Chapter 11 Gases
Energy order of the p p and s p orbitals changes across the period. Due to lower nuclear charge of B, C & N there is no s-p orbitals interaction Due to high nuclear charge of O, F& Ne there is an s-p orbitals interaction This is the correct order of molecular orbitals. The order given in the same slid of the presentation on Chapter 9 is wrong. O, F, Ne B, C, N
Topics Properties of gases The gas laws Avogadro The ideal gas equation Reactions with gaseous reactants and products Gas mixtures The kinetic molecular theory of gases Deviation from ideal behavior
11.1 Properties of Gases The properties of a gas are almost independent of its identity. (Gas molecules behave as if no other molecules are present.) Compressible (gas molcules are far apart from each other) Low density compared to liquids and solids Expand to fill a container (assume the shape and the volume of its container) Form homogeneous mixtures with one another in any proportion.
Gas pressure: definition Pressure: force per unit area newton (N): Unit of force pascal (Pa): Unit of pressure Standard pressure
Gas pressure: units Standard Atmospheric Pressure: 1 atm = 76 cm Hg = 760 mm Hg = 760 Torr = 101,35 Pa Very small unit, thus it is not commonly used
If a weatherman says that atmospheric pressure is 9.1 inches of mercury, what is it in torr?.54 cm10 mm 1torr 9.1 in 1in 1cm 1mm 739.6 torr Copyright McGraw-Hill 009 7
Example A. What is 475 mm Hg expressed in atm? 485 mm Hg x 1 atm = 0.65 atm 760 mm Hg B. The pressure of a tire is measured as 9.4 psi. What is this pressure in mm Hg? 9.4 psi x 1.00 atm x 760 mmhg = 1.5 x 10 3 mmhg 14.7 psi 1.00 atm
Calculation of atmospheric pressure Area 1 cm x 1 cm or 0.0001 m Mass: 1 kg where 9.80665 m/s is the gravitational constant.
Measurement of pressure Torricellian Barometer Barometer: an instrument used to measure atmospheric pressure 760 mmhg atmospheric pressure P = d g h d - density g - acc. of gravity h atmospheric pressure
Manometer Device for Measuring the Pressure of a confined gas Closed tube manometer Open tube manometer
11. The Gas Laws Gas laws empirical relationships among gas parameters Volume (V) Pressure (P) Temperature (T) Amount of a gas (n)
5. The Gas Laws of Boyle, Charles and Avogadro Boyle s Law: at constant n, T Charles Law: at constant n, P Avogadro s Law: at constant P, T PV = const V/T = const V/n = const
Boyle s law pressure-volume relationship at constant temperature P 1 V 1 =P V Copyright McGraw-Hill 009 14
1/Pressure (1/Torr) Pressure (Torr) 0.01 Boyle s Law Boyle's Law Slope= 1/k V p 1 0.01 0.008 800 700 0.006 600 Boyle's Law PV =k (at constant T and n) 0.004 500 400 0.00 300 P 1 V 1 = P V 0 00 100 0 0 50 100 150 00 Volume (L) 0 50 100 150 00 Volume (L)
Volume (L) 35 Charles s Law temperature-volume relationship at constant pressure Charles' Law V/T = b V = bt (constant P & n) V1 T 1 V T 30 5 0 15 10 5 0 0 100 00 300 400 500 Temperature (K)
Plots of V Versus T(Celsius) for Several Gases Volume of a gas Changes by 1 73 When the temp. Changes by 1 o C. I.e., at -73 o C, V=0???
All gases will solidify or liquefy before reaching zero volume.
Avogadro s law The volume of a gas sample is directly proportional to the number of moles in the sample at constant pressure and temperature V1 n 1 V n Copyright McGraw-Hill 009 19
Volume (L) Avogadro s Law Vn V = an (constant P& T) V1 V 10 Avogadro's Law n 1 n 100 80 60 40 0 0 0 1 3 4 5 moles
How will the volume of a given gas change if the quantity of gas, absolute temperature, and pressure, all double? Avogadro Charles Boyle 1/ 1 volume doubles 1
11.3 The Ideal Gas Law V 1 p VT Vn Boyle s law Charles's law Avogadro s law Universal gas constant V nt R( P ) PV nrt
The Ideal Gas Law PV = nrt V T P P T V V n P nr nr RT The Ideal Gas Law can be used to derive the gas laws as needed!
The value of R What is R for 1.00 mol of an ideal gas at STP (0 o C and 1.00 atm)?given that V of 1 mol of gas at STP=.4L PV nrt R PV nt R 1.00atmatm.4. LL 1.00mol (73K) R 0.081 mol. K R = 0.081 atm L mol -1 K -1
The Ideal Gas Law: Final and initial state problems PV T nr Constant PV 1 1 P V (Ideal gas equation) T 1 T
A steel cylinder with a volume of 68.0 L contains O at a pressure of 15,900 kpa at 3ºC. What is the volume of this gas at STP? P 1 15,900 kpa 1atm 101.3 kpa 157.0 atm P = 1 atm T 1 = 3 + 73 = 96 K T = 73 K V 1 = 68.0 L V =? PV PV = nrt nr constant T V PV T TP 1 1 1 157.0 atm68.0 L73 K 96 K1atm PV T PV T 1 1 1 9850 L
Copyright McGraw-Hill 009 7
Molar Volume At STP 4.0 g He 16.0 g CH 4 44.0 g CO 1 mole 1 mole 1mole (STP) (STP) (STP) V =.4 L V =.4 L V =.4 L
For an ideal gas, calculate the pressure of the gas if 0.15 mol occupies 338 ml at 3.0ºC. n = 0.15 mol V = 338 ml = 0.338 L T = 3.0 + 73.15 = 305.15 K P =? P PV = nrt L atm mol K 0.338 L 0.15 mol 0.0806 305.15 K P nrt V = 15.98 = 15.9 atm
Applications of the ideal gas equation Relation to density (d) density Relation to molar mass ( )
Molar mass of a gas P x V = n x R x T P x V = m x R x T M m = mass, in grams n m M M = molar mass, in g/mol Molar mass = m R T P V
Example A glass vessel weighs 40.1305 g when clean, dry and evacuated; it weighs 138.410 when filled with water at 5 C (d=0.9970 g cm -3 ) and 40.959 g when filled with propylene gas at 740.3 mm Hg and 4.0 C. What is the molar mass of polypropylene? PV nrt PV m M RT M mrt PV Volume of the vessel
m H V O flask = d H O 138.410 g 40.1305 g (0.9970 g cm -3 = 98.41 cm 3 = 0.09841 L ) m gas = m filled - m empty = (40.959 g 40.1305 g) = 0.1654 g
PV = nrt PV = m M RT M = m RT PV M = (0.6145 g)(0.0806 L atm mol -1 K -1 )(97. K) (0.9741 atm)(0.09841 L) M = 4.08 g/mol
Density Density (d) is mass divided by volume P x V = m x R x T M P = m x R x T V x M d = m P = V d x R x T M d PM RT
Example Calculate the density in g/l of O gas at STP. From STP, we know the P and T. P = 1.00 atm T = 73 K Rearrange the ideal gas equation for moles/l d = PXM R x T
d PXM RXT d (1.00 atm ) (0.081 X (3.0 L.atm mol.k g mol ) )X (73K) 1.43g / L The density of O gas at STP is 1.43 grams per liter
Example.00 g sample of SX 6 (g) has a volume of 39.5 Cm 3 at 1.00 atm and 0 o C. Identify the element X. Name the compound P= 1.00 atm V 1L 39.5 Cm3X 1000Cm3 T = 73+0 = 93K M = m RT PV 0.395L
MM L. atm (.00g)(0.081 mol. K (1.00atm)(0.395L) )(93K) = 146 g SX 6 /mol f Molar mass of (X 6 )= 146-3 = 114 g/mol Molar mass of X = (114 g/mol X 6 ) /6 = 19 X = with a molar mass of 19 = F The compound is SF 6
d PXM RXT d (1.00 atm ) (0.081 X (3.0 L.atm mol.k g mol ) )X (73K) 1.43g / L The density of O gas at STP is 1.43 grams per liter
Example.00 g sample of SX 6 (g) has a volume of 39.5 Cm 3 at 1.00 atm and 0 o C. Identify the element X. Name the compound P= 1.00 atm V 1L 39.5 Cm3X 1000Cm3 T = 73+0 = 93K M = m RT PV 0.395L
MM L. atm (.00g)(0.081 mol. K (1.00atm)(0.395L) )(93K) = 146 g SX 6 /mol f Molar mass of (X 6 )= 146-3 = 114 g/mol Molar mass of X = (114 g/mol X 6 ) /6 = 19 X = with a molar mass of 19 = F The compound is SF 6
11.4 Reactions with Gaseous Reactants and Products Amounts of gaseous reactants and products can be calculated by utilizing The ideal gas law to relate moles to T, P and V. Moles can be related to mass by the molar mass The coefficients in the balance equation to relate moles of reactants and products Standard Temperature and Pressure (STP): 0ºC and 1 atm 1 mole of gas occupies.4 L at STP.
Carbon monoxide reacts with oxygen to form carbon dioxide according to the equation: CO(g) + O (g) CO (g) What volume of O is required to completely react with 65.8 ml of CO at constant temperature and pressure? Use the fact that ml of reactant are proportional to moles of reactant. 65.8 ml of CO 1 ml of O ml of CO 3.9 ml O
Magnesium is an active metal that replaces hydrogen from an acid by the following reaction: Mg(s) + HCl(aq) MgCl (aq) + H (g) How many g of Mg are needed to produce 5.0 L of H at a temperature of 5 o C and a pressure of 745 mmhg? Mg(s) + HCl(aq) MgCl (aq) + H (g)? g 5.0 L Hint: find moles of H using PV = nrt then work as a stoichiometry problem. n = PV RT = 745 mmhg 5.0 L mol K 6.4 L mmhg 98 K n = 0.0 mol
Example 30. ml of 1.00 M HCl are reacted with excess FeS. What volume of gas is generated at STP? HCl + FeS FeCl + H S(g) # moles HCl = Vol (L) X M
HCl + FeS FeCl + H S 1.00mol HCl mol 0.081atm.030 1.00 mol HCl mol H 73.15 S 0.081atm L 0.030 L mol HCl mol 73. 15 K L mol HCl mol K 1.00atm PV = nrt V = nrt/p 0.339 L
Example The decomposition of sodium azide, NaN 3, at high temperatures produces N (g). What volume of N (g), measured at 735 mm Hg and 6 C, is produced when 70.0 g NaN 3 is decomposed. NaN 3 (s) Na(l) + 3 N (g)
NaN 3 (s) Na(l) + 3 N (g) Determine moles of N : n N = 70 g NaN 3 1 mol NaN 3 3 mol N X X = 1.6 mol N 65.01 g N 3 /mol N 3 mol NaN 3 Determine volume of N : nrt V = = P (1.6 mol)(0.0806 L atm mol -1 K -1 )(99 K) (735 mm Hg) 1.00 atm 760 mm Hg = 41.1 L
Relation of changes in pressure to moles in a reaction Example At constant temperature and volume Copyright McGraw-Hill 009 50
11.5 Gas Mixtures In gaseous mixtures, each gas behaves as if it occupies the container alone. Assuming no reaction between gases partial pressure (P i ): the pressure exerted by each gas in a gaseous mixture Dalton s law of partial pressure The total pressure is the sum of the partial pressures. P Total = P 1 + P + P 3 + P 4 + P 5... P t = SP i
Schematic of Dalton s Law P N P total = P N + P O Add O
P Total = n 1 RT + n RT + n 3 RT +... V V V In the same container R, T and V are the same. P Total = (n 1 + n + n 3 +...)RT V Thus, P Total = P 1 + P + P 3 + P 4 + P 5... P Total n Total ( RT V )
A 50.0 ml flask contains 1.00 mg of He and.00 mg of H at 5.0 o C. Calculate the total gas pressure in the flask in atmospheres. The total pressure is due to the partial pressures of each of these gases. so: P P P (n n total He H He H For He: 1.00 x 10-3 g He mol =.50 x 10-4 4.00 g For H : mol He.00 x 10-3 g H mol = 9.9 x 10-4 mol H.016 g ) RT V
The mole fraction Mole fraction: number of moles of one component in a mixture relative to the total number of moles in the mixture symbol is Greek letter chi c c 1 n 1 n Total = n 1 n n 1 n 3...
Mole fraction expressed in pressures... 3 1 1 1 n n n n n n Total c... ); ( ); ( 1 1 RT V P n RT V P n... ) ( ) ( ) ( ) ( 3 1 1 1 1 RT V P RT V P RT V P RT V P n n Total c...) )( ( ) ( 3 1 1 1 P P P RT V RT V P c P Total P P P P P 1 3 1 1 1...) ( c Total i Total i i P P n n c Total i i P P c
P Example A 1.00 L sample of dry air at 786 Torr and 5 o C contains 0.95 g N plus other gasses (such as O, Ar and CO.) a) What is the partial pressure of N? b) What is the mole fraction of N? mol 0.95g 0. 0330mol 8.0g 1 1 0.0330mol 0.081atm L mol K 98 K N c N P Total 1.00 L PV = nrt P = nrt/v 0.807 atm 760Torr 613Torr 0.807 atm 613Torr X N 0. 780 atm 786Torr
Collecting gas over water An insoluble gas is passed into a container of water, the gas rises because its density is much less than that of water and the water must be displaced KClO 3 O gas
Collection of Gases over Water Assuming the gas is saturated with water vapor, the partial pressure of the water vapor is the vapor pressure of the water. P total = P gas + P H O(g) P gas = P total P H O(g)
Oxygen was produced and collected over water at ºC and a pressure of 754 torr. KClO 3 (s) KCl(s) + 3 O (g) 35 ml of gas were collected and the vapor pressure of water at ºC is 1 torr. Calculate the number of moles of O and the mass of KClO 3 decomposed.
P total = P O + P H O = P O + 1 torr = 754 torr P O = 754 torr 1 torr = 733 torr = 733 / 760 atm V = 35 ml = 0.35 L T = ºC + 73 = 95 K n 733 atm0.35 L 760 1.910 mol O Latm 0.0806 (95 K) mol K O KClO 3 (s) KCl(s) + 3 O (g) n PV RT 1.910 mol O mol KClO 3 mol O 3 1.6 g KClO 3 1mol KClO 3 1.06 g KClO 3
A 50.0 ml flask contains 1.00 mg of He and and.00 mg of H at 5.0 o C. Calculate the total gas pressure in the flask in atmospheres. so: Ptotal PHe PH (nhe nh ) For He: 1.00 x 10-3 g He mol =.50 x 10-4 mol He 4.00 g For H :.00 x 10-3 g H mol = 9.9 x 10-4 mol H.016 g RT V And: P total = (.50 x 10-4 + 9.9 x 10-4 )(RT/V) = (0.0014 mol)(0.081 L atm)(5 + 73)K mol K (0.500 L) P total = 0.116 atm
11.6 The Kinetic Molecular Theory It explains why ideal gases behave the way they do. Postulates of the kinetic Theory: A gas is composed of particles that are separated by relatively large distances. The volume occupied by individual molecules is negligible. Gas molecules are constantly in random motion, moving in straight paths, colliding with the walls of their container and with one another in perfectly elastic collisions. Gas particles exert no attractive or repulsive forces on one another. The average kinetic energy of the particles is proportional to the absolute temperature.
Application of KMT to the gas laws KMT explains ideal gas laws Compressibility: gases are compressible because the gas molecules are separated by large distances. P&V: P = (nrt). (1/V) P 1/V (Boyle s law) When V decreases # collisions increases P & T: P = (nr/v).t P T When T increases, hits with walls become stronger and more frequent V & T: V=(nR/P).T V T (Charle s law) When T increases hits with walls become stronger and more frequent. To keep P constant, V must increase to compensate for particles speeds
V & n: V= (RT/P). N V n (Avogadro s law) To maintain constant P and T, as V increases n must increase When n increase P would increase if the volume is kept constant. V must increase to return P to its original value P total = P i (Dalton s law) Individual particles are independent of each other and their volumes are negligible. Thus identities of gas particles do not matter and would be treated as if they belong to one gas
Molecular speed Total kinetic energy of a mole of gas is 3/ RT 1 The average kinetic energy of one molecule is mu The average kinetic energy for one mole of gas molecules is (N A is Avogadro s number) 1 3 N A( mu ) ( RT ) avg Because m X N A = molar mass = M u 3RT M 3RT M Root mean square (rms) speed (u rms ) u Where M is the molar mass in kg/mole, and R has the units 8.3145 J/Kmol. The velocity will be in m/s
How fast do N molecules move at room temperature (5 o C)? u 3RT M u 3RT M 3 8.314 kgm s mol K 98 K 8.0 10-3 kg mol 515 m s 1150 mph 67 Copyright McGraw-Hill 009
For two gases 1 and u u rms rms 1 3RT M 1 3RT M M M 1 Thus, u u rms rms 1 M M 1
Effect of Molar Mass on Molecular Speed so smaller molar masses result in higher molecular speeds 69 Copyright McGraw-Hill 009
Molecular speed for same gas at two different temperatures u T 1 3 ( M RT 1 ) 1/ u T 3 ( RT M ) 1/ u u T T 1 T ( T 1 ) 1/
Molecular speed for two different gases at two different temperatures u T 1 3 ( M RT 1 1 ) 1/ u T 3 ( M RT ) 1/ u u T T 1 T M ( T M 1 ) 1/ 1
Effect of Temperature on Molecular Speed so higher temperatures result in higher molecular speeds 7 Copyright McGraw-Hill 009
Comparison of rms and other speed measurements Mean or average speed (u avg ) Most probable speed (u mp ) Rms speed (u rms ) Example: Assume five speeds:, 4, 4, 6 and 8 m/s u avg 4 4 6 8 4 5 5 4.8 m/s u mp 4.0 m/s u rms 4 4 6 8 136 5 5 5. m/s
Place the following gases in order of increasing r.m.s. speed at 300 K, H, CO, Ne, NH 3, Cl u Cl < u CO < u Ne < u NH3 < u H Which one has the highest average kinetic energy? At the same temperature, all have the same average kinetic energy.
Diffusion: Mixing of gases as a results of random motion and collisions. Open valve
Effusion The escape of a gas from a container to a region of vacuum 76 Copyright McGraw-Hill 009
Graham s Law The rate of diffusion or effusion is inversely proportional to the square root of the molar mass of particles. Rate of Rate of diffusion/ effusion for gas 1 diffusion/effusion for gas M M 1
11.7 Deviation from Ideal Behavior Real gases do not always behave ideally under certain conditions due to Gas molecules occupy significant volume (at high pressures) Gas molecules experience intermolecular forces of attraction and repulsion (at low temperatures) Effect of intermolecular forces on P
Van der Waal s equation corrects for Pressure deviations where a is a constant Volume effects where b is a constant
The ideal gas law becomes van der Waal s equation a and b have specific values for each gas 80 Copyright McGraw-Hill 009
81 Copyright McGraw-Hill 009
Properties of gases Gas pressure Units Calculation Measurement The gas laws Boyle s law Charles law Key Points 8 Copyright McGraw-Hill 009
Avogadro s law The ideal gas law Key Points Reactions with gaseous reactants and products Gas mixtures Dalton s law Mole fractions Partial pressures
Key Points The kinetic molecular theory Assumptions Application to the gas laws Molecular speed Diffusion and effusion Deviation from ideal behavior Factors causing deviation Van der Waal s equation 84 Copyright McGraw-Hill 009