Unit 9: The Gas Laws Chapters 13 and 14
Demo – Vacuum Pump • Can water boil at room temperature? Why/why not? • Water boils when vapor pressure = atmospheric pressure • Yes! If we lower the air/atmospheric pressure below the vapor pressure, the molecules don’t have to move as fast, bringing water to a boil at a much lower temp. • Can we inflate a balloon by taking out air in its surrounding? Why/why not? • Yes! In a fixed container, if you vacuum out air, the size of the balloon expands to make up for the missing air.
Recall….Kinetic Molecular Theory • Gases consist of tiny molecules • Gases are in constantrandom motion • Gases collide into each other and the walls of the container; this creates Gas Pressure • Collisions are elastic – KE is transferred • As the temperature increases so does the KE; gas molecules move faster!
Air Pressure • Force applied to an area • Pressure = force area • If force increases, pressure ? • If area increases, pressure ? Increases. Decreases.
Compressibility • Gases are easily compressed because of the large space between the particles in a gas. • Under pressure (added force), the particles in a gas are forced closer together, decreasing the volume.
Factors Affecting Gas Pressure • Amount of Gas – increasing # of particles will increase the pressure (and vv)
Factors Affecting Gas Pressure • Volume – • increasing the volume in which the gases can move will decrease the pressure (more room to move) • Decreasing the volume in which gases can move will increase the pressure.
Factors Affecting Gas Pressure • Temperature • Increasing the temperature will increase the pressure (more KE, more collisions with container) • Decreasing the temperature will decrease the pressure (less KE, less collisions with container)
The Gas Laws - Overview • Boyle’s Law – Pressure and Volume • Charles’ Law – Temperature and Volume • Gay – Lussac’s Law – Temperature and Pressure
Boyle’s Law • Relates Pressure (P) and Volume (V) • Measures the pressure and volume under one set of conditions (P1 and V1) and then changes the conditions (P2 and V2) • It is an inverse relationship meaning: • When the volume decreases, the pressure increases • When the volume increases, the pressure decreases. • Equation: P1V1 = P2V2
The Relationship between volume and pressure
Boyle Example #1 A balloon initially occupies 12.4 L at 1.00 atm. What will be the volume at 0.800 atm? 1. Write the equation: P1V1 = P2V2 • List the variables: V1 = 12.4 L P1 = 1.00 atm V2 = ? P2 = 0.800 atm • Plug in the numbers and solve! (1.00 atm) (12.4 L) = (0.800 atm) V2 V2 = 12.4 atm· L 0.800 atm = 15.5 L
Boyle Example #2 5.00 L of a gas is at 1.08 atm. What pressure is obtained when the volume is 10.0 L? 1. Write the equation: P1V1 = P2V2 2. List the variables: V1 = 5.00 L P1 = 1.08 atm V2 = 10.0 L P2 = ? 3. Plug in the numbers and solve! • (1.08 atm) (5.00 L) =(P2) (10.0 L) P2 = 0.54 atm (don’t forget the units!)
Hints • Know the units for Pressure • Know the units for Volume • Keep all Units • Standard Pressure = 1 atm = 760 mm Hg • BIG Hint: Peter V. Boyle likes to “Play” on “Vacation”!
Using the KMT… • Use the kinetic theory of gases to answer this question. If the volume decreases, why does the pressure increase? - Smaller volume means more collisions. More collisions will increase the pressure.
Charles’s Law • Relates Temperature (T) and Volume (V) • Temperature must be measured in Kelvin • Kelvin = °C + 273 • Equation: T1V2 = T2V1 notice the numbers!
Charles’s Law • Direct Relationship – as one variable goes up, the other goes UP • Temperature increased, the volume will increase and v.v.
Charles Example #1 Given 17 L of a gas at 358 K. What is its volume at 283 K? – work through the problem on your own first before proceeding 1. Write the equation: T1V2 = T2V1 2. List the variables: T1 = 358 Kelvin V1 = 17 L T2 = 283 Kelvin V2 = ? 3. Plug in the numbers and solve! T1V2 = T2V1 (358K)V2 = (283K)(17L) V2 = (283K)(17L) 358K V2 = 13.4 L
Given 200 mL of a gas at 364 K. What is the temperature when the volume is increased to 500 mL? work through the problem on your own first Write the equation: T1V2 = T2V1 2. List the variables: T1 = 364 Kelvin V1 = 200 mL T2 = ? V2 = 500 mL 3. Plug in the numbers and solve! T1V2 = T2V1 (364 K)(500 mL) = T2(200 mL) T2 = (364 K) (500 mL) 200 mL T2 = 910 K In degrees Celsius: 910 K – 273 = 637°C = T2 Charles’ Example #2
Hints • Temperature must be in Kelvin • Recall: K = °C + 273 • Volume is measured in: • L or mL 1L = 1000 mL • Always keep your units in the problem and of course the answer • Hint: “Charles in Charge” was on T.V.!
Using the KMT… • Use the kinetic theory of gases to answer this question. If the temperature increases, why does the volume also increase? – think and write your answer The particles are moving faster as the temperature increases and hit the walls of the container with more force causing the volume to increase
Gay-Lussac’s Law • Relates Temperature (T) and Pressure (P) • Temperature must be measured in Kelvin • Kelvin = °C + 273 • T1P2 = T2P1 - notice the numbers
Gay-Lussac’s Law • Direct Relationship – as one variable goes up, the other goes UP • Temperature increased, the pressure will increase • Draw a sample graph of pressure and temperature.
Example Problems A container filled with a gas at 1.00 atm at 273 K. What will be the new pressure if the temperature increases to 298 K? work through the problem on your own before proceeding • Write the equation: T1P2 = T2P1 • List the variables: T1 = 273 K P1 = 1.00 atm T2 = 298 K P2 = ? 3. Plug in the numbers and solve! T1P2 = T2P1 (273 K)P2 = (298 K)(1.00 atm) P2 = 1.09 atm
A gas has a pressure of 535 mm Hg at 40.0 °C. What is the temperature in Kelvin and degrees Celsius at 760 mm Hg? work through the problem on your own before proceeding 1. Write the equation: T1P2 = T2P1 2. List the variables: T1 = 40°C + 273 = 313 K P1 = 535 mm Hg P2 = 760 mm Hg T2 = ? in K and °C 3. Plug in the numbers and solve! T1P2 = T2P1 (313 K)(760 mm Hg) = T2(535 mm Hg) T2 = 445 K or °C = 444 – 273 = 171°C Example
Hints • Temperature must be in Kelvin • Always keep your units • Read the problem • Standard temperature is 273 K • Standard pressure is 1 atm = 760 mm Hg • Hint: Good Landscaping make Pretty Tulips!
Using the KMT… • Use the kinetic theory of gases to answer this question. • If the temperature decreases, why does the pressure also decrease? When the temperature decreases the particles are moving with less kinetic energy (slower). The collisions are less frequent and with less force, so the pressure decreases too!
Ideal Gas Law • This gas law is the only law that considers the number of gas molecules (in units of moles)
Formula • PV = nRT P = Pressure in atmospheres V = Volume in Liters n = number of moles R = Ideal Gas Constant 0.0821 L x atm K x mol T = Temperature in Kelvin
Example #1 At what pressure would 0.150 mole of nitrogen gas at 23.0 °C occupy 8.90 L? • Write the equation: PV = nRT 2. List the variables: P = ? V = 8.90 L n = 0.150 mol R = 0.0821 L*atm/K*mol T = 23 + 273 = 296 K 3. Plug in the numbers and solve! PV = nRT or P = nRT V P = (0.150 mol)(.0821L*atm/K*mol)(296 K) = 0.41 atm 8.90L
Example #2 How many moles of gas are contained in a 20.0 L cylinder at a pressure of 100.0 atm and a temperature of 35.0 °C? • Write the equation: PV = nRT 2. List the variables: P = 100.0 atm V = 20.0 L n = ? R = R = 0.0821 L*atm/K*mol T = 35.0 + 273 =308 K 3. Plug in the numbers and solve! PV = nRT or n = PV RT n = (100.0 atm) (20.0 L) = 79.1 mol (0.0821L*atm/K*mol) (308 K)
Hints • Rearrange the equation before you plug in your variables. Ex: Solve for Temperature PV = nRT T = P V n R now plug in the variables • Units! Keep them! • Memorize R = 0.0821 L *atm / mol * K
When you are asked to solve for grams… • Sometimes you will be asked to solve for grams. First solve for n = moles, then convert to grams using the molar mass. • Remember Molar Mass – multiply the number of atoms by the mass and add up all the elements in the compound. Ex. What is the molar mass of water? H2O = 2 ( 1.0g) + 1(16.0g) = 18.0 g H2O If you had 4.5 moles of water, how many grams would you have? 4.5 moles x 18.0 g = 81 g. of H2O 1mol