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

Part 1: What Is a Gas Law? C + 273 = __K 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: We will cover Boyle’s, Charles's and Gay- Lussac’s laws today. We will cover the Combined Gas Law and the Partial Pressures Law tomorrow. 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” Combined Gas Law P1V1 = P2V2 T1 T2 Boyle’s Law P1V1 = P2V2 Charles’s Law V1 = V2 T1 T2 Gay-Lussac Law P1 = P2 T1 T2 Partial Pressures Law PT = P1 + P2 + P3 C + 273 = __K

millimeters of mercury C + 273 = __K 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 (cm3). 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! Unit Abbr. STP value atmospheres atm 1 atm millimeters of mercury mmHg 760 mmHg pounds per square inch psi 14.7 psi kilopascals kPa 101.325 kPa

Steps for Solving ANY Gas Law Problem: 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 P1 and V1 indicate initial (or starting) conditions P2 and V2 indicate new (or final) conditions Steps for Solving ANY Gas Law Problem: Write out a column of information down the left-hand side. Make sure all of your variable’s units match (i.e. if P1 is in kPa, then P2 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). Write the original equation for the gas law you will be using. Rearrange the equation to solve for the variable you need. Boyle’s Law P1V1 = P2V2 P V

Steps for Solving ANY Gas Law Problem: Write out a column of information down the left-hand side. Make sure all of your variable’s units match (i.e. if P1 is in kPa, then P2 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). Write the original equation for the gas law you will be using. Rearrange the equation to solve for the variable you need. 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. Calculate, then box your answer! Ex1: Using 14.3 L of N2 as the initial volume, calculate the volume that would result if the pressure was raised from 150 kPa to 250 kPa.

P1V1 = P2V2 V2 = (150 kPa)(14.3 L) = ____ ____ 250 kPa P2 P2 Ex1: Using 14.3 L of N2 as the initial volume, calculate the volume that would result if the pressure was raised from 150 kPa to 250 kPa. P1 = __________ V1 = __________ P2 = __________ V2 = __________ 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 ? P1V1 = P2V2 ____ ____ P2 P2 V2 = (150 kPa)(14.3 L) = 250 kPa V2 = 150 14.3 ÷ 250 = V2 = P1V1 P2 V2 = 8.58 L Charles’s Law V1 = V2 T1 T2 V T

V1T2 = V2T1 ____ ____ V2 = (5.0 L)(316 K) = T1 T1 310 K 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? V1 = __________ T1 = ____C____K V2 = __________ T2 = ____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? Charles’s Law V1 = V2 T1 T2 5.0 L 37 ? 43 V1T2 = V2T1 ____ ____ T1 T1 V2 = (5.0 L)(316 K) = 310 K 310 316 V2 = 5.0 316 ÷ 310 = V2 = V1T2 T1 V2 = 5.10 L Gay-Lussac Law P1 = P2 T1 T2 P T

P1T2 = P2T1 ____ ____ T1 T1 P2 = (125.3 kPa)(1318 K) = 290 K P2 = P1T2 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? P1 = __________ T1 = ____C____K P2 = __________ T2 = ____C____K Gay-Lussac Law P1 = P2 T1 T2 P1T2 = P2T1 125.3 kpa 17 ? 1045 ____ ____ T1 T1 P2 = (125.3 kPa)(1318 K) = 290 K 290 1318 P2 = P1T2 T1 P2 = 125.3 1318 ÷ 290 = P2 = 569.47 kPa