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Unformatted text preview: CHAPTER FIVE GASES Pressure 21. 22. 23. 10 mm 1 atm 4.75 cm X = 47.5 mm Hg or 47.5 torr; 47.5 torr X = 6.25 X 10'2 atrn
cm ' a 760 torr 1.013 x105 Pa
atm 6.25 X 10'2 atm X = 6.33 X 103 Pa If the levels of Hg in each arm of the manometer are equal, then the pressure in the ﬂask is equal to
atmospheric pressure. When they are unequal, the difference in height in ﬁrm will be equal to the
difference in pressure in mm Hg between the ﬂask and the atmosphere. Which level is higher will
tell us whether the pressure in the ﬂask is less than or greater than atmospheric. a. Pﬂask < Pm; Pﬂask = 760.  118 = 642 torr 1 atm
760 torr = 0.845 atm 1.013 x105 Pa
atm 0.845 atm X = 8.56 X 104 Pa b. Pﬂask > Pm; PM; =1 760. torr + 215 torr = 975 torr 975 torr x 1 an“ = 1.28 atm
760 torr
s ’.
m =13o x‘105 pa 1.28 atm X
' atm c. Pﬂﬂk=635118=517torr; Pﬂgk=635+215=850.torr suppose we have a column of Hg 1.00 cm X 1.00 cm X 76.0 em = V = 76.0 cm3: 1 kg
1000 g mass = 76.0 cm3 X 13.59 g/cm3 = 1.03 X 103 g X = 1.03 kg
F =mg=1.03 kg X 9.81 m/s2= 10.1 kg m/s2 =10.1N 85‘ 86 CHAPTER 5 GASES 24. 25. 2
Force 10'1 N x( 100m) = 1.01 x 105 1 or 1.01 x 105 Pa 2 area ' cm 2 m m (Note: 76.0 cm Hg = 1 atm = 1.01 X 105 Pa.) To exert the same pressure, a column of water will have to contain the same mass as the 76.0 cm column of Hg. Thus, the column of water will have to be 13.59 times taller or 76.0 cm x 13.59 =
1.03 ><‘103 em = 10.3 m. r a. The pressure is proportional to the mass of the ﬂuid. The mass is proportional to the volume of the column of ﬂuid (or to the height of the column assuming the area of the column‘of ﬂuid is
constant). ' mass d = density = ; The volume of silicon oil is the same as the volume of Hg in Exercise
volume 522.
m m . m d . VH8=V°ﬂ; _Hg_: 011, mo“: Hg’oﬂ
d ng_ doil ng Since P is proportional to the mass of liquid: dHg 13.6 d .
Poi, = PH8[ “J = Pm [139) = 0.0956 PHg This conversion applies only to the column of silicon oil. 1 a. PM = 760. torr  (118 X 0.0956) torr = 760.  11.3 = 749 torr ' 5
1 atm = 0.986 atm; 0.986 atm x W 760 torr atm l b. PM= 760. torr + (215 X 0.0956) torr = 760. + 20.6= 781 ton 5
mm =1.03 atm; 1.03 atm x W
760 torr V = 9.99 X 104 Pa 749 torr X 781 torr >< = 1.04 X 105 Pa
atm ‘ b. If we are measuring the same pressure, the height of the silicon oil column would be 13.6 + 1.30 = 10.5 times the height of a mercury column. The advantage of using a less dense ﬂuid than mercury is in measuring small pressures. The height difference measured will be larger for the
less dense ﬂuid. Thus, the measurement will be more precise. a. 4.8amxm=36x1o3mmHg; b. 3.6X103mmHgX “0” =3.6><103torr
atm » _ MHg
5 ~ _' 
c. 4.8 atmx Maw 105 Pa; d. 4.8 atmx 14'7p51=71psi atm atm 28. 29. 30. 31. 32. 33. P v
For H2: P2= ‘ 1 =360.torr>< 2'00]? =240.torr V2 3.00L ,, PTO,r = PH? + PNZ, PNZ = PTOT — PHz = 320. torr  240. torr= 80. torr P V 
Foer: Pl= 2 2 =80. torrx 3‘001“ =240torr
V1 1.00L
. n T n T ‘
'PV =nRT, ﬂ = X = constant, ——1—1 = g; mol x molar mass =mass
P R P P l 2 n1 (molar mass) T1 112 (molar mass) T2 mass1 x T1 mass2 x T2 P1 P2 P1 P2
mass xT P 3 '
mas52= 1 1 2 =1.00x10 gx291Kx050.psr =309g
TZP1 299 K x 2050. ps1
P V P V V P T
PV=nRT,nisconstant. EX=nR=constanL 1 1: 2 2,V2= 112
T T1 T2 P2T1
V2=1.00L x 760' m“ x w =2.82 L; AV=2;82  1.00 =1.82L
220. torr (273 + 23) K ~ n RT 220 g x 1 mol x 0.08206 L atm x 300' K
_ C02 _ 44.01 g mol K  = 3.08 atm
4.00 L P=P c02‘ V With air present, the partial pressure of C02 will still be 3 .08 atm. The total pressure will be the s
of the partial pressures. . latm
P =P +P.=3.08 atm+ 740.torrx =3.08+0.974=4.05 atm
W C02 3“ _ [ 760 tort) 
‘ P V P V
ﬂ=nR=aconstang ————l 1= 2 2 v‘
T T1 T2 :
P v T 2 P2 = 1 1 2 =710torr>< _____5'0X10 In“ x——————(273 +820”; =5.1x104torr «‘
VZTI . 25 mL (273 + 30.) K D: g = p 135 atmx200.0L =Mlx103mol
RT 0.08206 La‘m x (273 + 24) K mol K CHAPTER 5 GASES 89 ForHe: 1.11x103m01x199513 =4.44 x 103 gHe
 . mo _ Foer: 1.11 x 103mol x Egg—€53 =2.24 x 103gH2
m0 I 060 g x 1 mol ) X 0.08206 L atm x (273 + 22) K
34. P ___ nRT : 32.00 g mol K = 0091 am
V, 5.0 L
35. As N02 is converted completely into N204, the moles of gas present will decrease by a factor of one
half (from the 2:1 mol ratio in the balanced equation). Using Avogadro’s law, V1_‘V2’V_V n2_ 1~ ———.——, 2— 1><———25.0.mLx——12.5mL n1 n2 n1 2 N204(g) will occupy onehalf the original volume of N02(g). I P V P V x V
36. PV=nRT, nis constant. £1] =nR=constant, 4—1—1 = 2 2; V2= 1.040 V1, so —1 = T T1 T2 V2 1.040
P V T
P2: 1 1 2:75.psix 1.000 x (273 +58)K=82psi
V T ‘ 1.040 (273 + 19)K 2 1 n T n T
37. PV =nRT, P is constant. ﬂ = E— = constant, —1—1 = —23
‘ g V R V1 V2
11 T V 3‘ 3 \ ‘ V
_2= 1 2 = 294Kx4.20x10 m =0'921 ’ . 11} Tzvl K X 103 m3 38. Wecan use the ideal gas law to calculate the partial pressure of each gas or to calculate the total
pressure. There will be less math if we calculate the total pressure ﬁom the ideal gas law.'
1 mol 02 no = 1.80 x 102 mg 02 x _1—g— x __ = 5.63 x 103 mol 02
.2 1000 mg 32.00 g‘O2 18 A 3 I
mm = 9.00 x 10 molecules NO x 1000 cm X ‘ 1 mol N0 = 1.49 x 10.2 mol NO/L
‘ cm3 , L . 6.022 x 1023 molecules NO nw = nNZ + n02 + nNO = 2.00 x 102 + 5.63 x 103 +1.49 x 102 = 4.05 x 102 mol 4.05 x 10‘2 mol x W x 273 K
RT mol K . . P = 0.907 atm
“"“ V ' 1.00 L 2.00 x 102 mol Nz P ax P = ————————— x 0.907 atm = 0.448 atm; x = mole fraction 0le
“2 “2 m‘ 4.05 x 10*2 mol total ' N2 2 92 CHAPTER 5 GASES “ 46. P X (molar mass) = dRT, d = mass , p x (molar mass) = mass x RT
volume V
0.800 x 00820? 1]; mm x 373 K
Molar mass = = M = ___—__.11£—_.—__— = 96.9 g/mol
PV (750. torr x' mm )x 0256 L
760 torr Mass of CHCl c 12.0 + 1.0 + 35.5 = 48.5; = 2.00; Molecular formula is C2H2C12. 47. IfBe“, the formula is Be(C5H702)3 and molar mass a 13.5 + 1502) + 21(1) + 6(16) = 311 g/mol. IfBe“, the formula is Be(C5}L02)2 and molar mass = 9.0 + 10(12) + 14(1) + 4( 16) = 207 g/mol. Data Set I (molar mass = dRT/P and d = massN):v
0.2022 g x 0.08206 L atm x 286 K
molar mass = M = m01 = 209 g/mol
PV (765.2 torr x 1 mm )x 22.6 x 10‘3 L
760 torr
Data set IL 0.2224 g x W x 290. K \
molar mass = M = ‘ “101 K = 202 g/mol
‘ PV _ (764.6 torr x 1 3"” )x 26.0 x 10‘3 L
760 torr . These results are close to the expected value of 207 g/mol for Be(C5}L02)2. Thus, we conclude from
these data that beryllium is a divalent element with an atomic weight (mass) of 9.0 g/mol. 48. We assume that 28.01 g/mol is the true value for the. molar mass of N2. The value of 28.15 g/mol is the average molar mass of the amount of N2 and Ar in air. Let x '= % of the number of moles that
are N2 molecules. Then 100  x = % of the number of moles that are Ar atoms. Solving: 2815 = ' 100 2815 =28.01x+ 3995  39.95 x, 11.94x = 1180. x=. 98.83% N2; % Ar=100‘.00 x= 1.17% Ar Ratio of moles of Ar to moles of N2 = 918' 1873 = 1.18 X 10". M20}; x W x 273.2 K
dRT 49. molar mass = ——  L (H101 Kv — ——————————————'— = 15.90 g/mol
P' 1.000 atm l,‘
i
x ,
.‘ 
r,
a.
G:
6..
V,
v.
. .
1‘ .i 5,.
_
.
., 96 59. 60. ' Molar mass: —— = PV=nRT,VandTlare constant. — =—3 or —— =— CHAPTER 5 GASES =0.510 molN 7379 g C x 1”“ = 6.144 mol'C; 7.14 gN x 1m]
. 12.011,g 14.01 g 109ng “‘01 =10.8molH; 82ng 1mm =0.51molO
1.008g 16.00 g Dividing all values by 0.51 gives an empirical formula “of CanNO. 4.02 g x 0.08206 L atm
dRT L mol K 1 atm
760 torr x 400. K,
= 392 g/_mol P 256 torr x Empirical formula'mass of ClenNO zl95 g/mol and 33—:— s 2. Thus, the molecular formula is C24H42N202. P V
1 1 =0500 atm x 2.00L
v 3.00L For NH3: P2 ; = 0.333 atm 2
PV‘
ForOZ: P2'= 1 1:1.50atmx1'00L V2 3.00 L After the stopcock is opened, V and T will be constant, so P °< n. The balanced equation requires: = 0.500 atm n02 = ——P°2 = 3 = 1.25 a 1’02 = 0.500 atm =‘150 . PM} 0.333 atm The actual ratio present is: The actual ratio is larger than the required ratio, so NH3 in the denominator is limiting. Since equal
mol of NO will be produced as NH3 reacted, the partial pressure of NO produced is 0.333 atm (the
same as PNH3 reacted). PP’Pn l l 1 n1 112 P2 n2 When and T are constant, then pressure is directly proportional to moles of gas present and ‘
pressure ratios are identical to mol ratios. At 25 °c: 2 H,(g) + 02(g) ,> 2 H200), H200) is produced. The balanced equation requires 2 mol H2 for every mol O2 reacted. The same ratio (2: 1) holds true
for pressure units. The actual pressure ratio present is 2 atm H2 to 3 atm 03, well below the required
2:1 ratio. Therefore H2 is the limiting reagent. The only gas present at 25 °C aﬁer the reaction goes
to completion will be the excess 02. . _ prawmvau.. .. ,2... , A. ,. ' CHAPTER 5 GASES ' 97 61. latmO
2 = 1.00 atm 02 P reacted =2.00 ath X ————
02( ) 2 2 atm H2  ' P02 (excess) = P02 (initial)  P02 (reacted) = 3.00 atm  1.00 atm = 2.00 atm 02 At 125 °C: 2 H2(g) + 02(g) + 2 H20(g), H20(g) is produced. The major difference in the problem is that gaseous water is now a product, which will increase the
total pressure. 2 atm H20
——————— = 2.00 atm H20
2 atm H2 Pm, = P02 (excess) + PH2O (produced) = 2.00 atm 02 + 2.00 atm H20 = 4.00 atm PH20 (produced) = 2.00 atm H2. x Rigid container (constant volume): As reactants are converted to products, the mol of gas particles
present decrease by onehalf. As 11 decreases, the pressure will decrease (by onehalt). Density is
the mass per unit volume. Mass is conserved in a chemical reaction, so the density of the gas will
not change since mass and volume do not change. Flexible container (constant pressure): Pressure is constant since the container changes volume in
order to keep a constant pressure. As the mol of gas particles decrease by a factor of 2, the volume
of the container will decrease (by onehalf). We haVe the same mass of gas in a smaller volume, so
the gas density will increase (is doubled). Kinetic Molecular Theory and Real Gases 62. a. Containers ii, iv, vi, and viii have volumes twice that of containers i, iii, v, and vii. Containers
iii, iv, vii, and viii have twice the number of molecules (mol) present as compared to containers
i, ii, v, and vi. The container with the lowest pressure will be the one which has the fewest mol
of gas present in the largest volume (containers ii and vi both have the lowest P). The smallest .
container with the most mol of gas present will have the highest pressure (containers iii and vii
both have the highest P). All the other containers (i, iv, v and viii) will have the same pressure
between the two extremes. The order is: ii = vi < i = iv = v = viii < iii = vii._ b. All have the same average kinetic energy since the temperature is the same in each container.
Only the temperature determines the average kinetic energy. c. The least dense gas will be container ii since it has the fewest of the lighter Ne atoms present
in the largest ‘volume. Container vii has the most dense gas since the largest number of the
‘ heavier Ar atoms are present in the smallest volume. To ﬁgure out the ordering for the other
containers, we will calculate the relative density of each. In the following table, ml equals the
mass of Ne in container i, Vl equals the volume of container i, and d1 equals the density of the gas in container i. l
'1'
l 98 63. ‘ 64. 65. 66. um mass, m m, 2V1
volume CHAPTER 5 GASES From the table, the order of gas density is: ii < i = iv = vi < iii = v = viii < vii d. uum = (3 RT/M)“; the root mean square velocity only depends on the temperature and the molar mass. Since T is constant, the heavier argon molecules will have the slower root mean square
velocity as compared to the neon molecules. The order is:
v=vi=vii=viii<i=ii=iii=iv. (KE),VE= 3/2 RT; KE depends only on temperature. At each temperature CH, and N2 will have the same average KE. For energy units of joules (J), use R = 8.3145 J mol'1 K". To determine average
KE per molecule, divide by Avogadro’s number, 6.022 x 1023 molecules/mol. at 273 K: (KB)avg = 3 x 83145 J x 273 K = 3.40 x 103 J/mol = 5.65 x 1021 J/molecule
2 mol K
at 546 K: (KE),VB= 3 x 83145 J x 546 K= 6.81 x 103 J/mol = 1.13 x 1020 J/molecule
2 mol K
 1/2 I
um= 2E1: _, where R= 83145 andM=molarmass inkg
M mol K
For CH4, M = 1.604 x 102 kg and for N2, M = 2.802 x 102 kg.
, ' U2
3 x 8'31145 J x 273 K
For CH4 at 273 K: um, = L = 652 m/s
11604 x 102 kg/mol ‘
. At 546 K, um for OH, is 921 m/s. ‘ For N2, um= 493 m/s at 273 K and 697 m/s at 546 K. No; The numbers calculated in Exercise 5.63 are the average kinetic energies at the various
temperatures. At each temperature, there is a distribution" of energies. Similarly, the numbers
calculated in Exercise 5.64 are a special kind of average velocity. There is a distribution of velo
cities as shown in Figures 5.155.17 of the text. Note that the major reason there is a distribution of kinetic energies is because there is a distribution of velocities for any gas sample at some
temperature. a. All the gases have the same average kinetic energy since they are all at the same temperature . [KEm = (3/2)RT] 100 \ CHAPTER 5 GASES
W Of the choices, the gas with the molar mass closest to 29.99 is NO. 1/2 ’
Rate M 12 17 1/2 12 16 1/2 '
70. 1:[ 2] , C 02(300) =1_02; c 0=[30.0) =1_04 Ratez M1 Izclso 29.0 Izclso 28.0 The relative rates of effusion of 12C160: 12C‘7O: 12C‘“O are 1.04: 1.02: 1.00.
Advantage: CO2 isn't as toxic as CO. Major disadvantages of using CO2 instead of CO: 1. Can get a mixture of oxygen isotopes in C02. ' 2. 'Some species, e.g., 12060130 and 120702, would effuse (gaseously diffuse) at about the
same rate since the masses are about equal. Thus, some species cannot be separated from ‘3: each other.
g 1/2
1 Rate M
9 71. l = ——3 ; Rate1 = 240.1111“, Rate2 = 47'8.mL, M2 = 16'04g and M1 = ? Rate2 Ml nun mm mol
1/2' ,
1 ﬂ .= ﬂ = 0.502, 16.04 = (0.502)2 x M1, M1 = ﬂ = 63'7 g
‘ 78 1  0.252 mol ‘ Ratel M2 172 g
‘ 72. = —— where M = molar mass; Let Gas (1) = He, Gas (2) = Cl2
Rate2 M1
1.0 L
 . 1/2
4.5 min = 70.90 ,. t . = 4.209, t=19 min
1.0 L 4.003 4.5 nnn
1:
73. a. PV = nRT
0.5000 mol x W x (25.0 + 273.2) K
P = “RT. = ‘ “101 K ‘ = 12.24 atm
V 1.0000 L
2 . . .
b. 'P + (V  nb) = nRT; For N2: a = 1.39 atm Lz/mol2 and b = 0.0391 L/mol
0.5000 2 ' ' I _
P + 1.39 1 0000 atm (1.0000 L  0.5000 X 0.0391 L) —— 12.24 L atm (P + 0.348 atm)(0.9805 L) = 12.24 L atm 12.24 L atm
0.9805 L P:  0.348 atm = 12.48  0.348 = 12.13 atm CHAPTER 5 GASES ' ~ ‘ 101 74. 75. 76. 77. .3. The ideal gas law is high by 0.11 atm or 1021113 x 100 = 0.91%..
a. PV = nRT
0.5000 mol x W 'x 298.2 K
P = ——“RT = m°1K = 1.224 atm
V 10.000 L
, 2 .
b. P + a( (v  nb) = nRT; For N2: a = 1.39 atm L2/m012 and b = 0.0391 L/mol
2 .
1 P + 1.39( 265833) atm] (10.000 L  0.5000 x 0.0391 L) = 12.24 L atm (P + 0.00348 atm)(10.000 L  0.0196 L): 12.24 L atm 12.24 L atm P + 0.00348 atm =
‘ 9.980 L = 1.226 atm, 'P = 1.226  0.00348 = 1.223 atm c. The results agree to :I: 0.001 atm (0.08%). d. In Exercise 5.73 the pressure is relatively high’and thereis a signiﬁcant disagreement. In
Exercise 5.74 the pressure is around 1 atm and both gas laws show better agreement. The ideal
gas law is valid at relatively low pressures. The kinetic molecular theory assumes that gas particles do not exert forces on each other and that
gas particles are volumeless. Real gas particles do exert attractive forces for each other, and real gas
particles do have volumes. A gas behaves most ideally at low pressures and high temperatures. The
effect of attractive forCes is minimized at high temperatures since the gas particles are moving very
rapidly. At low pressure, the container volume is relatively large (P and Vuare inversely related) so
the volume of the container taken up by the gas particles is negligible. a. At constant temperature, the average kinetic energy of the He gas sample will equal the average
kinetic energy of the C12 gas sample. In order for the average kinetic energy to be the same, the
smaller He atoms must move at a faster average velocity as compared to C12. Therefore, plot
A, with the slower. average velocity, would be for the C12 sample, and plot B would be for the
He sample. Note the average velocity in each plot is a little past the top peak. b. As temperature increases, the average velocity of a gas will increase. Plot A would be for 02(g)
at 273 K and plot B, with the faster average velocity, would be for 02(g) at 1273 K. Since a gas behaves more ideally at higher temperatures, 02(g) at 1273 K would behave most
ideally. 1 . ‘ I. . nRT n 2
Rearrangmg the Van der Waals' equatlon gives: P = V b — a[ —)
. — 1’1 102 ‘ CHAPTER 5 GASES —————————_—_—_——___—_ 78. 79. 80. 81. 82. 83. r[’P+EV£2_) (Vnb)=nRT, PV+ —an— P is the measured pressure and V is the volume of the container.
For NH3: a = 4.17 atm L2 /mol2 and b = 0.0371 L/mol For the ﬁrst experiment (assuming four signiﬁcant ﬁgures in all values): _ nRT n 2_ 1.000x0.08206x273.2 1.000 2
P —a — ———————_ — 4.17 —_
V —nb 172.1 — 0.0371 V 172.1 P = 0.1303  0.0001 = 0.1302 atm. In Example 5.1, PM = 0.1300 atm. The difference is less than
0.1%. The ideal gas law also gives 0.1302 atm. At low pressures, the van der Waals equation agrees
with the ideal gas law. ' For experiment 6, the measuredpressure is 1.000 atm. The ideal gas law gives P = 1.015 atm and '
the van der Waals equatiOn gives P = 1.017  0.086 = 1.008 atm. The van der Waals equation
accounts for approximawa onehalf of the deviation from ideal behavior. Note: As pressure
increases, deviation from the ideal gas law increases. 2 an 2V 3 2 3
a“ b =nRT; PV+31L —an —,a“ b
V2 V2 V V2 = nRT At lowP and high T, the molar volume of a gas will be relatively large. The an2N and an’b/V2 terms
become negligible because V is large. Since nb is the actual volume of the gas molecules.
themselves, then nb << V: and an is negligible compared to PV. Thus, PV = nRT. The corrected (ideal) volume is the volume accessible to the gas molecules. For a real gas,'this ’ V volume is less than the container volume because of the space occupied by the gas molecules. The van der Waals constant b is a measure of the size of the molecule. Thus, Cng should have the
largest value of b since it has the largest molar mass (size). ' 0.244 atm L 2 mol 2
Since a is a measure of intermolecular attractions, the attractions are greatest for C02. The values ofa are: H2, ; co,,3.59;iN2, 1.39; CH4, 2.25 3/2 2
ﬂu) = 47‘ m u 26 (—mu /2kBT)
21thT
(~mu2/2kBT) o , 2 ‘ . .
As 11 > 0, e —> e = 1; At small values of u, the u term causes the functlon to increase. At large values of u, the exponent term, mu2/2kBT, is a large negative number and e raised to a large
negative number causes the function to decrease. As 11 —> oo, e" —> 0. ' we) U2
3 8.3145 kg m2 ‘
1/2 s 2 mol K
28.02 x 10‘3 kg/mol J (227°C t 273)K
= 667 m/s C “Mm H».LJT£§J£.J1;L”—ni w. u; 2; was... l, M l v . A ‘1
‘ S
a
I: 106 CHAPTER 5 GASES 94. For benzene: 89.6X10'9gx “‘01 =1.15 x 10‘9mol benzene 78.11 g n R 1.15x10'9m01xWx296K
Vb“: hen = _.______m°1K =2.84 x 108L ,
P 748 torrx 1th 284 10'8L 760m ' Mixing ratio= #— x 106 a 9.47 x 103 ppmv
3.00L 
9 '8 “ppm: vol. ofXx 10 = 2.84x 10 L X 109:9.47 ppbv total v01. 3.00 L 1.15 x 10'9 mol benzene x l L x 6.022 x 1023 molecules L cm 3 H101 = 2.31 X 1011 molecules benzene/cm3
For toluene: 1 mol
92.13 g 153 X 10'9.g C7HB X = 1.66 X 10'9 mol toluene 1.66 x10'9 mol x W x 296 K nmlRT : mol K P 748 torr x 760 ‘0”
atm 106 = 1.37 X 10'2 ppmv (or 13.7 ppbv) \ . v.0,= =4.1o x 108 L ‘8
4.10x 10 L x Mixing ratio =
' 3.00 L 1.66 x 10‘9 mol toluene x 1 L x 6.022 x 1023 molecules = 3.33 X 10“ moleeules toluene/cm3 Additional Exercises
1.149gx 11110102 95. _ 0.050mLx = 1.8 x 103molo2 mL 32.0 g _
R 1.8 x10‘3 mol x W x 310.K
V? 1—T =—————————‘i‘°—————— =4.610'2L=46mL
P 1.0 atm ‘
_ , 12mL CZHSOH _ a
96. 750. mL Julce X ————————  90. mL C2H50H present 100 mL juice 0.79 g CZHSOH 1 mol CZHSOH 2 mol CO2 \
———————— x x——————=1.5molC02 90. mL CZHSOH x ———_——_ .
mL CZHSOH 46.07 g CZHSOH 2 mol CZHSOH CHAPTER 5 GASES 107 The COiwill occupy (825  750. =) 75 mL not occupied by the liquid (headspace).
0.08206 L atm = n€02 mol K
C02 V 75 x 10‘3 L Actually, enough CO2 will dissolve in the wine to lower the pressure of CO2 to a much more
reasonable value. XRT 1.5molx x298K P = 490 atm 97. Mn(s)+x HCl(g)~>.MnC1x(S)+ ’2‘—H2(g) __ PV _ 0.951 atmx3.22L n — —— — =0.100 molH2
x373K “2 RT 0.08V206Latm
molK mol CI in compound = mol HCl = 0.100 mol H2 X M}— = 0.200 mol Cl
% mol H2 molCl = 0.200mol c1 _ 0.200 mol c1 = 400 mol Mn 2.747 g Mn x 1 mol Mn 0.05000 mol Mn 54.94 g Mn The formula of compound is MnC14. 98. a. Volume ofhot air: V = gnr3 €719.50 m)3 = 65.4 m3 (Note: radius = diameter/2 = 5.00/2 = 2.50 m) 10dm 3 1L
X
m = 6.54 X 104 L 65.4 m3 X ( 745torrx 1am“ x6.54x104L
_PV_ 1 _ __ _ 760 torr
RT 0.08206 L atm
mol K =2.31X103 mol air x (273 + 65) K Mass ofhot air=2.31 X 103 mol X 29'0 g = 6,70 x 104 g
mol
PV ‘ 3.2% atmx 6.54 x104L
Air displaced: n = —¥ = 0 82 6L =2.66 X 103 mol air
R m x (273 + 21) K
mol K
~ . . _ 3 g __ 4
Mass ofair displaced — 2.66 X 10 mol X l — 7.71 X 10 g
mo 1 Lift = 7.71x104 g  6.70 x 104g: 1.01 x 104 g CHAPTER 5 GASES ‘ * , 113 110. a) Ideal Gas Equation: PV = nRT 1.0011101 x W x 310. K
P = LIST = “12315 = 25.4 atm (V ~nb) = nRT b) Van der Waals Equation: MW . 2 A
P = \IIIRTb " a ; For C02: a = 3.59 atin L2 mol'z; b = 00427 L/mOI
—n
1.00m01 X'WX3IO.K 2
P: mol K _ 3.59 atmL2 X (1.00 mol)
1.00 L  1.00 mol X 0.0427 L/mol mol 2 ‘ (11.00 L)2 P = 26.6 atm  3.59 atm = 23.0 atm nCH4 'ncm = 5.71 mol Ar; xcm = —— #0650 ~ 111. nN— 228g ———————
nCH4 + nAr new + (5.71) _ 39.95 g/mol 0.650 (nCH4 + 5.71) = new, 3.71 = 0.350 new, nCH4 =‘10.6 mol CH4; KEavg gRT for 1 mol So KEml = (10.6 + 5.71 mol) X 3/2 X 8.3145 J mol“1 K'1 x 298 K = 6.06 X .104 J = 60.6 k] . . 1/2
' 112. a) P—V— = a + .BP b) Mm“) = 2 mu 0< M[ = Nat constantT
n 1mpact M
(straight line, y = b + mx)
E
n Amu
' slope = m = B impact
OI yintercept = b = 01
P M nR(T.c + 273) c) TK='T.C +273; P= V = constant (T.C + 273) (straight line, y = mx + b) nR
V yintercept = b = slope = m = 273 nR
V 116 ~ ' CHAPTER 5 GASES Challenge Problems
120. a. The number of collisions of gas particles with the walls of the container is proportional to:
N T i
z o: __ _
" V M Where N = # of gas particles, V= volume of container, T = temperature (Kelvin),
and M = molar mass of gas particles in kg. Since both He samples are in separate containers of the same volume, then V and M are
constant. Since pressure and volume are constant, then P or nT (also 11 °< N). ‘ Thus zA o: Nﬁ N T
“F1 = 2; and and NIT] = NJ,
NW 21
So, we have: ,——
22 N1 2 2 1" T2
Thus,E=———=T—,—:—,2‘/TT=‘/'T_2
2 1 2 «T7 «T7 0‘
Solving: 4 T1 = T2, T1 = 1/4 T2; and since P °< nT, then 111 = 4 112 Although the number of collisions in container #1 is twice as high, the temperature is one
fourth that of container #2. This is because there are four times the number of moles of ’
helium gas in container #1. b. There are two times the number of collisions, but because the temperature is lower, the gas particles are hitting with less forceful collisions. Overall, the pressure is the same in each
container. 121. Cr(s) + 3 HCl(aq) «y CrC13(aq) + 3/2 H,(g); za(s) + 2 HCl(aq) —> ZnC12(aq) + H,(g) ( 750. torr x 1mm J x 0.225 L mol H2 produced = n = ﬂ = 760 tOFT RT W x(273 + 27)K = 9.02 x 103 mol H2 . . . ~mol K
9.02 X 10'3 mol H2 = mol H2 from Cr reaction + mol H2 from Zn reaction From the balanced equation: 9.02 x 10'3 mol H2 = mol Cr X (3/2) + mol Zn x 1, Let x = mass of Cr and y = mass of Zn, then: X+y=0362gand902 ><1o3='15x + y
52.00 65.38 We have two equations and two unknowns. Solving by simultaneous equations: ...
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 Fall '07
 Kerber
 Chemistry

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