"Gases The Gas Laws Labs 18 Molar Mass of a Volatile Liquid 19 Calcium Carbonate Analysis Molar Volume of"
Gases The Gas Laws Labs #18 Molar Mass of a Volatile Liquid #19 Calcium Carbonate Analysis: Molar Volume of Carbon Dioxide 2/16/2012 1 Gases have general characteristics • Expansion: Expand indefinitely to fill the space available to them • Indefinite shape: Fill all parts of container evenly, so have no definite shape of their own • Compressibility: Most compressible of states of matter • Mixing: Two or more gases will mix evenly and completely when confined to same container • Low density: Have much lower densities than liquids and solids (typically about 1/1000 those of liquids/solids) • Pressure: Exert pressure on their surroundings 2/16/2012 2 Vapors • Gas phase at temperature where same substance can also exist in liquid or solid state • Below critical temperature of substance (vapor can be condensed by increasing pressure without reducing temperature) 2/16/2012 3 Pressure: force per unit area (P= f/a) • Force generated by • Collision frequency collisions of gas particle dependent on w/container walls • Gas particle velocity • Related to velocity of gas • Distance to container walls. particles • Changing temperature • Total force = sum of forces changes collision force, as of all collisions each second per unit area well as collision frequency • Force = m x acceleration • Collision frequency • Pressure dependent on changed by altering size of • Gas particle velocity container • Collision frequency • Force of collisions is not affected 2/16/2012 4 (force of 1 newton exerted on one square meter of area) (29.92” Hg) (psi) 2/16/2012 5 Manometers • Used to measure pressure of enclosed gas • U-tube partially filled with liquid, typically Hg • One end connected to container of gas being measured • Other end sealed with vacuum existing above liquid, or open to atmosphere 2/16/2012 6 Closed-end manometer • Pressure is just difference between two levels (in mm of Hg)-indicates pressure of system attached to apparatus • Gas connected to one arm • Space above Hg in other arm is vacuum • Liquid in tube falls to height (directly proportional to pressure exerted by gas in the tube) • Since pressure of gas causes liquid levels to be different in height, it is this difference (h) that is measure of gas pressure in container • Pgas = Ph • If left to atmosphere, it measures atmospheric pressure-barometer 2/16/2012 7 Open-end manometer • Used to measure pressure of gas in container • Difference in Hg levels indicates pressure difference in gas pressure and atmospheric pressure • Atmospheric pressure pushes mercury in one direction • Gas in container pushes it in the other direction • Two ends connected to gases at different pressures • Closed end to gas in bulb (gas filled) • Open end to atmosphere • If pressure of gas higher than atmospheric, Hg level lower in arm connected to gas (Pgas = Pbarometric + h) • If pressure of gas lower than atmospheric, Hg level higher in arm connected to gas (Pgas = Pbarometric - h) • If levels are equal, gas at atmospheric pressure 2/16/2012 8 • If fluid other than mercury is used: • Difference in heights of liquid levels inversely proportional to density of liquid and represents the pressure • Greater density of liquid, smaller difference in height • High density of mercury (13.6 g/mL) allows relatively small monometers to be built • Readings must be corrected for relative densities of fluid used and of mercury (mm Hg = mm fluid) density fluid density Hg 2/16/2012 9 2/16/2012 10 • A sample of CH4 is confined in a water manometer. The temperature of the system is 30.0 °C and the atmospheric pressure is 98.70 kPa. What is the pressure of the methane gas, if the height of the water in the manometer is 30.0 mm higher on the confined gas side of the manometer than on the open to the atmosphere side. (Density of Hg is 13.534 g/mL). 2/16/2012 11 • 1) Convert 30.0 mm of H2O to equivalent mm of mercury: • (30.0 mm) (1.00 g/mL) = (x) (13.534 g/mL) x = 2.21664 mm (I will carry some guard digits.) • 2) Convert mmHg to kPa: • 2.21664 mmHg x (101.325 kPa/760.0 mmHg) = 0.29553 kPa • 3) Determine pressure of enclosed wet CH4: • At point A in the above graphic, we know this: Patmo. press. = Pwet CH4 + Pthe 30.0 mm water column • 98.70 kPa = x + 0.29553 kPa • x = 98.4045 kPa • 4) Determine pressure of dry CH4: • From Dalton's Law, we know this: Pwet CH4 = Pdry CH4 + Pwater vapor • (Water's vapor pressure at 30.0 °C is 31.8 mmHg. Convert it to kPa.) • 98.4045 kPa = x + 4.23965 kPa • x = 94.1648 kPa • Based on provided data, use three significant figures; so 94.2 kPa. 2/16/2012 12 Boyle’s Law X axis independent variable Y axis dependent variable Pressure-Volume Relationship Temperature/# molecules (n) constant 2/16/2012 13 • Inversely proportional (one goes up, other goes down) • V 1/P or • P = k/T where K is constant • PV = constant • P1V1 = P2V2 • Gas that strictly obeys Boyle’s Law is ideal gas (holds precisely for gases at very low temperatures) 2/16/2012 14 Pressure applied vs. volume measured • Shows inverse proportion • If pressure doubled, volume decreased by ½ 2/16/2012 15 Pressure vs. inverse of volume • Plot of V (P) against 1/P (1/V) gives straight line – Graph of equation P = k1 x 1/V • Same relationship holds whether gas is being expanded or contracted 2/16/2012 16 • A gas which has a pressure of 1.3 atm occupies a volume of 27 L. What volume will the gas occupy if the pressure is increased to 3.9 atm at constant pressure? • P1 = 1.3 atm • V1 = 27 L • P2 = 3.9 atm • V2 = ? • P1V1 = P2V2 • 1.3 atm (27 L) = 3.9 atm (X) = 9.0 L 2/16/2012 17 Charles’ Law Temperature-Volume Relationship Pressure/n constant 2/16/2012 18 • Volume of gas directly proportional to temperature (T V), and extrapolates to zero at zero Kelvin (convert Celsius to Kelvin) 2/16/2012 19 Used to determine absolute zero • From extrapolated line, determine T at which ideal gas would have zero volume • Since ideal gases have infinitely small atoms, only contribution to volume of gas is pressure exerted by moving atoms bumping against walls of container • If no volume, then no kinetic energy left • Absolute zero is T at which all KE has been removed • Does not mean all energy has been removed, merely all KE http://www.absorblearning.com/media/attachment.action?quick=10o&att=2629 2/16/2012 20 • If T↓, V↓ and vs. • V = kT • P is constant • k = proportionality constant • V/T = constant V1 V2 ( P constant) T1 T2 2/16/2012 21 • A gas at 30oC and 1.00 atm occupies a volume of 0.842 L. What volume will the gas occupy at 60.0oC and 1.00 atm? • V1 = 0.842 atm • T1 = 30oC = 303 K • V2 = ? • T2 = 60.0oC = 333 K • V1/T1 = V2/T2 • 0.842/303 K = X/333 K = 0.925 L 2/16/2012 22 Gay Lussac’s Law Pressure-Temperature Relationship Volume/n constant 2/16/2012 23 • When two gases react, do so in volume ratios always expressed as small whole numbers • When H burns in O, volume of H consumed is always exactly 2x volume of O • Direct relationship between pressure and temperature (P T) • P/T = constant • Pi/Ti = Pf/Tf 2/16/2012 24 Avogadro’s Law (at low pressures) Volume-Amount Relationship Temperature/Pressure constant 2/16/2012 25 • Equal volumes of gases, measured at same temperature and pressure, contain equal numbers of molecules • Avogadro's law predicts directly proportional relation between # moles of gas and its volume • Helped establish formulas of simple molecules when distinction between atoms and molecules was not clearly understood, particularly existence of diatomic molecules • Once shown that equal volumes of hydrogen and oxygen do not combine in manner depicted in (1), became clear that these elements exist as diatomic molecules and that formula of water must be H2O rather than HO as previously thought 2/16/2012 26 • V of gas proportional to # of moles present (V n) • At constant T/P, V of container must increase as moles of gas increase • V/n = constant • Vi/ni = Vf/nf 2/16/2012 27 • A 5.20 L sample at 18.0oC and 2.00 atm pressure contains 0.436 moles of a gas. If we add an additional 1.27 moles of the gas at the same temperature and pressure, what will the total volume occupied by the gas be? • V1 = 5.20 L • n1 = 0.436 mol • V2 = X • n2 = 0.436 + 1.27 = 1.70 mol • V1/n1 = V2/n2 • 5.20 L/0.436 mol = X/1.70 mol = 20.3 L 2/16/2012 28 Generalization applicable to most gases, at pressures up to about 10 atm, and at temperatures above 0°C. Ideal gas’s behavior agrees with that predicted by ideal gas law. Ideal Gas Law (Holds closely at P < 1 atm) 2/16/2012 29 Formulated from combination of Boyle’s law, Charles’s law, Gay-Lussac’s law, and Avogadro’s principle • Combined • If proportionality constant called R • Rearrange to form ideal gas equation 2/16/2012 30 • Experimentally observed relationship between these properties is called the ideal gas law: PV = nRT • Pressure (P) in atm • Volume (V) in L • Absolute temperature (T) in K (Charles’ Law) • Amount (number of moles, n) • R (universal ideal gas constant) • 0.08206 liter ∙ atm/mole ∙ K or 0.08206 L∙atm∙K–1·mol–1 • 8.31 liter ∙ kPa/mole ∙ K • 8.31 J/mole ∙ K • 8.31 V ∙ C/mole ∙ K • 8.31 x 10-7 g ∙ cm2/sec2 ∙ mole ∙ K • 6.24 x 104 L ∙ mm Hg/mol ∙ K • 1.99 cal/mol ∙ K 2/16/2012 31 • You can use ideal gas law equation for all problems • Given 3 of 4 variables and calculate 4th PiVi= niRTi PfVf nfRTf • Cancel all constants and R, make appropriate substitutions from given data to perform calculation 2/16/2012 32 • A sample containing 0.614 moles of a gas at 12.0oC occupies a volume of 12.9 L. What pressure does the gas exert? • PV = nRT • P (12.9 L) = (0.614 mol)(0.08206 L atm/K mol)(285 K) = 1.11 atm 2/16/2012 33 2/16/2012 34 Homework: Read 5.1-5.3, pp. 189-202 Q pp. 232-234, #29, 31-34, 44 2/16/2012 35 Standard Temperature and Pressure (STP) • STP = 0°C (273K) and 1.00 atm pressure (760 mm Hg) • One mole of gas at STP will occupy 22.42 L • Allows you to compare gases at STP to each other 2/16/2012 36 • What volume will 1.18 mole of O2 occupy at STP? • PV = nRT • (1atm)(X) = (1.18 mol)(0.08206 L atm/K mol)(273 K) = 26.4 L • Alternate way: • At STP, 1 mole occupies 22.4 L • Vi/ni = Vf/nf • 22.4 L/1 mol = X/1.18 mol = 26.4 L 2/16/2012 37 • A sample containing 15.0 g of dry ice, CO2(s), is put into a balloon and allowed to sublime according to the following equation: CO2(s) CO2(g) How big will the balloon be (what is the volume of the balloon) at 22oC and 1.04 atm after all of the dry ice has sublimed? • 15.0 g CO2 1 mol CO2 = 0.341 mol CO2 44.0 g CO2 • PV = nRT • (1.04 atm)(X) = (0.341 mol)(0.08206 L atm/K mol)(295K) = 7.94 L 2/16/2012 38 • 0.500 L of H2(g) are reacted with 0.600 L of O2(g) according to the equation 2H2(g) + O2(g) 2H2O(g). What volume will the H2O occupy at 1.00 atm and 350oC? • Find limiting reactant: • 0.500 L H2 1 mol H2 = 0.0223 mol H2/2 22.4 L • 0.600 L O2 1 mol O2 = 0.0268 mol O2/1 22.4 L • PV = nRT • (1atm)(X) = (0.0223 mol)(0.08206 L atm/K mol)(623 K) = 1.14 L 2/16/2012 39 2/16/2012 40 Molar Mass • Gives density which can be determined if P, T and molar mass are known • Density directly proportional to molar mass • Density increases as gas pressure increases • Density decreases as temperature increases 2/16/2012 41 • A gas at 34.0oC and 1.75 atm has a density of 3.40 g/L. Calculate the molar mass (M.M.) of the gas. • M.M. = dRT P • M.M. = (3.40 g/L)(0.08206 L atm/K mol)(307K) (1.75 atm) • M.M. = 48.9 g/mol 2/16/2012 42 Example • What are the expected densities of argon, neon and air at STP? • PV = nRT • PV = g (RT) molar mass • Density = g/V = (Molar mass)P/RT • density Ar = (39.95 g/mol)(1.00 atm)/(0.0821 L- atm/mol-K)(273 K) = 1.78 g/L • density Ne = 0.900 g/L • density air = 1.28 g/L • molar mass = (0.80)(28 gN2/mol) + (0/20)(32 g O2/mol) = 28.8 g/mol 2/16/2012 43 Molar Volume (V/n) • V = RT n P • (0.0821 L-atm/mol-K)(273K) = 22.4 L/mol 1.00 atm 2/16/2012 44 Homework: Read 5.4, pp. 203-206 Q pp. 234-235, #52, 56, 58, 60 2/16/2012 45 Dalton’s Partial Pressure 2/16/2012 46 • When two gases are mixed together, gas particles tend to act independently of each other • Each component gas of mixture of gases uniformly fills containing vessel • Each component exerts same pressure as it would if it occupied that volume alone • Total pressure of mixture is sum of individual pressures, called partial pressures, of each component • Pt = P A + P B + P C + … 2/16/2012 47 2/16/2012 48 • Since each component obeys ideal gas law (PV=nRT) and has same T and V, it follows that partial pressure of each gas in container is directly proportional to # moles of gas present • Pi/Pt = ni/nt • Pi = ni/nt x Pt • Pi = XiPt • Xi = mole fraction of gas component i 2/16/2012 49 • A volume of 2.0 L of He at 46oC and 1.2 atm pressure was added to a vessel that contained 4.5 L of N2 at STP. What is the total pressure and partial pressure of each gas at STP after the He is added? • Find # moles of He at original conditions-will lead us to finding partial pressure of He at STP. • PV = nRT • (1.2 atm)(2.0 L) = n(0.08206 L atm/k mol)(319 K) • n = 0.0917 mol He • When gases are combined under STP, partial pressure of He will change while N2 will remain the same since it is already at STP. • PHeV = nRT • (X)(4.5 L) = (0.091 mol)(0.08206 L atm/K mol)(273 K) • P = 0.457 atm • Total pressure = 1.00 + 0.46 = 1.46 atm 2/16/2012 50 • Mole fraction • Xi = ni = Pi ntotal Ptotal • You can use either the number of moles or the pressure of each component of your system to evaluate the mole fraction • 4.5 L N2 1 mol = 0.201 mol N2 22.4 L • XN2 = 0.201 mol = 0.69 1.00 atm = 0.69 0.293 mol 1.457 atm • X He = 0.0917 mol = 0.31 0.457 atm = 0.31 0.293 mol 1.457 atm 2/16/2012 51 • A mixture of gases consists of 3.00 moles of helium, 4.00 moles of argon, and 1.00 moles of neon. The total pressure of the mixture is 1,200 torr. • Calculate the mole fraction of each gas. • nt = 3.00 mol + 4.00 mol + 1.00 mol = 8.00 mol • Xi = ni/nt • XHe = 3.00 mol/8.00 mol = 0.375 • XAr = 4.00 mol/8.00 mol = 0.500 • XNe = 1.00 mol/8.00 mol = 0.125 • Calculate the partial pressure of each gas. • Pi = XiPi • PHe = 0.375 x 1,200 torr = 450. torr • PAr = 0.500 x 1,200 torr = 600. torr • PNe = 0.125 x 1,200 torr = 150. torr • He-1.2 L, 0.63 atm, 16oC Ne-3.4 L, 2.8 atm, 16oC 2/16/2012 52 • Whenever gases are collected by displacement of water, total gas pressure is sum of partial pressure of collected gas and partial pressure of water vapor • Pt = Pgas + PH2O • Consequently, partial pressure of collected gas is Pgas = Pt – PH2O, where partial pressure of water depends on temperature and corresponds to vapor pressure of water 2/16/2012 53 • When oxygen gas is collected over water at 30°C and the total pressure is 645 mm Hg: • What is the partial pressure of oxygen? Given the vapor pressure of water at 30°C is 31.8 mm Hg. • Pt = PO2 + PH2O • PO2 = Pt – PH2O = 645 mm Hg – 31.8 mm Hg PO2 = 613 mm Hg • What are the mole fractions of oxygen and water vapor? • Recall that the partial pressures of O2 and H2O are related to their mole fractions, PO2 = XO2Pt PH2O = XH2OPt • X O2 = 613/645 = 0.950, and XH2O = 31.8/645 = 0.049 • Note also that the sum of the mole fractions is 1.0, within the number of significant figures, given: X O2 + XH2O = 0.950 + 0.049 = 0.999 2/16/2012 54 • The vapor pressure of water in air at 28oC is 28.3 torr. Calculate the mole fraction of water in a sample of air at 28oC and 1.03 atm pressure. • XH2O = PH2O = 28.3 torr = 0.036 Pair 783 torr 2/16/2012 55 2/16/2012 56 2/16/2012 57 • The safety air bags in cars are inflated by nitrogen gas generated by the rapid decomposition of sodium azide. If an air bag has a volume of 36 L and is to be filled with nitrogen gas at a pressure of 1.15 atm at a temperature of 26.0oC, how many grams of NaN3 must be decomposed? 2NaN3(s) 2Na(s) + 3N2(g) • Gas data mol N2 mol NaN3 g NaN3 • n = PV/RT = (1.15 atm)(36 L) = 1.7 mol N2 (0.0821 L-atm/mol-K)(299K) • 1.7 mol N2 2 mol NaN3 65.0g NaN3 = 72 g NaN3 3 mol N2 1 mol NaN3 2/16/2012 58 Homework: Read 5.5, pp. 206-211 Q pg. 235, #64, 66, 68, 70, 72 2/16/2012 59 Kinetic Molecular Theory of Gases Describes the properties of gases 2/16/2012 60 Main assumptions that explain behavior of ideal gas: • Gases composed of tiny, invisible molecules that are widely separated from one another in empty space • Gas molecule behave as independent particles • Volume occupied by molecule considered negligible • Gas molecules are in constant motion, continuous, random and straight-line motion • Collisions of particles with walls of container cause pressure exerted by gas • Molecules collide with one another, but collisions are perfectly elastic (no net loss of energy-exert no force on each other) • Attractive forces between atoms and/or molecules in gas are negligible • Average kinetic energy of collection of gas particles assumed to be directly proportional to Kelvin temperature of gas 2/16/2012 61 Temperature is measure of average kinetic energy of gas • Equal #s of molecules of any gas have same average kinetic energy at same temperature • Greater temperature, greater fraction of molecules moving at higher speeds (higher average kinetic energy of gas molecules) • Individual molecules move at varying speeds • Momentum conserved in each collision, but one colliding molecule might be deflected off at high speed while another is nearly stopped • Result is that molecules at any instant have wide 2/16/2012range of speeds 62 Average KE, e, related to At higher temperatures, root mean square (rms) greater fraction of molecules speed u moving at higher speeds • 4 molecules in gas sample have speeds of 3.0, 4.5, 5.2, and 8.3 m/s. • Average speed • rms • Because mass of molecules does not increase, rms speed of molecules must increase with increasing temperature 2/16/2012 63 • Average kinetic energy of single gas molecule • KE = ½ mv2 • m = mass of molecule (kg) • v = speed of molecule (meters/sec) • KE measured in joules • Average translational kinetic energy of any kind of molecule in ideal gas is given by: 2/16/2012 64 2/16/2012 65 2/16/2012 66 J = kg· m2/s2 2/16/2012 67 • Based on understanding of pressure, molecular meaning of gas laws can be appreciated as follows: 2/16/2012 68 Boyle’s Law: inverse relationship between P & V: • Kinetic molecular theory agrees • If V of container decreased, gas particles strike walls of container more frequently, increasing pressure of gas • If V of container increased, fewer impacts of gas molecules with wall per second, decreasing pressure 2/16/2012 69 Charles’ Law: direct relationship between T & V: • Increasing temperature increases average kinetic energy and speed of gas molecules • Increases force of each collision w/container wall • Increases frequency of collisions w/container wall • Both increase pressure • If pressure of gas is to remain constant, volume must increase to decrease frequency of collisions with container walls • If one of the walls is a movable piston, then the volume of the container will expand until balanced by the external force 2/16/2012 70 Gay-Lussac‘s Law: direct relationship between T & P • Increase in temperature increases kinetic energy of gas particles • Force of each collision increases • Frequency of collisions with container walls increases • When gas is heated in container with fixed volume, gas molecules impact more forcefully with wall, increasing pressure 2/16/2012 71 Avogadro’s Law • As more gas molecules are added to the container, # of impacts per second with wall increases, pressure increases correspondingly • If V of container is not fixed, then V will increase 2/16/2012 72 Graham’s Law of Effusion Experiments show that molecules of a gas do not all move at same speed but are distributed over a range. 2/16/2012 73 Maxwell-Boltzmann distribution • Probability distribution that forms basis of kinetic theory of gases • Explains many fundamental gas properties, including pressure and diffusion 2/16/2012 74 Highest point on curve marks most probable speed-some move with speeds much less than most probable speed, while others move with speeds much greater than this speed As temp. increases, curve flattens out (more molecules have higher kinetic energies and therefore higher average speeds) (speed) 2/16/2012 75 • Mean free path Diffusion • Average distance molecule travels between collisions • Spread of gas • Factors affecting it: • Density (increasing density molecules decreases MFP) throughout volume • Radius of molecule (increasing • Much slower than size increases collision frequency, reducing MFP) average molecular • Pressure (increasing pressure speed because of increases collision frequency, collision between reducing MFP) molecules • Temperature (increasing T increases collision frequency, but does not affect MFP) 2/16/2012 76 Effusion • Escape of gas through small opening into vacuum • Rate (in mol/s) proportional to average speed u • More collisions gas has w/walls of container, higher probability it will hit pinhole and go through it 2/16/2012 77 • Average kinetic energy has same value for all gases at same temperature • Gas molecules with higher molar masses will have slower average speeds, while lighter molecules will have higher average speeds • Rates two gases effuse through a pinhole in a container are inversely related to the square roots of the molecular masses of gas particles • Use equation to find relative rates of diffusion of gases, whether evacuated or not) • All gases expand to fill container • Heavier gases diffuse more slowly than lighter ones 2/16/2012 78 • KE = cT = ½ mv2 • KE-average kinetic energy • c = constant that is same for all gases • T = temperature in Kelvin • M = mass of gas • v2 = average of the square of the velocities of the gas molecules • Since cT will be the same for all gases at the same temperature, the average KE of any two gases at the same T will be the same • KE1 = KE2 • ½ m1V21 = ½ m2v22 • root mean square velocity = vrms = √v2 • √m1/m2 = vrms2/vrms1 • Higher the T, higher vrms 2/16/2012 79 2/16/2012 80 Effusion: Rate of effusion for gas 1 M2 Rate of effusion for gas 2 M1 Diffusion: Distance traveled by gas 1 M2 Distance traveled by gas 2 M1 2/16/2012 81 • How many times faster than He would NO2 gas effuse? • MNO2 = 46.01 g/mol • MHe = 4.003 g/mol • √MNO2/MHe = rateHe/RateNO2 • √46.01/4.003 = 3.390 • So He would effuse 3.39 times faster as NO2 2/16/2012 82 Real Gases (Nonideal Gases) • Ideal gas law works very well for most gases, however, it does not work well for gases under high pressures or gases at very low temperature • Conditions used to condense gases • Generalized that any gas close to its BP (condensation point) will deviate significantly from ideal gas law • Kinetic molecular theory makes two assumptions • Gases have no volume • They exhibit no attractive or repulsive forces 2/16/2012 83 • As a real gas is cooled and/or compressed, distances between particles decreases dramatically, and these real volumes and forces can no longer be ignored • Cooling gas decreases average KE of its molecules • All gases when cooled enough will condense to liquid • Suggests that intermolecular forces of attraction exist between molecules of real gases • Condensation occurs when average KE is not great enough for molecules to break away from intermolecular attractive forces • When they stick together, there are fewer particles bouncing around and creating P, so real P in nonideal situation will be smaller than P predicted by ideal gas equation 2/16/2012 84 2/16/2012 85 Van der Waals Equation Developed modification of ideal gas law to deal with nonideal behavior of real gases 2/16/2012 86 Real Gases Must correct ideal gas behavior when at high pressure (smaller volume) and low temperature (attractive forces become important) 2/16/2012 87 correction for attractive forces correction for small/finite volume 2 [ Pobs a (n / V ) ] V nb nRT corrected pressure corrected volume Pideal Videal PidealVeffective = nRT Van der Waals (real gases) P = pressure of gas (atm) V = volume of gas (L) n = #moles of gas (mol) T = absolute temperature (K) R = gas constant, 0.0821 L-atm/mol-K a = constant, different for each gas, that takes into account the attractive forces between molecules b = constant, different for each gas, that takes into account the volume of each molecule 2/16/2012 88 • Calculate the pressure exerted by 0.3000 mol of He in a 0.2000 L container at -25oC • Using the ideal gas law • Using van der Waal’s equation • PV = nRT • (X)(0.2000L) = (0.3000 mol)().08206 L atm/K mol)(248) = 30.55 atm 2 [ Pobs a (n / V ) ] V nb nRT • From table 5.3, a = 0.0341 atm L2/mol2 and b = 0.0237 L mol • (P + 0.0341 x 0.30002/0.20002)(0.2000 – 0.3000 x 0.0237) = 0.3000(0.08206)(248)= 31.60 atm 2/16/2012 89 2/16/2012 90 Pneumatic trough used for collecting gas samples over water • A gas (H2, O2) which is not soluble (or only slightly soluble) in water can be collected over water • Pneumatic trough w/bottle full of water submerged • Gas from reaction bubbles into bottle, displacing water • Pressure of gas inside collecting bottle determined to solve ideal gas equation • When gases are collected over water, there is some water vapor collected • Pressure of water vapor depends on T only and is obtained from reference table • Pressure of gas that was generated is calculated from Dalton’s law of partial pressure Pgas = Patm - Pwater 2/16/2012 91 2/16/2012 92 Homework: Read 5.6-5.9, pp. 212-229 Q pp. 235-236, #74, 76, 78, 80, 82, 84, 86, 93 Do one additional exercise and one challenge problem. Submit quizzes by email to me: http://www.cengage.com/chemistry/book_content/0547125321_z umdahl/ace/launch_ace.html?folder_path=/chemistry/book_con tent/0547125321_zumdahl/ace&layer=act&src=ch05_ace1.xml http://www.cengage.com/chemistry/book_content/0547125321_z umdahl/ace/launch_ace.html?folder_path=/chemistry/book_con tent/0547125321_zumdahl/ace&layer=act&src=ch05_ace2.xml http://www.cengage.com/chemistry/book_content/0547125321_z umdahl/ace/launch_ace.html?folder_path=/chemistry/book_con tent/0547125321_zumdahl/ace&layer=act&src=ch05_ace3.xml 2/16/2012 93