pressure inside the container at 20.0 °C was at 3.00 atm. What is the pressure of the nitrogen after its temperature is increased? 3. A sample of gas at 3.00 x 103 mm Hg inside a steel tank is cooled from 500.0 °C to 0.00 °C. What is the final pressure of the gas in the steel tank? HOMEWORK = Practice problems with each law! FINISH LAB TO TURN IN AT END OF CLASS! Allow students to work out problems with guided help on their notes then show how to set up each problem via overhead transparency.