Answers to selected problems, Chapter 5
Review questions
14. All three parts of this question can be answered directly
from the ideal gas law, PV = nRT.
(a) When the pressure is increased on a fixed amount of a gas
at constant temperature, nRT will remain constant. This makes PV a constant, and
V will this vary inversely with P (Boyle's law). So if pressure increases,
volume decreases.
(b) When only temperature and volume are allowed to
vary, we get V/T = nR/P, where nR/P is a constant. Thus, V = (nR/P)T = CT, where
C is a constant. This makes volume vary directly (linearly) with T. So when
temperature decreases, volume decreases proportionately.
(c) Pressure decreases and temperature increases. Note that
intuitively, but these changes will increase the volume of a fixed amount of a
gas. To show that mathematically, solve for volume, and get V = nRT/P.
Increasing temperature (in the numerator) will increase the ratio (volume).
Decreasing pressure, in the denominator will also increase the ratio.
17. Since container A has twice the number of molecules as container B but twice the volume as well, it will have the same number of molecules per unit volume (same density of gas). They will then hit the walls of container A at the same rate as in container B, creating the same pressure in A as in B.
22. To compare the densities in containers A and B, we can again
use the ideal gas law, PV = nRT. By writing it as n/V = P/RT, we get a quantity
n/V that is very close to density.
(a) If containers A and B have the same volume and
temperature, the denominators of n/V = P/RT are equal for the two gases, and n
will vary with P. That means that density (n/V) will also vary with P. So if A
is at higher pressure, it contains more moles and therefore is at greater
density.
(b) This one is easy. If A and B have the same pressure and
temperature, then P/RT will be the same for both containers. But P/RT is
proportional to density, so the densities in A and B will be equal.
(c) A and B have the same pressure and volume, but the gas in
A has a higher temperature. This means that P/R is a constant and that density
(n/V) is inversely proportional to T. So gas A, with the higher temperature, has
the lower density.
Note: There are intuitive ways to answer these these parts as
well. You should address the problem this way, too.
Problems
31. (a) (0.985 atm)(706 mmHg/1 atm) = 749
mmHg
(b) (849 Torr)(1 atm/760 Torr) = 1.12 atm
(c) (642 Torr)(101.325 kPa/760 Torr) = 85.5 kPa
(d) (15.5 lb/in.2)(760 mmHg/14.696 in.2)
= 801 mmHg
35. (a) Since 1 atmosphere of pressure is equivalent to a height
of 760 mm of Hg, we can set up the simple proportion (H/4.36 atm) = (0.76 mHg/1
atm). This gives H = (4.36 atm)(0.76 mHg/atm) = 3.31 mHg.
(b) Since pressure = mass per unit area and mass = density x
volume, pressure = density x volume per unit area. Since volume per unit area
just equals height, this means that pressure = density x height. Thus if columns
of liquids 1 and 2 are to produce the same pressure, D1H1
= D2H2. This makes the height of a column of liquid needed
to produce a given pressure inversely proportional to its density. So H2
= H1(D1/D2) = (25.0 cm Hg)(13.3 g/mL Hg/1.59 g/mL
CCl4) = 2.14 m CCl4.
39. A closed-end manometer measures differences in pressures. Thus to measure a difference in pressure of 1 atmosphere by using mercury, a difference in height of nearly 1 m will be needed. (The absolute height of either side is unimportant.)
41. For Boyle's law, one need only use the simplified version of
the ideal gas law: P1V1 = P2V2. (You
should also practice using the full version of the law.)
(a) (1752 Torr)(521 mL) = (752 Torr)(V2) ®
V2
= (521 mL)(1752 Torr/752 Torr) = 1210 mL.
(b) 3.55 atm = (3.55 atm)(760 Torr/atm) = 2698 Torr.
(1752 Torr)(521 mL) = (2698 Torr)(V2)
® V2 = (521 mL)(1752 Torr/2698 Torr) = 338 mL.
(c) (125 kPa)(760 Torr/101.325 kPa) = 937.6 Torr.
(1752 Torr)(521 mL) = (937.6 Torr)(V2)
® V2 = (521 mL)(1752 Torr/937.6 Torr) = 974 mL.
45. Just set up Boyle's law algebraically. (Remember how I have
said to set up problems algebraically first? Here is a good example of how
useful this approach can be.)
(P)(56.0 mL) = (P + 100.0 Torr)(49.4 mL). Now it's just
algebra. 56.0 P = 49.4 P + 4940 ® 6.6 P = 4940 ®
P = 750 Torr.
49. For Charles's law, you may use the abbreviated form of the
ideal gas law: V1/T1 = V2/T2.
V2 = V1(T2/T1)
= (154 mL)[283.0 K/(273.2 K + 99.8 K)] = (154 mL)(283.0/373.0) = 117 mL.
53. Remember that Charles's law uses absolute, or kelvin, temperatures. So if we call the original conditions state 1 and the final conditions state 2, we have: T1 = T2(V1/V2) = (T1 + 15)(V1/1.1V1), which transforms to T1 = (T1 + 15)(1/1.1). To solve for T1, just write 1.1 T1 = T1 + 15, or 0.1 T1 = 15, which makes T1 = 150 K.
55. This problem is easy once we notice that it is for STP. That means that each mole of the neon gas will occupy 22.4 L. The mass in kg of 4.55x103 L of neon at STP will then be its number of moles x its molecular weight M (its molar mass) / 1000, or (4.55x103 L/22.4 L)(20.2 g Ne per mole)(10-3 kg/g) = 4.10 kg Ne.
59. Estimates which of these samples contains the greatest
number of molecules (the greatest number of moles of the substance).
(a) 5.0 g H2 is 2.5 moles. (b) 50 L of SF6
gas at STP is just over 2 moles. (c) 1.0x1024 molecules of CO2
is just under 2 moles (because 6x1023 molecules makes one mole). (d)
67 L of a gas at STP makes about 3 moles.
Therefore, (d) has the greatest number of molecules.
It is also possible to work this problem by using molecules
instead of moles. But that requires the user to keep too many larger numbers in
mind. Better to use moles.
61. To calculate the new pressure, just use the combined gas law PV/T = constant. Since the volume of the gas in the can remains constant, this equations simplifies to P/T = constant, or P1/T1 = P2/T2. If we let the heated state be state 2, we just solve for P2 and get P2 = P1(T2/T1) = (721 Torr)[(755 + 273)/(25 + 273)] = 2487 Torr = 2490 Torr to three significant figures.
65. This one offers a chance to use all three terms of the combined gas law. The question is to solve for volume in a new state, which we may call state 2: Since PV/T = constant, or P1V1/T1 = P2V2/T2, we solve for V2 and get V2 = V1(P1T2/P2T1). Another version, which you might prefer, is V2 = V1(P1/P2)(T2/T1). A third version is V2 = (P1V1/T1)/(P2/T2). Use whichever version helps you best understand how the two sets of conditions are being compared. Putting the numbers into the first version, we get V2 = (2.53 m3)[(191 Torr)(298 K)]/[(1142 Torr)(258 K)] = 0.489 m3.
67. For this problem, we use the old standby PV = nRT, with the
only tricks being the units and the corresponding values of the coefficient R
(see page 195).
(a) V = nRT/P = (1.12 mol)(0.0821 L-atm/mol-K)(335 K)/(1.38
atm) = 22.3 L.
(b) P = nRT/V = (125 g CO/28 g CO/mole)(0.0821 L-atm/mol-K)(302
K)/3.96 L) = 27.9 atm.
(c) Use the variant of the formula PV = mRT/M, where m is
mass (g) and M is molar mass (g). This will give an answer for m in grams that
needs to be converted to milligrams.
m = PVM/RT = [(725 Torr/760 Torr/atm)(0.0345
L)(2 g/mole)]/[(0.0821 L-atm/mol-K)(261 K)] = 0.00307 g = 3.07 mg
(d) Calculate the pressure in atmospheres and convert to kP.
P = nRT/V = (173 g/28 g/mol N2)(0.082
L-atm/mol-K)(273 K)/(8.35 L) = 16.58 atm
16.58 atm(101.325 kPa/atm) = 1680 kPa.
73. The tricky part here is to first convert 1.0x109
molecules per cubic meter to moles or moles per liter. Converting to moles means
dividing by the number of molecules per mole, which is just Avogadro's number,
6.023x1023 molecules per mole. Then use a volume of 1000 L = 1 m3.
P = nRT/V = (1.0x109 molecules/6.023x1023
molecules/mole)(0.0821 L-atm/mol-K)(298 K)/(103 L) = 4.1x10-17
atm.
77. This forbidding-looking problem can be broken into accessible parts. First realize that because the compound is a hydrocarbon, it is made up only of H and C. Since the problem tells you that it is 8.75% H by mass, you know that it is 91.25% C by mass ('cause that's all there is in a hydrocarbon, folks). These percentages allow you to get the molar ratio of H and C by dividing by the atomic masses: (8.75 H)/(1.008 g H/mol H) = 8.68 moles of H for (91.25 )/(12 g C/mol C) = 7.60 moles of C. That's 1 mole of C per 1.142 moles of H. Since we express molar ratios in terms of small whole numbers, the empirical formula will be C7H8. The only remaining question is to find the actual formula. We can do this by getting the molecular weight, which we can determine from the ideal gas law with m and M: PV = mRT/M. Solving algebraically for M gives M = mRT/PV = [(1.261 g)(0.0821 L-atm/mol-K)(388 K)]/[761 Torr/760 Torr/atm)(0.435 L)] = 92.2 g/mol. This corresponds to the formula weight of C7H8, which is 84 + 8 = 92. So the hydrocarbon is C7H8. Extra question for you: what is a possible structure for this compound?
79. For calculating densities of gases, use the ideal gas law in
the form D = m/V = PM/RT.
(a) Density of CO(g) at STP: (1 atm)(28 g/mol)/[0.0821 L-atm/mol-K)(273
K)] = 1.25 g L-1.
(b) Ar(g) at 1.26 atm and 325 °C: [(1.26 atm)(39.9
g/mol)]/[0.0821 L-atm/mol-K)(598 K)] = 1.02 g L-1.
83. This problem is a little like no. 77, except easier. You
find the molecular weight of sulfur vapor and see how many sulfur atoms it
requires to get that weight in one molecule.
Starting from PV = (m/M)RT, get M = mRT/PV = [(4.33 g)(0.0821
L-atm/mol-K)(718 K)]/[(755 mmHg/760 mmHg/atm)(1 L)] = 257 u. That corresponds to
8 atoms of sulfur (8x32 = 256), so the formula for vaporous sulfur is S8.
87. This one looks harder than it is. Note that it gives liters and asks for liters, so nothing need be converted to those pesky grams. You just determine the limiting reactant by considering the volumes of the two reactants, remembering that at equal temperatures and pressures, moles of gases are proportional to volumes (because the molar volumes of gases are very similar to one another). In the equation 2 SO2(g) + O2(g) ® 2 SO3(g), 2 moles of SO2 are needed to react with 1 mole of O2. Thus, SO2/O2 = 2/1 (in moles). The actual molar proportions are 1.15/0.65 = 1.77, which means there is less SO2 per mole of O2 than the 2/1 required by the reaction. Thus, SO2 is limiting. You can just look at this equation and see that since the coefficients of SO2 and SO3 are both 2, a given number of moles of SO2 (or liters of SO2) will produce the same number of moles of SO3 (or liters of SO3). Thus 1.15 L of SO2, the limiting reactant, will give 1.15 L of SO3 at the same temperature and pressure. [Recall how we earlier set up this kind of calculation in a proportion: (Coeff. of substance 1)/(Actual moles of 1) = (Coeff. 2)/(Actual moles of 2). Here this is 2/1.15 L = 2/x L, so that x = 1.15.]
91. Two basic steps: use the gas equation to get the moles of
hydrogen gas that are produced, then use that with the equation to get the moles
and the mass of Mg needed.
(a) Moles of hydrogen. Use the form of the ideal gas law that
is n = PV/RT = [(758 Torr/760 Torr/atm)(0.02850 L)]/[(0.0821 L-atm/mol-K)(299
K)] = 0.00116 mol H2(g).
(b) The molar proportion for this equation is (1 mole Mg)/(x
moles Mg) = (1 mole H2)/(0.00116 mol H2) ®
x =
(1/1)(0.00116 mol H2) = 0.00116 mol Mg = 0.028 g Mg = 28 mg Mg.
101.