CHAPTER 6 GASES PRACTICE EXAMPLES 1A
(E) The pressure measured by each liquid must be the same. They are related through P = g h d Thus, we have the following g hDEG dDEG = g hHg dHg. The g’s cancel; we substitute known values: 9.25 mDEG ×1.118 g/cm3DEG = hHg× 13.6 g/cm3Hg 1.118g/cm3 hHg 9.25 m 0.760 m Hg, P = 0.760 m Hg =760.mmHg 13.6 g/cm3
1B
(E) The solution is found through the expression relating density and height: hTEG d TEG hHg d Hg We substitute known values and solve for triethylene glycol’s density: 9.14 mTEG ×dTEG = 757 mmHg ×13.6 g/cm3Hg. Using unit conversions, we get 0.757 m dTEG= 13.6 g/cm3 1.13 g/cm3 9.14 m
2A
(E) We know that Pgas = Pbar + ΔP with Pbar = 748.2 mmHg. We are told that ΔP = 7.8 mmHg. Thus, Pgas = 748.2 mmHg + 7.8 mmHg = 756.0 mmHg.
2B
(M) The difference in pressure between the two levels must be the same, just expressed in different units. Hence, this problem is almost a repetition of Practice Example 6-1. hHg=748.2 mmHg – 739.6 mmHg=8.6 mmHg. Again we have g hg dg=g hHg dHg. This becomes hg × 1.26 g/cm3 glycerol = 8.6 mmHg ×13.6 g/cm3 Hg 13.6 g / cm3 Hg hg 8.6 mmHg 93 mm glycerol 126 . g / cm3glycerol
3A
(M) A = r2
(here r = ½(2.60 cm ×
1m ) = 0.0130 m) 100 cm
0.0130 m)2 = 5.31 × 10-4 m2 F = m × g = (1.000 kg)(9.81 m s-2) = 9.81 kg m s-2 = 9.81 N 9.81 N F P= = 18475 N m-2 or 1.85 × 104 Pa = A 5.31 10-4 m 2 P (torr) = 1.85 × 104 Pa × = 139 torr
208
Chapter 6: Gases
3B
4A
101, 325 Pa = 1.000 × 104 Pa 1013.25 mb The area of the cylinder is unchanged from that in Example 6-3, (1.32 × 10-3 m2). F F = 1.000 × 104 Pa = P= A 1.32×10-3 m 2 Solving for F, we find F = 13.2 (Pa)m2 = 13.2 (N m-2)m2 = 13.2 N F = m × g = 13.2 kg m s-2 = m 9.81 m s -2 F 13.2 kg m s -2 = 1.35 kg = total mass = mass of cylinder + mass added weight = m = g 9.81 m s -2 An additional 350 grams must be added to the top of the 1.000 kg (1000 g) red cylinder to increase the pressure to 100 mb. It is not necessary to add a mass with the same cross sectional area. The pressure will only be exerted over the area that is the base of the cylinder on the surface beneath it. (M) Final pressure = 100 mb. 100 mb
(M) The ideal gas equation is solved for volume. Conversions are made within the equation.
1 mol NH 3 0.08206 L atm 25 273 K 20.2 g NH 3 17.03g NH 3 mol K nRT 24.4 L NH 3 V 1atm P 752 mmHg 760 mmHg 4B
bg
(E) The amount of Cl 2 g is 0.193 mol Cl 2 and the pressure is 0.980 atm (0.993 barr × (1 atm/1.01325 barr) = 0.980 atm). This information is substituted into the ideal gas equation after it has been solved for temperature. 0.980 atm 7.50 L PV T 464 K nR 0.193mol 0.08206 L atm mol1 K 1
5A
(E) The ideal gas equation is solved for amount and the quantities are substituted. 10.5atm 5.00 L PV n 2.11mol He RT 0.08206 L atm 30.0 273.15 K mol K
5B
(M)
1 atm 1000L 6.67 10-7 Pa 3.45 m3 PV 101325 Pa 1m3 n= = = 9.28 × 10-10 moles of N2 -1 -1 RT (0.08206 L atm K mol ) (25 273.15)K
6.022 1023 molecules of N 2 1 mole N 2 14 molecules of N2 = 5.59 × 10 molecules N2 molecules of N2 = 9.28 × 10-10 mol N2 ×
209
Chapter 6: Gases
6A
(E) The general gas equation is solved for volume, after the constant amount in moles is cancelled. Temperatures are converted to kelvin. . mL 2.14 atm 37.8 + 273.2 K V PT 100 2.11 mL V2 1 1 2 1.02 atm 36.2 + 273.2 K P2 T1
b
b
6B
g
g
bg
(M) The flask has a volume of 1.00 L and initially contains O 2 g at STP. The mass of O 2 g that must be released is obtained from the difference in the amount of O 2 g at the two temperatures, 273 K and 373 K. We also could compute the masses separately and subtract them. We note that 1.00 bar is 0.987 atm. PV PV PV 1 1 MO mass released = n STP n100 C M O MO R 273 K 373 K R 273 K R373 K 0.987 atm 1.00 L 1 32.00 g 1 = = 0.378 g O 2 1 1 0.08206 L atm mol K 273 K 373 K 1 mol O 2
bg bg
2
2
7A
(M) The volume of the vessel is 0.09841 L. We substitute other values into the expression for molar mass. 0.08206 L atm 22.4 273.2 K 40.4868g 40.1305g mRT mol K 86.4 g/mol M PV 1atm 772 mmHg 0.09841L 760 mmHg
7B
(M) The gas’s molar mass is its mass (1.27 g) divided by the amount of the gas in moles. The amount can be determined from the ideal gas equation. 1 atm . L 737 mm Hg 107 760 mm Hg PV n 0.0424 mol gas 0.08206 L atm RT 25 273 K mol K 1.27 g M= = 30.0 g/mol 0.0424 mol This answer is in good agreement with the molar mass of NO, 30.006 g/mol.
FG H
8A
b
IJ K
2
g
(M) The molar mass of He is 4.003 g/mol. This is substituted into the expression for density. MP 4.003 g mol 1 0.987 atm d 0162 . g/L RT 0.08206 L atm mol -1K -1 298 K When compared to the density of air under the same conditions (1.16 g/L, based on the “average molar mass of air”=28.8g/mol) the density of He is only about one seventh as much. Thus, helium is less dense (“lighter”) than air.
210
Chapter 6: Gases
8B
(M) The suggested solution is a simple one; we merely need to solve for mass of gas from density and its moles from the ideal gas law. m(gas) D V 1.00 g/L 1.00 L 1.00 g
1atm 745 mmHg 1.00 L 760 mmHg PV n 0.0312 mol RT 0.08206 L atm K 1 382 K Therefore, the molar mass of the gas is as follows: g 1.00 g M 32.0 g/mol mol 0.0312 mol The molecular weight suggests that the gas is O2. 9A
(M) The balanced equation is 2 NaN 3 (s) 2 Na (l) 3 N 2 (g)
1atm 776 mmHg 20.0L 760 mmHg PV moles N 2 0.821mol N 2 0.08206 L atm RT (30.0 273.2) K mol K Now, solve the stoichiometry problem. 2 mol NaN 3 65.01 g NaN 3 mass NaN 3 = 0.821 mol N 2 = 35.6 g NaN 3 3 mol N 2 1 mol NaN 3 9B
(M) Here we are not dealing with gaseous reactants; the law of combining volumes cannot be used. From the ideal gas equation we determine the amount of N 2 g per liter under the specified conditions. Then we determine the amount of Na(l) produced simultaneously, and finally the mass of that Na(l). 1 atm Pressure: 1.0 barr 0.987 atm 1.01325 barr 0.987 atm 1.000 L 2 mol Na 22.99 g Na 0.619 g Na(l) mass of Na(l) 0.08206 L atm (25 273) K 3mol N 2 1mol Na mol K
bg
10A (E) The law of combining volumes permits us to use stoichiometric coefficients for volume ratios. 5 L O2 O 2 volume = 1.00 L NO g = 1.25 L O 2 g 4 L NO
bg
bg
211
Chapter 6: Gases
10B (E)The first task is to balance the chemical equation. There must be three moles of hydrogen for every mole of nitrogen in both products (because of the formula of NH 3 ) and reactants: N 2 g + 3 H 2 g 2 NH 3 g . The volumes of gaseous reactants and products are related by their stoichiometric coefficients, as long as all gases are at the same temperature and pressure. 2 L NH 3 g volume NH 3 g = 225 L H 2 g = 150. L NH 3 3 L H2 g
bg
bg
bg
bg
bg
bg bg
11A (M) We can work easily with the ideal gas equation, with a the new temperature of T = 55 + 273 K = 328 K . The amount of Ne added is readily computed. 1 mol Ne nNe 12.5 g Ne 0.619 mol Ne 20.18g Ne 0.08206 L atm (175 . 0.619) mol 328 K ntotal RT mol K P 13 atm 5.0 L V
b
g
11B (E) The total volume initially is 2.0 L + 8.0 L = 10.0 L . These two mixed ideal gases then obey the general gas equation as if they were one gas. . atm 10.0 L 298 K PV T 100 . atm P2 1 1 2 55 2.0 L 273K V2 T1 12A (M) The partial pressures are proportional to the mole fractions. nH O 0.00278 mol H 2 O 2.50 atm = 0.0348atm H 2 O(g) PH 2 O 2 Ptot . 0197 n tot mol CO 2 0.00278 mol H 2 O PCO2 Ptot PH 2 O 2.50 atm 0.0348 atm = 2.47 atmCO 2 (g) 12B (M) Expression (6.17) indicates that, in a mixture of gases, the mole percent equals the volume percent, which in turn equals the pressure percent. Thus, we can apply these volume percents—converted to fractions by dividing by 100—directly to the total pressure. N 2 pressure = 0.7808 748 mmHg = 584 mmHg,
O 2 pressure = 0.2095 748 mmHg = 157 mmHg, CO 2 pressure = 0.00036 748 mmHg = 0.27 mmHg, Ar pressure = 0.0093 748 mmHg
= 7.0 mmHg
bg
13A (M) First compute the moles of H 2 g , then use stoichiometry to convert to moles of HCl.
1atm (755 25.2) torr 0.0355 L 760 mmHg 6 mol HCI amount HCI = 0.00278 mol HCI 0.08206 L atm 3mol H 2 (26 273) K mol K
212
Chapter 6: Gases
bg
13B (M) The volume occupied by the O 2 g at its partial pressure is the same as the volume occupied by the mixed gases: water vapor and O 2 g . The partial pressure of O 2 g is found by difference. O 2 pressure = 749.2 total pressure 23.8 mmHg H 2 O pressure = 725.4 mmHg
bg
b
bg
g
1 atm 725.4 mm Hg 0.395 L 760 mm Hg PV 0.0154 mol O 2 mol O 2 R T 0.08206 L atm K 1 298 K 2 mol Ag 2 O 231.74 g Ag 2 O 7.14 g Ag 2 O 1 mol O 2 1 mol Ag 2 O Mass% Ag2O = 7.14/8.07 × 100 = 88.4% mass Ag 2O = 0.0154 mol O 2
The volume of the dry gas in turn is determined as follows: 0.0154 mol 0.08206 L atm mol1 K 1 298 K V 0.382 L 749.2 / 760 atm 14A (M) The gas with the smaller molar mass, NH 3 at 17.0 g/mol, has the greater root-mean-square speed
urms
3 RT 3 8.3145 kg m 2 s 2 mol1 K 1 298 K 661m/s M 0.0170 kg mol1
14B (D)
2180 mi 1h 5280 ft 12in. 2.54 cm 1m 974.5 m/s 1h 3600s 1mi 1ft 1in. 100 cm Solve the rms-speed equation (6.20) for temperature by first squaring both sides. bullet speed =
2
974.5 m 2.016 103 kg 2 urms M 1s 1mol H 2 3RT 2 76.75 K T urms 8.3145 kg m 2 M 3R 3 s 2 mol K We expected the temperature to be lower than 298 K. Note that the speed of the bullet is about half the speed of a H 2 molecule at 298 K. To halve the speed of a molecule, its temperature must be divided by four. 15A (M) The only difference is the gas’s molar mass. 2.2 104 mol N 2 effuses through the orifice in 105 s. M N2 ? mol O 2 28.014 g/mol 0.9357 4 M O2 2.2 10 mol N 2 31.999 g/mol
moles O 2 = 0.9357 2.2 104 = 2.1104 mol O 2
213
Chapter 6: Gases
15B (M) Rates of effusion are related by the square root of the ratio of the molar masses of the two gases. H 2 , effuses faster (by virtue of being lighter), and thus requires a shorter time for the same amount of gas to effuse. M H2 2.016 g H 2 /mol H 2 105 s 28.2 s timeH2 = time N2 M N2 28.014 g N 2 /mol N 2 16A (M) Effusion times are related as the square root of the molar mass. It requires 87.3 s for Kr to effuse. M unk M unk unknown time 131.3 s substitute in values 1.50 Kr time M Kr 87.3 s 83.80 g/mol
b
g
2
M unk 1.504 83.80 g / mol = 1.90 102 g / mol 16B (M) This problem is solved in virtually the same manner as Practice Example 18B. The lighter gas is ethane, with a molar mass of 30.07 g/mol. M (C2 H 6 ) 30.07 g C 2 H 6 / mol C 2 H 6 C 2 H 6 time = Kr time 87.3 s 52.3 s M ( Kr) 83.80 g Kr / mol Kr 17A (D) Because one mole of gas is being considered, the value of n 2 a is numerically the same as the value of a , and the value of nb is numerically the same as the value of b . 2
nRT na P 2 V nb V
1.00 mol
0.083145 L barr
mol K (2.00 0.0427) L
273 K
3.66 L2 barr mol2 (2.00 L) 2
11.59 barr 0.915 barr
= 10.68 barr CO 2 g compared with 10.03 barr for Cl2 g Pideal
nRT V
1.00 mol
bg
0.083145 L barr 273 K mol K 11.35 ~ 11.4 barr 2.00 L
bg
Cl 2 g shows a greater deviation from ideal gas behavior than does CO 2 g . 17B (M) Because one mole of gas is being considered, the value of n 2 a is numerically the same as the value of a , and the value of nb is numerically the same as the value of b . 0.083145 L barr 1.00 mol 273 K nRT n2a 1.47 L2 barr mol K 2 11.58 atm 0.368 atm P V nb V (2.00 0.0395) L (2.00 L) 2 = 11.2 barr CO g
bg
bg
compared to 10.03 barr for Cl 2 g , 11.2 barr for CO, and 11.35 barr for CO 2 g . Thus, Cl 2 g displays the greatest deviation from ideality, 11.4 barr.
bg
214
Chapter 6: Gases
INTEGRATIVE EXERCISE (M) First, convert the available data to easier units.
A.
101.3×103 Pa × (1 atm/1.013×105 Pa) = 1 atm, and 25 °C = 298 K. Then, assume that we have a 1 L container of this gas, and determine how many moles of gas are present: n=
PV (1.00 atm)(1.00 L) = = 0.0409 mol. RT 0.08206 L atm K 1 298 K
Knowing the density of the gas (1.637 g/L) and its volume (1 L) gives us the mass of 1 L of gas, or 1.637 g. Therefore, the molar mass of this gas is: (1.637 g / 0.0409 mol) = 40.03 g/mol Now, determine the number of moles of C and H to ascertain the empirical formula: mol C: 1.687 g CO 2 mol H: 0.4605 g H 2 O
1 mol CO 2 1 mol C 0.0383 mol 44.01 g CO 2 1 mol CO 2 1 mol H 2 O 2 mol H 0.05111 mol 18.02 g H 2 O 1 mol H 2 O
Dividing by the smallest value (0.0383 mol C), we get a H:C ratio of 1.33:1, or 4:3. Therefore, the empirical formula is C3H4, which has a molar mass of 40.07, which is essentially the same as the molar mass calculated. Therefore, the actual formula is also C3H4. Below are three possible Lewis structures: H
HC
H C
C
H
C H
H C
HC
CH2
215
C
CH3
Chapter 6: Gases
B.
(D) First, let us determine the amount of each element in the compound:
mol C: 151.2 103 g CO 2
1 mol CO 2 1 mol C 0.003435 mol C 44.01 g CO 2 1 mol CO 2
12.01 g C = 0.04126 g C 1 mol C 1 mol H 2 O 2 mol H 0.007727 mol H g H2O 18.02 g H 2 O 1 mol H 2 O
g C: 0.003435 mol C mol H: 69.62 103
1.01 g H = 0.007804 g H 1 mol H PV 1 atm 9.62 103 L 2 mol N 0.0008589 mol N mol N: = 1 RT 1 mol N 2 0.08206 L atm K (273 K)
g H: 0.007727 mol H
14.01 g N = 0.01203 g N 1 mol N Therefore, the mass of O is determined by subtracting the sum of the above masses from the mass of the compound: g N: 0.0008589 mol N
g O: 0.1023 (0.04126 0.007804 0.01203) 0.04121 g O 1 mol O mol O: 0.04121 g O = 0.002576 mol O 16.0 g O
To determine the empirical formula, all of the calculated moles above should be divided by the smallest value. Doing so will give the following ratios: C: 0.003435/0.0008589 = 4 H: 0.007727/0.0008589 = 9 N: 1 O: 0.002576/0.0008589 = 3 The empirical formula is C4H9NO3, and has a molar mass of 119.14 g/mol. To determine the actual formula, we have to calculate its molecular mass. We know the density at a given volume and therefore we need to find out the number of moles. Before that, we should convert the experimental conditions to more convenient units. T = 127 °C + 273 = 400 K, and P = 748 mm Hg/760 mm Hg = 0.9842 atm. n=
PV (0.9842 atm) 1 L = = 0.02999 mol RT 0.08205 L atm K 1 (400 K)
MM = g/mol = 3.57 g/0.02999 = 119 g/mol
Therefore, the empirical and molecular formulas are the same, being C4H9NO3.
216
Chapter 6: Gases
EXERCISES Pressure and Its Measurement 1.
(E) (a) (b) (c) (d)
2.
3.
P 736 mmHg
1 atm 0.968 atm 760 mmHg
1 atm 0.766 atm 1.01325 bar 1 atm P 892 torr 117 . atm 760 torr 1000 Pa 1 atm 2.22 atm P = 225 kPa 1 kPa 101,325 Pa P 0.776 bar
(E)
760 mmHg 748 mmHg 1 atm h 928 torr = 928 mmHg
(a)
h 0.984 atm
(c)
h 142 ft H 2 O
12 in 1 ft
(b)
2.54 cm 10 mmH 2 O 1 mmHg 1 mHg 3.18 mHg 1 in 1 cm H 2 O 13.6 mmH 2 O 1000 mmHg
(E)
We use: hbnz d bnz hHg d Hg
hbnz 0.970 atm 4.
0.760 m Hg 13.6 g/cm3 Hg 11.4 m benzene 1 atm 0.879 g/cm3 benzene
(E) We use: hgly d gly hCCl4 d CCl4 hgly 3.02 m CCl 4
1.59 g/cm3 CCl4 3.81 m glycerol 1.26 g/cm3 glycerol
5.
(E) P = Pbar – h1 = 740 mm Hg – 30 mm (h1) = 710 mm Hg
6.
(E) P = Pbar + h1 = 740 mm Hg + 30 mm (h1) = 770 mm Hg
217
Chapter 6: Gases
7.
F = m × g and 1 atm = 101325 Pa = 101325 kg m-1 s-2 = P =
F m 9.81 m s -2 = 1 m2 A
101325 kg m -1 s -2 1 m 2 = 10329 kg 9.81 m s -2 (Note:1 m2 = (100 cm)2 = 10,000 cm2) mass (per m2) =
P (kg cm-2) = 8.
m 10329 kg = 1.03 kg cm-2 = 2 A 10,000 cm
(E) Start by noting that 1 atm is defined as 101325 Pa. Pascals is the unit name used to describe SI units of force (F) per unit area, and is expressed as N/m2. Since F = ma, the MKS units for pressure are kg·m·s-2·m-2, or kg·m-1·s-2. The Imperial unit of measuring force is pounds-force, or lbf. Considering acceleration due to gravity, an object weighing 1 N has a mass of (1/9.8 m/s2) 0.102 kg. We now can convert the MKS units to psi as follows:
101325
N 0.102 kg 1 lb 1 m2 (2.54 cm) 2 = 14.7 psi m2 1N 0.453 kg (100 cm) 2 1 in 2
The Simple Gas Laws 9.
(E) (a) (b)
10.
11.
762 mmHg = 52.8 L 385 mmHg 762 mmHg V = 26.7 L = 7.27 L 760 mmHg 3.68 atm 1 atm
V = 26.7 L
(E) Apply Charles’s law: V kT . (a)
T 273 98 371 K
(b)
T 273 20 253 K
Ti 26 273 299 K 371 K V = 886 mL 1.10 103 mL 299 K 253 K V = 886 mL 7.50 102 mL 299 K
(E) Charles’ Law states that V1/T1 = V2/T2. Therefore,
3.0 L 1.50 L , and T2 = 225 K 450 K T2 12.
(E) Pf Pi
Vi 0.725 L 105 kPa 33.8 kPa (0.334 atm) 2.25 L Vf
218
Chapter 6: Gases
Vf
FG H
IJ K
(E) Pi Pf
14.
(E) We let P represent barometric pressure, and solve the Boyle’s Law expression below for P.
Vi
721 mmHg
35.8 L 1875 L 1 atm 50.6 atm 35.8L 760 mm H 2 O
13.
P 42.0 mL ( P 85 mmHg) 37.7 mL P 15.
42.0P 37.7 P 3.2 103
3.2 103 7.4 102 mmHg 42.0 37.7
(E) Combining Boyle’s and Charles’ Law, we get the following expression:
P1 V1 P2 V2 = T1 T2 Therefore,
0.340 atm 5.00 103 m3 300 K P2 V2 T1 T2 = = = 255 K P1 V1 1.000 atm 2.00 103 m3
16.
(E) First, convert temperatures from Celsius to the Kelvin scale. 22 C + 273.15 = 295.15 K -22 C = (-22 + 273.15) = 251.15 K
Volume decrease is proportional to the ratio of the temperatures in Kelvin. 251.15 K 0.851 (volume will be about 85% of its original volume). volume contraction = 295.15 K 17.
(E) STP: P = 1 barr and T = 273.15. P (1 barr) = 0.9869 atm 0.9869 atm 0.0750 L 39.948 g Ar mass Ar = = 0.132 g Ar L atm 1 mol Ar 0.08206 273.15 K K mol
18.
(M) STP: P = 1 barr (0.9869 atm) and T = 273.15 K (note: one mole of gas at STP occupies 22.698
L). 1 mol Cl2 L atm 250.0 g Cl2 70.906 g 0.08206 K mol 273.15 K nRT V= = = 80.08 L Cl2 P 0.9869 atm 1 mol Cl2 22.698 L at STP = 80.03 L Cl2 Alternatively V = n Vm = 250.0 g Cl2 70.906 g 1 mol
219
Chapter 6: Gases
19.
(M) (a)
Conversion pathway approach: mass = 27.6 mL
1L 1000 mL
1 mol PH 3
34.0 g PH 3 1000 mg PH 3 22.698 L STP 1 mol PH 3 1g
= 41.3 mg PH 3
Stepwise approach: 27.6 mL
1L
0.0276 L
1000 mL 1 mol PH 3 0.0276 L 0.001216 mol PH 3 22.698 L STP 34.0 g PH 3 0.001216 mol PH 3 0.0413 g PH 3 1 mol PH 3
0.0413 g PH 3
(b)
1000 mg PH 3 1g
41.3 mg PH 3
number of molecules of PH 3 = 0.001216 mol PH3
6.022 1023 molecules 1 mol PH 3
number of molecules of PH 3 = 7.32 1020 molecules
20.
(E) (a)
mass = 5.0 1017 atoms
1 mol Rn 222 g Rn 106 g 7 = 8.30 10 mol 1 mol 1g 6.022 1023 atoms
mass = 1.8 102 g Rn g
(b) 21.
volume = 8.30 107 mol
22.698 L 106 L = 19 L Rn g 1 mol 1L
(M) At the higher elevation of the mountains, the atmospheric pressure is lower than at the beach. However, the bag is virtually leak proof; no gas escapes. Thus, the gas inside the bag expands in the lower pressure until the bag is filled to near bursting. (It would have been difficult to predict this result. The temperature in the mountains is usually lower than at the beach. The lower temperature would decrease the pressure of the gas.)
220
Chapter 6: Gases
22.
(M) Based on densities, 1 mHg = 13.6 mH 2 O . For 30 m of water, 1 mHg 1000 mm 1 atm 30 mH 2 O = 2.2 mHg Pwater 2.2 mHg = 2.9 atm 13.6 mH 2 O 1m 760 mmHg To this we add the pressure of the atmosphere above the water: Ptotal 2.9 atm 1.0 atm 3.9 atm .
When the diver rises to the surface, she rises to a pressure of 1.0 atm. Since the pressure is about one fourth of the pressure below the surface, the gas in her lungs attempts to expand to four times the volume of her lungs. It is quite likely that her lungs would burst.
General Gas Equation 23.
24.
Pf V f Pi Vi nR is Ti Tf obtained from the ideal gas equation. This expression can be rearranged as follows. Vi Pi Tf 4.25 L 748 mmHg (273.2 26.8) K Vf 4.30 L Pf Ti 742 mmHg (273.2 25.6) K (E) Because the number of moles of gas does not change,
(E) We first compute the pressure at 25 C as a result of the additional gas. Of course, the gas pressure increases proportionally to the increase in the mass of gas, because the mass is proportional to the number of moles of gas. 12.5 g P= 762 mmHg = 953 mmHg 10.0 g Now we compute the pressure resulting from increasing the temperature. 62 + 273 K P= 953 mmHg = 1.07 103 mmHg 25 + 273 K
b b
25.
g g
(E) Volume and pressure are constant. Hence niTi =
nf ni
=
PV = nfTf R
Ti (21 273.15) K = = 0.609 (60.9 % of the gas remains) T f (210 273.15) K
Hence, 39.1% of the gas must be released. Mass of gas released = 12.5 g × 26.
39.1 = 4.89 g 100
(M) First determine the mass of O 2 in the cylinder under the final conditions. 115 . atm 34.0 L 32.0 g / mol PV . g O2 mass O 2 n M 517 M L atm RT 0.08206 (22 273) K mol K mass of O 2 to be released = 305 g 51.7 g = 253 g O 2
221
Chapter 6: Gases
Ideal Gas Equation 27.
bg
(M) Assume that the CO 2 g behaves ideally and use the ideal gas law: PV = nRT 1 mol CO 2 L atm 89.2 g 0.08206 (37 273.2) K 1000 mL nRT 44.01 g mol K 5.32 10 4 mL V
P
737 mmHg
28.
FG 358. g O nRT H (M) P
29.
(E) mass n M
V
PV RT
2
1 atm
1L
760 mmHg
IJ K
1 mol O 2 0.08206 L atm (46 273.2) K 32.00 g O 2 mol K 2.29 atm 12.8 L M
11.2 atm 18.5 L 83.80 g/mol 702 g Kr L atm 0.08206 (28.2+273.2) K mol K
30. (M)
Mol He = 7.41 g He × (1 mol He/4.003 g He) = 1.85 mol
T
PV nR
3.50 atm 72.8 L 1.68 103 K L atm 1.85 mol 0.08206 mol K
t (C) = (1.68 103 273) 1.41 103 C 31.
32.
1 mol gas 8.3 1015 mol gas 23 6.022 10 molecules gas We next determine the pressure that the gas exerts at 25 C in a cubic meter L atm 298.15 K 8.3 1015 mol gas 0.08206 101,325 Pa K mol P= = 2.1 1011 Pa 3 1 atm 10 dm 1 L 1 m3 3 1 m 1 dm (M) 1 mol CO 2 n CO2 1242 g CO 2 = 28.20 mol CO 2 T = 273.15 + (-25) = 248.15 K 44.010 g CO 2
(M) n gas 5.0 109 molecules gas
2
1000 L 0.250 m VCO2 = r 2 h = 3.1416 = 85.9 L 1.75 m 2 1 m3 L atm 101,325 Pa 248.15 K 28.20 mol CO 2 0.08206 1 atm K mol = 6.77 105 Pa P= 85.9 L 222
Chapter 6: Gases
33.
(E) The basic ideal law relationship applies here. Molar volume is the amount of volume that one mole of a gas occupies. If PV=nRT, then molar volume is V/n, and the relationship rearranges to: V R T = Vm = n P (a) Vm = (298 K·0.08206 L·atm·K-1)/1.00 atm = 24.4 L·mol-1 (b) Patm = 748 mmHg / 760 mmHg = 0.978 atm Vm = (373 K·0.08206 L·atm·K-1)/0.978 atm = 31.3 L·mol-1
34.
(E) The molar volume formula given above can be rearranged to solve for T. T = (P/R)·Vm = (2.5 atm/0.08205 L·atm·K-1)×22.4 L·mol-1 = 682 K
Determining Molar Mass 35.
(M) Use the ideal gas law to determine the amount in moles of the given quantity of gas. M
mRT PV
0.418 g 0.08206
L atm
339.5 K mol K = 104 g mol-1 1 atm 743 mmHg 760 mmHg 0.115 L
Alternatively 1 atm 1L 743 mmHg 115 mL PV 760 mmHg 1000 mL n 0.00404 mol gas M = RT
36.
L atm
mol K
(273.2 66.3) K
(M) Use the ideal gas law to determine the amount in moles in 1 L of gas. n
37.
0.08206
0.418 g = 103 g / mol 0.00404 mol
PV RT
1 atm 725 mmHg 760 mmHg 1 L 0.02801 mol gas 0.08206
L atm
mol K
415 K
M =
mass
moles
=
0.841 g = 30.0 g mol-1 0.02801 mol
(M) First we determine the empirical formula for the sulfur fluoride. Assume a 100 g sample of SxFy. 1 mol S 1 mol F moles S = 29.6 g S = 0.923 mol S moles F = 70.4 g F = 3.706 mol F 32.064 g S 18.9984 g F Dividing the number of moles of each element by 0.923 moles gives the empirical formula SF4. To find the molar mass we use the relationship: dRT 4.5 g L-1 0.08206 L atm K -1 mol-1 293 K Molar mass = = = 108 g mol-1 P 1.0 atm molecular formula mass 108 g mol-1 Thus, molecular formula empirical formula SF4 = SF4 = -1 empirical formula mass 108.06 g mol
223
Chapter 6: Gases
38.
(M) We first determine the molar mass of the gas.
T = 24.3 + 273.2 = 297.5 K
P = 742 mmHg
1 atm = 0.976 atm 760 mmHg
mRT 2.650 g 0.08206 L atm mol1 K 1 297.5 K 155 g/mol 1L PV 0.976 atm 428 mL 1000 mL Then we determine the empirical formula of the gas, based on a 100.0-g sample. 1 mol C mol C = 15.5 g C = 1.29 mol C 0.649 1.99 mol C 12.01 g C 1 mol Cl mol Cl = 23.0 g Cl = 0.649 mol Cl 0.649 1.00 mol Cl 35.45 g Cl 1 mol F mol F = 61.5 g F = 3.24 mol F 0.649 4.99 mol F 19.00 g F Thus, the empirical formula is C2 ClF5 , which has a molar mass of 154.5 g/mol. This is the same as the experimentally determined molar mass. Hence, the molecular formula is C 2 ClF5 .
M=
39.
(M) L atm (23 273) K mRT mol K (a) M 55.8 g/mol PV 1 atm 1L 749 mmHg 760 mmHg 102 mL 1000 mL (b) The formula contains 4 atoms of carbon. (5 atoms of carbon gives a molar mass of at least 60―too high―and 3 C atoms gives a molar mass of 36―too low to be made up by adding H’s.) To produce a molar mass of 56 with 4 carbons requires the inclusion of 8 atoms of H in the formula of the compound. Thus the formula is C4 H 8 . 0.231 g 0.08206
40.
(M) First, we obtain the mass of acetylene, and then acetylene’s molar mass. mass of acetylene = 56.2445 g 56.1035 g = 0.1410 g acetylene L atm 0.1410 g 0.08206 (20.02 273.15) K mRT mol K 26.04 g/mol M PV 1 atm 1L 749.3 mmHg 132.10 mL 760 mmHg 1000 mL The formula contains 2 atoms of carbon (3 atoms of carbon gives a molar mass of at least 36-too high and 1C atom gives a molar mass of 12-too low to be made up by adding H’s.) To produce a molar mass of 26 with 2 carbons requires the inclusion of 2 atoms of H in the formula. Thus the formula is C2H2.
224
Chapter 6: Gases
Gas Densities 41.
(E) d
MP RT
P
dRT M
L atm (32 273) K 760 mmHg mol K 28.0 g/mol 1 atm
1.80 g/L 0.08206
P 1.21 103 mmHg 1L 28.0 g N 2 Molar volume N2 15.56 L/mol 1.8 g 1 mol N 2
L atm (22.8 273.2) K mol K 62.5 g / mol 1 atm 756 mmHg 760 mmHg 62.5 g 1L Molar volume 24.4 L/mol mol 2.56 g
dRT 42. (E) M M
43.
(M) (a)
(b)
44.
45.
2.56 g / L 0.08206
d
MP RT
28.96 g/mol 1.00 atm 1.18 g/L air L atm 0.08206 (273 25)K mol K
44.0 g / mol CO 2 100 . atm MP 180 . g / L CO 2 RT 0.08206 L atm (273 25) K mol K Since this density is greater than that of air, the balloon will not rise in air when filled with CO 2 at 25 C ; instead, it will sink!
d
44.0 g / mol 100 . atm MP MP becomes T 454 K = 181 C RT RT 0.08206 L atm 118 . g/L mol K L atm 2.64 g/L 0.08206 (310+273)K MP dRT mol K becomes M 124 g/mol (E) d 1 atm RT P 775 mmHg 760 mmHg Since the atomic mass of phosphorus is 31.0, the formula of phosphorus molecules in the vapor must be P4 . 4 atoms / molecule 31.0 = 124 (E) d
b
g
225
Chapter 6: Gases
46.
(M) We first determine the molar mass of the gas, then its empirical formula. These two pieces of information are combined to obtain the molecular formula of the gas. L atm 2.33 g/L 0.08206 296 K dRT mol K M 57.7 g/mol 1 atm P 746 mmHg 760 mmHg 1 mol C mol C = 82.7 g C = 6.89 mol C 6.89 1.00 mol C 12.0 g C 1 mol H mol H = 17.3 g H = 17.1 mol H 6.89 2.48 mol H 1.01 g H Multiply both of these mole numbers by 2 to obtain the empirical formula, C2 H 5 , which has an empirical molar mass of 29.0 g/mol. Since the molar mass (calculated as 57.7 g/mol above) is twice the empirical molar mass, twice the empirical formula is the molecular formula, namely, C4 H10 .
Gases in Chemical Reactions 47.
(E) Balanced equation: C3 H8 g + 5 O 2 g 3 CO 2 g + 4 H 2 O l
Use the law of combining volumes. O 2 volume = 75.6 L C 3 H 8 48.
5 L O2 = 378 L O 2 1 L C3H 8
(E) Each mole of gas occupies 22.7 L at STP.
H 2 STP volume 1.000 g Al
1 mol Al 3 mol H 2 (g) 22.7 L H 2 (g) at STP 26.98 g Al 2 mol Al(s) 1 mol H 2
= 1.262 L H 2 g 49.
bg
(M) Determine the moles of SO 2 g produced and then use the ideal gas equation.
Conversion pathway approach: 3.28 kg S 1000 g S 1 mol S 1 mol SO 2 mol SO 2 = 1.2 106 kg coal 100.00 kg coal 1 kg S 32.1 g S 1 mol S 1.23 106 mol SO 2 nRT V P L atm 1.23 106 mol SO 2 0.08206 296 K mol K V 3.1 107 L SO 2 1 atm 738 mmHg 760 mmHg
226
Chapter 6: Gases
Stepwise approach: 3.28 kg S 3.94 104 kg S 100.00 kg coal 1000 g S 3.94 104 kg S 3.94 107 g S 1 kg S 1 mol S 3.94 107 g S 1.23 106 mol S 32.1 g S 1 mol SO 2 1.23 106 mol S = 1.23 106 mol SO 2 1 mol S
1.2 106 kg coal
V
50.
nRT P
L atm 296 K mol K 3.1 107 L SO 2 1 atm 738 mmHg 760 mmHg
1.23 106 mol SO 2 0.08206
bg
(M) Determine first the amount of CO 2 g that can be removed. Then use the ideal gas law.
mol CO 2 = 1.00 kg LiOH
V
51.
nRT P
1 mol CO 2 1000 g 1 mol LiOH = 20.9 mol CO 2 1 kg 23.95 g LiOH 2 mol LiOH
L atm (25.9 273.2) K mol K 519 L CO 2 (g) 1 atm 751 mmHg 760 mmHg
20.9 mol 0.08206
(M) Determine the moles of O2, and then the mass of KClO3 that produced this amount of O2. 1 atm 1L 738 mmHg 119 mL 760 mmHg 1000 mL mol O 2 0.00476 mol O 2 L atm 0.08206 (22.4 273.2)K mol K
mass KClO 3 = 0.00476 mol O 2 % KClO 3 = 52.
2 mol KClO 3 122.6 g KClO 3 = 0.389 g KClO 3 3 mol O 2 1 mol KClO 3
0.389 g KClO 3 100% = 10.9% KClO 3 357 . g sample
b g
bg
bg
(M) Determine the moles and volume of O 2 liberated. 2 H 2 O 2 aq 2 H 2 O 1 + O 2 g 1 mol O 2 1.01 g 0.0300 g H 2 O 2 1 mol H 2 O 2 mol O 2 = 10.0 mL soln 1 mL 1 g soln 34.0 g H 2 O 2 2 mol H 2 O 2
= 0.00446 mol O 2
227
Chapter 6: Gases
V
L atm (22 273) K 1000 mL mol K 109 mL O 2 1 atm 1L 752 mmHg 760 mmHg
0.00446 mol O 2 0.08206
53.
(M) First we need to find the number of moles of CO(g) Reaction is 3 CO g 7 H 2 g C 3 H 8 g 3 H 2 O l 1 atm 28.5 L 760 torr PV 760 tor = 1.27 moles CO n CO = = L atm RT 0.08206 273.15 K K mol 7 mol H 2 L atm n H2 RT 1.27 mol CO 3 mol CO 0.08206 K mol 299 K VH2 (required) = = = 73.7 LH 2 1 atm P 751 mmHg 760 mmHg
54.
(M) (a) In this case, H2 is the limiting reagent. 2 L NH 3 Volume NH 3 = 313 L H 2 = 209 L NH 3 3 L H2 (b) Moles of NH3 (@ 315 C and 5.25 atm) 2 mol NH 3 5.25 atm 313 L = 2.27 101 mol NH 3 3.41 101 mol H 2 L atm 3 mol H 2 0.08206 (315 273)K mol K L atm 2.27 101 mol NH 3 0.08206 298 K mol K = 5.80 102 L NH 3 V (@25 C, 727mmHg) 1 atm 727 mmHg 760 mmHg
af
af
af
af
Mixtures of Gases 55.
(M) Determine the total amount of gas; then use the ideal gas law, assuming that the gases behave ideally. 1 mol Ne 1 mol Ar moles gas = 15.2 g Ne + 34.8 g Ar 20.18 g Ne 39.95 g Ar
= 0.753 mol Ne + 0.871 mol Ar = 1.624 mol gas L atm . mol 0.08206 (26.7 273.2) K nRT 1624 mol K V= = = 5.59 L gas P 7.15 atm
228
Chapter 6: Gases
56.
(M) 2.24 L H 2 (g) at STP is 0.100 mol H 2 (g). After 0.10 mol He is added, the container holds 0.20 mol gas. At STP, pressure is in barr; 1 barr is 0.987 atm L atm 0.20 mol 0.08206 (273 100)K nRT mol K V 6.2 L gas P 0.987 atm
57.
(M) The two pressures are related, as are the number of moles of N 2 g to the total number of moles of gas. 28.2 atm 53.7 L PV 61.7 mol N 2 moles N 2 L atm RT (26 273) K 0.08206 mol K 75.0 atm total moles of gas = 61.7 mol N 2 = 164 mol gas 28.2 atm 20.18 g Ne mass Ne = 164 mol total 61.7 mol N 2 = 2.06 103 g Ne 1 mol Ne
58.
(M) Solve a Boyle’s law problem for each gas and add the resulting partial pressures. 2.35 L 3.17 L PH2 762 mmHg 324 mmHg PHe 728 mmHg 418 mmHg 5.52 L 5.52 L Ptotal PH 2 PHe 324 mmHg + 418 mmHg 742 mmHg
bg
59. (M) Initial pressure of the cylinder 1 mol O 2 (1.60 g O 2 )(0.08206 L atm K -1 mol-1 )(273.15 K) nRT 31.998 g O 2 = P= = 0.500 atm V 2.24 L
We need to quadruple the pressure from 0.500 atm to 2.00 atm. The mass of O2 needs to quadruple as well from 1.60 g 6.40 g or add 4.80 g O2 (this answer eliminates answer (a) and (b) as being correct). One could also increase the pressure by adding the same number of another gas (e.g. He) mass of He = nHe × MMHe 1 mol O 2 = 0.150 moles = 0.150 moles of He) 31.998 g O 2 4.0026 g He mass of He = 0.150 moles × = 0.600 g He ((d) is correct, add 0.600 g of He) 1 mol He
(Note: moles of O2 needed = 4.80 g ×
229
Chapter 6: Gases
60.
(M) (a) First determine the moles of each gas, then the total moles, and finally the total pressure. 1 mol H 2 1 mol He moles H 2 = 4.0 g H 2 moles He = 10.0 g He 2.02 g H 2 4.00 g He = 2.0 mol H 2
nRT P V (b) 61.
= 2.50 mol He
(2.0 2.50) mol 0.08206
PH 2 23 atm
4.3 L
L atm 273 K mol K 23 atm
2.0 mol H 2 10 atm 4.5 mol total
PHe 23 atm 10 atm = 13 atm
(M)
(a)
1 mol C6 H 6 L atm (35 273) K 0.728 g 0.08206 78.11 g C6 H 6 mol K 760 mmHg nRT Pben 2.00 L 1 atm V = 89.5 mmHg Ptotal = 89.5 mmHg C6 H 6 g + 752 mmHg Ar g = 842 mmHg
bg
(b) 62.
(D) (a)
Pbenzene 89.5 mmHg
bg
PAr 752 mmHg
The %CO 2 in ordinary air is 0.036%, while from the data of this problem, the %CO 2 in expired air is 3.8%. 38% P {CO 2 expired air} . CO 2 11 . 102 CO 2 (expired air to ordinary air) P {CO 2 ordinary air} 0.036% CO 2
(b/c) Density should be related to average molar mass. We expect the average molar mass of air to be between the molar masses of its two principal constituents, N 2 (28.0 g/mol) and O 2 (32.0 g/mol). The average molar mass of normal air is approximately 28.9 g/mol. Expired air would be made more dense by the presence of more CO 2 (44.0 g/mol) and less dense by the presence of more H 2 O (18.0 g/mol). The change might be minimal. In fact, it is, as the following calculation shows.
28.013 g N 2 31.999 g O 2 M exp.air 0.742 mol N 2 + 0.152 mol O 2 1 mol N 2 1 mol O 2 44.01 g CO 2 18.02 g H 2 O + 0.038 mol CO 2 + 0.059 mol H 2 O 1 mol CO 2 1 mol H 2 O 39.9 g Ar + 0.009 mol Ar = 28.7 g/mol of expired air 1 mol Ar 230
Chapter 6: Gases
Since the average molar mass of expired air is less than the average molar mass of ordinary air, expired air is less dense than ordinary air. Calculating the densities: (28.7 g/mol)(1.00 atm) d(expired air) = 1.13 g/L (0.08206 L atm/K mol)(310 K) d(ordinary air) 1.14 g/L 63.
(E) 1.00 g H2 0.50 mol H2 1.00 g He 0.25 mol He Adding 1.00 g of He to a vessel that only contains 1.00 g of H2 results in the number of moles of gas being increased by 50%. Situation (b) best represents the resulting mixture, as the volume has increased by 50%
64.
(E) The answer is (b), because the volume at 275 K with 1.5 times as many atoms is roughly 4/3 larger than the drawing.
65.
(M) In this problem, you don’t need to explicitly solve for moles of gas, since you are looking at the relationship between pressure and volume. 4.0 atm 1.0 L = 4.0 PV mol O 2 = = RT RT RT 2.0 atm 2.0 L = 4.0 PV mol N 2 = = RT RT RT total mol. of gas = 8.0/RT
Therefore, nRT 8.0 RT P= = = 4.0 atm V RT 2.0 66.
(M) Like the above problem, you also don’t need to explicitly solve for moles of each gas to get the total moles, because you are ultimately looking at pressure/volume relationships:
0.75 atm 1.0 L = 0.75 PV = RT RT RT 0.45 atm 2.5 L = 1.125 PV mol Xe = = RT RT RT 1.20 atm 1.0 L = 1.2 PV mol Ar = = RT RT RT total mol. of gas = 3.075/RT mol He =
The total volume after the opening of the valves is the sum of the volumes of the flasks (VFlask) and the tubes (VTube). Therefore,
231
Chapter 6: Gases
P=
nRT 3.075 RT = = 0.675 atm VFlask VTube RT 4.5 VTube
Solving for VTube, we get a volume of 0.055 L.
Collecting Gases over Liquids 67.
bg
(M) The pressure of the liberated H 2 g is 744 mmHg 23.8 mmHg = 720. mmHg
1 mol Al 3 mol H 2 L atm (273 25)K 1.65 g Al 0.08206 26.98 g 2 mol Al mol K nRT V 2.37 L H 2 (g) 1 atm P 720. mmHg 760 mmHg This is the total volume of both gases, each with a different partial pressure. 68.
(D) (a)
bg
The total pressure is the sum of the partial pressures of O 2 g and the vapor pressure of water. Ptotal PO2 PH2O 756 mmHg = PO2 19 mmHg
PO2 (756 19) mmHg = 737 mmHg (b)
(c)
The volume percent is equal to the pressure percent. VO PO 737 mmHg of O 2 %O 2 (g) by volume 2 100% 2 100% 100% 97.5% Vtotal Ptotal 756 mm Hg total Determine the mass of O 2 g collected by multiplying the amount of O 2 collected, in moles, by the molar mass of O 2 g .
bg
mass O2
69.
bg
1 atm 1L 737. mmHg 89.3 mL 760 mmHg 1000 mL 32.00 g O 2 PV 0.115 g O 2 L atm RT 1 mol O 2 0.08206 (21.3 273.2)K mol K
(M) We first determine the pressure of the gas collected. This would be its “dry gas” pressure and, when added to 22.4 mmHg, gives the barometric pressure. 1 mol O 2 L atm 146 0.08206 297 K . g 32.0 g O 2 mol K nRT 760 mmHg 729 mmHg P V 116 . L 1 atm barometric pressure = 729 mm Hg + 22.4 mmHg = 751 mmHg
FG H
IJ K
232
Chapter 6: Gases
70.
(M) We first determine the “dry gas” pressure of helium. This pressure, subtracted from the barometric pressure of 738.6 mmHg, gives the vapor pressure of hexane at 25 C . 1 mol He L atm 1072 . g 0.08206 298.2 K 4.003 g He mol K nRT 760 mmHg P 589.7 mmHg V 8.446 L 1 atm vapor pressure of hexane = 738.6 589.7 = 148.9 mmHg
FG H
71.
IJ K
(M) The first step is to balance the equation:
2NaClO3 2NaCl + 3O2 The pressure of O2 is determined by subtracting the known vapor pressure of water at the given temperature from the measured total pressure. PO2 = PTOT – PH2O = 734 torr – 21.07 torr = 713 torr Patm = 713 mmHg / 760 mmHg = 0.938 atm mol O 2 =
0.938 atm 0.0572 L = 0.00221 mol PV = RT 0.08206 L atm K 1 (296 K)
Mass of NaClO3 is then determined as follows: 0.00221 mol O 2 %NaClO3 =
72.
2 mol NaClO3 106.44 g = 0.1568 g NaClO3 3 mol O 2 1 mol NaClO3
0.1568 g 100 17.9% 0.8765 g
(M) The work for this problem is nearly identical to the above problem.
2KClO3 2KCl + 3O2 The pressure of O2 is determined by subtracting the known vapor pressure of water at the given temperature from the measured total pressure. PO2 = PTOT – PH2O = 323 torr – 25.22 torr = 298 torr Patm = 298 mmHg / 760 mmHg = 0.392 atm
233
Chapter 6: Gases
mol O 2 =
0.392 atm 0.229 L PV = = 0.003657 mol RT 0.08206 L atm K 1 (299 K)
Mass of KClO3 is then determined as follows: 0.003657 mol O 2 %KClO3 =
2 mol KClO3 122.54 g = 0.299 g KClO3 3 mol O 2 1 mol KClO3
0.299 g 100 72% 0.415 g
Kinetic-Molecular Theory 73.
(M) Recall that 1 J = kg m2 s-2 urms
74.
3RT M
J 1 kg m 2 s -2 303 K mol K 1J 326 m/s 70.91 103 kg Cl 2 1 mol Cl2
3 8.3145
(M) 2
3RT1 M = 3RT2 M
T1 T2
T1 273 K
Square both sides and solve for T1 .
T1 = 4 273 K = 1092 K
Alternatively, recall that 1 J = kg m2 s-2 3RT Solve this equation for temperature with urms doubled. 184 . 103 m/s M Murms 2 2.016 103 kg/mol (2 1.84 103 m/s) 2 T 1.09 103 K 3R J 1 kg m 2 s-2 3 8.3145 mol K 1J urms
75.
(M) J 298 K 3RT mol K M 0.00783 kg / mol = 7.83 g/mol. or 7.83 u. 2 (urms ) 2 mi 1 hr 5280 ft 12 in. 1m 2180 hr 3600 sec 1 mi 1 ft 39.37 in. 3 8.3145
76.
(E) A noble gas with molecules having urms at 25 C greater than that of a rifle bullet will have a molar mass less than 7.8 g/mol. Helium is the only possibility. A noble gas with a slower urms will have a molar mass greater than 7.8 g/mol; any one of the other noble gases will have a slower urms .
77.
(E) We equate the two expressions for root mean square speed, cancel the common factors, and solve for the temperature of Ne. Note that the units of molar masses do not 234
Chapter 6: Gases
have to be in kg/mol in this calculation; they simply must be expressed in the same units. T 300 K 3RT 3R 300 K 3R TNe Square both sides: Ne 4.003 20.18 . M 4.003 2018 20.18 Solve for TNe: TNe 300 K 1.51103 K 4.003 78.
(E) um , the modal speed, is the speed that occurs most often, 55 mi/h 38 + 44 + 45 + 48 + 50 + 55 + 55 + 57 + 58 + 60 average speed = = 51.0 mi/h = u 10 382 + 442 + 452 + 482 + 502 + 552 + 552 + 57 2 + 582 + 602 26472 miles urms = = 51.5 10 10 h
79.
(D) The greatest pitfall of this type of problem is using improper units. Therefore, convert everything to SI units.
MM O2 = 32.0 g/mol = 32.0×10-3 kg/mol 32.0 103 kg 1 mol Mass of O 2 molecule: = 5.314 1026 kg 23 mol 6.022 10 molec. R =8.3145 J·mol-1·K-1, or kg·m2/(s2·mol·K) Now, we must determine the urms first to determine kinetic energy: 3RT 3 8.3145 298 urms 482 m/s M 32.0 103 Kinetic energy of an O2 molecule is as follows:
ek 80.
1 1 2 m urms = (5.314 1026 kg)(482 m/s)2 = 6.17 1021 J/molecule 2 2
(D) To calculate the total kinetic energy for a certain quantity of gas molecules, we must first calculate the (average) kinetic energy for one molecule and then apply it to the bulk.
MM N2 = 28.0 g/mol = 28.0×10-3 kg/mol 28.0 103 kg 1 mol Mass of N 2 molecule: = 4.650 1026 kg 23 mol 6.022 10 molec. R =8.3145 J·mol-1·K-1, or kg·m2/(s2·mol·K) Now determine the urms first to determine kinetic energy: 3RT 3 8.3145 298 urms 515 m/s M 28.0 103 Kinetic energy of an N2 molecule is as follows: 235
Chapter 6: Gases
ek
1 1 2 m urms = (4.650 1026 kg)(515 m/s)2 = 6.17 1021 J/molecule 2 2
Therefore, the kinetic energy of 155 g of N2 is: 1 mol N 2 6.022 1023 molec. 6.17 1021 J ek : 155 g N 2 = 4.11104 J 14.0 g N 2 1 mol N 2 1 molec.
Diffusion and Effusion of Gases 81.
82.
(M) rate (NO 2 ) M (N 2 O) 44.02 x mol NO 2 / t 0.9781 rate (N 2 O) 46.01 0.00484 mol N 2 O/ t M (NO 2 ) mol NO 2 = 0.00484 mol 0.9781 = 0.00473 mol NO 2 (M) rate (N 2 ) mol (N 2 )/38 s 64s M (unknown) 1.68 M ( N2 ) rate (unknown) mol (unknown)/64s 38s
M (unknown)=(1.68)2 M (N 2 ) (1.68)2 (28.01 g/mol )=79 g/mol 83.
84.
(M) (a)
rate (N 2 ) rate (O 2 )
(c)
rate (14 CO 2 ) rate (12 CO 2 )
(M) rate of effusion O 2
M (O 2 ) 32.00 107 . M (N 2 ) 28.01 M (12 CO 2 ) M (14 CO 2 )
M(SO 2 )
(b) rate (H 2 O) rate (D 2 O)
44.0 0.978 46.0
M (D 2 O) 20.0 105 . M (H 2 O) 18.02
235 238 (d) rate ( 238 UF6 ) M (235 UF6 ) 352 1004 .
rate (
UF6 )
M(
UF6 )
349
64 = 2 = 1.4 rate of effusion SO 2 32 M(O 2 ) O2 will effuse at 1.4 times the rate of effusion of SO2. The situation depicted in (c) best represents the distribution of the molecules. If 5 molecules of SO2 effuse, we would expect that 1.4 times as many O2 molecules effuse over the same period of time =
=
(1.4 × 5 = 7 molecules of O2). This is depicted in situation (c). 85.
(M) For ideal gases, the effusion rate is inversely proportional to their molecular mass. As such, the rate of effusion of a known gas can be determined if the rate of effusion for another gas is known: rate of effusion of Ne rate of effusion of He
MM He MM Ne
236
Chapter 6: Gases
Since effusion is loosely defined as movement of a fixed number of atoms per unit time, and since in this problem we are looking at the time it takes for the same number of moles of both Ne and He to effuse, the above equation can be rearranged as follows: time He MM He molNe molHe = timeNe timeHe time Ne MM Ne x = 22 h
4.00 20.18
Solving for x, x = 22 4.00 / 20.18 = 9.80 h 86.
(M) The same reasoning discussed above works in this problem: time Hg MM Hg molRn molHg = timeRn timeHg time Rn MM Rn 1 = 1.082
200.59 MM Rn
Solving for MM Rn gives a value of 235 g/mol.
Nonideal Gases 87.
nRT n2a 2 V nb V Pideal = 11.2 atm, off by 1.3 atm or + 13%
(M) For Cl2 (g), n 2 a 6.49 L2 atm and nb 0.0562 L. Pvdw =
At 0ºC, Pvdw = 9.90 atm and
1.00 mol
(a)
(b)
(c)
nRT V 0.08206 L atm T 1.00 mol mol K 6.49 L2 atm 0.0422 T atm 1.62 atm 2 2.00 0.0562 L 2.00 L
At 100 C Pideal
Pvdw
0.08206 L atm 373K mol K 15.3atm 2.00 L
=0.0422 373K 1.62 14.1atm Pideal is off by 1.2 atm or +8.5% 0.08206 L atm 473 K 1.00 mol nRT mol K At 200 C Pideal 19.4 atm V 2.00 L Pvdw =0.04222 473K 1.623 18.35 atm Pideal is off by 1.0 atm or +5.5% 0.08206 L atm 673 K 1.00 mol nRT mol K At 400 C Pideal 27.6 atm V 2.00 L Pvdw =0.0422 673K 1.62 26.8atm Pideal is off by 0.8 atm or +3.0%
237
Chapter 6: Gases
88.
(M) (a)
V = 100.0 L, Pvdw
2
nRT na 2 V nb V
0.083145 L barr 298 K 1.502 7.857 L2 barr mol K (100.0 1.50 0.0879)L (100.0 L) 2
1.50 mol
= 0.3721 0.0018 barr SO 2 g 0.3704 atm
0.083145 L barr 298 K mol K 0.372 barr Pideal 100.0 L The two pressures are almost equal. 1.50 mol
89.
(b)
V = 50.0 L, Pvdw = 0.739 barr; Pideal = 0.744 barr. Here the two pressures agree within a few percent.
(c)
V = 20.0 L, Pvdw = 1.82 barr; Pideal = 1.86 barr
(d)
V = 10.0 L, Pvdw = 3.60 barr; Pideal = 3.72 barr
(E) The van der Waals parameter b is defined as the excluded volume per mole, or the volume that is taken up by 1 mole of gas once converted to a liquid.
From Table 6-5, bHe = 0.0238 L/mol. Therefore, the volume of a single He atom is:
Conversion pathway approach:
11012 pm 0.0238 L 1 mol He 1 m3 mol He 6.022 1023 atoms 1000 L 1 m3
3
= 3.95 107 pm3 / He atom
Step-wise approach: 0.0238 L 1 mol He 3.95 1026 L/atom 23 mol He 6.022 10 atoms 1 m3 3.95 1026 L/atom = 3.95 1029 m3 /atom 1000 L
110
12
3.95 1029 m3 /atom
pm
1 m3
3
= 3.95 107 pm3 / He atom
V = (4/3) π r3. Rearranging to solve for r gives r= 3 3V 4π . Solving for r gives an atomic radius of 211.3 pm.
238
Chapter 6: Gases
90.
(M) (a) The process is very similar to the above example. From Table 6-5, bCH4 = 0.0431 L/mol. Therefore, the volume of a single CH4 molecule is:
3
11012 pm 1 mol CH 4 0.0431 L 1 m3 = 7.16 107 pm3 / CH 4 molec. 23 3 mol CH 4 6.022 10 molec. 1000 L 1m 3 V = (4/3) π r . Solving for r gives a molecular radius of 258 pm. This value is greater than r=228 pm, because our calculated value is based on b, which is the volume taken up by a mole of gas that has been condensed into a liquid. (b) Compression ratio is given as follows: 1 mL 1L 16.043 g Vm = 2.43 103 L mol1 66.02 g 1000 mL mol PVm (100 bar)(2.43 103 L mol1 ) Z= = 9.0 103 1 1 RT 0.083145 bar L mol K 325 K
Integrative and Advanced Exercises 91. (E) A millimeter of mercury is 1/760 of an atmosphere of pressure at sea level. The density of mercury (13.5951 g/cm3) needs to be included in this definition and, because the density of mercury, like that of other liquids, varies somewhat with temperature, the pressure will vary if the density varies. The acceleration due to gravity (9.80665 m/s2) needs to be included because pressure is a force per unit area, not just a mass per unit area, and this force depends on the acceleration applied to the given mass, i.e., force = mass × acceleration. 92. (E) Of course, we express the temperatures in Kelvin:
0 C 273 K, - 100 C 173 K, – 200 C 73 K, – 250 C 23 K, and – 270 C 3 K. (A)
V 10.0 cm3 k 0.0250 cm3 K 1 T 400 K V 0.0250 cm 3K 1 273 K = 6.83 cm3 V 0.0250 cm 3 K 1 173 K 4.33 cm 3 V 0.0250 cm 3 K 1 23 K 0.58 cm 3
239
V 0.0250 cm 3 K 1 73 K 1.8 cm 3 V 0.0250 cm 3 K 1 3 K 0.08 cm 3
Chapter 6: Gases
(B)
(C)
20.0 cm 3 V k 0.0500 cm 3 K 1 V 0.0500 cm 3 K 1 273 K 13.7 cm 3 400 K T
V 0.0500 cm 3 K 1 173K 8.65 cm 3
V 0.0500 cm 3 K 1 73 K 3.7 cm 3
V 0.0500cm 3 K 1 23K 1.2 cm 3
V 0.0500 cm 3 K 1 3 K 0.2 cm 3
40.0 cm 3 V k 0.100 cm 3 K 1 T 400 K
V 0.100 cm 3 K 1 273 K 27.3 cm 3
V 0.100 cm 3 K 1 173K 17.3 cm 3
V 0.100 cm 3 K 1 73 K 7.3 cm 3
V 0.100 cm 3 K 1 23K 2.3 cm 3
V 0.100 cm 3 K 1 3 K 0.3 cm 3
As expected, in all three cases the volume of each gas goes to zero at 0 K. 93. (M) Initial sketch shows 1 mole of gas at STP with a volume of ~22.4 L.
Velocity of particles = vi. Note that velocity increases as
T (i.e., quadrupling temperature,
results in a doubling of the velocity of the particles). a)
Pressure drops to 1/3 of its original value (1 atm 0.333 atm). The number of moles and temperature are constant. Volume of the gas should increase from 22.4 L 67 L while the velocity remains unchanged at vi.
b) Pressure stays constant at its original value (1 atm torr). The number of moles is unchanged, while the temperature drops to half of its value (273 K 137 K). Volume of the gas should decrease from 22.4 L 11.2 L while the velocity drops to about 71% to 0.71vi( T 0.5 T ). c)
Pressure drops to 1/2 of its original value (1 atm 0.5 atm). The number of moles remains constant, while the temperature is increased to twice it original value (273 K 546 K). Volume of the gas should quadruple, increasing the volume from 22.4 L 90 L. Velocity of the molecules increases by 41% to 1.41vi( T 2 T ).
d) Pressure increases to 2.25 times its original value (2.25 atm) owing to a temperature increase of 50% (273 K 410 K) and a 50% increase in the number of particles (1 mole 1.5 moles). The volume of the gas should remain unchanged at 22.4 L, while the velocity increases by 22% to 1.22vi( T 1.5 T ).
240
Chapter 6: Gases
n = 1 mol P = 1 atm V = 22.4 L T = 273 K
(initial)
v = vi n = 1 mol P = 0.333 atm V = 11.2 L T = 273 K
(a)
v = vi n = 1 mol P = 1 atm V = 11.2 L T = 137 K v = 0.71vi n = 1 mol P = 0.500 atm V = 11.2 L T = 273 K
(b)
(c)
v = vi n = 1 mol P = 2.25 atm V = 22.4 L T = 273 K
(d)
v = vi
94.
(M) We know that the sum of the moles of gas in the two bulbs is 1.00 moles, and that both bulbs have the same volume, and are at the same pressure because they are connected. Therefore, n1T1 =n 2 T2 n1 T2 350 K = = 1.556 n 2 T1 225 K n1 =1.556 n 2
Therefore, n2(1.556) + n2 = 1.00. Solving the equation gives (Flask 2) n2=0.391, and (Flask 1) n1 = 0.609. 95.
(M) First, calculate the number of moles of gas: PV (1.00 atm)(6.30 L) n= = 0.2377 mol RT 0.082058 L atm K 1 (323 K)
Molecular mass (MM) then can be calculated: MM = (10.00 g)/(0.2377 mol) = 42.069 g/mol Now we must determine the mole ratio of C to H: 85.6 g C × (1 mol C/12.01 g C) = 7.13 mol 14.4 g H × (1 mol H/1.01 g H) = 14.3 mol
241
Chapter 6: Gases
Now, divide both by the smallest number: C:C mole ratio = 1 H:C mole ratio = 2 Therefore, the empirical formula is CH2, with a formula unit molar mass of ~14.03 g/mol. To determine molecular formula, divide MM by formula unit MM: 42.069/14.03 = 3.00. Therefore, the molecular formula is C3H6. 96.
(M) First we determine the molar mass of the hydrocarbon. L atm 0.7178g 0.08206 (65.0 273.2) K mRT mol K M 52.0 g/mol 0.00987 atm PV 99.2 kPa 0.3907 L 1 kPa Now determine the empirical formula. A hydrocarbon contains just hydrogen and carbon. 1 mol CO 2 1 mol C 0.05514 mol C 0.05514 1.000 mol C amount C 2.4267 g CO 2 44.01 g CO 1 mol CO 2 2 1 mol H O 2 2 mol H 0.05513 mol H 0.05514 1.000 mol H amount H 0.4967 g H O 2 18.02 g H O 1 mol H O 2 2 The empirical formula is CH. This gives an empirical molar mass of 13 g/mol, almost precisely one-fourth of the experimental molar mass. The molecular formula is therefore C4H4.
97.
(M) Stepwise approach: Note that three moles of gas are produced for each mole of NH4NO3 that decomposes. 1 mol NH 4 NO3 3 moles of gas amount of gas = 3.05 g NH 4 NO3× × = 0.114 mol gas 80.04 g NH 4 NO3 1 mol NH 4 NO3 T = 250 + 273 = 523 K P=
nRT 0.114 mol × 0.08206 L atm mol-1 K -1 × 523 K = = 2.25 atm V 2.18 L
Conversion pathway: nRT P= V 3.05 g NH 4 NO3× P=
1 mol NH 4 NO3 3 moles of gas × × 0.08206 L atm mol-1 K -1 × 523 K 80.04 g NH 4 NO3 1 mol NH 4 NO3 2.18 L
= 2.25 atm
242
Chapter 6: Gases
98.
(M) First, let’s convert the given units to those easier used: P = 101 kPa × (1 barr/101 kPa) × (1 atm/1.01 barr) = 0.9901 atm T = 819 °C + 273 K = 1092 K 1 mol NH 4 NO 2 mol NH 4 NO 2 128 g NH 4 NO 2 = 1.998 mol 64.052 g 3 mol gas mol gas = 1.998 mol NH 4 NO 2 = 5.994 1 mol NH 4 NO 2
V=
99.
5.994 mol 0.082058 L atm K 1 1092 K nRT = = 542 L P 0.9901 atm
(M) (a) nH2 = 1.00 g/2.02 g/mol = 0.495 mol H2; nO2= 8.60 g/32.0 g/mol = 0.269 mol O2 ntotal RT 0.769 mol 0.08206 L atm mol1 K 1 298 K 12.5 atm Ptotal PO2 PH 2 V 1.50 0L The limiting reagent in the production of water is H 2 (0.495 mol) with O2 (0.269 mol) in excess. 2 H 2 (g)
O 2 (g)
2 H 2 O(l)
(initial)
0.495 mol
0.269 mol
0
after reaction
~0
0.022 mol
0.495 mol
so Ptotal PO2 PH 2O PO2 Ptotal
ntotal RT 0.022 mol 0.08206 L atm mol1 K 1 298 K 0.359 atm 1.50 L V 23.8 mm Hg 0.359 atm 0.39 atm 760 mm Hg/atm
100. (M) 4.0 L air 1000 mL 3.8 mL CO 2 1 mL O 2 76 mL O 2 /min 1 min 1L 100 mL air 2 mL CO 2 (b) We first determine the amount of O2 produced per minute, then the rate of consumption of Na2O2.
(a) mL O 2 /min
1 atm 0.967 atm T 25 273 298 K 760 mmHg 0.967 atm 0.076 L PV n 0.0030 mol O 2 RT 0.08206 L atm mol1 K 1 298 K 0.0030 mol O 2 2 mol Na 2 O 2 77.98 g Na 2 O 2 60 min 28 g Na 2 O 2 /h rate 1 min 1 mol O 2 1 mol Na 2 O 2 1h
P 735 mmHg
243
Chapter 6: Gases
101. (E) Determine relative numbers of moles, and the mole fractions of the 3 gases in the 100.0 g gaseous mixture.
N 2 46.5 g N 2
amount
1 mol N 2 1.66 mol N 2 28.01 g N 2
2.86 mol 0.580
1 mol Ne 0.629 mol Ne 2.86 mol 0.220 20.18 g Ne 1 mol Cl2 0.575 mol Cl2 2.86 mol 0.201 amount Cl 2 40.8 g Cl2 70.91 g Cl2 Ne 12.7 g Ne
amount
total amount 1.66 mol N 2 0.629 mol Ne 0.575 mol Cl2 2.86 mol Since the total pressure of the mixture is 1 atm, the partial pressure of each gas is numerically very close to its mole fraction. Thus, the partial pressure of Cl2 is 0.201 atm or 153 mmHg 102. (M) Let us first determine the molar mass of this mixture.
dRT 0.518 g L1 0.08206 L atm mol 1 K 1 298 K M 13.4 g/mol 1 atm P 721 mmHg
760 mmHg
Then we let x be the mole fraction of He in the mixture. 13.4 g/mol x 4.003 g/mol (1.000 x) 32.00 g/mol 4.003x 32.00 32.00x x
32.00 13.4 32.00 4.003
0.664
Thus, one mole of the mixture contains 0.664 mol He. We determine the mass of that He and then the % He by mass in the mixture. mass He 0.664 mol He
4.003 g He 1 mol He
2.66 g He
%He
2.66 g He 13.4 g mixture
100% 19.9% He
103. (M) The volume percents in a mixture of gases also equal mole percents, which can be converted to mole fractions by dividing by 100. M
air
0.7808 mol N 2
28.013 g N 2 1 mol N 2
0.00036 mol CO 2
0.2095 mol O 2
44.010 g CO 2 1 mol CO 2
31.999
g
1 mol O 2
O2
39.948 g Ar 0.0093 mol Ar 1 mol Ar =28.96 g/mol
244
Chapter 6: Gases
104. (M) (a) PTOT = PO2 +PN2O = 154 torr + 612 torr = 766 torr
We know that PO2 = χ O2 PTOT . Therefore, χ O2 = 154/766 = 0.20 To determine the wt% of O2 and N2O, we have to use the mole fractions and determine the mass of each species:
mass O 2 : 0.20 mol O 2
32.0 g O 2 = 6.432 g 1 mol O 2
mass N 2O: 0.80 mol N 2 O
44.0 g N 2 O = 35.20 g 1 mol N 2 O
6.432 100 = 15.4% 6.432 35.20 wt% N 2 O =100 15.4 84.6% wt% O 2 =
(b) Apparent molar mass of the mixture: (32.00 g/mol)(154 torr/766 torr) + (44.02 g/mol)(612 torr/766 torr) = 41.6 g/mol 105. (M) The amount of N2(g) plus the amount of He(g) equals the total amount of gas. We use this equality, and substitute with the ideal gas law, letting V symbolize the volume of cylinder B. n N 2 + n He = n total becomes PN2 VN 2 + PHe VHe = Ptotal Vtotal
since all temperatures are equal. Substitution gives (8.35 atm × 48.2 L) + (9.50 atm × V) = 8.71 atm (48.2 L + V) 420 402 402 L atm 9.50 V 420 L atm 8.71V V 23 L 9.50 8.71 106. (M) To construct a closed-end monometer, fill the tube with mercury, making sure to remove any air bubbles. Inverting the tube results in a column of mercury with a vacuum in the space directly above the column. vacuum gas vacuum
difference in height of columns of mercury is the pressure of the gas
pressure equals difference in heights of the two columns of mercury
When the container is evacuated, the heights of both mercury columns are equal. As the gas pressure in the container increases, the heights of the mercury columns will change. Because the system is closed, Pbar is not needed. One drawback of this system is that gas pressure much larger than one atmosphere cannot easily be handled
245
Chapter 6: Gases
unless a very long column is used. (> 1 meter). Otherwise, the column of mercury would be pushed up to the closed end of the monometer at a critical pressure, and remain unchanged with increasing pressure. An open-ended monometer, would be better suited to measuring higher pressure than lower pressure. A gas pressure of one atmosphere results in the columns of mercury being equal. A sidearm of 760 mm pressure would allow one to measure pressures up to 2 atmospheres. 107. (D) g 23.2 g 17.7 g 1.1 g 50.5 g 44.01 + 28.01 + 2.016 + 16.043 + 28.01 100 mol 100 mol 100 mol 100 mol 100 mol 1 atm g 763 mmHg× 24.56 PM 760 mmHg mol g Mav = 24.56 g/mol density = = =1.015 RT L L atm 0.08206 296 K K mol
(a) Mav =
8.0
(b) Pco = Ptotal ×
V% 23.2 % = 763 mmHg × = 177 mmHg or 0.233 atm 100 % 100 %
(c) CO(g) + ½ O2(g) CO2(g) H2(g)+ ½ O2(g) H2O(g) CH4(g)+ 2 O2(g) CO2(g) + 2 H2O(g)
Use the fact that volume is directly proportional to moles when the pressure and temperature are constant. 1000 L of producer gas contains: 232 L CO which requires 116 L O2 88.5 L O2 177 L H2 which requires 22 L O2 11 L CH4 which requires 226.5 L O2 (Note: air is 20.95 % O2 by volume) Thus, the reaction requires
226.5 L =1.08×103 L . 0.2095
108. (M) First, balance the equation: C20H32O2 + 27 O2 → 20 CO2 + 16 H2O
mol O 2 needed: 2000 g C20 H 32 O 2
1 mol C20 H 32 O 2 27 mol O 2 304.52 g C20 H 32 O 2 1 mol C20 H 32 O 2
= 177.33 mol O 2 Using the ideal gas law, we can determine the volume of 177.33 mol of O2: 1 nRT 177.33 mol 0.082058 L atm K 298 K vol O 2 = = = 4336.30 L O 2 P 1.00 atm
246
Chapter 6: Gases
To determine the volume of air needed, we note that O2 represents 20.9% of air by volume: x(0.205) = 4336.30 L. Solving for x gives 20698 L, or 2.070×104 L. 109. (M) First recognize that 3 moles of gas are produced from every 2 moles of water, and compute the number of moles of gas produced. Then determine the partial pressure these gases would exert. amount of gas 1.32 g H 2 O
1 mol H 2 O 3 mol gas 0.110 mol gas 18.02 g H 2 O 2 mol H 2 O
nRT 0.110 mol 0.08206 L atm mol1 K 1 303 K 760 mmHg P 717 mmHg V 2.90 L 1 atm
Then the vapor pressure (partial pressure) of water is determined by difference.
Pwater = 748 mmHg - 717 mmHg = 31 mmHg 110. (D) First, balance the equations: Fe + 2 HCl → 2 FeCl2 + H2 2 Al + 6 HCl → 2 AlCl3 + 3 H2 Then, we calculate the partial pressure of O2: PO2 =PTOT - PH 2O =841 - 16.5 = 824.5 torr
mol H 2 =
1.0849 atm 0.159 L = 0.007192 mol 0.082058 L atm K 1 292 K
Now, we will try to express the calculated moles of H2 using the mole relationship between the metals and H2: 1 mol H 2 3 mol H 2 mol H 2 = mol Fe + mol Al = 0.007198 mol 1 mol Fe 2 mol Al 1 mol Fe 1 mol Al 1.0 mass Fe + 1.5 0.1924 mass Fe 55.85 g Fe 26.98 g Al
0.007192 0.017905 mass Fe 0.01070 0.055597 mass Fe Solving for massFe yields 0.09307 g of Fe. %Fe = (0.09307/0.1924) ×100 = 48.4%
247
Chapter 6: Gases
111. (M) The total pressure of the mixture of O2 and H2O is 737 mmHg, and the partial pressure of H2O is 25.2 mmHg. (a) The percent of water vapor by volume equals its percent pressure.
% H2O
25.2 mmHg 100% 3.42% H 2 O by volume 737 mmHg
(b) The % water vapor by number of molecules equals its percent pressure, 3.42% by number. (c) One mole of the combined gases contains 0.0342 mol H2O and 0.9658 mol O2. molar mass = 0.0342 mol H 2 O×
18.02 g H 2 O 31.999 g O 2 + 0.9658 mol O 2 × 1 mol H 2 O 1 mol O 2
=0.616 g H 2 O + 30.90 g O 2 = 31.52 g % H 2 O=
0.616 g H 2 O ×100% = 1.95 % H 2 O by mass 31.52 g
112. (D) (a) 1 mol of the mixture at STP occupies a volume of 22.414 L. It contains 0.79 mol He and 0.21 mol O 2 . 4.003 g He 32.0 g O 2 0.79 mol He 0.21 mol O 2 mass 1 mol He 1 mol O 2 STP density = = = 0.44 g / L volume 22.414 L
25 C is a temperature higher than STP. This condition increases the 1.00-L volume containing 0.44 g of the mixture at STP. We calculate the expanded volume with the combined gas law.
Vfinal 1.00 L
b25 + 273.2g K = 1.09 L 273.2 K
final density =
0.44 g = 0.40 g / L 1.09 L
We determine the apparent molar masses of each mixture by multiplying the mole fraction (numerically equal to the volume fraction) of each gas by its molar mass, and then summing these products for all gases in the mixture. 28.01 g N 2 32.00 g O 2 39.95 g Ar M air 0.78084 + 0.20946 + 0.00934 1 mol N 2 1 mol O 2 1 mol Ar
= 21.87 g N 2 + 6.703 g O 2 + 0.373 g Ar = 28.95 g / mol air 4.003 g He 32.00 g O 2 9.9 g mixture M mix 0.79 = 3.2 g He + 6.7 g O 2 = + 0.21 1 mol He 1 mol O 2 mol
248
Chapter 6: Gases
(b) In order to prepare two gases with the same density, the volume of the gas of smaller molar mass must be smaller by a factor equal to the ratio of the molar masses. According to Boyle’s law, this means that the pressure on the less dense gas must be larger by a factor equal to a ratio of molar masses.
Pmix
28.95 1.00 atm = 2.9 atm 9.9
113. (M) First, determine the moles of Cl2 and NaClO. Then, determine the limiting reagent 4.66 atm 1.0 L mol Cl2 = = 0.2007 mol 0.082058 L atm K 1 283 K
mol NaClO = (0.750 L)(2.00 M) = 1.50 mol 4 mol NaClO 0.2007 mol Cl 2 0.8028 mol NaClO needed 1 mol Cl2 Therefore, NaClO is the excess reagent. Now, we must calculate the theoretical yield of ClO2: 2 mol ClO 2 67.45 g ClO 2 0.2007 mol Cl2 27.07 g ClO 2 1 mol Cl2 1 mol ClO 2 %yield =
25.9 100 95.7% 27.07
114. (M) First, determine the moles of Na2S2O3, then use the chemical equations given to determine the moles of O3 in the mixture:
mol Na 2S2 O3 = 0.0262 L 0.1359 M = 0.003561 mol 1 mol I3 1 mol O3 mol O3 : 0.003561 mol Na 2S2 O3 0.001780 mol O3 2 mol Na 2S2 O3 1 mol I3 Using the ideal gas law, we can determine the moles of gas: 0.993 atm 53.2 L moles of gas = = 2.2123 mol 0.082058 L atm K 1 291 K
O 3
115.
0.001780 mol = 8.046 104 2.2123 mol total
(D) First compute the pressure of the water vapor at 30.1°C. 1 mol H 2 O 0.08206 L atm (30.1 273.2) K 0.1052 g 18.015 g H O mol K 760 mmHg 2 PH2O 13.72 mmHg 8.050 L
249
1 atm
Chapter 6: Gases
The water vapor is kept in the 8.050-L container, which means that its pressure is proportional to the absolute temperature in the container. Thus, for each of the six temperatures, we need to calculate two numbers: (1) the pressure due to this water (because gas pressure varies with temperature), and (2) 80% of the vapor pressure. The temperature we are seeking is where the two numbers agree. (T 273.2) K (T ) 13.72 mmHg P water (30.1 273.2) K (20. 273.2) K 13.3 mmHg P(20C) 13.72 mmHg (30.1 273.2) K For example, T 20. °C 19. °C 18. °C 17. °C 16. °C 15. °C Pwater, mmHg 13.3 13.2 13.2 13.1 13.1 13.0 80.0% v.p., 14.0 13.2 12.4 11.6 10.9 10.2 mmHg At approximately 19°C, the relative humidity of the air will be 80.0%. 116. (D) 1/A 1/V (cm-1) 0.0358 0.0328 0.0394 0.0437 0.0493 0.0562 0.0658 0.0787 0.0980 0.1316
3500
-1
Plot of Pressure versus A
3000
Pressure (mmHg)
Pgas (mmHg) 810.8 739.8 896.8 996.8 1123 1278 1494 1796 2216 2986
2500 2000 1500
Equation of the line P = 22670(A-1) + 2.5377
1000 500 0 0
0.05
0.1
0.15
A-1(cm -1)
Factors that would affect the slope of this straight line are related to deviations real gases exhibit from ideality. At higher pressures, real gases tend to interact more, exerting forces of attraction and repulsion that Boyle’s Law does not take into account. 117. (D) (a) T = 10 C = 283 K kg m R = 8.314472 2 s K mol
M = 28.96 g mol-1 (from question 99) or 0.02896 kg mol-1 m g = 9.80665 2 h = 14494 ft× 12in × 2.54 cm × 1m 4417.8 m s 1ft 1in 100 cm
0.02896
P = Po × 10
Mgh 2.303RT
760 mmHg × 10
kg m 9.80665 2 4417.8 m mol s
2.303 8.314472
250
kg m 283K s2 K mol
760 100.2314 = 446 mmHg
Chapter 6: Gases
(b) h = 900 ft×
12in 2.54 cm 1m × × 274.3m 1ft 1in 100 cm
0.02896
P = Po 10
kg m 9.80665 2 274.3 m mol s
2.303 8.314472
kg m 283K s K mol 2
= 0.967455 Po =
29 Po 30
or
1 smaller 30
118. (M) (a) urms is determined as follows:
3 8.3145 J mol1 K 1 298 K 3RT 482 m/s M 32.00 103 kg
u rms
(b)
M Fu 4 2 RT
3/2
u 2 exp Mu 2 / 2RT
32.00 103 Fu 4 2 (8.3145)(298)
3/ 2
32.00 103 (498) 2 1.92 103 (498) exp 2(8.3145)(298) 2
119. (D) Potential energy of an object is highest when the kinetic energy of the object is zero and the object has attained its maximum height. Therefore, we must determine the kinetic energy. But first, we have to determine the velocity of the N2 molecule. u rms
3 8.3145 J mol1 K 1 300 K 3RT 517 m/s M 28.00 103 kg
2 1 1 2 mu rms 28.00 103 kg 517 m/s 3742 J 2 2 e k e p m g h 28.00 103 kg 9.8 m/s 2 h = 3742 J
ek
Solving for h, the altitude reached by an N2 molecule is 13637 m or 13.6 km. 120. (D)
251
Chapter 6: Gases
121. (D) (a) First multiply out the left-hand side of the equation. an 2 an 2 abn 3 P 2 (V nb) nRT PV Pnb 2 V V V Now multiply the entire equation through by V2, and collect all terms on the right-hand side. 0 nRTV 2 PV 3 PnbV 2 an 2V abn 3 Finally, divide the entire equation by P and collect terms with the same power of V, to obtain: 2 n 3 ab RT bP 2 n a 3 0 V n V V P 0 P P (b) 1 mol CO 2 n 185 g CO 2 4.20 mol CO 2 44.0 g CO 2 (0.08206 L atm mol1 K 1 286 K) (0.0429 L/mol 12.5 atm) V2 0 V 3 4.20 mol 12.5 atm (4.20 mol) 2 3.61 L2 atm mol2 (4.20 mol)3 3.61 L2 atm mol2 0.0429 L/mol V 12.5 atm 12.5 atm 3 2 V 8.07 V 5.09 V 0.918
We can solve this equation by the method of successive approximations. As a first value, we use the volume obtained from the ideal gas equation: nRT 4.20 mol 0.08206 L atm mol 1 K 1 286 K 7.89 L V 12.5 atm P
252
Chapter 6: Gases
V = 7.89 L (7.89)3 - 8.06 (7.89)2 + (5.07 × 7.89) - 0.909 = 28.5 Try V = 7.00 L (7.00)3 - 8.06 (7.00)2 + (5.07 × 7.00) - 0.909 = -17.4 Try V = 7.40 L (7.40)3 - 8.06 (7.40)2 + (5.07 × 7.40) - 0.909 = 0.47 Try V = 7.38 L (7.38)3 - 8.06 (7.38)2 + (5.07 × 7.38) - 0.909 = -0.53 Try V = 7.39 L (7.39)3 - 8.06 (7.39)2 + (5.07 × 7.39) - 0.909 = -0.03 The volume of CO2 is very close to 7.39 L.
>0 <0 >0 <0 <0
A second way is to simply disregard the last term and to solve the equation 0 = V3 - 8.06 V2 + 5.07 V This equation simplifies to the following quadratic equation: 0 = V2 - 8.06 V + 5.07, which is solved with the quadratic formula. 8.06 (8.06) 2 4 5.07 8.06 6.68 14.74 V 7.37 L 2 2 2 The other root, 0.69 L does not appear to be reasonable, due to its small size. 122. (D) (a) The van der Waals gas equation is used to determine the actual pressure. n 2a P 2 V nb nRT V
2 nRT n 2 a 1 0.083145 280 1 1.382 P 96.5 bar = 9.65 MPa V nb V 2 0.2168 1 0.0319 0.2168 2
%error =
9.65-10 100 3.5%
10 (b) The volume using the ideal gas law is given as follows: V = nRT/P = (1 mol)(8.3145 kPa·L·K-1·mol-1)(280 K)/(10×103 kPa) = 0.233 L %error = (0.233 – 0.2168)/0.2168 × 100 = 7.47%
123. (M) (a)
B
V
PV = RT 1+
P 0.500
+
C
V = 500 cm 3 mol-1
B = -21.89 cm 3 mol-1
2 V T= 273 K
C = 1230 cm 6 mol-2
-21.89 cm 3 mol-1 1230 cm 6 mol-2 L L atm = 0.08206 273 K + 1+ 2 3 -1 mol K mol 500 cm mol 500 cm 3 mol-1
L atm L P 0.500 0.961 = 22.40 mol mol
253
22.4 P=
L atm mol
0.961
0.500 L mol
= 43.1 atm
Chapter 6: Gases
(b) The result is consistent with Figure 6-20. In Figure 6-20, the compressibility factor is slightly below 1 at pressures between 0 and 400 atm. At a pressure of ~50 atm, the compressibility factor is just slightly below 1. Using the data provided, a pressure of 43.1 atm has a compressibility factor of 0.961, which is slightly below 1. The data is consistent with the information provided in Fig 620. 124. (M)
1 atm 0.9618 atm 760 mm Hg 0.9618 atm 0.202 L PV mol H 2 7.99 103 mol 1 1 RT 0.08206 L atm mol K 296 K
PH 2 752 mm Hg 21.07 mm Hg
n H2 = n H2 (from Al) + n H2 (from Mg) Let x = mass of Al in g, therefore 0.156 x = mass Mg in grams 1 mol H 2 1 mol Al 3 mol H 2 1 mol Mg n H2 = 7.99 10-3 mol H 2 = x (0.156 x) 26.98 g Al 2 mol Al 24.305 g Mg 1 mol Mg 7.99 10-3 mol H 2 = 0.0556x + 0.006418 0.04114x 0.001572 = 0.01446x x = 0.1087 g Al 0.1087 g Al mass % Al = 100% = 69.7 % Al 0.156 g mixture mass % Mg = 100% 69.7 % = 30.3 % Mg
FEATURE PROBLEMS 125. (M) Boyle’s Law relates P and V, i.e., P × V = constant. If V is proportional to the value of A, and Pgas = Pbar + PHg (i.e. the pressure of the gas equals the sum of the barometric pressure and the pressure exerted by the mercury column), then a comparison of individual A × P products should show a consistent result or a constant. A (cm) 27.9 30.5 25.4 22.9 20.3 17.8 15.2 12.7 10.2 7.6
Pbar (atm) 739.8 739.8 739.8 739.8 739.8 739.8 739.8 739.8 739.8 739.8
PHg (mmHg) 71 0 157 257 383 538 754 1056 1476 2246
Pgas (mmHg) 810.8 739.8 896.8 996.8 1123 1278 1494 1796 2216 2986 254
A × Pgas (cm mmHg) 22621 22564 22779 22827 22793 22745 22706 22807 22601 22692
Chapter 6: Gases
Since consistent A × Pgas results are observed, these data conform reasonably well (within experimental uncertainty) to Boyle’s Law. 49.4 ) = 32.1 u of X 100 32.7 ) = 16.0 u of X Nitrosyl Fluoride 49.01 u ( 100 18.6 ) = 16.0 u of X Thionyl Fluoride 86.07 u ( 100 31.4 Sulfuryl Fluoride 102.07 u ( ) = 32.0 u of X 100 The atomic mass of X is 16 u which corresponds to the element oxygen. The number of atoms of X (oxygen) in each compound is given below: Nitryl Fluoride = 2 atoms of O Nitrosyl Fluoride = 1 atom of O Thionyl Fluoride = 1 atom of O Sulfuryl Fluoride = 2 atoms of O
126. (E) Nitryl Fluoride
65.01 u (
127. (M) (a) The N 2 g extracted from liquid air has some Ar(g) mixed in. Only O 2 g was removed from liquid air in the oxygen-related experiments. (b) Because of the presence of Ar(g) [39.95 g/mol], the N 2 g [28.01 g/mol] from liquid air will have a greater density than N 2 g from nitrogen compounds.
bg
bg
bg
(c)
bg
Magnesium will react with molecular nitrogen 3 Mg s + N 2 g Mg 3 N 2 s but not with Ar. Thus, magnesium
reacts with all the nitrogen in the mixture, but leaves the relatively inert Ar(g) unreacted. (d)
The “nitrogen” remaining after oxygen is extracted from each mole of air (Rayleigh’s mixture) contains 0.78084 + 0.00934 = 0.79018 mol and has the mass calculated below. mass of gaseous mixture = 0.78084 28.013 g/mol N 2 + 0.00934 39.948 g/mol Ar mass of gaseous mixture = 21.874 g N 2 + 0.373 g Ar = 22.247 g mixture.
Then, the molar mass of the mixture can be computed: 22.247 g mixture / 0.79018 mol = 28.154 g/mol. Since the STP molar volume of an ideal gas is 22.414 L, we can compute the two densities.
b g
d N2 =
28.013 g / mol 28.154 g/mol = 1.2561 g/mol = 1.2498 g / mol d mixture = 22.414 L / mol 22.414 L/mol
These densities differ by 0.50%.
255
Chapter 6: Gases
128. (M) (a)
First convert pressures from mmHg to atm:
Density (g/L) 1.428962 1.071485 0.714154 0.356985 (b)
Pressure (atm) 1.0000 0.75000 0.50000 0.25000
Density/Pressure (g/L.atm) 1.428962 1.4290 1.428647 1.4286 1.428308 1.4283 1.42794 1.4279 average = 1.4285 g/L.atm
d M O2 = 1.4285 g/L atm 0.082058 L atm/mol K 273.15 K RT P = 32.0175 g/mol Thus, the atomic mass of O2 = M O2 / 2 = 16.0086
M O2 =
M O2 This compares favorably with the value of 15.9994 given in the front of the textbook. 129.
(D) Total mass = mass of payload + mass of balloon + mass of H2 mass(m) RT Use ideal gas law to calculate mass of H2: PV = nRT = Molar mass(M) 3 3 2.54 cm 1×10-3 L g 3 12in 1atm 120ft × × × 2.016 3 3 3 1cm mol 1ft 1in PVM = m= = 306 g RT L atm 0.08206 273K K mol Total mass = 1200 g + 1700 g + 306 g 3200 g
We know at the maximum height, the balloon will be 25 ft in diameter. Need to find out what mass of air is displaced. We need to make one assumption – the volume percent of air is unchanged with altitude. Hence we use an apparent molar mass for air of 29 g mol-1 (question 99). Using the data provided, we find the altitude at which the balloon displaces 3200 g of air. 25 4 4 Note: balloon radius = = 12.5 ft. volume = ()r3 = (3.1416)(12.5)3 = 8181 ft3 2 3 3 3 3 -3 12in × 2.54 cm × 1×10 L = 231,660 L Convert to liters: 8181ft 3 × 3 3 1cm3 1ft 1in 1atm g 2.7×10 2 mb× (231,660 L) 29 1013.25 mb PVM mol = At 10 km: m = = 97,827 g RT L atm 0.08206 K mol 223K 1atm g 5.5×101mb× (231,660 L) 29 1013.25 mb PVM mol = = 20,478 g At 20 km: m = RT L atm 0.08206 K mol 217K
256
Chapter 6: Gases
1atm g 1.2×101mb× (231,660 L) 29 1013.25 mb PVM mol = At 30 km: m = = 4,215g RT L atm 0.08206 K mol 230K 1atm g 2.9×100 mb× (231,660 L) 29 1013.25 mb PVM mol = = 937 g At 40 km: m = RT L atm 0.08206 K mol 250K The lifting power of the balloon will allow it to rise to an altitude of just over 30 km.
SELF-ASSESSMENT EXERCISES 130. (E) (a) atm: Pressure exerted by the atmosphere at sea level per unit area, or 760 mm Hg (b) STP: Standard temperature and pressure, defined as a pressure of 1 barr at 273 K (c) R: Gas constant, which is the ratio between the product of the molar volume of a gas and pressure, and temperature. (d) Partial pressure: The ratio between the pressure of a gas in a container and the total pressure of the container (e) urms: Root mean squared velocity of gas molecules in a sample
131. (E) (a) Absolute zero: the lowest theoretical temperature, and the temperature at which all molecular/atomic vibrations cease. (b) Collection of gas over water: The process of isolating a gas (which does not react with water) generated by a reaction by bubbling it through a bottle and measuring its volume by the displacement of water. (c) Effusion of a gas: the escape of gas molecules from their container through a tiny orifice or pinhole. (d) Law of combining volumes: the volume ratio of gases consumed and generated is the same as their mole ratio, provided that temperature and pressure are kept constant throughout. 132. (E) (a) A barometer measures atmospheric pressure, whereas a manometer measures the pressure of a generated gas in a closed vessel. (b) Celsius temperature uses the freezing and boiling points of water to generate a temperature scale, whereas the Kelvin scale uses the temperature below which no molecular vibrations can occur as the zero point. (c) The ideal gas equation states the relationship between pressure, volume, moles, and temperature in an ideal gas. The general gas equation uses the ideal gas equation to set up a linear relationship between P, V, n, and T (P1V1/n1T1 = P2V2/n2T2).
257
Chapter 6: Gases
(d) An ideal gas is one in which the gas molecules themselves don’t occupy a volume, and that their interaction with each other is negligible. A real gas takes into account molecular volume of the gas and intermolecular/interatomic interactions. 133. (E) The answer is (d). The following shows the values for each: (a) P = g·h·d = (9.8 m/s2)(0.75 m)(13600 kg/m3) = 99960 Pa (b) Just for a rough approximation, we assume the density of air to be the same as that of nitrogen (this underestimates it a bit, but is close enough). The density of N2 is 0.02802 g/22.7 L = 1.234 g/L = 1.234 kg/m3. P = (9.8 m/s2)(16093 m)(1.234 kg/m3) = 194616 Pa. (c) P = g·h·d = (9.8 m/s2)(5.0 m)(1590 kg/m3) = 77910 Pa (d) P = nRT/V = (10.00 g H2 × 1 mol/2.02 g)(0.083145 L·barr·K-1)(273 K)/(22.7 L) = 4.95 barr = 495032 Pa 134. (E) The answer is (c), because the temperature decreases from 100 °C (373 K) to 200 K. 135. (M) P1/T1 = P2/T2. To calculate T2, rearrange the formula: T2 = 2.0 barr / (1.0 barr × 273 K) = 546 K 136. (M) The answer is (d). P1 V1 P2 V2 = T1 T2
V2
1 atm 22.4 L 298 K = 16.3 L 273 K 1.5 atm
137. (M) The answer is (b). Since the same number of moles of all ideal gases occupy the same volume, density is driven by the molar mass of the gas. Therefore, Kr has the highest density because it has a molar mass of 83 g/mol. 138. (E) The answer is (a). Using the formula urms = (3RT/M)1/2, increases T by a factor of 2 increase urms by a factor of 2 . 139. (M) (a) False. They both have the same kinetic energy (b) True. All else being equal, the heavier molecule is slower. (c) False. The formula PV=nRT can be used to confirm this. The answer is 24.4 L (d) True. There is ~1.0 mole of each gas present. All else being equal, the same number of moles of any ideal gas occupies the same volume. (e) False. Total pressure is the sum of the partial pressures. So long as there is nothing else but H2 and O2, the total pressure is equal to the sum of the individual partial pressures.
258
Chapter 6: Gases
140. (E) The answer is (c). Partial pressures are additive, so: PTOT =PH2O +PO2 PO2 (atm)=
PTOT PH2O 760 mm Hg
751 21 0.96 atm 760
141. (M) The answer is (a). First, determine the # moles of NH3 using the ideal gas law relationship: n = PV/RT = (0.500 atm × 4.48 L) / (0.08206 L·atm K-1 × 273 K) = 0.100 mol If 1.0 mole of a substance has 6.022×1023 atoms, 0.100 moles has 6.022×1022 atoms. 142. (M) The answer is (b). Since PV = nRT, the number of moles of O2 needed to satisfy the conditions of the problem is:
n = (2.00 atm × 2.24 L) / (0.08206 L·atm K-1 × 273 K) = 0.200 moles 1 mol O2 0.050 mol O 2 32.0 g O 2 To have an additional 0.150 mol in the system, we would add 0.6 g of He to the container.
The amount of O2 available = 1.60 g O 2
143. (E) The volumes of both gases were measured at the same temperature and pressure. Therefore, the proportionality constant between volume and moles for both gases is the same (that is, volume can essentially replace moles in the following calculations):
25.0 L H 2
3 L CO = 10.7 L CO needed 7 L H2
So, all of the H2 and 10.7 L of CO are consumed, and 1.3 L of CO remain. 144. (M) The answer is (a), that is, the partial pressure of H2 in the container is less than SO2. The reasoning can be derived from the kinetic molecular theory of gases. Gas molecules with smaller molar masses travel faster, and as such can escape faster from an orifice. This is expressed in Graham’s Law in section 6.8. 145. (M) The answer is (c). Gases behave more ideally at high temperatures and low pressures. 146. (M) (a) He or Ne: Ne has higher a and b values. (b) CH4 or C3H8: C3H8 has higher a and b values (c) H2 or Cl2: Cl2 has higher a and b values
259
Chapter 6: Gases
147. (D) We know that pressure is force per unit area, that is:
F mg A A Using the fact that area A=πr 2 =π(D/2) 2 and that mass m=d V , and that volume of a cylindrical tube is V=A h (where h is the height of the liquid in the tube), we can express the pressure formula as follows: F d V g d A hg P= = =d h g A A A Therefore, dVg d h g, and, A d d h= A π D/2 2 P=
As we can see, the height, h is inversely proportional to D. That is, the larger the diameter of the tube, the shorter the height of the liquid. 148. (D) First, convert the given information to more useful units. The pressure (752 torr) is equivalent to 0.9895 atm (752 mm/760 mm Hg), and the temperature is 298 K. Then, use the ideal gas relationship to determine how many moles of gas are present, assuming 1 L of gas:
PV = nRT n=
0.9895 atm 1 L PV 0.04046 mol RT 0.08206 L atm K 1 298 K
From the problem, we know the mass of the gas per 1 L, as expressed in the density. Mass of 1 L of this gas is 2.35 g. The molar mass of this substance is therefore:
MM = g/mol = 2.35 g/0.04046 mol. = 58.082 g/mol Now, let’s calculate the empirical formula for the hydrocarbon gas and see how it compares to the molar mass: 1 mol C mol C = 82.7 g C = 6.89 mol 12.01 g C 1 mol H mol H = 17.3 g H = 17.1 mol 1.01 g H H:C ratio = 17.1 mol / 6.89 mol = 2.5 The ratio between H and C is 2.5:1 or 5:2, making the empirical formula C2H5. The mass of this formula unit is 29.07 g/mol. Comparing to the calculated molar mass of the gas, it is smaller by a factor of 2. Therefore, the gas in question in C4H10 or butane.
260
Chapter 6: Gases
149. (E) N2 comprises 78.084% of atmosphere, oxygen 20.946%, argon 0.934%, and CO2 0.0379%. To graphically show the scale of this difference, divide all values by the smallest one (CO2). Therefore, for every single mark representing CO2, we need 2060 marks for N2, 553 marks for O2, and 25 for Ar. 150. (M) To construct a concept map, one must first start with the most general concepts. These concepts are not defined by or in terms of other concepts discussed in those sections. In this chapter, pressure is the overarching concept (6-1). The topics that fall under the rubric of pressure are liquid and gas pressure, and measuring pressure. Simple gas laws are derived from the concept of pressure (6-2). Simple gas laws include Boyle’s Law, Charles’ Law and Avogadro’s Law. These laws combine to form another subtopic, the Ideal Gas Law and the General Gas Equation. Take a look at the subsection headings and problems for more refining of the general and specific concepts.
261
