GASES Chapter 5 A Gas - Uniformly fills any container. - Mixes completely with any other gas - Exerts pressure on its surroundings. 05_47 Vacuum h = 760mm Hg for standard atmosphere Simple barometer invented by Evangelista Torricelli Pressure - is equal to force/unit area Newton meter 2 - SI units = = 1 Pascal (Pa) - 1 standard atmosphere = 101,325 Pa - 1 standard atmosphere = 1 atm = 760 mm Hg = 760 torr Pressure Unit Conversions The pressure of a tire is measured to be 28 psi. What would the pressure in atmospheres, torr, and pascals. 28 psi 1.000 atm = 1.9 atm 1 14.69 psi 28 psi 1.000 atm 760.0 torr 3 = 1.4 x 10 torr 1 14.69 psi 1.000 atm 28 psi 1.000 atm 101,325 Pa = 1.9 x 105 Pa 1 14.69 psi 1.000 atm Atmospheric pressure (Patm) 05_48 Atmospheric pressure (Patm) h Gas pressure (Pgas) less than atmospheric pressure (Pgas) = (Patm) - h (a) h Gas pressure (Pgas) greater than atmospheric pressure (Pgas) = (Patm) + h (b) Simple Manometer 05_1541 Pext Pext Volume is decreased Volume of a gas decreases as pressure increases at constant temperature 05_50 40 slope = k V(in3) P (in Hg) 100 50 20 P P 2 0 0 20 40 60 0 0.01 0.02 0.03 1/P (in Hg) V 2V V(in3) (a) P vs V (b) V vs 1/P BOYLE’S LAW DATA Boyle’s Law* (Pressure)( Volume) = Constant (T = constant) P1V1 = P2V2 V 1 P (T = constant) (T = constant) (*Holds precisely only at very low pressures.) Boyle’s Law Calculations A 1.5-L sample of gaseous CCl2F2 has a pressure of 56 torr. If the pressure is changed to 150 torr, will the volume of the gas increase or decrease? What will the new volume be? Decrease P1 = 56 torr P2 = 150 torr V1 = 1.5 L V2 = ? V1P1 = V2 P2 V1P1 V2 = P2 (1.5 L)(56 torr) V2 = (150 torr) V2 = 0.56 L Boyle’s Law Calculations In an automobile engine the initial cylinder volume is 0.725 L. After the piston moves up, the volume is 0.075 L. The mixture is 1.00 atm, what is the final pressure? P1 = 1.00 atm V1P1 = V2 P2 P2 = ? V1P1 P2 = V2 V1 = 0.725 L V2 = 0.075 L (0.725 L)(1.00 atm) P2 = (0.075 L) P2 = 9.7 atm Is this answer reasonable? A gas that strictly obeys Boyle’s Law is called an ideal gas. 05_51 Ideal 22.45 Ne PV (L • atm) 22.40 O2 22.35 CO2 22.30 22.25 0 0.25 0.50 0.75 1.00 P (atm) Plot of PV vs. P for several gases at pressures below 1 atm. He 05_53 6 5 CH4 V(L) 4 3 H2O 2 H2 1 N2O -300 -200 -273.2 ºC -100 0 100 200 300 T(ºC) Plot of V vs. T(oC) for several gases 05_1543 Pext Pext Energy (heat) added Volume of a gas increases as heat is added when pressure is held constant. Charles’s Law The volume of a gas is directly proportional to temperature, and extrapolates to zero at zero Kelvin. V = bT (P = constant) b = a proportionality constant Charles’s Law V1 V2 T1 T2 ( P constant) Charles’s Law Calculations Consider a gas with a a volume of 0.675 L at 35 oC and 1 atm pressure. What is the temperature (in Co) of the gas when its volume is 0.535 L at 1 atm V1 T1 pressure? V1 = 0.675 L V2 = 0.535 L T1 = 35 oC + 273 = 308 K T2 = ? V2 = T2 T1V2 T2 = V1 (308 K)(0.535 L) T2 = (0.675 L) T2 = 244 K -273 T2 = - 29 o C 05_1544 Pext Pext Gas cylinder Moles of gas increases Pext Increase volume to return to original pressure At constant temperature and pressure, increasing the moles of a gas increases its volume. Avogadro’s Law For a gas at constant temperature and pressure, the volume is directly proportional to the number of moles of gas (at low pressures). V = an a = proportionality constant V = volume of the gas n = number of moles of gas AVOGADRO’S LAW V1 n1 V2 n2 AVOGADRO’S LAW A 12.2 L sample containing 0.50 mol of oxygen gas, O2, at a pressure of 1.00 atm and a temperature of 25 oC is converted to ozone, O3, at the same temperature and pressure, what will be the volume of the ozone? 3 O2(g) ---> 2 O3(g) 0.50 mol O2 2 mol O3 = 0.33 mol O3 1 3 mol O2 V1 = 12.2 L V1 n = 1 V2 n2 V2 = ? V2 = n1 = 0.50 mol V2 = n2 = 0.33 mol V1 n 2 n1 (12.2 L)(0.33 mol) (0.50 mol) V2 = 8.1 L COMBINED GAS LAW V1 P2 T1 V2 P1 T2 COMBINED GAS LAW What will be the new volume of a gas under the following conditions? V1 P2 T1 V2 P1 T2 V1 = 3.48 L V2 = ? V1P1T2 P1 = 0.454 atm V2 = P2 = 0.616 atm P2 T1 T1 = - 15 oC + 273 (309 K)(0.454 atm)(3.48 L) V2 = = 258 K (258 K)(0.616 atm) o T2 = 36 C + 273 V2 = 3.07 L T2= 309 K 05_1542 Pext Pext Temperature is increased Pressure exerted by a gas increases as temperature increases provided volume remains constant. If the volume of a gas is held V constant, then 1 = 1. V2 Therefore: P1 T1 P2 T2 Ideal Gas Law - An equation of state for a gas. - “state” is the condition of the gas at a given time. PV = nRT IDEAL GAS 1. Molecules are infinitely far apart. 2. Zero attractive forces exist between the molecules. 3. Molecules are infinitely small--zero molecular volume. What is an example of an ideal gas? REAL GAS 1. Molecules are relatively far apart compared to their size. 2. Very small attractive forces exist between molecules. 3. The volume of the molecule is small compared to the distance between molecules. What is an example of a real gas? Ideal Gas Law PV = nRT R = proportionality constant = 0.08206 L atm mol P = pressure in atm V = volume in liters n = moles T = temperature in Kelvins Holds closely at P < 1 atm Ideal Gas Law Calculations A 1.5 mol sample of radon gas has a volume of 21.0 L at 33 oC. What is the pressure of the gas? p=? pV = nRT p = nRT V V = 21.0 L n = 1.5 mol L atm )(306K) mol K (21.0L) (1.5mol)(0.08206 T = 33 oC + 273 p = T = 306 K R = 0.08206 Latm molK p = 1.8 atm Ideal Gas Law Calculations A sample of hydrogen gas, H2, has a volume of 8.56 L at a temperature of O oC and a pressure of 1.5 atm. Calculate the number of moles of hydrogen present. p = 1.5 atm pV = nRT V = 8.56 L L atm R = 0.08206 mol K n=? T = O oC + 273 T = 273K n= pV RT (1.5 atm)(8.56 L) n= atm*L (0.08206 )(273K) mol*K n = 0.57 mol Standard Temperature and Pressure “STP” P = 1 atmosphere T = C The molar volume of an ideal gas is 22.42 L at STP GAS STOICHIOMETRY 1. Mass-Volume 2. Volume-Volume Molar Volume pV = nRT V = nRT p V V = 22.4 L 0.08296 L* atm 1 mole 273K mole * K 1.00 atm Gas Stoichiometry Not at STP (Continued) p = 1.00 atm T = 25 oC + 273 = 298 K V=? n = 1.28 x 10-1 mol 0.0821atm*L R= mole*K pV = nRT nRT V= P 0.082 atm * L 1.28 10 mole 298.0 K mole * K V= 1atm -1 V = 3.13 L O2 Gases at STP A sample of nitrogen gas has a volume of 1.75 L at STP. How many moles of N2 are present? 1mole -2 (1.75 L N 2 )( ) = 7.81 10 mol N 2 22.4 Liters Gas Stoichiometry at STP Quicklime, CaO, is produced by heating calcium carbonate, CaCO3. Calculate the volume of CO2 produced at STP from the decomposition of 152 g of CaCO3. CaCO3(s) CaO(s) + CO2(g) 152g CaCO3 1 mol 1mol CO 2 22.4L = 34.1 L CO2 1 100.1g 1mol CaCO3 1mol Note: This method only works when the gas is at STP!!!!! Volume-Volume If 25.0 L of hydrogen reacts with an excess of nitrogen gas, how much ammonia gas will be produced? All gases are measured at the same temperature and pressure. N2(g) + 3H2(g) ----> 2NH3(g) 25.0 L H 2 2 mol NH3 1 3 mol H 2 = 16.7 L NH3 MOLAR MASS OF A GAS m n= M n = number of moles m = mass M = molar mass MOLAR MASS OF A GAS mRT P= VM or DRT P= M therefore: M = DRT P Molar Mass Calculations M=? D= 1.95 g/L T = 27 oC + 273 p = 1.50 atm T = 300. K L atm R = 0.0821 mol K DRT M= p g L atm 1.95 0.08206 300. K L mol K M= (1.5 atm) g M = 32.0 mol Dalton’s Law of Partial Pressures For a mixture of gases in a container, PTotal = P1 + P2 + P3 + . . . Dalton’s Law of Partial Pressures Calculations A mixture of nitrogen gas at a pressure of 1.25 atm, oxygen at 2.55 atm, and carbon dioxide at .33 atm would have what total pressure? PTotal = P1 + P2 + P3 PTotal = 1.25 atm + 2.55 atm + .33 atm Ptotal = 4.13 atm Water Vapor Pressure 2KClO3(s) → 2KCl(s) + 3O2(g) When a sample of potassium chlorate is decomposed and the oxygen produced collected by water displacement, the oxygen has a volume of 0.650 L at a temperature of 22 oC. The combined pressure of the oxygen and water vapor is 754 torr (water vapor pressure at 22 oC is 21 torr). How many moles of oxygen are produced? Pox = Ptotal - PHOH Pox = 754 torr - 21 torr pox = 733 torr Water Vapor Pressure Continued 733 torr 1 atm p= 0.964 atm 1 760 torr V = 0.650 L pV = nRT pV n= RT T = 22 oC + 273 (0.964 atm)(0.650 L) T = 295 K n= L atm L atm 0.08206 (295K) R = 0.0821 mol K mol K n=? n = 2.59 x 10-2 mol MOLE FRACTION -- the ratio of the number of moles of a given component in a mixture to the total number of moles of the mixture. n1 1 = n total V1 1 = Vtotal P1 1 = Ptotal (volume & temperature constant) Kinetic Molecular Theory 1. Volume of individual particles is zero. 2. Collisions of particles with container walls cause pressure exerted by gas. 3. Particles exert no forces on each other. 4. Average kinetic energy Kelvin temperature of a gas. Relative number of O2 molecules with given velocity 05_58 0 4 x 102 8 x102 Molecular velocity (m/s) Plot of relative number of oxygen molecules with a given velocity at STP (Boltzmann Distribution). Relative number of N2 molecules with given velocity 05_59 273 K 1273 K 2273 K 0 1000 2000 Velocity (m/s) 3000 Plot of relative number of nitrogen molecules with a given velocity at three different temperatures. The Meaning of Temperature (KE) avg 3 RT 2 Kelvin temperature is an index of the random motions of gas particles (higher T means greater motion.) rms 3RT M Diffusion: describes the mixing of gases. The rate of diffusion is the rate of gas mixing. Effusion: describes the passage of gas into an evacuated chamber. 05_60 Pinhole Gas Vacuum Effusion of a gas into an evacuated chamber Effusion: Rate of effusion for gas 1 Rate of effusion for gas 2 M2 M1 Diffusion: Distance traveled by gas 1 Distance traveled by gas 2 M2 M1 Real Gases Must correct ideal gas behavior when at high pressure (smaller volume) and low temperature (attractive forces become important). 05_62 CH4 N2 2.0 H2 PV nRT CO2 1.0 Ideal gas 0 0 200 400 600 800 1000 P(atm) PV nRT Plots of vs. P for several gases at 200 K. Note the significant deviation from ideal behavior. Real Gases [ Pobs n 2 a ( ) ] V nb nRT V corrected pressure corrected volume Pideal Videal 05_68 Molecules of unburned fuel (petroleum) 0.4 Other pollutants 0.3 NO2 0.2 O3 NO 6:00 4:00 2:00 Noon 10:00 8:00 0 6:00 0.1 4:00 Concentration (ppm) 0.5 Time of day Concentration for some smog components vs. time of day NO2(g)= NO(g) + O(g) O(g) + O2(g) = O3(g) NO(g) 1 + O2(g) = NO2(g) 2 __________________________ 3 O2(g) = O3(g) 2 What substances represent intermediates? Which substance represents the catalyst? Chemistry is a gas!!