4 Physical Characteristics of Gases Gases assume the volume and shape of their containers.Gases are the most compressible state of matter.Gases will mix evenly and completely when confined to the same container.Gases have much lower densities than liquids and solids.10.1
5 (force = mass x acceleration) AreaBarometerPressure =(force = mass x acceleration)Units of Pressure1 pascal (Pa) = 1 N/m21 bar = 105 Pascals1 atm = 760 mmHg = 760 torr1 atm = 101,325 Pa1 atm = kPa10.2
6 10 miles 0.2 atm 4 miles 0.5 atm Sea level 1 atm 10.2
8 Apparatus for Studying the Relationship Between Pressure and Volume of a GasAs P (h) increasesV decreases10.3
9 Boyle’s LawThe volume of a fixed quantity of gas at constant temperature is inversely proportional to the pressure.10.3
10 Boyle’s Law P a 1/V Constant temperature P x V = constant Constant amount of gasP x V = constantP1 x V1 = P2 x V210.3
11 P x V = constant P1 x V1 = P2 x V2 P1 = 726 mmHg P2 = ? V1 = 946 mL A sample of chlorine gas occupies a volume of 946 mL at a pressure of 726 mmHg. What is the pressure of the gas (in mmHg) if the volume is reduced at constant temperature to 154 mL?P x V = constantP1 x V1 = P2 x V2P1 = 726 mmHgP2 = ?V1 = 946 mLV2 = 154 mLP1 x V1V2726 mmHg x 946 mL154 mL=P2 == 4460 mmHg10.3
13 Variation of gas volume with temperature at constant pressure. Charles’ & Gay-Lussac’s LawV a TTemperature must bein KelvinV = constant x TV1/T1 = V2 /T2T (K) = t (0C)10.3
14 A plot of V versus T will be a straight line. Charles’s LawThe volume of a fixed amount of gas at constant pressure is directly proportional to its absolute temperature.i.e.,VT= kA plot of V versus T will be a straight line.10.4
15 A sample of carbon monoxide gas occupies 3. 20 L at 125 0C A sample of carbon monoxide gas occupies 3.20 L at 125 0C. At what temperature will the gas occupy a volume of 1.54 L if the pressure remains constant?V1 /T1 = V2 /T2V1 = 3.20 LV2 = 1.54 LT1 = KT2 = ?T1 = 125 (0C) (K) = KV2 x T1V11.54 L x K3.20 L=T2 == 192 K10.3
16 Mathematically, this means Avogadro’s LawThe volume of a gas at constant temperature and pressure is directly proportional to the number of moles of the gas.Mathematically, this meansV = kn10.3
17 Avogadro’s Law V a number of moles (n) V = constant x n Constant temperatureConstant pressureV = constant x nV1 / n1 = V2 / n210.3
18 4NH3 + 5O2 4NO + 6H2O 1 mole NH3 1 mole NO At constant T and P Ammonia burns in oxygen to form nitric oxide (NO) and water vapor. How many volumes of NO are obtained from one volume of ammonia at the same temperature and pressure?4NH3 + 5O NO + 6H2O1 mole NH mole NOAt constant T and P1 volume NH volume NO10.3
19 Final and Initial State Problems Use Combined Gas LawList all variables (P,V,n,T) in two columns… one for initial state, one for final state.Substitute and solve ( make sure units cancel)
20 Final and Initial State Problems A sample of gas occupies 355 ml and 15oC and 755 torr of pressure. What temperature will the gas have at the same pressure if its volume increases to 453 ml?
21 Final and Initial State Problems A sample of gas at 30.0o has a pressure of 1.23 bar and a volume of m3 . If the volume of the gas is compressed to m3 at the same temperature, what is its pressure at this volume?
22 Final and Initial State Problems A Sample of gas at 27.0 oC has a volume of 2.08 L and a pressure of mm Hg. If the gas is in a sealed container, what is its pressure in bar when the temperature (in oC) doubles?
23 Ideal-Gas Equation V nT P So far we’ve seen that V 1/P (Boyle’s law)V T (Charles’s law)V n (Avogadro’s law)Combining these, we getV nTP10.3
27 Ideal-Gas Equation PV = nRT nT P V nT P V = R or The relationship then becomesorPV = nRT10.4
28 Ideal-Gas EquationThe constant of proportionality is known as R, the ideal gas constant.10.4
29 Ideal Gas Equation 1 Boyle’s law: V a (at constant n and T) P Charles’ law: V a T (at constant n and P)Avogadro’s law: V a n (at constant P and T)V anTPV = constant x = RnTPR is the gas constantPV = nRT10.4
30 PV = nRT PV (1 atm)(22.414L) R = = nT (1 mol)(273.15 K) The conditions 0 0C and 1 atm are called standard temperature and pressure (STP).Experiments show that at STP, 1 mole of an ideal gas occupies L.PV = nRTR =PVnT=(1 atm)(22.414L)(1 mol)( K)R = L • atm / (mol • K)10.4
31 PV = nRT nRT V = P 1.37 mol x 0.0821 x 273.15 K V = 1 atm V = 30.6 L What is the volume (in liters) occupied by 49.8 g of HCl at STP?T = 0 0C = KP = 1 atmPV = nRTn = 49.8 g x1 mol HCl36.45 g HCl= 1.37 molV =nRTPV =1 atm1.37 mol x x KL•atmmol•KV = 30.6 L10.4
32 PV = nRT n, V and R are constant nR V = P T = constant P1 T1 P2 T2 = Argon is an inert gas used in lightbulbs to retard the vaporization of the filament. A certain lightbulb containing argon at 1.20 atm and 18 0C is heated to 85 0C at constant volume. What is the final pressure of argon in the lightbulb (in atm)?PV = nRTn, V and R are constantnRV=PT= constantP1 = 1.20 atmT1 = 291 KP2 = ?T2 = 358 KP1T1P2T2=P2 = P1 xT2T1= 1.20 atm x358 K291 K= 1.48 atm10.4
33 Single State ProblemsUse Ideal gas Law – you’re given three for the four variables and R - solve for the unknown.PV = nRT
34 Single State ProblemsHow many grams of oxygen gas in a 10.0 L container will exert a pressure of 712 torr at a temperature of 25.0 o C?
35 Single State ProblemsAt what temperature will grams of sulfur dioxide in a ml container exert a pressure of atm?
36 Other calculations involving the Ideal Gas Law We can often determine the molar mass of a gas by measuring its mass at a given volume, temperature and pressure.Ideal Gas Law can also be used to determine the density of a gas at given conditions of temperature and pressure.
37 Densities of Gases n P V = RT If we divide both sides of the ideal-gas equation by V and by RT, we getnVPRT=10.4
38 So multiplying both sides by the molecular mass ( ) gives Densities of GasesWe know thatmoles molecular mass = massn = mSo multiplying both sides by the molecular mass ( ) givesPRTmV=10.4
39 Densities of Gases P RT m V = d = Mass volume = densitySo,PRTmV=d =Note: One only needs to know the molecular mass, the pressure, and the temperature to calculate the density of a gas.10.4
40 Molecular Mass P d = RT dRT P = We can manipulate the density equation to enable us to find the molecular mass of a gas:PRTd =BecomesdRTP =10.4
41 d is the density of the gas in g/L Density (d) Calculationsm is the mass of the gas in gmV=PMRTd =M is the molar mass of the gasMolar Mass (M ) of a Gaseous SubstancedRTPM =d is the density of the gas in g/L10.4
42 dRT P M = d = m V = = 2.21 1 atm x 0.0821 x 300.15 K M = M = A 2.10-L vessel contains 4.65 g of a gas at 1.00 atm and C. What is the molar mass of the gas?dRTPM =d =mV4.65 g2.10 L== 2.21gL2.21gL1 atmx x KL•atmmol•KM =M =54.6 g/mol10.4
43 When 4. 93 grams of carbon tetrachloride gas are in a 1 When 4.93 grams of carbon tetrachloride gas are in a 1.oo liter container at K, the gas exerts a pressure of torr. What is the molar mass of carbon tetrachloride?
44 A gaseous compound with a mass of 3 A gaseous compound with a mass of grams has a volume of 2236 mL at 27.0oC and 735 mm Hg. Calculate the molar mass of the compound.
45 Calculate the density of ammonia at 0oC and 1 atm of pressure.
46 Calculate the molar mass of a compound with a density of 0 Calculate the molar mass of a compound with a density of g/L at 25.0oC and torr pressure.
47 C6H12O6 (s) + 6O2 (g) 6CO2 (g) + 6H2O (l) Gas StoichiometryWhat is the volume of CO2 produced at 37 0C and 1.00 atm when 5.60 g of glucose are used up in the reaction:C6H12O6 (s) + 6O2 (g) CO2 (g) + 6H2O (l)g C6H12O mol C6H12O mol CO V CO21 mol C6H12O6180 g C6H12O6x6 mol CO21 mol C6H12O6x5.60 g C6H12O6= mol CO20.187 mol x x KL•atmmol•K1.00 atm=nRTPV == 4.76 L10.5
48 Dalton’s Law of Partial Pressures V and T are constantP1P2Ptotal = P1 + P25.6
49 PA = nART V PB = nBRT V XA = nA nA + nB XB = nB nA + nB PT = PA + PB Consider a case in which two gases, A and B, are in a container of volume V.PA =nARTVnA is the number of moles of APB =nBRTVnB is the number of moles of BXA =nAnA + nBXB =nBnA + nBPT = PA + PBPA = XA PTPB = XB PTPi = Xi PTmole fraction (Xi) =ninT5.6
50 Pi = Xi PT PT = 1.37 atm 0.116 8.24 + 0.421 + 0.116 Xpropane = A sample of natural gas contains 8.24 moles of CH4, moles of C2H6, and moles of C3H8. If the total pressure of the gases is 1.37 atm, what is the partial pressure of propane (C3H8)?Pi = Xi PTPT = 1.37 atm0.116Xpropane ==Ppropane = x 1.37 atm= atm5.6
51 PT = PO + PH O 2KClO3 (s) 2KCl (s) + 3O2 (g) 5.6 Bottle full of oxygen gas and water vapor2KClO3 (s) KCl (s) + 3O2 (g)PT = PO + PH O25.6
53 Scuba Diving and the Gas Laws Chemistry in Action:Scuba Diving and the Gas LawsDepth (ft)Pressure (atm)1332663PV5.6
54 Kinetic Molecular Theory of Gases A gas is composed of molecules that are separated from each other by distances far greater than their own dimensions. The molecules can be considered to be points; that is, they possess mass but have negligible volume.Gas molecules are in constant motion in random directions, and they frequently collide with one another. Collisions among molecules are perfectly elastic.Gas molecules exert neither attractive nor repulsive forces on one another.The average kinetic energy of the molecules is proportional to the temperature of the gas in kelvins. Any two gases at the same temperature will have the same average kinetic energyKE = ½ mu25.7
55 Kinetic theory of gases and … Compressibility of GasesBoyle’s LawP a collision rate with wallCollision rate a number densityNumber density a 1/VP a 1/VCharles’ LawCollision rate a average kinetic energy of gas moleculesAverage kinetic energy a TP a T5.7
56 Kinetic theory of gases and … Avogadro’s LawP a collision rate with wallCollision rate a number densityNumber density a nP a nDalton’s Law of Partial PressuresMolecules do not attract or repel one anotherP exerted by one type of molecule is unaffected by the presence of another gasPtotal = SPi5.7
57 Apparatus for studying molecular speed distribution 5.7
58 3RT urms = M The distribution of speeds of three different gases at the same temperatureThe distribution of speedsfor nitrogen gas moleculesat three different temperaturesurms =3RTM5.7
59 Chemistry in Action: Super Cold Atoms Gaseous Rb Atoms1.7 x 10-7 KBose-Einstein Condensate
60 M2 M1 NH4Cl NH3 17 g/mol HCl 36 g/mol Gas diffusion is the gradual mixing of molecules of one gas with molecules of another by virtue of their kinetic properties.M2M1r1r2=NH4ClNH317 g/molHCl36 g/mol5.7
61 Gas effusion is the is the process by which gas under pressure escapes from one compartment of a container to another by passing through a small opening.M2M1r1r2t2t1==Nickel forms a gaseous compound of the formula Ni(CO)x What is the value of x given that under the same conditions methane (CH4) effuses 3.3 times faster than the compound?M2 =r1r2()2x M1r1 = 3.3 x r2= (3.3)2 x 16 = 174.2M1 = 16 g/molx • 28 = 174.2x = 4.1 ~ 45.7
62 Deviations from Ideal Behavior 1 mole of ideal gasRepulsive ForcesPV = nRTn =PVRT= 1.0Attractive Forces5.8
63 Effect of intermolecular forces on the pressure exerted by a gas. 5.8
64 ( ) } } Van der Waals equation nonideal gas an2 P + (V – nb) = nRT V2 correctedpressure}correctedvolume5.8