1. April 24, 2013Perseverance Aficionado: a devotee; a fan; an enthusiastic person about a sport or hobby Do Now: Quad Card Topic: Air pressure

2. Basic Gas Laws (Boyle’s, Charles’s & Gay-Lussac’s)

3. What makes a hot air balloon inflate? HINT: look at the name!

4. Part 1: What Is a Gas Law? the gas laws are simple, mathematical relationships between the pressure (P), volume (V), temperature (T), and moles (n), of a gas. Basic gas laws involve P, V, and T only. the 5 basic gas laws: gas laws use the Kelvin temperature scale (K). Why? The Celsius scale (  C) has negative values and a zero value (the Kelvin scale does not). If we used the Celsius scale, we might calculate a zero or negative volume/pressure from it, which can’t exist! to convert a Celsius temp into Kelvin, just add 273 notice that Kelvin temps do not have a degree sign, just a “K”  C + 273 = __K Boyle’s Law P 1 V 1 = P 2 V 2 Boyle’s Law P 1 V 1 = P 2 V 2 Charles’s Law V 1 = V 2 T 1 T 2 Charles’s Law V 1 = V 2 T 1 T 2 Gay-Lussac Law P 1 = P 2 T 1 T 2 Gay-Lussac Law P 1 = P 2 T 1 T 2 Combined Gas Law P 1 V 1 = P 2 V 2 T 1 T 2 Combined Gas Law P 1 V 1 = P 2 V 2 T 1 T 2 Partial Pressures Law P T = P 1 + P 2 + P 3 Partial Pressures Law P T = P 1 + P 2 + P 3

5. to convert a Celsius temp into Kelvin, just add 273 notice that Kelvin temps do not have a degree sign, just a “K” there is a certain temperature that is considered “standard,” as well as a standard pressure. The values for standard temp and pressure (STP) are 273 K and 1 atm. all the gas laws (except Charles’s law) involve pressure. Most people are not familiar with the many units pressure can be measured in (except maybe psi). So here they are: remember: units of volume = milliliters (mL), liters (L), and cubic centimeters (cm 3 ).  C + 273 = __K UnitAbbr.STP value atmospheresatm1 atm millimeters of mercurymmHg760 mmHg pounds per square inchpsi14.7 psi kilopascalskPa101.325 kPa NOTE: the STP values shown here can be used to convert one pressure unit to another, which will need to be done quite a bit in your calculations!

6. Part 2: Boyle’s Law (1662) Boyle’s Law states that the pressure of a fixed mass of gas varies inversely with the volume at a constant temperature. this means if you compare the initial volume and pressure of a gas with the new conditions of the gas, you will get an inverse relationship every time P 1 and V 1 indicate initial (or starting) conditions P 2 and V 2 indicate new (or final) conditions Steps for Solving ANY Gas Law Problem: 1.Write out a column of information down the left-hand side. Make sure all of your variable’s units match (i.e. if P 1 is in kPa, then P 2 must be in kPa as well). If one doesn’t match, convert it to match the other, using a conversion table. Put a question mark in the space for the variable you are trying to solve for (what you DON’T have). 2.Write the original equation for the gas law you will be using. 3.Rearrange the equation to solve for the variable you need. Boyle’s Law P 1 V 1 = P 2 V 2 Boyle’s Law P 1 V 1 = P 2 V 2 P V

7. Steps for Solving ANY Gas Law Problem: 1.Write out a column of information down the left-hand side. Make sure all of your variable’s units match (i.e. if P 1 is in kPa, then P 2 must be in kPa as well). If one doesn’t match, convert it to match the other, using a conversion table. Put a question mark in the space for the variable you are trying to solve for (what you DON’T have). 2.Write the original equation for the gas law you will be using. 3.Rearrange the equation to solve for the variable you need. 4.Plug in the values and units you have in to the rearranged equation, and make sure all your units will cancel except for one. This will be the unit for your answer. 5.Calculate, then box your answer! Ex1: Using 14.3 L of N 2 as the initial volume, calculate the volume that would result if the pressure was raised from 150 kPa to 250 kPa.

8. P 1 = __________ V 1 = __________ P 2 = __________ V 2 = __________ Part 3: Charles’s Law (1787) Charles’s Law states that the volume of a fixed mass of gas varies directly with the temperature at a constant pressure. this means that as the volume of gas increases, so does the temperature Ex2: A sample of gas occupied a volume of 5.0L at a temp of 37.0  C. If the temp were to increase by 6  C, what would be the volume of the gas under this new condition? 150 kPa 14.3 L 250 kPa ? P 1 V 1 = P 2 V 2 ____ P 2 P 2 V 2 = P 1 V 1 P 2 V 2 = (150 kPa)(14.3 L) = 250 kPa V 2 = 8.58 L V T Charles’s Law V 1 = V 2 T 1 T 2 Charles’s Law V 1 = V 2 T 1 T 2 V 2 = 150  14.3 ÷ 250 =

9. Ex2: A sample of gas occupied a volume of 5.0L at a temp of 37.0  C. If the temp were to increase by 6  C, what would be the volume of the gas under this new condition? V 1 = __________ T 1 = ____  C  ____K V 2 = __________ T 2 = ____  C  ____K Part 4: Gay-Lussac’s Law (1802) Gay-Lussac’s Law states that the pressure of a fixed mass of gas varies directly with the temperature at a constant volume. this means that as the pressure of gas increases, so does the temperature Ex3: The gas left in a used aerosol can is at a pressure of 125.3 kPa at 17  C. If the can is thrown into a fire, what will the pressure be inside the can at 1045  C? 5.0 L 37 ? 43 V 1 T 2 = V 2 T 1 ____ T 1 T 1 V 2 = V 1 T 2 T 1 V 2 = (5.0 L)(316 K) = 310 K V 2 = 5.10 L Charles’s Law V 1 = V 2 T 1 T 2 Charles’s Law V 1 = V 2 T 1 T 2 V 2 = 5.0  316 ÷ 310 = 310 316 Gay-Lussac Law P 1 = P 2 T 1 T 2 Gay-Lussac Law P 1 = P 2 T 1 T 2 P T

10. Ex3: The gas left in a used aerosol can is at a pressure of 125.3 kPa at 17  C. If the can is thrown into a fire, what will the pressure be inside the can at 1045  C? P 1 = __________ T 1 = ____  C  ____K P 2 = __________ T 2 = ____  C  ____K Gay-Lussac Law P 1 = P 2 T 1 T 2 Gay-Lussac Law P 1 = P 2 T 1 T 2 125.3 kpa 17 ? 1045 ____ T 1 T 1 290 1318 P 1 T 2 = P 2 T 1 P 2 = P 1 T 2 T 1 P 2 = (125.3 kPa)(1318 K) = 290 K P 2 = 569.47 kPa P 2 = 125.3  1318 ÷ 290 =