1.4: Gas Properties, Basic Gas Laws 1/17/17

can expand and contract Part 1: Gas Properties gases have many properties that are very different from solids and liquids: The first three rows in the table above sum up what we covered in our notes on Phases of Matter (1.3). The last two rows are explained next: unlike liquids and solids, gases have what is known as an “indefinite” volume. this means that the volume of a gas is dependent upon more than just how many molecules are present in the sample of gas. Property Solids & Liquids Gases particle spacing particle movement kinetic energy volume exhibit pressure? close together vibrational or fluid low to medium definite amount no very spread out very fast, random high to very high can expand and contract yes

this means that the volume of a gas is dependent upon more than just how many molecules are present in the sample of gas. since gas particles move very quickly and spread out in all directions (a process called diffusion), gases can occupy many different volumes for the same amount of gas particles. They will take the shape and volume of any container they are placed in. the two other variables that determine the volume of a gas are temperature (T) and pressure (P). pressure = the amount of force exerted per unit of area. according to the Kinetic Molecular Theory (KMT), gas particles are continuously moving and colliding with any object in their path, including the sides of their containers. these collisions result in pressure inside containers, such as a balloon. The arrows in the balloon show these forces.

pressure = the amount of force exerted per unit of area. according to the Kinetic Molecular Theory (KMT), gas particles are continuously moving and colliding with any object in their path, including the sides of their containers. these collisions result in pressure inside containers, such as a balloon. The arrows in the balloon show these forces. expandable containers, like balloons, relieve pressure by increasing in size. Fixed-volume containers, such as spray cans, decrease internal pressure as gas molecules are released. pressure is measure in Pascals (Pa), or kilopascals (kPa). gas laws help show the effect of pressure on the volume and temperature of a gas.

millimeters of mercury Part 1I: What Is a Gas Law? the gas laws are simple, mathematical relationships between the pressure (P), volume (V), and temperature (T) of a gas. the 3 basic gas laws: Boyle’s law and Gay-Lussac’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). Charles’s Law V1 = V2 T1 T2 Gay-Lussac Law P1 = P2 T1 T2 Boyle’s Law P1V1 = P2V2 Unit Abbr. atmospheres atm millimeters of mercury mmHg pounds per square inch psi kilopascals kPa

Variables involved Which gas law? Prediction of answer Part III: 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 for these problems, this is what’s important: Boyle’s Law P1V1 = P2V2 P V Variables involved Which gas law? Prediction of answer

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 = __________ Ex2: A 5.6 L weather balloon rises into the atmosphere where it volume increases to 7.9 L. If the initial pressure in the balloon was 13.5 psi, what is the new pressure inside the balloon? 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 P2 = (13.5 psi)(5.6 L) = 7.9 L 13.5 psi 5.6 L ? 7.9 L ____ ____ V2 V2 P1V1 = P2V2 P2 = 13.5 5.6 ÷ 7.9 = P2 = P1V1 V2 P2 = 9.57 psi

V1T2 = V2T1 ____ ____ V2 = (5.0 L)(316 K) = T1 T1 310 K Part IV: 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 Ex3: 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 Charles’s Law V1 = V2 T1 T2 V T 5.0 L 37 ? 43 ____ ____ T1 T1 V1T2 = V2T1 Cross-multiplied version of Charles’s Law (easier to rearrange than original) V2 = (5.0 L)(316 K) = 310 K 310 316 V2 = 5.0 316 ÷ 310 = V2 = V1T2 T1 V2 = 5.10 L

V2 = V1T2 V2 = (4000 L)(379 K) = T1 312 K V2 = 4000 379 ÷ 312 = Ex4: The gas in a 4000 L hot-air balloon is heated from 39C to 106C. What new volume does this hot-air balloon have after heating? V1 = __________ T1 = ____C____K V2 = __________ T2 = ____C____K Part V: 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 Ex5: 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 V1T2 = V2T1 4000 L 39 ? 106 V2 = V1T2 T1 V2 = (4000 L)(379 K) = 312 K 312 379 V2 = 4000 379 ÷ 312 = V2 = 4858.97 L Gay-Lussac Law P1 = P2 T1 T2 P T

P1T2 = P2T1 ____ ____ T1 T1 P2 = (125.3 kPa)(1318 K) = 290 K P2 = P1T2 Ex5: 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 Ex6: What temperature is required to raise the pressure in a helium gas tank from 3200 mmHg to 9800 mmHg if the original temp. of the gas was 28C? P2 = _________ Gay-Lussac Law P1 T2= P2T1 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 3200 mmHg 28 9800 mmHg ? P1T2 = P2T1 T2 = (9800 mmHg)(301 K) = 3200 mmHg ____ ____ P1 P1 301 ? T2 = 9800 301 ÷ 3200 = T2 = P2T1 P1 T2 = 921.81 K = 648.81C