Chapter 9
Gases: Their Properties and Behavior

9.1 1.00 atm = 14.7 psi

1.00 mm Hg x = 1.93 x 10^-2 psi

9.2 1.00 atmosphere pressure can support a column of Hg 0.760 m high. Because the density of H₂O is 1.00 g/mL and that of Hg is 13.6 g/mL, 1.00 atmosphere pressure can support a column of H₂O 13.6 times higher than that of Hg. The column of H₂O supported by 1.00 atmosphere will be (0.760 m)(13.6) = 10.3 m.

9.3 The pressure in the flask is less than 0.975 atm because the liquid level is higher on the side connected to the flask. The 24.7 cm of Hg is the difference between the two pressures.

Pressure difference = 24.7 cm Hg x = 0.325 atm

Pressure in flask = 0.975 atm - 0.325 atm = 0.650 atm

9.4 The pressure in the flask is greater than 750 mm Hg because the liquid level is lower on the side connected to the flask.
Pressure difference = 25 cm Hg x = 250 mm Hg
Pressure in flask = 750 mm Hg + 250 mm Hg = 1000 mm Hg

9.5 (a) Assume an initial volume of 1.00 L.
First consider the volume change resulting from a change in the number of moles with the pressure and temperature constant.
; V_f = = 0.75 L
Now consider the volume change from 0.75 L as a result of a change in temperature with the number of moles and the pressure constant.
; V_f = = 1.0 L
There is no net change in the volume as a result of the decrease in the number of moles of gas and a temperature increase.

(b) Assume an initial volume of 1.00 L.
First consider the volume change resulting from a change in the number of moles with the pressure and temperature constant.
; V_f = = 0.75 L
Now consider the volume change from 0.75 L as a result of a change in temperature with the number of moles and the pressure constant.
; V_f = = 0.5 L
The volume would be cut in half as a result of the decrease in the number of moles of gas and a temperature decrease.

9.6 n = = 4.461 x 10³ mol CH₄
CH₄, 16.04 amu; mass CH₄ = (4.461 x 10³ mol) = 7.155 x 10⁴ g CH₄
9.7 C₃H₈, 44.10 amu; V = 350 mL = 0.350 L; T = 20^oC = 293 K
n = 3.2 g x = 0.073 mol C₃H₈
P = = = 5.0 atm
9.8 P = 1.51 x 10⁴ kPa x = 149 atm; T = 25.0^oC = 298 K
n = = 267 mol He

9.9 The volume and number of moles of gas remain constant.
; T_f = = 301 K = 28^oC

9.10 CaCO₃(s) + 2 HCl(aq) CaCl₂(aq) + CO₂(g) + H₂O(l)
CaCO₃, 100.1 amu; CO₂, 44.01 amu
mole CO₂ = 33.7 g CaCO₃ x = 0.337 mol CO₂
mass CO₂ = 0.337 mol CO₂ x = 14.8 g CO₂
V = = 7.55 L

9.11 C₃H₈(g) + 5 O₂(g) 3 CO₂(g) + 4 H₂O(l)
n_propane = = 2.76 mol C₃H₈
2.76 mol C₃H₈ x = 8.28 mol CO₂
V = = 186 L = 190 L

9.12 n = = 0.0446 mol
molar mass = = 34.1 g/mol; molecular mass = 34.1 amu
Na₂S(aq) + 2 HCl(aq) ® H₂S(g) + 2 NaCl(aq)
The foul-smelling gas is H₂S, hydrogen sulfide.
9.13 12.45 g H₂ x = 6.176 mol H₂
60.67 g N₂ x = 2.166 mol N₂
2.38 g NH₃ x = 0.140 mol NH₃
n_total = = 6.176 mol + 2.166 mol + 0.140 mol = 8.482 mol
= 0.7281; = 0.2554; = 0.0165

9.14 (from Problem 9.13). T = 90^oC = 363 K
= 25.27 atm
= (0.7281)(25.27 atm) = 18.4 atm
= (0.2554)(25.27 atm) = 6.45 atm
= (0.0165)(25.27 atm) = 0.417 atm

9.15 P_Total = (0.0287)(0.977 atm) = 0.0280 atm

9.16 The number of moles of each gas is proportional to the number of each of the different gas molecules in the container
n_total = n_red + n_yellow + n_green = 6 + 2 + 4 = 12
X_red = = 0.500; X_yellow = = 0.167; X_green = = 0.333
P_red = X_red P_total = (0.500)(600 mm Hg) = 300 mm Hg
P_yellow = X_yellow P_total = (0.167)(600 mm Hg) = 100 mm Hg
P_green = X_green P_total = (0.333)(600 mm Hg) = 200 mm Hg

9.17 , M = molar mass, R = 8.314 J/(K mol), 1 J = 1 kg m²/s²
at 37^oC = 310 K,
at -25^oC = 248 K,

9.18 (a) ; = 1.62
O₂ diffuses 1.62 times faster than Kr.

(b) ; = 1.04
C₂H₂ diffuses 1.04 times faster than N₂.

9.19 ;
Thus, the relative rates of diffusion are .

9.20 P = = 20.5 atm

P = = 20.3 atm

9.21 The amount of ozone is assumed to be constant.
Therefore nR =
Because V h, then where h is the thickness of the O₃ layer.
h_f = = 3.8 x 10^-5 m
(Actually, V = 4r²h, where r = the radius of the earth. When you go out 30 km to get to the ozone layer, the change in r² is less than 1%. Therefore you can neglect the change in r² and assume that V is proportional to h.)

9.22 For ether, the MAC = x 100% = 2.0%

9.23 (a) Let X = partial pressure of chloroform.
MAC = x 100% = 0.77%
Solve for X. X = 760 mm Hg x = 5.9 mm Hg

(b) CHCl₃, 119.4 amu
PV = nRT; n = = 0.00347 mol CHCl₃
mass CHCl₃ = 0.00347 mol CHCl₃ x = 0.41 g CHCl₃

Understanding Key Concepts

9.24 (a) The volume of a gas is proportional to the kelvin temperature at constant pressure. As the temperature increases from 300 K to 450 K, the volume will increase by a factor of 1.5.

(b) The volume of a gas is inversely proportional to pressure at constant temperature. As the pressure increases from 1 atm to 2 atm, the volume will decrease by a factor of 2.

(c) PV = nRT; The amount of gas (n) is constant.

Therefore nR = .

Assume V_i = 1 L and solve for V_f.

There is no change in volume.

9.25 If the sample remains a gas at 150 K, then drawing (c) represents the gas at this temperature. The gas molecules still fill the container.

9.26 The two gases should mix randomly and homogeneously and this is best represented by drawing (c).

9.27 The two gases will be equally distributed among the three flasks.

9.28 The gas pressure in the bulb in mm Hg is equal to the difference in the height of the Hg in the two arms of the manometer.

9.29 A

When stopcock A is opened, the pressure in the flask will equal the external pressure, and the level of mercury will be the same in both arms of the manometer.

9.30 (a) Because there are more yellow gas molecules than there are blue, the yellow gas molecules have the higher average speed.
(b) Each rate is proportional to the number of effused gas molecules of each type.
M_yellow = 25 amu

; ;

M_blue = = 36 amu

9.31 (a) The temperature has increased by about 10% (from 300 K to 325 K) while the amount and the pressure are unchanged. Thus, the volume should increase by about 10%.

(b) The temperature has increased by a factor of 1.5 (from 300 K to 450 K) and the pressure has increased by a factor of 3 (from 0.9 atm to 2.7 atm) while the amount is unchanged. Thus, the volume should decrease by half (1.5/3 = 0.5).

(c) Both the amount and the pressure have increased by a factor of 3 (from 0.075 mol to 0.22 mol and from 0.9 atm to 2.7 atm) while the temperature is unchanged. Thus, the volume is unchanged.

Additional Problems

Gases and Gas Pressure

9.32 Temperature is a measure of the average kinetic energy of gas particles.

9.33 Gases are much more compressible than solids or liquids because there is a large amount of empty space between individual gas molecules.

9.34 P = 480 mm Hg x = 0.632 atm

P = 480 mm Hg x = 6.40 x 10⁴ Pa

9.35 P = 352 torr x = 46.9 kPa

P = 0.255 atm x = 194 mm Hg

P = 0.0382 mm Hg x = 5.09 Pa

9.36 P_flask > 754.3 mm Hg; P_flask = 754.3 mm Hg + 176 mm Hg = 930 mm Hg

9.37 P_flask < 1.021 atm (see Figure 9.4)

P_difference = 28.3 cm Hg x = 0.372 atm

P_flask = 1.021 atm - P_difference = 1.021 atm - 0.372 atm = 0.649 atm

9.38 P_flask > 752.3 mm Hg (see Figure 9.4)

If the pressure in the flask can support a column of ethyl alcohol (d = 0.7893 g/mL) 55.1 cm high, then it can only support a column of Hg that is much shorter because of the higher density of Hg.

55.1 cm x = 3.21 cm Hg = 32.1 mm Hg

P_flask = 752.3 mm Hg + 32.1 mm Hg = 784.4 mm Hg

P_flask = 784.4 mm Hg x = 1.046 x 10⁵ Pa

9.39 Compute the height of a column of CHCl₃ that 1.00 atm can support.
760 mm Hg x = 6941 mm CHCl₃; therefore 1.00 atm = 6941 mm CHCl₃
The pressure in the flask is less than atmospheric pressure.
P_atm - P_flask = 0.849 atm - 0.788 atm = 0.061 atm
0.061 atm x = 423 mm CHCl₃
The chloroform will be 423 mm higher in the manometer arm connected to the flask.

9.40

The % volume for a particular gas is proportional to the number of molecules of that gas in a mixture of gases.
Average molecular mass of air
= (0.7808)(mol. mass N₂) + (0.2095)(mol. mass O₂)
+ (0.0093)(at. mass Ar) + (0.000 37)(mol. mass CO₂)
= (0.7808)(28.01 amu) + (0.2095)(32.00 amu)
+ (0.0093)(39.95 amu) + (0.000 37)(44.01 amu) = 28.96 amu

9.41 The % volume for a particular gas is proportional to the number of molecules of that gas in a mixture of gases.
Average molecular mass of a diving-gas = (0.020)(mol. mass O₂) + (0.980)(at. mass He)
= (0.020)(32.00 amu) + (0.980)(4.00 amu) = 4.56 amu

The Gas Laws

9.42 (a) ; = P_f

Let P_i = 1 atm, T_i = 100 K, T_f = 300 K

P_f = = 3 atm
The pressure would triple.
(b) ; = P_f
Let P_i = 1 atm, n_i = 3 mol, n_f = 1 mol
P_f = = atm
The pressure would be 1/3  the initial pressure.

(c) nRT = P_iV_i = P_fV_f; = P_f
Let P_i = 1 atm, V_i = 1 L, V_f = 1 - 0.45 L = 0.55 L
P_f = = 1.8 atm
The pressure would increase by 1.8 times.

(d) nR = ; = P_f
Let P_i = 1 atm, V_i = 1 L, T_i = 200 K, V_f = 3 L, T_i = 100 K
P_f = = 0.17 atm
The pressure would be 0.17 times the initial pressure.

9.43 (a) ;
Let V_i = 1 L, T_i = 400 K, T_f = 200 K
= 0.5 L
The volume would be halved.

(b) ;
Let V_i = 1 L, n_i = 4 mol, n_f = 5 mol
= 1.25 L
The volume would increase by 1/4.

(c) nRT = P_i V_i = P_f V_f;
Let V_i = 1 L, P_i = 4 atm, P_f = 1 atm
= 4 L
The volume would increase by a factor of 4.

(d) ;
Let V_i = 1 L, T_i = 200 K, T_f = 400 K, P_i = 1 atm, P_f = 2atm
= 1 L
There is no volume change.

9.44 They all contain the same number of gas molecules.

9.45 For air, T = 50^oC = 323 K.
n = = 0.0931 mol air
For CO₂, T = -10^oC = 263 K
n = = 0.101 mol CO₂
Because the number of moles of CO₂ is larger than the number of moles of air, the CO₂ sample contains more molecules.

9.46 n and T are constant; therefore nRT =
V_f = = 7210 L
n and P are constant; therefore
V_f = = 51.5 L

9.47 T_i = 20^oC = 293 K; nR =
V_f = = 1.0 x 10³ L

9.48 15.0 g CO₂ x = 0.341 mol CO₂
P = = 27.98 atm
27.98 atm x = 2.1 x 10⁴ mm Hg

9.49 20.0 g N₂ x = 0.714 mol N₂
T = = 41 K


9.50 = 1.7 x 10^-21 mol H/L
P = = 1.4 x 10^-20 atm
P = 1.4 x 10^-20 atm x = 1 x 10^-17 mm Hg

9.51 CH₄, 16.04 amu; 5.54 kg = 5.54 x 10³ g; T = 20^oC = 293 K
P = = 189.6 atm
P = 189.6 atm x = 1.92 x 10⁴ kPa

9.52 n = = 308.9 mol
mass Ar = 308.9 mol x = 12340 g = 1.23 x 10⁴ g

9.53 P = 13,800 kPa x = 136.2 atm
n and T are constant; therefore nRT = P_iV_i = P_fV_f
V_f = = 250.6 L
250.6 L x = 167 balloons

Gas Stoichiometry

9.54 For steam, T = 123.0^oC = 396 K
n = = 0.43 mol steam
For ice, H₂O, 18.02 amu; n = 10.5 g x = 0.583 mol ice
Because the number of moles of ice is larger than the number of moles of steam, the ice contains more H₂O molecules.

9.55 T = 85.0^oC = 358 K
n_Ar = = 0.156 mol Ar
Cl₂, 70.91 amu; = 0.156 mol Cl₂
There are equal numbers of Ar atoms and Cl₂ molecules in their respective samples.

9.56 The containers are identical. Both containers contain the same number of gas molecules. Weigh the containers. Because the molecular mass for O₂ is greater than the molecular mass for H₂, the heavier container contains O₂.

9.57 Assuming that you can see through the flask, Cl₂ gas is greenish and He is colorless.

9.58 room volume = 4.0 m x 5.0 m x 2.5 m = 50 m³
room volume = 50 m³ x = 5.0 x 10⁴ L
n_total = = 2.23 x 10³ mol
= (0.2095)n_total = (0.2095)(2.23 x 10³ mol) = 467 mol O₂
mass O₂ = 467 mol x = 1.5 x 10⁴ g O₂

9.59 0.25 g O₂ x = 7.8 x 10^-3 mol O₂
V = = 0.198 L = 0.200 L = 200 mL O₂

9.60 (a) CH₄, 16.04 amu; d = = 0.716 g/L
(b) CO₂, 44.01 amu; d = = 1.96 g/L
(c) O₂, 32.00 amu; d = = 1.43 g/L
(d) UF₆, 352.0 amu; d = = 15.7 g/L

9.61 Average molar mass = (0.270)(molar mass F₂) + (0.730)(molar mass He)
= (0.270)(38.00 g/mol) + (0.730)(4.003 g/mol) = 13.18 g/mol
Assume 1.00 mole of the gas mixture. T = 27.5^oC = 300.6 K
V = = 26.3 L
d = = 0.501 g/L

9.62 n = = 0.0290 mol
molar mass = = 34.0 g/mol; molecular mass = 34.0 amu

9.63 (a) Assume 1.000 L gas sample
n = = 0.0446 mol
molar mass = = 30.1 g/mol; molecular mass = 30.1 amu

(b) Assume 1.000 L gas sample
n = = 0.0405 mol
molar mass = = 26.0 g/mol; molecular mass = 26.0 amu

9.64 2 HgO(s) ® 2 Hg(l) + O₂(g); HgO, 216.59 amu
10.57 g HgO x = 0.024 40 mol O₂
V = = 0.5469 L

9.65 2 HgO(s) ®  2 Hg(l) + O₂(g); HgO, 216.59 amu
mass HgO = 0.0155 mol O₂ x = 6.71 g HgO

9.66 Zn(s) + 2 HCl(aq) ® ZnCl₂(aq) + H₂(g)
(a) 25.5 g Zn x = 0.390 mol H₂
V = = 9.44 L

(b) n = = 0.092 56 mol H₂
0.092 56 mol H₂ x = 6.05 g Zn

9.67 2 NH₄NO₃(s) ®  2 N₂(g) + 4 H₂O(g) + O₂(g); NH₄NO₃, 80.04 amu
Total moles of gas = 450 g NH₄NO₃ x = 19.68 mol
T = 450^oC = 723 K
V = = 1.17 x 10³ L

9.68 (a) V_24h = (4.50 L/min)(60 min/h)(24 h/day) = 6480 L
= (0.034)V_24h = (0.034)(6480 L) = 220 L
n = = 8.70 mol CO₂
8.70 mol CO₂ x = 383 g = 380 g CO₂

(b) 2 Na₂O₂(s) + 2 CO₂(g) ® 2 Na₂CO₃(s) + O₂(g); Na₂O₂, 77.98 amu
3.65 kg = 3650 g
3650 g Na₂O₂ x = 5.4 days

9.69 2 TiCl₄(g) + H₂(g) ®  2 TiCl₃(s) + 2 HCl(g); TiCl, 189.69 amu
(a) T = 435^oC = 708 K
= 2.79 mol H₂
2.79 mol H₂ x = 1058 g = 1060 g TiCl₄

(b) n_HCl = 2.79 mol H₂ x = 5.58 mol HCl
V = = 125 L HCl

Dalton's Law and Mole Fraction

9.70 Because of Avogadro's Law (V µ n), the % volumes are also % moles.

In decimal form, % mole = mole fraction.
= (0.7808)(1.000 atm) = 0.7808 atm
= (0.2095)(1.000 atm) = 0.2095 atm
= (0.0093)(1.000 atm) = 0.0093 atm
= (0.000 37)(1.000 atm) = 0.000 37 atm
Pressures of the rest are negligible.

9.71

= (0.94)(1.48 atm) = 1.4 atm
= (0.040)(1.48 atm) = 0.059 atm
= (0.015)(1.48 atm) = 0.022 atm
= (0.0050)(1.48 atm) = 0.0074 atm

9.72 Assume a 100.0 g sample. g CO₂ = 1.00 g and g O₂ = 99.0 g
mol CO₂ = 1.00 g CO₂ x = 0.0227 mol CO₂
mol O₂ = 99.0 g O₂ x = 3.094 mol O₂
n_total = 3.094 mol + 0.0227 mol = 3.117 mol
= 0.993 = 0.007 28
= (0.993)(0.977 atm) = 0.970 atm
= (0.007 28)(0.977 atm) = 0.007 11 atm

9.73 From Problem 9.74: X_HCI = 0.026,
P_HCI = X_HCl P_total = (0.026)(13,800 kPa) = 3.6 x 10² kPa
= (0.094)(13,800 kPa) = 1.3 x 10³ kPa
P_Ne = X_Ne P_total = (0.88)(13,800) kPa) = 1.2 x 10⁴ kPa

9.74 Assume a 100.0 g sample.
g HCl = (0.0500)(100.0 g) = 5.00 g; 5.00 g HCl x = 0.137 mol HCl
g H₂ = (0.0100)(100.0 g) = 1.00 g; 1.00 g H₂ x = 0.496 mol H₂
g Ne = (0.94)(100.0 g) = 94 g; 94 g Ne x = 4.66 mol Ne
n_total = 0.137 + 0.496 + 4.66 = 5.3 mol
= 0.026 = 0.094 = 0.88

9.75 Assume a 1.000 L gas sample.
n = = 0.044 61 mol
average molar mass = = 31.67 g/mol
31.67 =
Solve for x: x = 0.3064, 1 - x = 0.6936
The mixture contains 30.64% Ar and 69.36% N₂.
Assume 100 moles of gas.
= 0.3064 = 0.6936

9.76 P_total = ; = 747 mm Hg - 23.8 mm Hg = 723 mm Hg

n = = 0.1384 mol H₂
0.1384 mol H₂ x = 3.36 g Mg

9.77 P_total = = 755 mm Hg
= 755 mm Hg - 28.7 mm Hg = 726.3 mm Hg
(a) = 0.962

(b) NaCl, 58.44 amu
0.0232 mol Cl₂ x = 2.71 g NaCl

Kinetic-Molecular Theory and Graham's Law

9.78 The kinetic-molecular theory is based on the following assumptions:
1. A gas consists of tiny particles, either atoms or molecules, moving about at random.
2. The volume of the particles themselves is negligible compared with the total volume of the gas; most of the volume of a gas is empty space.
3. The gas particles act independently; there are no attractive or repulsive forces between particles.
4. Collisions of the gas particles, either with other particles or with the walls of the container, are elastic; that is, the total kinetic energy of the gas particles is constant at constant T.
5. The average kinetic energy of the gas particles is proportional to the Kelvin temperature of the sample.

9.79 Diffusion -- The mixing of different gases by random molecular motion and with frequent collisions.
Effusion -- The process in which gas molecules escape through a tiny hole in a membrane without collisions.

9.80 Heat is the energy transferred from one object to another as the result of a temperature difference between them.
Temperature is a measure of the kinetic energy of molecular motion.

9.81 The atomic mass of He is much less than the molecular mass of the major components of air (N₂ and O₂). The rate of effusion of He through the balloon skin is much faster.

9.82 = 443 m/s

9.83 For Br₂: = 214 m/s
For Xe: u =
Square both sides of the equation and solve for T.
45796 m²/s² =
T = 241 K = -32^oC

9.84 For H₂, = 1360 m/s
For He, = 2010 m/s
He at 375^oC has the higher average speed.

9.85 UF₆, 352.02 amu; T = 25^oC = 298 K
= 145 m/s
Ferrari:
145 = 64.8 m/s
The UF₆ molecule has the higher average speed.

9.86 ; ;
M_X = = 17.2 g/mol; molecular mass = 17.2 amu

9.87 ; ; ;
M_Z = 37.9 g/mol; molecular mass = 37.9 amu; The gas could be F₂.

9.88 HCl, 36.5 amu; F₂, 38.0 amu; Ar, 39.9 amu
= 1.05 = 1.02
The relative rates of diffusion are HCl(1.05) > F₂(1.02) > Ar(1.00).

9.89 Because CO and N₂ have the same mass, they will have the same diffusion rates.

9.90 u =
Square both sides of the equation and solve for T.

T = 0.325 K = -272.83^oC (near absolute zero)

9.91 230 km/h x = 63.9 m/s
u = 63.9 m/s =
Square both sides of the equation and solve for T.
4083 m²/s² =
T = 5.24 K = -268^oC

General Problems

9.92 = 1.03
= 1.01
The relative rates of diffusion are .

9.93 Average molecular mass of air = 28.96 amu; CO₂, 44.01 amu
P = 760 mm Hg x = 1155 mm Hg

9.94 = 1.1 L

9.95 1 atm = 1033.228 g/cm²
column height = (1033.228 g/cm²)(1 cm³/0.89g) = 1200 cm = 12 m

9.96 n = = 0.657 mol Ar
0.657 mol Ar x = 26.2 g Ar
m_total = 478.1 g + 26.2 g = 504.3 g

9.97 This is initially a Boyle's Law problem, because only P and V are changing while n and T remain fixed. The initial volume for each gas is the volume of their individual bulbs. The final volume for each gas is the total volume of the three bulbs.

nRT = P_iV_i = P_fV_f; V_f = 1.50 + 1.00 + 2.00 = 4.50 L
For CO₂: = 0.710 atm
For H₂: = 0.191 atm
For Ar: = 0.511 atm
From Dalton's Law, P_total =
P_total = 0.710 atm + 0.191 atm + 0.511 atm = 1.412 atm

9.98 (a) Bulb A contains CO₂(g) and N₂(g); Bulb B contains CO₂(g), N₂(g), and H₂O(s).
(b) Initial moles of gas = n =
Initial moles of gas = 0.030 35 mol
mol gas in Bulb A = n = = 0.011 78 mol
mol gas in Bulb B = n = = 0.017 29 mol
= n_initial - n_A - n_B = 0.030 35 - 0.011 78 - 0.017 29 = 0.001 28 mol = 0.0013 mol H₂O

(c) Bulb A contains N₂(g).
Bulb B contains N₂(g) and H₂O(s).
Bulb C contains N₂(g) and CO₂(s).

(d) n_A = = 0.001 803 mol
n_B = = 0.002 646 mol
n_C = = 0.006 472 mol
= n_A + n_B + n_C = 0.001 803 + 0.002 646 + 0.006 472 = 0.010 92 mol N₂

(e) = n_initial - - = 0.030 35 - 0.0013 - 0.010 92 = 0.0181 mol CO₂

9.99 C₃H₅N₃O₉, 227.1 amu
(a) moles C₃H₅N₃O₉ = 1.00 g x = 0.004 40 mol
n_air = = 0.0208 mol air

(b) moles gas from C₃H₅N₃O₉ = 0.004 40 mol x
moles gas from C₃H₅N₃O₉ = 0.0319 mol gas from C₃H₅N₃O₉
n_total = 0.0319 mol + 0.0208 mol = 0.0527 mol

(c) P = = 6.04 atm

9.100 NH₃, 17.03 amu; mol NH₃ = 45.0 g x = 2.64 mol
or      

(a) At T = 0^oC = 273 K
P = = 59.1 atm
P =
P = 65.6 atm - 29.1 atm = 36.5 atm

(b) At T = 50^oC = 323 K
P = = 70.0 atm
P =
P = 77.6 atm - 29.1 atm = 48.5 atm

(c) At T = 100^oC = 373 K
P = = 80.8 atm
P =
P = 89.6 atm - 29.1 atm = 60.5 atm
At the three temperatures, the van der Waals equation predicts a much lower pressure than does the ideal gas law. This is likely due to the fact that NH₃ can hydrogen bond leading to strong intermolecular forces.

9.101 (a) n_total = = 0.007 06 mol
(b) n_B = = 0.004 71 moles
(c) d = = 0.872 g/L
(d) molar mass = = 46.3 g/mol, NO₂; mol. mass = 46.3 amu
(e) Hg₂CO₃(s) + 6 HNO₃(aq) ®  2 Hg(NO₃)₂(aq) + 3 H₂O(l) + CO₂(g) + 2 NO₂(g)

9.102 CO₂, 44.01 amu
mol CO₂ = 500.0 g CO₂ x = 11.36 mol CO₂
PV = nRT
= = 816 atm

9.103 (a) Let x = mol C_nH_{2n + 2} in reaction mixture.
Combustion of C_nH_{2n + 2} nCO₂ + (n +1)H₂O needs mol O₂
Balanced equation is: C_nH_{2n + 2}(g) + O₂(g) ® nCO₂(g) + (n + 1)H₂O(g)
In going from reactants to products, the increase in the number of moles is
[n + (n + 1)] - = per mol of C_nH_{2n + 2} reacted.
Before reaction: total mol = = 0.032 70 mol
After reaction: total mol = = 0.036 52 mol
Difference = 0.032 70 mol - 0.036 52 mol = 0.003 82 mol
Increase in number of mol = x = 0.003 82 mol; x =
Also x =
So = ; 0.148 n - 0.148 = 0.107 n + 0.0154
0.041 n = 0.163; n = = 4.0
C_nH_{2n + 2} is C₄H₁₀ (butane); molar mass = (4)(12.01) + (10)(1.008) = 58.12 g/mol

(b) 0.148 g C₄H₁₀ x = 0.002 55 mol C₄H₁₀
mol O₂ initially = total mol - mol C₄H₁₀ = 0.032 70 mol - 0.002 55 mol = 0.030 15 mol O₂
= 0.156 atm
= 1.844 atm

(c) C₄H₁₀(g) +   O₂(g) ® 4 CO₂(g) + 5 H₂O(g)
0.002 55 mol C₄H₁₀ x = 0.0102 mol CO₂
0.002 55 mol C₄H₁₀ x = 0.012 75 mol H₂O
mol O₂ unreacted = total mol after reaction - mol CO₂ - mol H₂O
= 0.03652 mol - 0.0102 mol - 0.01275 = 0.01357 mol O₂
= 0.833 atm

= 1.041 atm

= 1.108 atm

9.104 (a) average molecular mass for natural gas
= (0.915)(16.04 amu) + (0.085)(30.07 amu) = 17.2 amu
total moles of gas = 15.50 g x = 0.901 mol gas

(b) P = = 1.44 atm
(c) = (1.44 atm)(0.915) = 1.32 atm
= (1.44 atm)(0.085) = 0.12 atm

(d) H_combustion(CH₄) = -802.3 kJ/mol and H_combustion(C₂H₆) = -1427.7 kJ/mol
Heat liberated = (0.915)(0.901 mol)(-802.3 kJ/mol)
+ (0.085)(0.901)(-1427.7 kJ/mol) = -771 kJ

Multi-Concept Problems

9.105 X + 3 O₂ 2 CO₂ + 3 H₂O

(a) X = C₂H₆O

C₂H₆O + 3 O₂ 2 CO₂ + 3 H₂O
(b) It is an empirical formula because it is the smallest whole number ratio of atoms. It is also a molecular formula because any higher multiple such as C₄H₁₂O₂ does not correspond to a stable electron-dot structure.
(c)
(d) C₂H₆O, 46.07 amu
mol C₂H₆O = 5.000 g C₂H₆O x = 0.1085 mol C₂H₆O
H_combustion = = -1328.6 kJ/mol
H_combustion = [2 H^o_f(CO₂) + 3 H^o_f(H₂O)] - H^o_f(C₂H₆O)
H^o_f(C₂H₆O) = [2 H^o_f(CO₂) + 3 H^o_f(H₂O)] - H_combustion
= [(2mol)(-393.5 kJ/mol) + (3 mol)(-241.8 kJ/mol)] - (-1328.6 kJ)
= -183.8 kJ/mol

9.106 (a) 2 C₈H₁₈(l) + 25 O₂(g) ®  16 CO₂(g) + 18 H₂O(g)
(b) 4.6 x 10¹⁰ L C₈H₁₈ x = 3.64 x 10¹³ g C₈H₁₈
3.64 x 10¹³ g C₈H₁₈ x = 2.55 x 10¹² mol CO₂
2.55 x 10¹² mol CO₂ x = 1.1 x 10¹¹ kg CO₂
(c) = 5.7 x 10¹³ L of CO₂

(d) 12.5 moles of O₂ are needed for each mole of isooctane (from part a).
12.5 mol O₂ = (0.210)(n_air); n_air = = 59.5 mol air
= 1.33 x 10³ L

9.107 (a) Freezing point of H₂O on the Rankine scale is (9/5)(273.15) = 492^oR.
(b)
(c) P =
P = 204.2 atm - 88.0 atm = 116 atm

9.108 n = = 7.25 mol of all gases
(a) 0.004 00 mol "nitro" x = 0.0290 mol hot gases

(b) n = = 0.0190 mol B + C + D
n_A = n_total - n_(B+C+D) = 0.0290 - 0.0190 = 0.0100 mol A; A = H₂O

(c) n = = 0.007 00 mol C + D
n_B = n_(B+C+D) - n_(C+D) = 0.0190 - 0.007 00 = 0.0120 mol B; B = CO₂

(d) n = = 0.006 00 mol D

n_C = n_(C+D) - n_D = 0.007 00 - 0.006 00 = 0.001 00 mol C; C = O₂

molar mass D = = 28.0 g/mol; D = N₂

(e) 0.004 C₃H₅N₃O₉(l) ®  0.0100 H₂O(g) + 0.012 CO₂(g) + 0.001 O₂(g) + 0.006 N₂(g)
Multiply each coefficient by 1000 to obtain integers.
4 C₃H₅N₃O₉(l)  ® 10 H₂O(g) + 12 CO₂(g) + O₂(g) + 6 N₂(g