CHAPTER 10 Problems: 22, 24, 34, 40, 44, 48, 50, 56, 62, 66, 70, 81, 82, 88, 94, 106,
128
22) At 46. C a sample of ammonia gas exerts a pressure of 5.3 atm. What is the pressure
when the volume of the gas is reduced to one-fourth of the original value at the same
temperature?
Since n ant T are held constant we can use the relationship based on Boyle’s law
piVi = pfVf
or
pf = pi (Vi/Vf) = (5.3 atm) (4) = 21.2 atm.
24) A sample of air occupies 3.8 L when the pressure is 1.2 atm. Assuming constant
temperature
a) What volume does it occupy for p = 6.6 atm?
Since n and T are constant
piVi = pfVf
Vf = Vi (pi/pf) = (3.8 L) (1.2 atm/6.6 atm) = 0.69 L
b) What pressure is required to compress it to 0.075 L?
We may use the equation found in problem 22, which applies, since n and T are
kept constant
pf = pi (Vi/Vf) = (1.2 atm) (3.8 L/0.075 L) = 61. Atm
34) A sample of nitrogen in a 4.5 L container at a temperature of 27. C exerts a pressure
of 4.1 atm. Calculate the number of moles of gas in the sample.
pV = nRT
so
T = 27. C = 300. K
n = pV =
(4.1 atm) (4.5 L)
RT (0.08206 L.atm/mol.K) (300. K)
= 0.75 mol
40) An ideal gas originally at 0.85 atm and 66. C was allowed to expand until its final
volume, pressure, and temperature were 94. mL, 0.60 atm, and 45. C. What was the
initial volume?
Since n is kept constant, we may say
piVi = pfVf
Ti
Tf
Ti = 66. C = 339. K Tf = 45. C = 318. K
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Or
Vi = Vf (pf/pi) (Ti/Tf) = (94. mL) (0.60 atm/0.85 atm) (339. K/318. K)
= 71. mL
44) At STP, 0.280 L of a gas weighs 0.400 g. Calculate the molar mass of the gas.
STP means p = 1.000 atm, T = 0.0 C = 273.2 K. Using the ideal gas law we can
derive an equation for n (see problem 34)
n = pV =
(1.00 atm) (0.280 L)
= 0.01249 mol
RT (0.08206 L.atm/mol.K) (273.2 K)
M = m/n = (0.400 g)/(0.01249 mol) = 32.0 g/mol
48) A 2.10 L vessel contains 4.65 g of a gas at 1.00 atm and 27.0 C.
a) Calculate the density of the gas in g/L.
d = m/V = (4.65 g)/(2.10 L) = 2.214 g/L
b) What is the molar mass of the gas?
From the ideal gas law (see problem 34)
T = 27.0 C = 300.2 K
n = pV =
(1.00 atm) (2.10 L)
= 0.08525 mol
RT (0.08206 L.atm/mol.K) (300.2 K)
M = m/n = (4.65 g)/(0.08525 mol) = 54.5 g/mol
50) A certain anesthetic contains 64.9 % C, 13.5 % H, and 21.6 % O by mass. At 120. C
and 750. mm Hg, 1.00 L of the gaseous compound weighs 2.30 g. What is the molecular
formula of the compound?
Let us first find the empirical formula. Assume 100.0 g compound. Then
64.9 g C
1. mol = 5.404 mol C  1.350 = 4.00  4
12.01 g
13.5 g H 1. mol = 13.39 mol H  1.350 = 9.92  10
1.008 g
21.6 g O 1. mol = 1.350 mol O 1.350 = 1.00
16.00 g
Empirical formula is C4H10O (FW = 74.12 g/mol)
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We can now find the molecular mass of the compound.
120. C = 393. K
750. mmHg (1atm/760 mmHg) = 0.9868 atm
n = pV =
(0.9868 atm) (1.00 L)
RT (0.08206 L.atm/mol.K) (393. K)
= 0.03060 mol
M = m/n = (2.30 g)/(0.03060 mol) = 75.2 g/mol
The number of formula units in the molecule is 75.2/74.12 = 0.99  1
So the molecular formula is C4H10O (the same as the empirical formula)
56) When coal is burned the sulfur present is converted into sulfur dioxide (SO2), which
is partly responsible for acid rain.
S(s) + O2(g)  SO2(g)
If 3.15 kg of S reacts with oxygen, calculate the volume of SO2 gas (in mL) formed at
30.5 C and 1.04 atm.
moles S = 3150. g 1 mol = 98.25 mol S
32.06 g
Based on the stoichiometry of the reaction, 1 mole of S leads to the formation of 1 mole
of SO2, and so moles SO2 = 98.25 mol
pV = nRT
V = nRT = (98.25 mol) (0.08206 L.atm/mol.K) (303.6 K) = 2354. L
p
1.04 atm
62) Calculate the mass, in grams of hydrogen chloride produced when 5.6 L of molecular
hydrogen (H2), measured at STP, reacts with an excess of chlorine gas (Cl2).
The reaction taking place is
H2(g) + Cl2(g)  2 HCl(g)
We can interpret the above reaction in terms of volumes of reactants and products as long
as we assume the same temperature and pressure throughout and look only at substances
in the gas phase
So
V(HCl) = 5.6 L H2 2 L HCl = 11.2 L HCl
1 L H2
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The moles of HCl is then
n = pV =
(1.000 atm)(11.2 L)
= 0.500 mol
RT (0.08206 L.atm/mol.K)(273.2 K)
The mass of HCl (M = 36.46 g/mol) is then
n = (0.500 mol) (36.46 g/mol) = 18.2 g HCl
66) A mixture of gases contains 0.31 mol CH4, 0.25 mol C2H6, and 0.29 mol C3H8. The
total pressure is 1.50 atm. Calculate the partial pressures of each gas.
ntotal = 0.31 mol + 0.25 mol + 0.29 mol = 0.85 mol
X(CH4) = 0.31 mol/0.85 mol = 0.365
X(C2H6) = 0.25 mol/0.85 mol = 0.294
X(C3H8) = 0.29 mol/0.85 mol = 0.341
By Dalton’s law
pi = Xi ptotal
so
p(CH4) = (0.365) (1.50 atm) = 0.55 atm
p(C2H6) = (0.294) (1.50 atm) = 0.44 atm
p(C3H8) = (0.341) (1.50 atm) = 0.51 atm
70) A piece of sodium metal reacts completely with water as follows:
2 Na(s) + 2 H2O(l)  H2(g) + 2 NaOH(aq)
The hydrogen gas generated is collected over water at 25.0 C. The volume of the gas is
246. mL measured at 1.00 atm. Calculate the number of grams of sodium used in the
reaction. (At T = 25.0 C the vapor pressure of water is 0.0313 atm).
We have sufficient information to find the moles of H2 generated. Note that
because the gas was collected over water (in an apparatus similar to the one shown in fig
10.14)
ptotal = pH2 + pH2O
T = 25.0 C = 298.2 K
pH2 = ptotal – pH2O = 1.00 atm – 0.0313 atm = 0.9687 atm
4
n = pV =
(0.9687 atm) (0.246 L)
RT (0.08206 L.atm/mol.K) (298.2 K)
= 0.009738 mol
Using the balanced chemical equation, we may say
n(Na) = (0.09738 mol H2) 2 mol Na =0.01948 mol Na
1 mol H2
g Na = 0.01948 mol Na 22.99 g Na = 0.448 g Na
mol Na
81) What is the difference between gas effusion and gas diffusion?
Diffusion is the mixing of gases at the same temperature and pressure due to the
random motion of the particles making up the gas. Effusion is the movement of a gas
through a small pinhole into vacuum.
The rate for both diffusion and effusion is ~ (1/M)1/2, and so at a particular
temperature light gases diffuse and effuse more rapidly than heavy gases.
If you compare the rate of diffusion or effusion for two gases at the same
temperature, you may say (as a result of the above) that
Rate(gas 2) = (M1/M2)1/2
Rate(gas 1)
So, for example, the rate of diffusion or effusion for helium (M = 4.0 g/mol) is about
twice the rate for methane (CH4, M = 16.0 g/mol)
82) Compare the root mean square speed of O2 and UF6 at T = 65. C.
urms = (3RT/M)1/2
M(O2) = 32.00 g/mol
T = 65. C = 338. K
M(UF6) = 352.0 g/mol
urms(O2) = [3(8.314 J/mol.K)(338. K) / (32.00 x 10-3 kg/mol) ]1/2 = 513. m/s
urms(UF6) = [3(8.314 J/mol.K)(338. K) / (352.0 x 10-3 kg/mol) ]1/2 = 155. m/s
Notice we converted temperature into Kelvin and molecular mass into kg/mol.
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88) The average distance traveled by a molecule between successive collision is called
the mean free path. For a given amount of gas, how does the mean free path depend on
the following:
a) density – Decreases as density increases, as there are more molecules per unit
volume and so the gas is more crowded.
b) temperature, at contant volume – The same, independent of temperatrure, since
the density is the same. The gas molecules move faster on average at higher temperature,
but the average distance moved between collisions doesn’t change.
c) pressure, at constant temperature – Increased pressure will decrease the volume
and so increase the density. Therefore the gas is more crowded, and so the mean free
path decreases.
d) volume, at constant temperature – If the volume decreases the density
increases, and so the gas is more crowded. This will make the mean free path decrease.
e) size of the gas molecules – The bigger the gas molecule the more likely it will
collide with another gas molecule, and so the smaller the mean free path.
94) Use the data in Table 10.6 to calculate the pressure exerted by 2.50 mol of CO2
confined in a volume of 5.00 L at 450. K. Compare the presure to that calculated by the
ideal gas law. Which pressure is correct?
For the van der Waals equation
a(CO2) = 3.59 L2.atm/mol2
b(CO2) = 0.0427 L/mol
p = nRT – an2
(V – nb) V2
= (2.50 mol)(0.08206 L.atm/mol.K)(450. K) - (3.59 L2.atm/mol2)(2.50 mol)2
5.00 L - (2.50 mol)(0.0427 L/mol)
(5.00 L)2
= 18.866 atm – 0.898 atm = 17.97 atm
For the ideal gas law
P = nRT = (2.50 mol)(0.08206 L.atm/mol.K)(450. K) =18.46 atm
V
(5.00 L)
Neither pressure is likely correct – however, we would expect the pressure calculated
using the van der Waals equation to be closer to the actual pressure than the pressure
calculated using the ideal gas law.
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106) Consider the following apparatus. Calculate the partial pressures of helium and
neon after the stopcock is opened. The temperature remains constant at 16. C.
If we are clever we can do this problem without a lot of calculations. Since T is
kept constant we can use Boyle’s law
piVi = pfVf
pf = pi(Vi/Vf) Vtotal = 1.2 L + 3.4 L = 4.6 L
For He
p = (0.63 atm)(1.2 L/4.6 L) = 0.164 atm
For Ne
p = (2.8 atm)(3.4 L/4.6 L) = 2.07 atm
128) Nitrogen forms several gaseous oxides. One of them has a density of 1.33 g/L when
measured at 764. mm Hg and 150. C. What is the formula of the compound?
Assume 1 liter of gas. Then the mass of gas is 1.33 g.
To find the moles of gas we can use the ideal gas law
764. mm Hg = 1.0053 atm
150. C = 423.2 K
n = pV =
(1.0053 atm)(1.000 L)
= 0.02895 mol
RT (0.08206 L.atm/mol.K)(423.2 K)
M = m/n = (1.33 g/0.02895 mol) = 45.9 g/mol
O3 = 48.0 g/mol
NO2 = 46.0 g/mol
N2O = 44.0 g/mol
So the gas is most likely NO2.
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