Gas Laws Gases

Kelvin Temperature Scale Used in gas law equations 0 K is referred to as absolute zero – the point where all particle motion stops. Conversions: K = 0C + 273 0C = K – 273 Convert 230 C to Kelvin 273 + 23 = 296 K

The boiling point of water is 1000 C The boiling point of water is 1000 C. What is the boiling point in degrees Kelvin? 100 + 273 = 373 K

Variables Affecting Gas Behavior Number of gas particles Temperature Pressure Volume

STP Standard Temperature and Pressure Temp: 0 degrees Celsius or 273 degrees K Pressure: 1 atm Equals 760 mm Hg, 101.3 kPa, 14.7 psi

Boyle’s Law Volume of a given amount of gas at a constant temperature varies inversely with the pressure When one goes up, the other goes down (P1)(V1) = (P2)(V2)

Boyle’s Law in action

Example P1 = 210 kPa V1 = 4.0L P2 = X V2 = 2.5L A sample of He in a balloon is compressed from 4.0L to 2.5L at a constant temp. If the pressure of the gas is initially 210 kPa, what will the pressure be at 2.5L? P1 = 210 kPa V1 = 4.0L P2 = X V2 = 2.5L (210)(4) = (X)(2.5) X = 336 kPa

Example Air trapped in a cylinder with piston occupies 145.7 mL at 1.08 atm. What is the new volume of air when the pressure is increased to 1.43 atm by applying force to the piston? P1 = 1.08atm V1 = 145.7 mL P2 = 1.43atm V2 = X (1.08)(145.7) = (1.43)(X) X = 110.0 mL

Charles’s Law Volume of a given mass of gas is directly proportional to its temp (K) at constant pressure When one goes up, the other goes up V1/T1 = V2/T2

Charles Law in Action

Example A gas at 40.00C occupies a volume of 2.32L at constant pressure. If the temp is raised to 75.00C, what will the volume be? V1 = 2.32L T1 = 40+273 = 313K V2 = X T2 = 75+273 = 348K (2.32)/(313) = (X)/(348) X = 2.6L

Example A volume of air in a balloon occupies .620L at 250C and constant pressure. If the volume of the balloon decreases to .570L, what was the temp change in the room? V1 = .620L T1 = 25+273 = 298K V2 = .570L T2 = X (.620)/(298) = (.570)/(X) X = 273K

Lussac’s Law P1/T1 = P2/T2 As temperature goes up, pressure goes up Pressure of a given mass of gas varies directly with the temp (K) when volume stays constant When one goes up, the other does too P1/T1 = P2/T2 As temperature goes up, pressure goes up

Example The pressure of a gas in a tank is 3.20 atm at 22.00C. If the temp rises to 60.00C, what will be the gas pressure in the tank? P1 = 3.20 atm T1 = 295K P2 = X T2 = 333K (3.20)/(295) = (X)/(333) X = 3.6 atm

Combined Gas Law States relationship between P, V, T for a fixed amount of gas P is inversely related to V, and directly related to T V is directly proportional to T P1V1 = P2V2 T1 T2

Example A gas at 110kPa and 30.00C fills a container with an initial volume of 2.00L. If the temp is raised to 80.00C and the pressure to 400kPa, what is the new volume? P1 = 110kPa V1 = 2.0L T1 = 303K P2 = 400kPa V2 = X T2 = 353K (110)(2)/(303) = (400)(X)/(353) X = .58L

Example At 00C and 1.00 atm, a sample of gas occupies 30.0mL. If the temp is increased to 30.00C, and the entire sample transferred to a 20.0mL container, what will be the pressure inside the container? P1 = 1 atm V1 = 30.0mL T1 = 273K P2 = X V2 = 20.0mL T2 = 303K (1)(30)/(273) = (X)(20)/(303) X = 1.66 atm

Avogadro’s Principle Equal volumes of gas at the same temp and pressure contain equal numbers of particles How many gas particles are in .166mol He? .166 mol (6.02 x 1023 particles/mol) = 9.99 x 1022 How many gas particles are in .166mol Ne?

Molar Volume The volume that 1mol of gas occupies at STP (00C and 1atm) 1 mol ANY gas at STP = 22.4L 1 mol of Ar 1 mol of He 1 mol of Xe

Example Calculate the volume that 2000g of methane (CH4) will occupy at STP. 2000g CH4 x 1mol CH4 x 22.4L CH4 = 2792.7L CH4 16.042g CH4 1 mol CH4

Example Calculate the volume that .881mol Ar will occupy at STP. .881 mol Ar x 22.4 L = 19.7 L Ar 1mol Ar

Ideal Gas Law Describes the behavior of an ideal gas in terms of P, V, T, and # moles (n) of gas present Ideal Gas One whose particles take up no space and have no intermolecular attractive forces. Follows gas laws under all conditions of T and P. Real Gas Do not always follow all of the gas laws

Ideal Gas Law Ideal Gas Constant (R) – experimentally determined constant that depends on the units of pressure PV = nRT P units may vary V units are L n = moles T unit is K

Ideal Gas Constant Units of R Numerical value of R Units of Pressure L • atm mol • 0K 0.0821 atm L • kPa 8.314 kPa L • mmHg 62.4 mm Hg

Example Calculate the number of moles of gas contained in a 3.0L vessel at 300.0K with a pressure of 1.50atm. P=1.50atm V=3.0L n=x R=.0821 T=300 (1.50)(3) = n(.0821)(300) n = .18 mol

Example Determine the Celsius temp of 2.59mol of gas contained in a 1.00L vessel at a pressure of 143kPa. P=143kPa V=1L n=2.59 R=8.314 T=x (143)(1) = (2.59)(8.314)(x) X = 6.64K 6.64K-273 = -2660C

Gas Stoichiometry CH4(g) + 2O2(g)  CO2(g) + 2H2O(g) 1 mol = 1 volume 1 vol CH4 2 vol O2 1 vol CO2 2 vol O2 1 vol CO2 2 vol H2O

Example CH4(g) + 2O2(g)  CO2(g) + 2H2O(g) If you have 6.7L of CH4, how much water vapor can be produced? 6.7L CH4 x 2 vol H2O = 13.4L H2O 1 vol CH4

Example CH4(g) + 2O2(g)  CO2(g) + 2H2O(g) Calculate the volume of oxygen needed to completely react with 74.3g CH4 74.3g CH4 x 1mol CH4 x 2 mol O2 x 22.4L O2 = 207.5L O2 16.042g CH4 1mol CH4 1 mol O2