Example A 1.00-L sample of dry air at 25 ⁰C contains 0.0319 mol N2, 0.00856 mol O2, 0.000381 mol Ar, and 0.00002 mol CO2. Calculate the partial pressure of N2(g) in the mixture. nN2RT PN2 = ——— V =0.0319 mol (0.08206 L- atm/mol-K) 298 K 1.00 L = 0.780 atm

Partial Pressure = Mole fraction x Total Pressure The mole fraction (x) is the ratio of the number of moles of a given component in a mixture to the total number of moles in the mixture n1 x1 = —— ntotal P1 = —— Ptotal P1 = x1 Ptotal Partial Pressure = Mole fraction x Total Pressure

Ex. The partial pressure of oxygen gas was observed to be 156 torr in air with an atmospheric pressure of 743 torr. Calculate the mole fraction of O2 present. X1 = P1 = 156torr Ptotal 743torr = 2.10x10-1 (no unit)

Ex. The mole fraction of nitrogen in the air is 0. 7808 Ex. The mole fraction of nitrogen in the air is 0.7808. Calculate the partial pressure (in torr) of N2 at STP. .7808 = P1 760 torr P1 = 593torr

Collection of Gases over Water If O2 is being generated … As (essentially insoluble) gas is bubbled into the container for collection, the water is displaced. The gas collected is usually saturated with water vapor. Assuming the gas is saturated with water vapor, the partial pressure of the water vapor is the vapor pressure of the water. Pgas = Ptotal – PH2O(g) = Pbar – PH2O(g) Pbar = barometric pressure … what TWO gases are present here?

Vapor Pressure of Water as a Function of Temp Temp Vapor Pressure of Water as a Function of Temp Temp.(C) Pressure (mmHg) 15 12.8 16 13.6 17 14.5 18 15.5 19 16.5 20 17.5 21 18.7 22 19.8 23 21.1

Example: Hydrogen produced in the following reaction is collected over water at 23⁰C when the barometric pressure is 742 Torr: 2 Al(s) + 6 HCl(aq)  2 AlCl3(aq) + 3 H2(g) What volume of hydrogen gas will be collected in the reaction of 1.50 g Al(s) with excess HCl(aq)? Ptotal = PH2 + PH2O 742 Torr = PH2 + 21.1 Torr PH2 = 721 Torr =.949atm 1.50 g Al x 1 mol Al x 3 mol H2 = 0.0834 mol H2 26.98 g Al 2 mol Al V = nRT =0.0834 mol (0.08206 L-atm/mol-K)(273 + 23 K) P .949 atm = 2.13 L

Gas molecules do not all move at the same speed. From this postulate we expect a distribution of velocities, as seen in the diagram below. Gas molecules do not all move at the same speed. The particle in a gas have a wide range of velocities: some may be nearly still, while others move at great speed. Thus there is a wide range of kinetic energies in any gas, and an expected distribution of velocities is seen below:

stop and add pogil here

Which gases move slower? What might account for this?

Diffusion: The mixing of gases. At a fixed temperature, molecules of higher mass (M) move more slowly than molecules of lower mass. Effusion Describes the passage of a gas through a tiny orifice into an evacuated chamber. Rate of effusion measures the speed at which the gas is transferred into the chamber.

Gases with lower molecular masses will diffuse and/or effuse faster than gases with higher molecular masses.

Effusion Copyright © Cengage Learning. All rights reserved

Diffusion Copyright © Cengage Learning. All rights reserved

SO2, Ne, NH4 SO2 = 64.1 g NH4 = 18.0 g Ne = 20.2 g Concept Check: Arrange the following gases in order of increasing speed of the particles at the same temperature: SO2, NH4, Ne SO2 = 64.1 g SO2, Ne, NH4 NH4 = 18.0 g Ne = 20.2 g

Graham’s Law of Effusion M1 and M2 represent the molar masses of the gases. Copyright © Cengage Learning. All rights reserved

Which gas effuses faster, Helium or Nitrogen? How much faster? Example: Which gas effuses faster, Helium or Nitrogen? How much faster? Helium is 2.6 times faster than N2; Nitrogen is .38 times slower than He

Example: Which gas effuses faster, Oxygen or Carbon Dioxide? How much faster? Oxygen effuses 1.17 times faster than CO2; CO2 effuses .855 times slower than O2

Since Rate is a unit per time (1/t), Graham’s law can be rearranged to solve for time instead of rate assuming the same amount of each gas.

Example: It takes 354 s for 1.00 mL of Xe to effuse through a small hole. Under the same conditions, how long will it take for 1.00 mL of nitrogen to effuse? Solution A: x = 164 s

Example: It takes 354 s for 1.00 mL of Xe to effuse through a small hole. Under the same conditions, how long will it take for 1.00 mL of nitrogen to effuse? Solution B: x = 164 s

Real Gases and van der Waals Equation An ideal gas is a hypothetical concept. No gas exactly follows the ideal gas law. Real gas molecules do have volume, and the molecules of real gases do experience attractive forces. Gases behave most like ideal gases at low pressure, high temperature. Copyright © Cengage Learning. All rights reserved

Real Gases (van der Waals Equation) n2 a (P + —— ) (V-nb) = nRT V2 Corrected pressure Corrected volume ‘a’ and ‘b’ are constants derived from experimental results a has units of L2atm/mol2 b has units of L/mol. Copyright © Cengage Learning. All rights reserved

n2 a (P + —— )(V-nb) = nRT V2 Two corrective Factors: 1. n2a/V2 alters the pressure It accounts for the intermolecular attractive forces between gas molecules A low value for a reflects weak IM forces; a high value reflects stronger IM forces. 2. nb alters the volume It accounts for the volume occupied by the gas molecules The value of b is generally much lower than a The value of b generally increases with the size of the molecule

Example: 1. 00 mol of carbon dioxide gas at 373 K occupies 536 mL Example: 1.00 mol of carbon dioxide gas at 373 K occupies 536 mL. What is the calculated value of the pressure in atm using: (a) Ideal gas equation (b) Van der Waals equation? Van der Waals constants for carbon dioxide: a = 3.61 L2 atm mol-2; b = 0.0428 L mol-1 . Use the ideal gas law equation! PV=nRT (P) (.536L) = (1.00mol)(0.08206)(373K) P = 57.1atm

Use Van der Waals equation! Example: 1.00 mol of carbon dioxide gas at 373 K occupies 536 mL. What is the calculated value of the pressure in atm using: (a) Ideal gas equation (b) Van der Waals equation? Van der Waals constants for carbon dioxide: a = 3.61 L2 atm mol-2; b = 0.0428 L mol-1 . n2 a (P + —— )(V-nb) = nRT V2 Use Van der Waals equation! (1.00)2(3.61) [P + -----------------][0.536–(1.00)(0.0428)]=(1.00)(0.08206)(373) (.536)2 (P + 12.6)(0.493) = 30.6 P + 12.6 = 62.1 P = 49.5 atm

Predict which substance has the largest "b" constant: NH3, N2, CH2Cl2, Cl2, CCl4 The value of b constant is merely the actual volume of the gas. A larger molecular volume results in a larger b value. From the compound in the list, CCl4 is the largest so it will have the greatest b constant.

If ever need to add in mean root velcity

Root Mean Square Velocity Gas molecules do not all move at the same speed. The average velocity of the gas is known as the Root Mean Square Velocity. R = 8.315 L-Kpa/mol-K **Must use this R so units cancel! T = temperature of gas (in K) M = mass of a mole of gas in kg Final units are in m/s. Copyright © Cengage Learning. All rights reserved

Example: Calculate the root mean square velocity for the atoms in a sample of oxygen gas at 32°C. * M must be in Kg 32.00 g ×1 kg = .03200 kg 1000g * T must be in K T= 32 + 273 = 305 K = 488m/s Answer is 487.57m/s