Pressure The molecules in a gas are in constant motion. The purple balls represent gaseous atoms that collide with each other and the walls of the container. Gas molecules are in constant motion. "Pressure" is a measure of the collisions of the atoms with the container.
Pressure Force per unit area Equation: P = F/A F = force A = area Next
Formulas For Surface Area Square: 6s² rectangle: 2ab + 2bc + 2ac Cylinder: 2πr²+ 2 πrh sphere: 4πr²
Units of Measurement N/m 2 Newton per square meter N/cm 2 Newton per square centimeter Pa Pascal kPa kilopascal Torr mmHg millimeters of mercury lb/in 2 pound per square inch
Calculating Pressure Using P = F/A Suppose that a woman weighing 135 lb and wearing high-heeled shoes momentarily places all her weight on the heel of one foot. If the area of the heel is 0.50 in²., calculate the pressure exerted on the underlying surface. P = F/A = 135 lb/0.50 in² P = 270 lb/in²
Liquid Pressure The pressure depends height of liquid column and density of liquid. Ρ = dgh = density x acceleration due to gravity x height
Calculating Pressure Exerted by a Liquid Calculate the height of a mercury column, in feet, that is equal to a column of water that is 110 ft. high. Known values d water = 1.0 g/cm 3 d mercury = 13.6 g/cm 3 g = 32 ft/s 2 Ρ = dgh P water = P mercury dgh = dgh (1.0 g/cm 3 )(32ft/s 2 )(110 ft) = (13.6 g/cm 3 )(32 ft/s 2 )(h) h = 8.1 ft. Hg
Barometer Barometer is a device used to measure the pressure exerted by the atmosphere. Height of mercury varies with atmospheric conditions and with altitude.
Standard Atmosphere (atm) Pressure exerted by a mercury column of exactly 760 mm in height the density of Hg = 13.5951 g/cm 3 (0 o C) The acceleration due to gravity (g) is 9.80665 m/s 2 exactly. 1 atm = 760 mmHg 760 Torr 14.696 lb/in 2 101.325 kPa Check textbook for more values
Converting Pressure to an Equivalent Pressure A gas is at a pressure of 1.50 atm. Convert this pressure to a.Kilopascals 1 atm = 101.325 kPa 1.50 atm x 101.325 kPa = 152 kPa 1 atm b.mmHg 1.50 atm x 760 mmHg = 1140 mmHg 1 atm
Manometers Used to compare the gas pressure with the barometric pressure. Next
Types of Manometers Closed-end manometer The gas pressure is equal to the difference in height (Dh) of the mercury column in the two arms of the manometer
2.Gas pressure is greater than the barometric pressure. P > 0 P gas = P bar + P
3.Gas pressure is less than the barometric pressure. P < 0 P gas = P bar + P
Standard Temperature & Pressure (STP) Temperature = 0ºC Pressure = 1 atm
The Simple Gas Laws Boyles Law Charles Law Gay-Lussacs Law Combined Gas Law
Temperature & Gas Laws Three temperature scales Fahrenheit (ºF) Celsius (ºC) Kelvin K (no degree symbol) Always use K when performing calculations with the gas law equations. K = (ºC) + 273 Example: ºC = 20º K = 293 Next
Absolute Zero of Temperature Temperature at which the volume of a hypothetical gas becomes zero -273.14 ºC Next
Absolute or Kelvin Scale Temperature scale that has -273.15 ºC as its zero. Temperature interval of one Kelvin equals one degree celsius. 1 K = 1 ºC
Boyles Law For a fixed amount of gas at constant temperature, gas volume is inversely proportional to gas pressure. P 1 V 1 = P 2 V 2
Example. Charles Law A 4.50-L sample of gas is warmed at constant pressure from 300 K to 350 K. What will its final volume be? Given: V 1 = 4.50 L T 1 = 300. K T 2 = 350. K V 2 = ? Equation: V 1 = V 2 T 1 T 2 or V 1 T 2 = V 2 T 1 (4.50 L)(350. K) = V 2 (300. K) V 2 = 5.25 L
Gay-Lussacs Law The pressure of a sample of gas is directly proportional to the absolute temperature when volume remains constant. P 1 = P 2 T 1 T 2 or P 1 T 2 = P 2 T 1
On the next slide (43) The amount of gas and its volume are the same in either case, but if the gas in the ice bath (0 ºC) exerts a pressure of 1 atm, the gas in the boiling-water bath (100 ºC) exerts a pressure of 1.37 atm. The frequency and the force of the molecular collisions with the container walls are greater at the higher temperature.
Combined Gas Law Pressure and volume are inversely proportional to each other and directly proportional to temperature. P 1 V 1 = P 2 V 2 T 1 T 2 or P 1 V 1 T 2 = P 2 V 2 T 1
Example. Combined Gas Law A sample of gas is pumped from a 12.0 L vessel at 27ºC and 760 Torr pressure to a 3.5-L vessel at 52ºC. What is the final pressure? Given: P 1 = 760 Torr P 2 = ? V 1 = 12.0 L V 2 = 3.5 L T 1 = 300 K T 2 = 325 K Equation: P 1 V 1 = P 2 V 2 T 1 T 2 or P 1 V 1 T 2 = P 2 V 2 T 1 (760 Torr)(12.0 L)(325 K) = ( P 2) (3.5 L)(300 K) P 2 = 2.8 x 10³ Torr
Avogadros Explanation of Gay-Lussacs Law When the gases are measured at the same temperature and pressure, each of the identical flasks contains the same number of molecules. Notice how the combining ratio: 2 volumes H 2 to 1 volume O 2 to 2 volumes H 2 O leads to a result in which all the atoms present initially are accounted for in the product.
Avogadros Hypothesis Different gases compared at same temperature and pressure a. Equal - equal number of Volumes molecules b. Equal number of – equal moleculess volumes
Avogadros Law At a fixed temperature and pressure, the volume of a gas is directly proportional to the amount of gas. V = c · n V = volume c = constant n= # of moles Doubling the number of moles will cause the volume to double if T and P are constant.
More STP Values For Gases 1 mol of a gas = 22.4 L Number of molecules contained in 22.4 L of a gas is 6.022 x 10 23
Equation Includes all four gas variables: Volume Pressure Temperature Amount of gas Next
PV = nRT Gas that obeys this equation if said to be an ideal gas (or perfect gas). No gas exactly follows the ideal gas law, although many gases come very close at low pressure and/or high temperatures. Ideal gas constant, R, is R = PV nT = 1 atm x 22.4 L 1 mol x 273.15 K R = 0.082058 L·atm/mol· K
Applications of the Ideal Gas Law Molar Mass Density
Molar Mass Determination n = number of moles Moles = mass of sample m Molar mass M Ideal Gas Equation: n = PV RT Substituting: m = PV M RT M = mRT PV
Example: Molar Mass The density of carbon tetrachloride vapor at 714 torr and 125ºC is 4.43 g/L. What is its molar mass? M = dRT P = (4.43 g/L)(0.0821 L-atm/mol-k)(398 K) (714 torr x 1 atm/760 torr) M = 154 g/mol
Gas Density (d) d = m/V so V = m/d Ideal Gas Equation: PV = nRT Substituting P m = mRT d M m cancels out d = PM RT
Example. Vapor Density The mean molar mass of the atmosphere at the surface of Titan, Saturns largest moon, is 28.6 g/mol. The surface temperature is 95 K, and the pressue is 1.6 Earth atm. Assuming ideal behavior, calcuate the density of Titans atmosphere. d = PM RT = (1.6)(1 atm)(28.6 g/mol) (0.0821 L-atm/mol-K)(95 K) d = 5.9 g/L
Gases in Chemical Reactions Ideal Gas Law & Balanced Chemical Equation
Example. Reaction Stoichiometry How many moles of nitric acid can be prepared using 450 L of NO 2 at a pressure of 5.00 atm and a temperature of 295 K? Step 1: Use Ideal Gas law equation n = PV (5.00 atm)(450 L) RT (0.0821 L-atm/mol-K)(295 K) n = 92.9 mol Next ---- >
Step 2 3NO 2 (g) + H 2 O --- > 2HNO 3(aq) + NO (g) 92.9 mol NO 2 x 2 mol HNO 3 3 mol NO 2 = 61.9 mol HNO 3
Daltons Law of Partial Pressure Mixture of Gases
Total Pressure The total pressure of a mixture of gases is the sum of the partial pressures of the components of the mixture. P total = P A + P B + ……
Mole Fraction of Compound in Mixture Fraction of all molecules in the mixture contributed by that component. Sum of all the mole fractions in a mixture is 1. Expression for mole fraction of a substance in terms of P and V n a = P a = V a n tot P tot V tot
Example: Gas Mixtures & Partial Pressure A gaseous mixture made from 6.00 g O 2 and 9.00 g CH 4 is placed in a 15.0 L vessel at 0ºC. What is the partial pressure of each gas, and what is the total pressure in the vessel? Step 1: n O2 = 6.00 g O 2 x 1 mol O 2 32 g O 2 = 0.188 mol O 2 Next ---- >
n CH4 = 9.00 g CH 4 x 1 mol CH 4 16.0 g CH 4 = 0.563 mol CH 4 Step 2: Calculate pressure exerted by each P O2 = nRT V = (0.188 mol O 2 )(0.0821 L-atm/mol-K)(273 K) 15.0 L = 0.281 atm
P CH4 = (0.563 mol)(0.0821 L-atm/mol-K)(273 K) 15.0 L = 0.841 atm Step 3: Add pressures P total = P O2 + P CH4 = 0.281 atm + 0.841 atm P total = 1.122 atm
Mole fraction: n a/ n total n O2 = 0.188 mol (0.188 + 0.563) mol n O2 = 0.250 n CH4 = 0.563 mol (0.188 + 0.563) mol n CH4 = 0.750 0.250 + 0.750 = 1.000
Volume of each Gas P O2 = V O2 P tot V tot 0.281 atm = V O2 (0.281 + 0.841) atm = 15 L V O2 = 3.8 L V CH4 = 15 L – 3.8 L = 11 L
Gas Collected Over Water Wet Gas: mixture of the desired gas and water vapor P bar = P gas + P water P gas = P bar – P water Next
P bar = P gas + P water P gas = P bar - P water
Gas Collected Over Water A sample of KClO 3 is partially decomposed, producing O 2 gas that is collected over water. The volume of gas collected is 0.250 L at 26ºC and 765 torr total pressure. 2KClO 3(s) ----- > 2 KCl (s) + 3O 2(g) Next ---- >
a.How many moles of O 2 are collected? Step 1: Calculate pressure of O 2 in mixture P O2 = P tot – P water vapor = 765 torr – 25 torr P O2 = 740 torr Next ---- >
n O2 = PV RT = (740 torr)(1atm/760 torr)(0.250 L) (0.0821 L-atm/mol-K)(299 K) n O2 = 9.92 x 10 -3 mol Next ----- >
b.How many grams of KClO 3 were decomposed? (9.92 x 10 -3 mol O 2 ) x 2 mol KClO 3 x 123 g KCLO 3 3 mol O 2 1 mol KClO 3 = 0.811 g KClO 3
c. When dry, what volume would the collected O 2 gas occupy at the same temperature and pressure? Remove water vapor. P 2 = 765 torr Temperature is the same. P 1 V 1 = P 2 V 2 V 2 = (740 torr)(0.250 L) (765 torr) V 2 = 0.242 L
Assumptions The molecules of gases are in rapid random motion. Their average velocities (speeds) are proportional to the absolute (Kelvin) temperature. At the same temperature, the average kinetic energies of the molecules of different gases are equal. Next
Relationship Between Temperature and Average Kinetic Energy Higher temperature means greater motion (KE) av = 3 RT 2 Obtained from PV = RT = 2 (KE) av n 3
PV = RT = 2 (KE) av = ½ mv² n 3 Apply to all gases whether alone or mixed with other gases 1.Constant V a. T changes – KE changes b. T increases – pressure increases c. KE changes – speed of molecules changes 2.Constant temperature a. V increases, P decreases b. KE doesnt change – T didnt change c. Speed of molecules doesnt change-KE didnt change
Root-Mean Square Speed (u 2 ) u 2 represents the average of the squares of the particles velocities. The square root of u 2 is called the root mean square velocity. The unit of measurement is m/s. u 2 = 3RT Units: R = 8.3145 J K -1 mol -1 M or 8.3145 kgm 2 /s 2 K -1 mol -1 Joule = kg·m 2 /s 2 Joule is a unit of measurement for energy.
Example How is the root-mean square speed (rms) of F 2 molecules in a gas sample changed by a)An increase in temperature. b)An increase in volume c)Mixing with a sample of Ar at the same temperature
Example. Root-Mean Square Speed Calculate the root mean square speed for the atoms in a sample of helium gas at 25ºC. Given: T = 25ºC = 298 K Known: R = 8.3145 J/K·mol or 8.3145 kgm 2 /s 2 / K·mol M He : Change g to kg 4.00 g/mol = 4.00 x 10 -3 kg/mol Equation: u 2 = 3RT M u 2 = 3 (8.3145 kgm 2 /s 2 /K·mol)(298K) 4.00 x 10 -3 kg/mol u = 1.36 x 10 3 m/s
Effusion The escape of gas molecules from their container through a tiny orifice or pin hole.
Diffusion Diffusion is the moving or mixing of molecules of different substances as a result of random molecular motion.
Factors That Influence Effusion and Diffusion The greater the temperature the faster the ga will move or vise versa/ If the mass increases, the kinetic energy will also increase if the speed remains constant. If the mass increases, the speed decreases if the kinetic energy remains constant.
Example. Rate of Effusion Calculate the ratio of the effusion rates of hydrogen gas (H 2 ) and uranium hexafluoride (UF 6 ), a gas used in the enrichment process fuel for nuclear reactors. Known: Molar Masses H 2 = 2.016 g/mol UF 6 = 352.02 g/mol (Rate of effusion)² = M U compound M H gas = 352.02 2.016 Rate of effusion = 13.21
Grahams Law The rate of effusion (or diffusion) of two different gases are inversely proportional to the square roots of their molar masses. (rate of effusion of A) 2 = M B (rate of effusion of B) 2 M A
Ratios of Rates the square root of two molar masses is also equal to the ratio Molecular speeds Effusion times Distance traveled by molecules Amount of gas effused
Based on Assumptions of Kinetic Molecular Theory Rates of diffusion of different gases are inversely proportional to the square roots of their densities. Rates of diffusion are inversely proportional to the square roots of their molar masses.
Ideal vs. Nonideal (Real) Gases Ideal High pressure- compressible, volume approaches zero Force of collisions with container wall is great Nonideal High pressure – molecules are practically incompressible Force of collision with container wall is less due to attractive force among the molecules.
Behavior of Gases Behave ideally at High temperatures Low pressures Behave nonideally at Low temperatures High pressures
Equation corrects for volume and intermolecular forces (P + n²a/V²)(V-nb) = nRT n²a/V² = related to intermolecular forces of attraction n²a/V² is added to P = measured pressure is lower than expected a & b have specific values for particular gases V - nb = free volume within the gas
Intermolecular Force of Attraction Attractive forces of the orange molecules for the purple molecule cause the purple molecule to exert less force when it collides with the wall than it would if these attractive forces did not exist.
Source of Sketches/problems http://cwx.prenhall.com/bookbind/pu bbooks/hillchem3/medialib/media_po rtfolio/04.htmlhttp://cwx.prenhall.com/bookbind/pu bbooks/hillchem3/medialib/media_po rtfolio/04.html Chemistry, 7 th ed. Brown, LeMay & Bursten, Prentice Hall