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Chapter 5. Gases Student Objectives 5.1 Breathing: Putting Pressure to Work  Define pressure and understand how differences in pressure lead to the ...
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Chapter 5. Gases Student Objectives 5.1 Breathing: Putting Pressure to Work 

Define pressure and understand how differences in pressure lead to the act of breathing.

5.2 Pressure: The Result of Molecular Collisions    

Understand pressure from a molecular point of view. Understand examples of pressure: wind and pressure imbalance on the eardrum. Understand why pressure can be measured in mm of Hg, and convert between the different pressure units. Understand the origin of the two numbers for blood pressure.

5.3 The Simple Gas Laws: Boyle’s Law, Charles’s Law, and Avogadro’s Law      

Know and be able to rationalize Boyle’s law, the inverse relationship between volume and pressure. Use the inverse mathematical relationship between pressure and volume to solve initial and final states problems at constant temperature and amount. Know and be able to rationalize Charles’s law, the direct relationship between volume and temperature. Use the direct mathematical relationship between volume and temperature to solve initial and final states problems at constant pressure and amount. Know and be able to rationalize Avogadro’s law, the direct relationship between volume and amount (moles). Use the direct mathematical relationship between volume and amount to solve initial and final states problems at constant pressure and temperature.

5.4 The Ideal Gas Law  

Know and understand how the ideal gas law combines the three simple gas laws into one equation. Calculate using the ideal gas law and the gas constant, R, with the appropriate value and units.

5.5 Applications of the Ideal Gas Law: Molar Volume, Density, and Molar Mass of a Gas   

Define standard temperature and pressure and molar volume of an ideal gas. Know and understand the relationship between molar volume, molar mass, and density. Calculate using density, molar mass, and molar volume.

5.6 Mixtures of Gases and Partial Pressures     

Define and understand partial pressure of a gaseous component in a mixture. Define and determine mole fraction of a component in a mixture. Know and understand that Dalton’s law of partial pressures relates to mole fraction of a gas to the partial pressure of the gas. Understand how the total pressure affects the partial pressures of gases in blood, especially during deep‐sea diving. Know and understand the technique of collecting gases over water.

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Chapter 5. Gases 5.7 Gases in Chemical Reactions: Stoichiometry Revisited  

Understand how stoichiometry applies to gases via the number of moles in the ideal gas law. Understand how stoichiometry relates to molar volume.

5.8 Kinetic Molecular Theory: A Model for Gases      

Define kinetic molecular theory for gases. Understand each of the three postulates/assumptions of the kinetic molecular theory. Understand how the kinetic molecular theory explains Boyle’s, Charles’s, Avogadro’s, and Dalton’s laws. Follow the derivation of the ideal gas law from the kinetic molecular theory. Understand that all gases at the same temperature have the same kinetic energy, and understand the relationship between speed and molar mass. Understand the graphical representation of the distribution of molecular speeds.

5.9 Mean Free Path, Diffusion, and Effusion of Gases  

Define and understand mean free path. Define diffusion and effusion and understand how they are related to the kinetic molecular theory.

5.10 Real Gases: The Effects of Size and Intermolecular Forces   

Understand that the ideal gas law is an approximation that works well under certain circumstances and not so well at low temperature and/or high pressure. Understand that nonideal behavior arises from the finite volume of gas particles and the intermolecular forces between particles. Recognize and identify the components of the van der Waals equation.

Section Summaries Lecture Outline  

Terms, Concepts, Relationships, Skills Figures, Tables, and Solved Examples

Teaching Tips  

Suggestions and Examples Misconceptions and Pitfalls



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Chapter 5. Gases Lecture Outline Terms, Concepts, Relationships, Skills 5.1 Breathing: Putting Pressure to Work  Pressure o practical definition o breathing as a result of pressure difference 5.2 Pressure: The Result of Molecular Collisions  Pressure o molecular view o units: millimeters of mercury, torr, atmosphere, pascal, pounds per square inch o measurements  mercury barometer  manometer o blood pressure



Figures, Tables, and Solved Examples  

Intro figure: pressure in a champagne bottle Figure 5.1 Gas Pressure



unnumbered figure: regions of high and low atmospheric pressure Figure 5.2 Pressure and Particle Density Figure 5.3 Pressure Imbalance Figure 5.4 The Mercury Barometer Table 5.1 Common Units of Pressure Example 5.1 Converting between Pressure Units Figure 5.5 The Manometer Chemistry and Medicine: Blood Pressure

      

5.3 The Simple Gas Laws: Boyle’s Law, Charles’s Law, and Avogadro’s Law  Simple gas laws o Boyle’s law: volume and pressure  V  1/P  P 1 V 1 = P 2 V 2 o Charles’s law: volume and temperature  V  T  V 1 /T 1 = V 2 /T 2 o Avogadro’s law: volume and amount  V  n  V 1 /n 1 = V 2 /n 2

            

5.4 The Ideal Gas Law  Derived from simple gas laws: PV = nRT o units: Kelvin for T, moles for n o gas constant R: function of P, V units

   

Figure 5.6 The J‐Tube Figure 5.7 Volume versus Pressure Figure 5.8 Molecular Interpretation of Boyle’s Law Figure 5.9 Increase in Pressure with Depth Example 5.2 Boyle’s Law Figure 5.10 Volume versus Temperature Chemistry in Your Day: Extra‐long Snorkels Figure 5.11 Molecular Interpretation of Charles’s Law unnumbered figure: hot‐air balloon unnumbered figure: balloon in liquid N 2 Example 5.3 Charles’s Law Figure 5.12 Volume versus Number of Moles Example 5.4 Avogadro’s Law

unnumbered figure: relationship between ideal gas law and simple gas laws unnumbered figure: label on aerosol can Example 5.5 Ideal Gas Law I Example 5.6 Ideal Gas Law II



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Chapter 5. Gases Teaching Tips Suggestions and Examples 5.1 Breathing: Putting Pressure to Work  All students should have experience with a bottle of carbonated liquid, though it is more likely with soda pop than champagne. 5.2 Pressure: The Result of Molecular Collisions   Recognizing pressure as related to force will likely help most students. The difference can be explained by talking about snowshoes and ice skates.  Large numbers of people have experienced their ears “popping” in an airplane or upon another ascent in altitude.  Tornados, hurricanes, and cyclones are extreme examples of differential pressures in weather.  Blood pressure is an interesting alternative since the previous examples deal with gases. 5.3 The Simple Gas Laws: Boyle’s Law, Charles’s Law, and  Avogadro’s Law  The easiest example of Boyle’s law is a cylinder with a sliding piston. A plastic syringe works  well as long as one has a convenient connection to a gauge.  Limited numbers of students will have personal  experience with diving, though most students likely will have experience swimming underwater.  Charles’s law can be demonstrated using balloons  and dry ice/acetone or liquid nitrogen.  Conceptual Connection 5.1 Boyle’s Law and Charles’s Law  Avogadro’s law can be demonstrated by producing a gas over water to show displacement. An interesting challenge question comes from burning a candle within a sealed container connected to a pressure gauge. Oxygen is consumed, but carbon dioxide is produced. 5.4 The Ideal Gas Law  A series of R values is possible with different units especially for pressure.  Common problems include using mixtures of unit conversions: oC to K, mL to L, mmHg to atm.

Misconceptions and Pitfalls

Barometers and manometers combine liquids and gases. The end of the tube can be open, under a vacuum, or filled with air. The outcomes are different and can be confusing.

Breathing at increased depths of water becomes more difficult but is possible even to substantial depths. Hot‐air and helium balloons rise because of decreased density compared with the surrounding air. Some students memorize the different equations for initial and final states problems, but all of them can be derived easily from the ideal gas law. In initial and final states problems with T, the ratio of the values must be in Kelvin and not in degrees Celsius.

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Chapter 5. Gases Lecture Outline Terms, Concepts, Relationships, Skills 5.5 Applications of the Ideal Gas Law: Molar Volume, Density, and Molar Mass of a Gas  Standard temperature and pressure o molar volume  Density: molar volume/molar mass o

d

Figures, Tables, and Solved Examples   

unnumbered figure: molar volumes of different gases at STP Example 5.7 Density Example 5.8 Molar Mass of a Gas

PM , M  molar mass RT

 Finding the molar mass of a gas 5.6 Mixtures of Gases and Partial Pressures  Mixtures of gases o partial pressure o mole fraction o Dalton’s law  Deep‐sea diving o hypoxia o oxygen toxicity o nitrogen narcosis  Collecting gases over water o vapor pressure of water 5.7 Gases in Chemical Reactions: Stoichiometry Revisited  Calculations: moles from the ideal gas law o P, V, T of gas A  moles A  moles B  P, V, T of gas B  Calculations: molar volume o V of gas A  moles A  moles B  mass of gas B

       

 

Table 5.3 Composition of Dry Air Example 5.9 Total Pressure and Partial Pressures Figure 5.13 Oxygen Partial Pressure Limits unnumbered figure: gas pressure in lungs Example 5.10 Partial Pressures and Mole Fractions Figure 5.14 Collecting a Gas over Water Table 5.4 Vapor Pressure of Water versus Temperature Example 5.11 Collecting Gases over Water

Example 5.12 Gases in Chemical Reactions Example 5.13 Using Molar Volume in Gas Stoichiometric Calculations

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Chapter 5. Gases Teaching Tips Suggestions and Examples 5.5 Applications of the Ideal Gas Law: Molar Volume, Density, and Molar Mass of a Gas  Conceptual Connection 5.2 Molar Volume  Density can be calculated from molar volume and molar mass or from P, T and M.  Conceptual Connection 5.3 Density of a Gas 5.6 Mixtures of Gases and Partial Pressures  The partial pressure of oxygen in air is important in physiology. Oxygen available in the lungs for transfer to the blood can be changed by a change in the total air pressure or by a change in the percent oxygen in air.  Conceptual Connection 5.4 Partial pressures  Partial pressure of oxygen plays a role in diving and in climbing at high altitude. Too little or too much can be dangerous.  The vapor pressure of water must be considered when collecting gases over water, a common laboratory technique. The solubility of gases in water is generally small except for CO 2 .  The vapor pressure of water is a function of temperature and is covered in more detail in Chapter 11. 5.7 Gases in Chemical Reactions: Stoichiometry Revisited  The ideal gas law is used to obtain moles rather than using mass or volume and concentration of an aqueous solution.  At the same pressure and temperature, the mole ratio from a balanced reaction is the same as the volume ratio for gaseous components.  At the same volume and temperature, the mole ratio from a balanced reaction is the same as the pressure ratio for gaseous components.  Conceptual Connection 5.5 Pressure and Number of Moles

 

 



Misconceptions and Pitfalls Ideal gases have a common molar volume but not a common mass. Standard temperature for gases is not the same standard as in thermochemistry and thermodynamics.

That gases behave ideally, independent of identity, gives rise to partial pressures. The sum of the mole fractions is 1.

Stoichiometry problems involving gases are inherently the same as other stoichiometry problems. Gases simply present another means of obtaining moles of substances.

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Chapter 5. Gases Lecture Outline Terms, Concepts, Relationships, Skills 5.8 Kinetic Molecular Theory: A Model for Gases  Kinetic molecular theory o postulates/assumptions  particles negligibly small  average kinetic energy proportional to T  elastic collisions  Relevance to gas laws o pressure as a force per unit area o Boyle’s law o Charles’s law o Avogadro’s law o Dalton’s law o ideal gas law  Temperature and molecular speeds





Figures, Tables, and Solved Examples      

Figure 5.15 A Model for Gas Behavior Figure 5.16 Elastic versus Inelastic Collsions Figure 5.17 The Pressure on the Wall of a Container Figure 5.18 Velocity Distribution for Several Gases at 25 oC Figure 5.19 Velocity Distribution for Nitrogen at Several Temperatures Example 5.14 Root Mean Square Velocity

1 2 3 mv  RT 2 2 3 RT  M

o

KE 

o

urms

Distribution of molecular speeds o dependence on T o dependence on M

5.9 Mean Free Path, Diffusion, and Effusion of Gases  Mean free path  Diffusion  Effusion o

Graham’s law:

rateA  rateB

  

Figure 5.20 Mean Free Path Figure 5.21 Effusion Example 5.15 Graham’s Law of Effusion

MB MA



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Chapter 5. Gases Teaching Tips Suggestions and Examples 5.8 Kinetic Molecular Theory: A Model for Gases  The physics in the kinetic molecular theory is remarkably simple given all of the behaviors it explains. The ideal gas law can be derived from the definition of pressure as force per unit area.  Two of the three assumptions in the kinetic molecular theory give rise to nonideal behavior.  Conceptual Connection 5.6‐5.7 Kinetic Molecular Theory I & II 5.9 Mean Free Path, Diffusion, and Effusion of Gases  Diffusion is easy for students to understand, as they have experience with odors (especially bad ones) spreading throughout a room.  An analogy for diffusion is thermal equilibration, as they both are illustrations of the second law of thermodynamics.









Misconceptions and Pitfalls One of the main uses of going through a derivation is pointing out that knowing where an equation comes from and the assumptions required to derive it gives clues concerning its range of applicability. The analogy of billiard balls to atoms in an elastic collision is partially flawed since the balls lose some energy to friction with the table. Students are likely to have difficulty with distributions of velocities within a collection of molecules.

Most students are surprised by the high speeds of atoms and molecules at room temperature.



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Chapter 5. Gases Lecture Outline Terms, Concepts, Relationships, Skills 5.10 Real Gases: The Effects of Size and Intermolecular Forces  Real gases o high pressures: finite particle volume o low temperatures: intermolecular forces  van der Waals equation o

2  n   P  a    V  nb  nRT  V   

Figures, Tables, and Solved Examples      

Figure 5.22 Molar Volumes of Real Gases Figure 5.23 Particle Volume and Ideal Behavior Figure 5.24 The Effect of Particle Volume Table 5.5 Van der Waals Constants for Common Gases Figure 5.25 The Effect of Intermolecular Forces Figure 5.26 Real versus Ideal Behavior



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Chapter 5. Gases Teaching Tips Suggestions and Examples 5.10 Real Gases: The Effects of Size and Intermolecular Forces  Two of the three assumptions from the kinetic molecular theory break down and give rise to nonideal behavior.  Each of the effects can be demonstrated independently.  Conceptual Connection 5.8 Real Gases





Misconceptions and Pitfalls Why do we use the ideal gas law if we know it is “wrong”? For some gases under some conditions, the corrections are unnecessary.

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Chapter 5. Gases Additional Problem for Converting between Pressure Units (Example 5.1)

A normal systolic blood pressure value is 110. mmHg. What is the pressure in pascal?

Sort

Given 110 mmHg

The problem gives a pressure in mmHg and asks you to find the equivalent in pascal.

Find Pa

Strategize

Conceptual Plan

Since Table 5.1 does not have a direct conversion factor between mmHg and Pa, but does provide relationships between both of these units and atmospheres, you can convert to atm as an intermediate step.

mmHg



atm

1 atm 760 mmHg



Pa

101,325 Pa 1 atm

Relationships Used 1 atm = 760 mmHg 101,325 Pa = 1 atm (both from Table 5.1) Solve

Solution

Follow the conceptual plan to solve the problem. Begin with 110 mmHg and use the conversion factors to arrive at the pressure in Pa.

110 mmHg ×

1 atm 101,325 Pa = 14,700 Pa × 760 mmHg 1 atm

The units (Pa) are correct. The magnitude of the answer (14,700) makes physical sense since the blood pressure is a fraction of one atm.

Check

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Chapter 5. Gases

Additional Problem for Ideal Gas Law (Example 5.5)

Calculate the volume in L occupied by 1.43 mol of oxygen gas at a pressure of 3.25 atm and a temperature of 298 K.

Sort

Given n = 1.43 mol

The problem gives you the number of moles of oxygen gas, the pressure, and the temperature. You are asked to find the volume in L.

P = 3.25 atm T = 298 K Find V in L

Strategize

Conceptual Plan

You are given three of the four variables (n, P, T) in the ideal gas law and asked to find the fourth (V). The conceptual plan shows how the ideal gas law provides the relationship between the given quantities and the quantity to be found.

n, P, T PV



V

 nRT

Relationships Used PV = nRT (ideal gas law)

Solve

Solution

To solve the problem, first solve the ideal gas law for V.

PV  nRT nRT V  P

Then substitute the given quantities to compute V.

1.43 mol  V

=

0.08206 L atm  298 K mol K 3.25 atm

= 10.8 L

The units of the answer are correct and the magnitude makes sense.

Check

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Chapter 5. Gases Additional Problem for Density (Example 5.7)

Calculate the density of oxygen gas at 37.4 oC and a pressure of 720 mmHg.

Sort

Given T = 37.4 oC

The problem gives you the temperature and pressure of a gas and asks you to find its density. The problem also states that the gas is oxygen.

Find d

Strategize

Conceptual Plan

Equation 5.6 provides the relationship between the density of a gas and its temperature, pressure, and molar mass. The temperature and pressure are given, and you can compute the molar mass from the formula of the gas, which we know is O 2 .

P = 720 mmHg

P, T, 

 d

d

P RT

Relationships Used d

P RT

molar mass O 2 = 32.00 Solve

Solution

To solve the problem, you must gather each of the required quantities in the correct units. Convert the temperature to kelvins and the pressure to atmospheres.

T(K) = 37.4 oC + 273 = 310.4 K

Now substitute the quantities into the equation to compute density.

P  720 mmHg 

d

1 atm = 0.947 atm 760 mmHg

P RT

 32.00 g  0.947 atm    1 mol  = 0.08206 L atm  310.4 K mol K

= 1.19 g/L

The units of the answer (g/L) are correct. The magnitude of the answer (1.19) makes physical sense compared to the density of gases calculated earlier.

Check

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Chapter 5. Gases Additional Problem for Molar Mass of a Gas (Example 5.8)

A sample of gas has a mass of 0.205 g. Its volume is 0.112 L at a temperature of 25 oC and a pressure of 740 mmHg. Find its molar mass.

Sort The problem gives you the mass of a gas sample, along with its volume, temperature, and pressure. You are asked to find the molar mass.

Given       m  =  0.205 g                    V  =  0.112 L              T(oC)  = 25 oC                    P  =  740 mmHg 

Strategize

Conceptual Plan

The conceptual plan has two parts:

     P, V, T               n                    PV  nRT         n, m                  molar mass       molar mass = mass (m )            

Find molar mass (g/mol)

1) Use the ideal gas law to find the number of moles of gas.

 

moles (n )

2) Use the definition of molar mass to find the molar mass.

Relationships Used PV = nRT molar mass =

mass (m ) moles (n )

Solve

Solution

Convert the temperature to K and pressure to atm. Then find the number of moles, solving the ideal gas law for n.

T(K)  =  25 oC +  273  =  298 K 

Use the number of moles (n) and the given mass (m) to find the molar mass.

P

 740 mmHg 

1 atm = 0.974 atm   760 mmHg

PV  nRT   PV  n RT 0.974 atm  0.112 L = 4.46  103 mol n = 0.08206 L atm  298 K mol K mass (m ) molar mass = moles (n ) 0.205 g = = 46.0 g/mol 4.46  10 3 mol

The units (g/mol) are correct. The magnitude of the answer (46.0) seems to make sense for a gas. For example, it could be NO 2 .

Check

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Chapter 5. Gases Additional Problem for Total Pressure and Partial Pressures (Example 5.9)

A 1.50-L mixture of helium, neon, and argon has a total pressure of 754 mmHg at 310 K. If the partial pressure of helium is 431 mmHg and partial pressure of neon is 211 mmHg, what mass of argon is present in the mixture?

Sort

Given P He

= 431 mmHg

P Ne

= 211 mmHg

The problem gives you the partial pressures of two of the three components in a gas mixture, along with the total pressure, the volume, and the temperature, and asks you to find the mass of the third component.

P total = 754 mmHg V

= 1.50 L

T

= 310 K

Find m Ar Strategize

Conceptual Plan

You can find the mass of argon from the number of moles of argon, which you can calculate from the partial pressure of argon and the ideal gas law.

   P total , P He , P Ne                P Ar                          Ptotal  PHe  PNe  PAr      P Ar , V, T              n Ar                      PV  nRT     n Ar                  m Ar               1 mol Ar            

Begin by finding the partial pressure of argon from Dalton’s law of partial pressures. Then use the partial pressure of argon together with the volume of the sample and the temperature to find the number of moles of argon.

39.95 g Ar

Relationships Used P total = P He + P Ne + P Ar (Dalton’s law)

Finally, use the molar mass of argon to compute the mass of argon from the number of moles of argon.

PV = nRT (ideal gas law) molar mass Ar = 39.95 g/mol

Solve

Solution

Follow the conceptual plan. To find the partial pressure of argon, solve the equation for P Ar , and substitute the values of the other partial pressures.

P total   =  P He   +  P Ne   +  P Ar        P Ar   =  P total  ‐  P He   ‐  P Ne                  =  754 mmHg    431 mmHg    211 mmHg              =  112 mmHg  1 atm 112 mmHg  = 0.147 atm  

Convert the partial pressure from mmHg to atm and use it in the ideal gas law to compute the amount of argon in moles.

760 mmHg

Use the molar mass of argon to convert from amount of argon in moles to mass of argon.

PV 0.147 atm 1.50 L    = 8.67  10 3 mol RT 0.08206 L atm  310 K mol K 39.94 g Ar 8.67  10 3 mol Ar  = 0.346 g Ar   mol Ar

Check

The units of the answer are correct and the magnitude makes sense.

n

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Chapter 5. Gases Additional Problem for Partial Pressures and Mole Fractions (Example 5.10)

A 10-L scuba diving tank is willed with a mixture of helium and oxygen containing 20.4 g He and 3.52 g O 2 at 298 K. Calculate the mole fraction and partial pressure of each component in the mixture and calculate the total pressure.

Sort

Given  m He   =  20.4 g                mO2 =  3.52 g 

The problem gives the masses of two gases in a mixture and the volume and temperature of the mixture. You are asked to find the mole fraction and partial pressure of each component, as well as the total pressure.

                V  =  10 L                  T  = 298 K    Find  He ,  O , P He , PO , P total 2 2

Strategize

Conceptual Plan

The conceptual plan has several parts:

   m He                n He                m O2                 nO2  

1) To calculate the mole fraction of each component, first find the number of moles of each component. To accomplish this, convert masses to moles.

            1 mol He                              

2) Compute the mole fraction of each component using the mole fraction definition. Calculate the total pressure from the sum of the moles of both components. Alternatively, you could calculate the partial pressure of each component and sum them.

4.00 g He

1 mol O2 32.00 g O2

 

n n O He 2  = ;  = He O n +n n + n 2 He O He O 2 2 (nHe + nO2 )  RT   Ptotal  V PHe = He Ptotal ; PO2 =  O2 Ptotal

  

 

Relationships Used

3) Use the mole fractions of each component and the total pressure to calculate the partial pressure of each component.

 a   = n a  / n total   (mole fraction definition)  P total V  =  n total RT  (ideal gas law)        P a   =   a  P total    

Solve

Solution

Follow the conceptual plan to solve the problem. Begin by converting each of the masses to amounts in moles. Compute each of the mole fractions. Compute the total pressure. Compute the partial pressure of each component.

20.4 g He 

1 mol He = 5.10 mol He 4.00 g He

1 mol O2 3.52 g O2  32.00 g O2

  

He = O = 2

Ptotal 

PHe

 

= 0.110 mol O2  

nHe 5.10 = = 0.979 nHe + nO2 5.10 + 0.110 nO2 nHe + nO2

=

 

0.110 = 0.0211 0.110 + 5.10

(nHe + nO2 )  RT V

0.08206 L atm (5.10 + 0.110) mol   298 K mol K = 10.0 L = He Ptotal = 0.979  12.7 atm = 12.5 atm

  = 12.7 atm

PO2 =  O2 Ptotal = 0.021 12.7 atm = 0.267 atm

Check

The units of the answer are correct and the magnitudes are reasonable.

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Chapter 5. Gases

Additional Problem for Gases in Chemical Reactions (Example 5.12)

Sort You are given the mass of carbon monoxide, a reactant of a chemical reaction. You are asked to find the volume of the product produced (methanol) at a specified temperature and pressure.

Methanol (CH 3 OH) can be synthesized by the reaction:     CO(g)  +  2 H 2 (g)      CH 3 OH(g)  What volume (in liters) of methanol gas, measured at a  temperature of 473 K and a pressure of 820 mmHg, is  produced from 100.0 g of carbon monoxide CO?  Given  100.0 g CO              T  =  473 K              P  =  820 mmHg  Find V methanol

Strategize

Conceptual Plan

The volume of methanol produced can be calculated from the number of moles of methanol, which can be obtained from the number of moles of carbon monoxide and the stoichiometry of the reaction.

   g CO              mol CO             mol CH 3 OH                     1 mol CO             1 mol CH3OH  

Find the number of moles of carbon monoxide from its mass by using the molar mass. Use the stoichiometric relationship from the balanced equation to find the number of moles of methanol formed from that quantity of carbon monoxide.

1 mol CO

28.01 g CO

   n (mol CH 3 OH), P, T               V methanol                                       PV  nRT                    Relationships Used PV = nRT (ideal gas law) 1 mol CO : 1 mol CH 3 OH (from balanced equation) molar mass CO = 28.01 g

Substitute the moles of methanol, together with the pressure and temperature into the ideal gas law to find the volume of hydrogen. Solve

Solution

Follow the conceptual plan to solve the problem. Begin by using the mass of carbon monoxide to get the number of moles of CO.

100.0 g CO 

Convert the number of moles of carbon monoxide to moles of methanol. Convert the pressure to atmospheres and use the ideal gas law to find the volume of methanol.

1 mol CO = 3.570 mol CO 28.01 g CO

3.570 mol CO  = 3.570 mol CH 3 OH  P = 820 mmHg 

       VCH3OH =

1 atm = 1.08 atm 760 mmHg

nCH3OHRT

 

P

0.08206 L atm  473 K mol K = = 128 L 1.08 atm   The units of the answer are correct. The magnitude makes sense. You start with about 3.5 moles of reactant. If the reaction was conducted at STP, you would expect about 80 L. Since the temperature is much higher than 298 K and the pressure is somewhat higher, the result should be greater.pressure is somewhat higher, the result should be greater. 3.570 mol 

Check

 

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Chapter 5. Gases

Additional Problem for Molar Volume and Stoichiometry (Example 5.13)

How many grams of water form when 2.41 L of oxygen gas at STP completely react with an appropriate amount of H 2 ? 2 H 2 (g) + O 2 (g)  2 H 2 O(g)

Sort

Given 2.41 L O 2

You are given the volume of oxygen gas (a reactant) at STP and asked to determine the mass of water that forms upon complete reaction.

Find g H 2 O

Strategize

Conceptual Plan

Since the reaction occurs under standard conditions, you can convert directly from the volume (in L) of hydrogen gas to the amount in moles. Then use the stoichiometric relationship from the balanced equation to find the number of moles of water formed. Finally, use the molar mass of water to obtain the mass of water formed.

  L O 2           mol O 2             mol H 2 O           g H 2 O           1 mol O2            2 mol H2O              18.02 g H2O     22.4 L O2

1 mol O2

1 mol H2O

Relationships Used 1 mol = 22.4 L (at STP) 1 mol O 2 : 2 mol H 2 O (from balanced equation) molar mass H 2 O = 18.02 g/mol

Solve

Solution

Follow the conceptual plan to solve the problem.

Check

2.41 L O2 

1 mol O2 22.4 L O2



2 mol H2O 1 mol O2



18.02 g H2O = 3.88 g 1 mol H2O

The units of the answer are correct. The magnitude makes sense since you start with about one-tenth of a molar volume (22.4  10 = 2.2) and make twice as many moles of water as oxygen. The mass of water is about one-fifth of a molar mass (18  5 = 3.6).

83 Copyright © 2014 by Pearson Education, Inc.

Chapter 5. Gases

Additional Problem for Root Mean Square Velocity (Example 5.14)

Calculate the root mean square velocity of oxygen molecules at 100 oC.

Sort

Given  O 2              T  =  100 oC 

You are given the kind of molecule and the temperature and asked to find the root mean square velocity.

Find u rms

Strategize

Conceptual Plan

The conceptual plan for this problem simply shows how the molar mass of oxygen and the temperature (in kelvins) can be used with the equation that defines the root mean square velocity.

   , T               u rms                u = 3 RT         rms



Relationships Used

1 J  =  1 kg m2 / s2     3 RT   

urms =

Solve

Solution

The quantities must be in the correct units—temperature in K and molar mass in kg/mol.

T      =  100 oC +  273  =  373 K  

Substitute the quantities into the equation.

=

urms =

32.00 g O2 1 mol O2



1 kg = 3.200  102 kg/mol O2   1000 g

3 RT  8.314 J  373 K mol K 3.200  10 2 kg/ mol O2 3

=

  kg m2

=

2.91 105

J = kg

2.91 105

s2 kg

= 538 m/s The units of the answer (m/s) are correct, and the magnitude seems reasonable compared with the velocity previously calculated for 25 oC.

Check

84 Copyright © 2014 by Pearson Education, Inc.

Chapter 5. Gases

Additional Problem for Graham’s Law of Effusion (Example 5.15)

An unknown gas effuses at a rate that is 0.578 times that of nitrogen gas at the same temperature. Calculate the molar mass of the unknown gas in g/mol.

Sort

Given

You are given the ratio of effusion rates for the unknown gas and nitrogen and asked to find the molar mass of the unknown gas.

Rateunk = 0.578 RateN2

Find  unk Strategize

Conceptual Plan

The conceptual plan uses Graham’s law of effusion. You are given the ratio of rates and you know the molar mass of nitrogen. You can use Graham’s law to find the molar mass of the unknown gas.

rate ratio,  nitrogen



Rateunk = RateN2

 unk  N2  unk

Relationships Used Rateunk = RateN2

Solve

Solution

Solve the equation for  unk and substitute the correct values to compute it.

Rateunk = RateN2  unk =

=

 N2  unk

(Graham’s law)

 N2  unk  N2 2

 Rateunk     RateN  2   28.01 g/mol

 0.578 

2

= 83.8 g/mol

The units of the answer (g/mol) are correct, and the magnitude makes sense. The unknown gas appears to be krypton.

Check

85 Copyright © 2014 by Pearson Education, Inc.

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