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Chapter 11
Gases
Gases
11
11.1 Properties of Gases
11.2 The Kinetic Molecular Theory of Gases
Molecular Speed
Diffusion and Effusion
11.3 Gas Pressure
Definition and Units of Pressure
Calculation of Pressure
Measurement of Pressure
11.4 The Gas Laws
Boyle’s Law: The Pressure-Volume Relationship
Charles’s and Gay-Lussac’s Law: The Temperature-Volume Relationship
Avogadro’s Law: The Amount-Volume Relationship
The Gas Laws and Kinetic Molecular Theory
The Combined Gas Law: The Pressure-Temperature-Amount-Volume Relationship
11.5 The Ideal Gas Equation
Applications of the Ideal Gas Equation
Gases
11
11.6 Real Gases
Factors That Cause Deviation from Ideal Behavior
The van der Waals Equation
van der Waals Constants
11.7 Gas Mixtures
Dalton’s Law of Partial Pressures
Mole Fractions
11.6 Reactions with Gaseous Reactants and Products
Calculating the Required Volume of a Gaseous Reactant
Determining the Amount of Reactant Consumed Using Change in Pressure
Using Partial Pressures to Solve Problems
Properties of Gases
Gases differ from solids and liquids in the following ways:
A sample of gas assumes both the shape and volume of the container.
Gases are compressible.
The densities of gases are much smaller than those of liquids and solids and are highly variable depending on temperature and pressure.
Gases form homogeneous mixtures (solutions) with one another in any proportion.
11.1
The Kinetic Molecular Theory
The kinetic molecular theory explains how the molecular nature of gases gives rise to their macroscopic properties.
The basic assumptions of the kinetic molecular theory are as follows:
A gas is composed of particles that are separated by large distances. The volume occupied by individual molecules is negligible.
Gas molecules are constantly in random motion, moving in straight paths, colliding with perfectly elastic collisions.
Gas molecules do not exert attractive or repulsive forces on one another.
The average kinetic energy of a gas molecules in a sample is proportional to the absolute temperature:
11.2
The Kinetic Molecular Theory
Gases are compressible because molecules in the gas phase are separated by large distances (assumption 1).
Pressure is the result of the collisions of gas molecules with the walls of their container (assumption 2).
Decreasing volume increases the frequency of collisions.
Pressure increases as collision frequency increases.
The Kinetic Molecular Theory
Heating a sample of gas increases its average kinetic energy (assumption 4).
Gas molecules must move faster.
Faster molecules collide more frequently and at a greater speed.
Pressure increases as collision frequency increases.
The Kinetic Molecular Theory
The total kinetic energy of a mole of gas is equal to:
The average kinetic energy of one molecule is:
For one mole of gas:
m is the mass
is the mean square speed
rearrange and
Take the square root (m x NA = M)
The Kinetic Molecular Theory
The root-mean-square (rms) speed (urms) is the speed of a molecule with the average kinetic energy in a gas sample.
urms is directly proportional to temperature
The Kinetic Molecular Theory
The root-mean-square (rms) speed (urms) is the speed of a molecule with the average kinetic energy in a gas sample.
urms is inversely proportional to the square root of M.
The Kinetic Molecular Theory
When two gases are at the same temperature, it is possible to compare the the urms values of the different gases.
Worked Example 11.1
Strategy Use equation and the molar masses of He and CO2 to determine the ratio of their root-mean-square speeds. When solving a problem such as this, it is generally best to label the lighter of the two molecules as molecule 1 and the heavier as molecule 2. This ensures that the result will be greater than 1, which is relatively easy to interpret.
Determine how much faster a helium atom moves, on average, than a carbon dioxide molecule at the same temperature.
Solution The molar masses of He and CO2 are 4.003 and 44.02 g/mol, respectively.
On average, He atoms move 3.316 times as fast as CO2 molecules at the same temperature.
44.02 g
1 mol
= 3.316
4.003 g
1 mol
urms(He)
urms(CO2)
=
Think About It Remember that the relationship between molar mass and molecular speed is reciprocal. A CO2 molecule has approximately 10 times the mass of an He atom. Therefore, we should expect an He atom, on average, to be moving approximately √10 times (~3.2 times) as fast as a CO2 molecule.
The Kinetic Molecular Theory
Diffusion is the mixing of gases as the result of random motion and frequent collisions.扩散
Effusion is the escape of gas molecules from a container to a region of vacuum.溢散
The Kinetic Molecular Theory
Graham’s law states that the rate of diffusion or effusion of a gas is inversely proportional to the square root of its molar mass.
The Kinetic Molecular Theory
Determine the molar mass and identity of a diatomic gas that moves 4.67 times as fast as CO2.
Solution:
Step 1: Use the equation below to determine the molar mass of the unknown gas:
urms(unknown gas) = 4.67 x urms(CO2)
The gas must be H2.
Gas Pressure
Pressure is defined as the force applied per unit area:
The SI unit of force is the newton (N), where
The SI unit of pressure is the pascal (Pa), defined as 1 newton per square meter.
1 N = 1 kg m/s2
1 Pa = 1 N/m2
11.3
Gas Pressure
Gas Pressure
A barometer is an instrument that is used to measure atmospheric pressure.
Standard atmospheric pressure (1 atm) was originally defined as the pressure that would support a column of mercury exactly 760 mm high.
h = height in m
d = is the density in kg/m3
g = is the gravitational constant (9.80665 m/s2)
Gas Pressure
A barometer is an instrument that is used to measure atmospheric pressure.
1 atm*= 101,325 Pa
= 760 mmHg*
= 760 torr*
= 1.01325 bar
= 14.7 psi
* Represents an exact number
Gas Pressure
A manometer is a device used to measure pressures other than atmospheric pressure.
Worked Example 11.2
Strategy Use P = hdg to calculate pressure. Remember that the height must be expressed in meters and density must be expressed in kg/m3.
Calculate the pressure exerted by a column of mercury 70.0 cm high. Express the pressure in pascals, in atmospheres, and in bars. The density of mercury is 13.5951 g/cm3.
Solution
h = 70.0 cm × = 0.700 m
d =
g = 9.80665 m/s2
1 m
100 cm
13.5951 g
cm3
×
1 kg
1000 g
×
100 cm
1 m
3
= 1.35951×104 kg/m3
Worked Example 11.2 (cont.)
Solution
pressure = 0.700 m × × = 9.33×104 kg/m s2
9.33×104 Pa
9.33×104 Pa × = 0.921 atm
0.921 atm × = 0.933 bar
1.35951×104 kg
m3
9.80665 m
s2
1 atm
101,325 Pa
1.01325 bar
1 atm
Think About It Make sure your units cancel properly in this type of mon errors include forgetting to express height in meters and density in kg/m3. You can avoid these errors by becoming familiar with the value of atmospheric pressure in the various units. A column of mercury slightly less than 760 mm is equivalent to slightly less than 101,325 Pa, slightly less than 1 atm, and slightly less than 1 bar.
The Gas Laws
Boyle’s law states that the pressure of a fixed amount of gas at a constant temperature is inversely proportional to the volume of the gas.
11.4
(a) (b) (c)
P (mmHg) 760 1520 2280
V (mL) 100 50 33
P1V1 = P2V2
at constant temperature
The Gas Laws
Calculate the volume of a sample of gas at 5.75 atm if it occupies 5.14 L at 2.49 atm. (Assume constant temperature.)
Solution:
Step 1: Use the relationship below to solve for V2:
P1V1 = P2V2
Worked Example 11.3
Strategy Use P1V1 = P2V2 to solve for V2.
If a skin diver takes a breath at the surface, filling his lungs with 5.82 L of air, what volume will the air in his lungs occupy when he dives to a depth where the pressure of 1.92 atm (Assume constant temperature and that the pressure at the surface is exactly 1 atm.)
Solution P1 = 1.00 atm, V1 = 5.82 L, and P2 = 1.92 atm.
V2 = = = 3.03 L
P1 × V1
P2
1.00 atm × 5.82 L
1.92 atm
Think About It At higher pressure, the volume should be smaller. Therefore, the answer makes sense.
The Gas Laws
Charles’s and Gay-Lussac’s law, (or simply Charles’s Law) states that the volume of a gas maintained at constant pressure is directly proportional to the absolute temperature of the gas.
Higher temperature
Lower temperature
The Gas Laws
Charles’s and Gay-Lussac’s law, (or simply Charles’s Law) states that the volume of a gas maintained at constant pressure is directly proportional to the absolute temperature of the gas.
at constant pressure
Worked Example 11.4
Strategy Use V1/T1 = V2/T2 to solve for V2. Remember that temperatures must be expressed in kelvin.
A sample of argon gas that originally occupied 14.6 L at 25°C was heated to 50.0°C at constant pressure. What is its new volume
Solution T1 = 298.15 K, V1 = 14.6 L, and T2 = 323.15 K.
V2 = = = 15.8 L
V1 × T2
T1
14.6 L × 323.15 K
298.15 K
Think About It When temperature increases at constant pressure, the volume of a gas sample increases.
The Gas Laws
Avogadro’s law states that the volume of a sample of gas is directly proportional to the number of moles in the sample at constant temperature and pressure.
at constant temperature and pressure
Worked Example 11.5
Strategy Apply Avogadro’s law to determine the volume of a gaseous product.
If we combine 3.0 L of NO and 1.5 L of O2, and they react according to the balanced equation 2NO(g) + O2(g) → 2NO2(g), what volume of NO2 will be produced (Assume that the reactants and products are all at the same temperature and pressure.)
Solution Because volume is proportional to the number of moles, the balanced equation determines in what volume ratio the reactants combine and the ratio of product volume to reactant volume. The amounts of reactants given are stoichiometric amounts.
According to the balanced equation, the volume of NO2 formed will be equal to the volume of NO that reacts. Therefore, 3.0 L of NO2 will form.
Think About It When temperature increases at constant pressure, the volume of a gas sample increases.
The Gas Laws
A sample of gas originally occupies 29.1 L at 0.0°C. What is its new volume when it is heated to 15.0°C (Assume constant pressure.)
Solution:
Step 1: Use the relationship below to solve for V2: (Remember that temperatures must be expressed in kelvin.
The Gas Laws
What volume in liters of water vapor will be produced when 34 L of H2 and 17 L of O2 react according to the equation below:
2H2(g) + O2(g) → 2H2O(g)
Assume constant pressure and temperature.
Solution:
Step 1: Because volume is proportional to the number of moles, the balanced equation determines in what volume ratio the reactants combine and the ratio of product volume to reactant volume. The amounts of reactants given are stoichiometeric amounts.
34 L of H2O will form.
The Gas Laws
Heating at constant volume: pressure increases
Cooling at constant volume:
pressure decreases
The Gas Laws
Heating at constant pressure: volume increases
Cooling at constant pressure:
volume decreases
The Gas Laws
The presence of additional molecules causes an increase in pressure.
The Gas Laws
The combined gas law can be used to solve problems where any or all of the variables changes.
The Gas Laws
The volume of a bubble starting at the bottom of a lake at 4.55°C increases by a factor of 10 as it rises to the surface where the temperature is 18.45°C and the air pressure is 0.965 atm. Assume the density of the lake water is 1.00 g/mL. Determine the depth of the lake.
Solution:
Step 1: Use the combined gas law to find the pressure at the bottom of the lake; assume constant moles of gas.
The Gas Laws
The volume of a bubble starting at the bottom of a lake at 4.55°C increases by a factor of 10 as it rises to the surface where the temperature is 18.45°C and the air pressure is 0.965 atm. Assume the density of the lake water is 1.00 g/mL. Determine the depth of the lake.
Solution:
Step 2: Use the equation P = hdg to determine the depth of the lake. The pressure represents the difference in pressure from the surface to the bottom of the lake.
Pressure exerted by the lake = Pbottom of lake – Pair
P = 9.19 atm – 0.965 atm = 8.225 atm
The Gas Laws
The volume of a bubble starting at the bottom of a lake at 4.55°C increases by a factor of 10 as it rises to the surface where the temperature is 18.45°C and the air pressure is 0.965 atm. Assume the density of the lake water is 1.00 g/mL. Determine the depth of the lake.
Solution:
Step 2 (Cont): Convert pressure to pascals and density to kg/m3.
P = hdg
833,398 Pa = h(1000 kg/m3)(9.81 m/s2)
h = 85.0 m
Worked Example 11.6
Strategy In this case, because there is a fixed amount of gas, we use P1V1/T1 = P2V2/T2. The only value we don’t know is V2. Temperatures must be expressed in kelvins. We can use any units of pressure, as long as we are consistent.
If a child releases a 6.25-L helium balloon in the parking lot of an amusement park where the temperature is 28.50°C and the air pressure is 757.2 mmHg, what will the volume of the balloon be when it has risen to an altitude where the temperature is -34.35°C and the air pressure is 366.4 mmHg
Think About It Note that the solution is essentially multiplying the original volume by the ratio of P1 and P2, and by the ratio of T2 to T1. The effect of decreasing external pressure is to increase the balloon volume. The effect of decreasing temperature is to decrease the volume. In this case, the effect of decreasing pressure predominates and the balloon volume increases significantly.
Solution T1 = 301.65 K, T2 = 238.80 K.
V2 = = = 10.2 L
P1T2V1
P2T1
757.2 mmHg × 238.80 K × 6.25 L
366.4 mmHg × 301.65 K
The Ideal Gas Equation
The gas laws can be combined into a general equation that describes the physical behavior of all gases.
11.5
Boyle’s law
Avogadro’s law
Charles’s law
PV = nRT
rearrangement
R is the proportionality constant, called the gas constant.
The Ideal Gas Equation
The ideal gas equation (below) describes the relationship among the four variables P, V, n, and T.
PV = nRT
An ideal gas is a hypothetical sample of gas whose pressure-volume-temperature behavior is predicted accurately by the ideal gas equation.
The Ideal Gas Equation
The gas constant (R) is the proportionality constant and its value and units depend on the units in which P and V are expressed.
PV = nRT
The Ideal Gas Equation
Standard Temperature and Pressure (STP) are a special set of conditions where:
Pressure is 1 atm
Temperature is 0°C (273.15 K)
The volume occupied by one mole of an ideal gas is then 22.41 L:
PV = nRT
Worked Example 11.7
Strategy Convert the temperature in °C to kelvins, and use the ideal gas equation to solve for the unknown volume.
Calculate the volume of a mole of ideal gas at room temperature (25°C) and 1 atm.
Think About It With the pressure held constant, we should expect the volume to increase with increased temperature. Room temperature is higher than the standard temperature for gases (0°C), so the molar volume at room temperature (25°C) should be higher than the molar volume at 0°C–and it is.
Solution The data given are n = 1 mol, T = 298.15 K, and P = 1.00 atm. Because the pressure is expressed in atmospheres, we use R = 0.08206 L atm/K mol in order to solve for volume in liters.
V = = = 24.5 L
nRT
P
(1 mol)(0.08206 L atm/K mol)(298.15 K)
1 atm
Molar Volume of Some Common Gases at STP
(00C and 1 atm)
Gas
Molar Volume
(L/mol)
Condensation Point
(0C)
He
H2
Ne
Ideal gas
Ar
N2
O2
CO
Cl2
NH3
22.435
22.432
22.422
22.414
22.397
22.396
22.390
22.388
22.184
22.079
-268.9
-252.8
-246.1
---
-185.9
-195.8
-183.0
-191.5
-34.0
-33.4
The behavior of several real gases with increasing external pressure.
The Ideal Gas Equation
Using algebraic manipulation, it is possible to solve for variables other than those that appear explicitly in the ideal gas equation.
PV = nRT
d is the density (in g/L)
M is the molar mass (in g/mol)
The Ideal Gas Equation
What pressure would be required for helium at 25°C to have the same density as carbon dioxide at 25°C and 1.00 atm
Solution:
Step 1: Use the equation below to calculate the density of CO2 at
25°C and 1 atm.
The Ideal Gas Equation
What pressure would be required for helium at 25°C to have the same density as carbon dioxide at 25°C and 1 atm
Solution:
Step 2: Use the density of CO2 found in step 1 to calculate the
pressure for He at 25°C.
P = 11.0 atm
Worked Example 11.8
Strategy Use d = PM/RT to solve for density. Because the pressure is expressed in atm, we should use R = 0.08206 L atm/K mol. Remember the express temperature in kelvins.
Carbon dioxide is effective in fire extinguishers partly because its density is greater than that of air, so CO2 can smother the flames by depriving them of oxygen. (Air has a density of approximately 1.2 g/L at room temperature and 1 atm.) Calculate the density of CO2 at room temperature (25°C) and 1.0 atm.
Think About It The calculated density of CO2 is greater than that of air under the same conditions (as expected). Although it may seem tedious, it is a good idea to write units for each and every entry in a problem such as this. Unit cancellation is very useful for detecting errors in your reasoning or your solution setup.
Solution The molar mass of CO2 is 44.01 g/mol.
d = =
PM
RT
(1 atm)(44.01 g/mol)
0.08206 L atm/K mol)(298.15 K)
= 24.5 L
Real Gases
The van der Waals equation is useful for gases that do not behave ideally.
11.6
Experimentally measured pressure
corrected
pressure term
corrected
volume term
Container volume
Real Gases
The van der Waals equation is useful for gases that do not behave ideally.
Real Gases
Calculate the pressure exerted by 0.35 mole of oxygen gas in a volume of 6.50 L at 32°C using (a) the ideal gas equation and(b) the van der Waals equation.
Solution:
Step 1: Use the ideal gas equation to calculate the pressure of O2.
PV = nRT
Real Gases
Calculate the pressure exerted by 0.35 mole of oxygen gas in a volume of 6.50 L at 32°C using (a) the ideal gas equation and(b) the van der Waals equation.
Solution:
Step 2: Use table 11.6 to find the values of a and b for O2.
Real Gases
Calculate the pressure exerted by 0.35 mole of oxygen gas in a volume of 6.50 L at 32°C using (a) the ideal gas equation and (b) the van der Waals equation.
Solution
Step 3: Use the van der Waals equation to calculate P.
P = 1.3 atm
Worked Example 11.10
Strategy (a) Use the ideal gas equation, PV = nRT. (b) Use (P + an2/V2)(V – nb) = nRT and a and b values for NH3 from Table 11.5
A sample of 3.50 moles of NH3 gas occupies 5.20 L at 47°C. Calculate the pressure of the gas (in atm) using (a) the ideal gas equation and (b) the van der Waals equation.
Solution T = 320.15 K, a = 4.17 atm L/mol2, and b = 0.0371 L/mol.
(a) P = =
(b) Evaluating the correction terms in the van der Waals equation, we get
=
(3.50 mol)(0.08206 L atm/K mol)(320.15 K)
5.20 L
nRT
V
= 17.7 atm
an2
V2
(4.17 atm L/mol2)(3.50 mol)2
(5.20 L)2
= 1.89 atm
0.0371 L
mol
= 0.130 L
nb = (3.50 mol)
Worked Example 11.10 (cont.)
Solution Finally, substituting these results into the van der Waals equation, we get
(P + 1.89 atm)(5.20 L – 0.130 L) = (3.50 mol)(0.08206 L atm/K mol)(320.15 K)
P = 16.2 atm
Think About It As is often the case, the pressure exerted by the real gas is lower than predicted by the ideal gas equation.
Worked Example 11.11
Strategy (a) Use the formula for volume of a sphere to calculate the volume of a sphere of radius 2r. This is excluded volume defined by two molecules. (b) Multiply the excluded volume per molecule by Avogardo’s number to determine excluded volume per mole.
Consider a gas sample consisting of molecules with radius r. (a) Determine the excluded volume defined by two molecules and (b) calculate the excluded volume per mole (b) for the pare the excluded volume per mole with the volume actually occupied by a mole of the molecules.
Solution The equation for volume of a sphere is πr3; NA = 6.022×1023.
(a) Excluded volume defined by two molecules = π(2r)3 = 8( πr3).
(b) Excluded volume per mole = × =
The volume actually occupied by a mole of molecules with a radius r is 4NA( πr3).
8( πr3)
2 molecules
4
3
4
3
4
3
4
3
NA molecules
1 mole
4NA( πr3)
1 mole
4
3
4
3
Think About It The excluded volume per mole is four times the volume actually occupied by a mole of molecules.
Gas Mixtures
When two or more gases are placed in a container, each gas behaves as though it occupies the container alone.
1.00 mole of N2 in a 5.00 L container at 0°C exerts a pressure of 4.48 atm.
Addition of 1.00 mole of O2 in the same container exerts an additional 4.48 atm of pressure.
The total pressure of the mixture is the sum of the partial pressures (Pi):
Ptotal = PN2 + PO2 = 4.48 atm + 4.48 atm = 8.96 atm
11.7
Gas Mixtures
Dalton’s law of partial pressure states that the total pressure exerted by a gas mixture is the sum of the partial pressures exerted by each component of the mixture:
Gas Mixtures
Determine the partial pressures and the total pressure in a 2.50-L vessel containing the following mixture of gases at 15.8°C: 0.0194 mol He, 0.0411 mol H2, and 0.169 mol Ne.
Solution:
Step 1: Since each gas behaves independently, calculate the partial pressure of each using the ideal gas equation:
Gas Mixtures
Determine the partial pressures and the total pressure in a 2.50-L vessel containing the following mixture of gases at 15.8°C: 0.0194 mol He, 0.0411 mol H2, and 0.169 mol Ne.
Solution:
Step 2: Use the equation below to calculate total pressure.
Ptotal = 0.184 atm + 0.390 atm + 1.60 atm = 2.17 atm
Gas Mixtures
Each component of a gas mixture exerts a pressure independent of the other components. The total pressure is the sum of the partial pressures.
Worked Example 11.12
Strategy Use the ideal gas equation to find the partial pressure of each component of the mixture, and sum the two partial pressures to find the total pressure.
A 1.00-L vessel contains 0.215 mole of N2 gas and 0.0118 mole of H2 gas at 25.5°C. Determine the partial pressure of each component and the total pressure in the vessel.
Solution T = 298.65 K
Ptotal = PN2 + PH2 = 5.27 atm + 0.289 atm = 5.56 atm
(0.215 mol)(0.08206 L atm/K mol)(298.65 K)
1.00 L
= 5.27 atm
PN2 =
(0.0118 mol)(0.08206 L atm/K mol)(298.65 K)
1.00 L
= 0.289 atm
PH2 =
Think About It The total pressure in the vessel can also be determined by summing the number of moles of mixture components (0.215 + 0.0118 = 0.227 mol) and solving the ideal gas equation for Ptotal:
(0.227 mol)(0.08206 L atm/K mol)(298.65 K)
1.00 L
= 5.56 atm
Ptotal =
Gas Mixtures
The relative amounts of the components in a gas mixture can be specified using mole fractions.
There are three things to remember about mole fractions:
The mole fraction of a mixture component is always less than 1.
The sum of mole fractions for all components of a mixture is always 1.
Mole fractions are dimensionless.
Χi is the mole fraction.
ni is the moles of a certain component
ntotal is the total number of moles.
Worked Example 11.13
Strategy Use the ideal gas equation to calculate the total number of moles in the cylinder. Subtract moles of N2 from the total to determine moles of NO. Divide moles NO by total moles to get mole fraction.
In 1999, the FDA approved the use of nitric oxide (NO) to treat and prevent lung disease, which occurs commonly in premature infants. The nitric oxide used in this therapy is supplied to hospitals in the form of a N2/NO mixture. Calculate the mole fraction of NO in a 10.00-L gas cylinder at room temperature (25°C) that contains 6.022 mol N2 and in which the total pressure is 14.75 atm.
Solution The temperature is 298.15 K.
total moles = =
mol NO = total moles – N2 = 6.029 – 6.022 = 0.007 mol NO
χNO = = = 0.001
(14.75 atm)(10.00 L)
(0.08206 L atm/K mol)(298.15 K)
= 6.029 mol
PV
RT
nNO
ntotal
0.007 mol NO
6.029 mol
Think About It To check your work, determine χN2 by subtracting χNO from 1. Using each mole fraction and the total pressure, calculate the partial pressure of each component using χi = Pi/Ptotal and verify that they sum to the total pressure.
Reactions with Gaseous Reactants and Products
What mass (in grams) of Na2O2 is necessary to consume 1.00 L of CO2 at STP
2Na2O2(s) + 2CO2(g) → 2Na2CO3(s) + O2(g)
Solution:
Step 1: Convert 1.00 L of CO2 at STP to moles using the ideal gas equation.
11.8
PV = nRT
(1 atm)(1.00 L) = n(0.08206 Latm/mol K)(273.15 K)
n = 0.04461 moles CO2
Reactions with Gaseous Reactants and Products
What mass (in grams) of Na2O2 is necessary to consume 1.00 L of CO2 at STP
2Na2O2(s) + 2CO2(g) → 2Na2CO3(s) + O2(g)
Solution:
Step 2: Determine the stoichiometric amount of Na2O2.
Worked Example 11.14
Strategy Convert the given mass of Na2O2 to moles, use the balanced equation to determine the stoichiometric amount of CO2, and then use the ideal gas equation to convert moles of CO2 to liters.
Sodium peroxide (Na2O2) is used to remove carbon dioxide from (and add oxygen to) the air supply in spacecrafts. It works by reacting with CO2 in the air to produce sodium carbonate (Na2CO3) and O2.
2Na2O2(s) + 2CO2(g) → 2Na2CO3(s) + O2(g)
What volume (in liters) of CO2 (at STP) will react with a kilogram of Na2O2
Worked Example 11.14 (cont.)
Solution The molar mass of Na2O2 is 77.98 g/mol (1 kg = 1000 g). (Treat the specified mass of NaO2 as an exact number.)
1000 g Na2O2 ×
12.82 mol Na2O2 ×
VCO2 =
= 12.82 mol Na2O2
1 mol Na2O2
77.98 g Na2O2
2 mol CO2
2 mol Na2O2
= 12.82 mol Na2O2
(12.82 mol CO2)(0.08206 L atm/K mol)(273.15 K)
1 atm
= 287.4 L CO2
Think About It The answer seems like an enormous volume of CO2. If you check the cancellation of units carefully in ideal gas equation problems, however, with practice you will develop a sense of whether such a calculated volume is reasonable.
Reactions with Gaseous Reactants and Products
Although there are no empirical gas laws that focus on the relationship between n and P, it is possible to rearrange the ideal gas equation to find the relationship.
PV = nRT
The change in pressure in a reaction vessel can be used to determine how many moles of gaseous reactant are consumed:
rearrangement
Worked Example 11.15
Strategy Use Δn = ΔP×(V/RT) to determine Δn, the number of moles CO2 consumed.
Another air-purification method for enclosed spaces involves the use of “scrubbers” containing aqueous lithium hydroxide, which react with carbon dioxide to produce lithium carbonate and water:
2LiOH(aq) + CO2(g) → Li2CO3(s) + H2O(l)
Consider the air supply in a submarine with a total volume of 2.5×105 L. The pressure of 0.9970 atm, and the temperature 25°C. If the pressure in the submarine drops to 0.9891 atm as the result of carbon dioxide being consumed by an aqueous lithium hydroxide scrubber, how many moles of CO2 are consumed.
Solution ΔP = 0.9970 atm – 0.9891 atm = 7.9×10-3 atm, V = 2.5×105 L, and T = 298.15 K.
ΔnCO2 = 7.9×10-3 atm ×
2.5×105 L
(0.08206 L atm/K mol) × (298.15 K)
= 81 moles CO2
consumed
Think About It Careful cancellation of units is essential. Note that this amount of CO2 corresponds to 162 moles or 3.9 kg of LiOH. (It’s a good idea to verify this yourself.
Reactions with Gaseous Reactants and Products
The volume of gas produced by a chemical reaction can be measured using an apparatus like the one shown below.
When gas is collected over water in this manner, the total pressure is the sum of two partial pressures:
Ptotal = Pcollected gas + PH2O
Reactions with Gaseous Reactants and Products
The vapor pressure of water is known at various temperatures.
Gas Mixtures
Calculate the mass of O2 produced by the decomposition of KClO3 when 821 mL of O2 is collected over water at 30.0°C and 1.015 atm.
Solution:
Step 1: Use Table 11.5 to determine the vapor pressure of water at 30.0°C.
Reactions with Gaseous Reactants and Products
Calculate the mass of O2 produced by the decomposition of KClO3 when 821 mL of O2 is collected over water at 30.0°C and 1.015 atm.
Solution:
Step 2: Convert the vapor pressure of water at 30.0°C to atm and then use Dalton’s law to calculate the partial pressure of O2.
Ptotal = PO2 + PH2O
PO2 = Ptotal – PH2O
PO2 = 1.015 atm – 0.041842 atm = 0.973158 atm
Reactions with Gaseous Reactants and Products
Calculate the mass of O2 produced by the decomposition of KClO3 when 821 mL of O2 is collected over water at 30.0°C and 1.015 atm.
Solution:
Step 3: Convert to moles of O2 using the ideal gas equation and then find mass.
PV = nRT
Worked Example 11.16
Strategy Use Dalton’s law of partial pressures to determine the partial pressure of H2, use the ideal gas equation to determine moles of H2, and then use the molar mass of H2 to convert to mass. (Pay careful attention to units. Atmospheric pressure is given in atmospheres, whereas the vapor pressure of water is tabulated in torr.)
Calcium metal reacts with water to produce hydrogen gas:
Ca(s) + H2O(l) → Ca(OH)2(aq) + H2(g)
Determine the mass of H2 produced at 25°Cand 0.967 atm when 525 mL of the gas is collected over water as shown in Figure 11.23.
Solution PH2 = Ptotal – PH2O = 0.967 atm – 0.0313 atm = 0.936 atm
moles of H2 =
moles of H2 = (2.008×10-2 mol)(2.016 g/mol) = 0.0405 g H2
(0.9357 atm)(0.525 L)
(0.08206 L atm/K mol)(298.15 K)
= 2.01×10-2 mol
Think About It Check unit cancellation carefully, and remember that the densities of gases are relatively low. The mass of approximately half a liter of hydrogen at or near room temperature and 1 atm should be a very small number.
Chapter Summary: Key Points
11
Characteristics of Gases
Gas Pressure: Definition and Units
Calculation of Pressure
Measurement of Pressure
Boyle’s Law: The Pressure-Volume Relationship
Charles’s and Gay-Lussac’s Law: The Temperature-Volume Relationship
Avogadro’s Law: The Amount-Volume Relationship
Deriving the Ideal Gas Equation from the Empirical Gas Laws
Applications of the Ideal Gas Equation
Calculating the Required Volume of a Gaseous Reactant
Determining the Amount of Reactant Consumed Using Change in Pressure
Predicting the Volume of a Gaseous Product
Dalton’s Law of Partial Pressures
Mole Fractions
Using Partial Pressures to Solve Problems
Application of the Gas Laws
Molecular Speed
Diffusion and Effusion
Factors That Cause Deviation from Ideal Behavior
The van der Waals Equation

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