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Gas Laws. Properties of Gases. Particles far apart Particles move freely Indefinite shape Indefinite volume Easily compressed Motion of particles is constant and random. Gas Pressure. Gas pressure is the result of collisions of particles with their container.
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Properties of Gases • Particles far apart • Particles move freely • Indefinite shape • Indefinite volume • Easily compressed • Motion of particles is constant and random
Gas Pressure • Gas pressure is the result of collisions of particles with their container. • More collisions = more pressure • Less collisions = less pressure • Unit = kPa or atm (also torr or mmHg)
Units of Pressure • 1 atm = 101.3 kPa =760 torr = 760 mmHg • 1 atm = 101,325 Pa • 1 atm = 14.70 lb/in2 • 1 bar = 100,000 Pa = 0.9869 atm atm = atmosphere
Amount of Gas • If you add gas, then you increase the number of particles • This will increase the number of collisions… increasing gas pressure • Unit = mole
Volume • Decreasing the volume of a container results in more collisions with the side of the container and therefore an increase in gas pressure • Unit = L
Temperature • If the temp. of a gas increases, then the kinetic energy of the particles increase. • Increasing KE makes the particles move faster. • Faster moving particles hit the sides of the container more and increase gas pressure. • Unit = Kelvin (K) (K = °C + 273)
STP • Standard Temperature and Pressure Standard Temp = 273K Standard Pressure = 1 atm (101.3kPa, 760torr, 760mmHg)
Drill • No drill today • But… • It’s March Madness • So • Move back one seat
Objectives • iWBAT • solve Charles, Gay-Lussac, Combined and Ideal gas law problems.
Gas Laws • Boyle’s Law • Charles’s Law • Gay-Lussac’s Law • Avogadro’s Law • Combined Gas Law • Ideal Gas Law • Dalton’s Law of Partial Pressures
Boyle’s Law • As pressure of a gas increases, the volume decreases (if the temp is constant). • Inverse relationship P1V 1= P2V2
Charles’s Law • As temperature of a gas increases, the volume increases (if pressure is constant). • Direct relationship V 1= V2 T 1 T2
Gay-Lussac’s Law • As temperature of a gas increases, the pressure increases (if volume is constant). • Direct relationship P 1 = P2 T 1 T2
Combined Gas Law P1V1 = P2V2 T 1 T2
Avogadro’s Law • Equal volumes of gases at the same temperature and pressure contain an equal number of particles V 1 = V2 n 1 n2
Dalton’s Law of Partial Pressure • The sum of the partial pressures of all the components in a gas mixture is equal to the total pressure of the gas in a mixture. • In other words…all the individual pressures in a gas mixture add up to the total pressure. Ptotal = P1 + P2 + P3 + …
Dalton’s Law of Partial Pressure is used whenever a gas is obtained over water because there will also be water vapor present. • You will have to look up the pressure of water at a given temperature.
Homework • Pg 367 #1-5 (use Table A8 on pg 859 for #5) • Pg 375 #1-6 (use the Kinetic Molecular Theory assumption #2 and #5 on pg. 329 for #1 & #6 respectively)
Ideal Gas Law • The most important gas law!
Ideal Gas Law • An Ideal Gas does not exist, but the concept is used to model gas behavior • A Real Gas exists, has intermolecular forces and particle volume, and can change states.
Ideal Gas Law PV = nRT P = Pressure (atm) V = Volume (L) n = # of particles (mol) T = Temperature (K) R = Ideal gas constant 8.31 (kPa∙L) or 0.0821 (atm∙L) (mol∙K) (mol∙K)
At what temperature would 4.0 moles of hydrogen gas in a 100 liter container exert a pressure of 1.00 atm? Ideal Gas Law PV = nRT T = PV/nR = (1.00atm)(100L) (4.0mol)(.0821atm∙L/mol∙K) = 304.5 K 300K Use Ideal Gas Law when you don’t have more than one of any variable
Extensions! • PV = nRT n is moles. If we know the chemical formula for the gas we can convert moles to mass or to particles using Dimensional Analysis!
We could also use the fact that: moles = mass or n = m molar mass MM Plugging this in, we have PV = mRT MM This can be rearranged to solve for Density which is m/V m = P∙MM or D = P∙MM V R∙T R∙T
What is the density of water vapor at STP? D = P∙MM D = (1 atm)(18.02g/mol) R∙T (.0821atm∙L)(273K) mol∙K D = 0.804 g/L
Practice • Try problems # 5 & 6 on the Ideal Gas Law WS
Solve it! • What volume does 0.0881 moles of gas occupy at 110.4 kPa and 29°C?
Problem #1 • Oxygen occupies a volume of 66L at 6.0atm. What volume will it occupy at 920kPa?
Problem #2 • At 25°C a gas has a volume of 6.5mL. What volume will the gas have at 50.°C?
Problem #3 Initially you have gas at 640mmHg, 2.5L, and 22°C. What is the new temperature at 750mmHg and 5L?