example 11.1
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Solution Map :. Solution :. SKILLBUILDER 11.1. Converting between Pressure Units. Convert a pressure of 173 in. Hg into pounds per square inch. EXAMPLE 11.1. Converting between Pressure Units. - PowerPoint PPT PresentationTRANSCRIPT
Copyright ©2009 by Pearson Education, Inc.Upper Saddle River, New Jersey 07458
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Introductory Chemistry, Third EditionBy Nivaldo J. Tro
You are given a pressure in psi and asked to convert it to mm Hg. Find the required conversion factors in Table 11.1.
EXAMPLE 11.1 Converting between Pressure Units
Given: 125 psi Find: mm HgConversion Factors:
1 atm = 14.7 psi 760 mm Hg = 1 atm
A high-performance road bicycle tire is inflated to a total pressure of 125 psi. What is this pressure in millimeters of mercury?
Begin the solution map with the given units of psi. Use the conversion factors to convert first to atm and then to mm Hg.
Convert a pressure of 173 in. Hg into pounds per square inch.
SKILLBUILDER 11.1 Converting between Pressure Units
Follow the solution map to solve the problem.
Solution Map:
Solution:
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Introductory Chemistry, Third EditionBy Nivaldo J. Tro
EXAMPLE 11.1 Converting between Pressure Units
Continued
Convert a pressure of 23.8 in. Hg into kilopascals.
SKILLBUILDER PLUS
FOR MORE PRACTICE Example 11.13; Problems 27, 28, 29, 30, 33, 34, 35, 36.
Copyright ©2009 by Pearson Education, Inc.Upper Saddle River, New Jersey 07458
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Introductory Chemistry, Third EditionBy Nivaldo J. Tro
You are given an initial pressure, an initial volume, and final pressure. You are asked to find the final volume.
This problem requires the use of Boyle's law.
EXAMPLE 11.2 Boyle’s Law
Given:P1 = 4.0 atmV1 = 6.0 LV1 = 6.0 L
Find: V2
Equation: P1 V1 = P2 V2
A cylinder equipped with a moveable piston has an applied pressure of 4.0 atm and a volume of 6.0 L. What is the volume of the cylinder if the applied pressure is decreased to 1.0 atm?
Draw a solution map beginning with the given quantities. Boyle’s law shows the relationship necessary to get to the find quantity.
Solve the equation for the quantity you are trying to find (V2)and then substitute the numerical quantities into the equation to compute the answer.
Solution Map:
Solution:
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Introductory Chemistry, Third EditionBy Nivaldo J. Tro
EXAMPLE 11.2 Boyle’s Law
Continued
FOR MORE PRACTICE Example 11.14; Problems 37, 38, 39, 40.
A snorkeler takes a syringe filled with 16 mL of air from the surface, where the pressure is 1.0 atm, to an unknown depth. The volume of the air in the syringe at this depth is 7.5 mL. What is the pressure at this depth? If the pressure increases by an additional 1 atm for every 10 m of depth, how deep is the snorkeler?
SKILLBUILDER 11.2 Boyle’s Law
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Introductory Chemistry, Third EditionBy Nivaldo J. Tro
You are given an initial volume, a final volume, and a final temperature. You are asked to find the intitial temperature in both kelvins (T1) and degrees Celsius (t1).
EXAMPLE 11.3 Charles’s Law
Given:V1 = 2.80 LV2 = 2.57 t2 = 0 °C
A sample of gas has a volume of 2.80 L at an unknown temperature. When the sample is submerged in ice water att = 0 °C its volume decreases to 2.57 L. What was its initial temperature (in kelvins and in Celsius)? Assume a constant pressure. (To distinguish between the two temperature scales, use t for temperature in °C and T for temperature in K.)
This problem requires the use of Charles's law.
Draw a solution map beginning with the given quantities. Charles’s law shows the relationship necessary to get to the find quantity.
Find: T1 and t1
Equation:
Solution Map:
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Introductory Chemistry, Third EditionBy Nivaldo J. Tro
Solve the equation for the quantity you aretrying to find (T1).
Before you substitute in the numerical values, you must convert the temperature to kelvins. Remember, gas law problems must always be worked using Kelvin temperatures. Once you have converted the temperature to kelvins, substitute into the equation to find Convert the temperature to degrees Celsius to find t1.
EXAMPLE 11.3 Charles’s Law
Continued
Solution:
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Introductory Chemistry, Third EditionBy Nivaldo J. Tro
EXAMPLE 11.3 Charles’s Law
Continued
FOR MORE PRACTICE Problems 43, 44, 45, 46.
A gas in a cylinder with a moveable piston with an initial volume of 88.2 mL is heated from 35°C to 155°C. What is the final volume of the gas in milliliters?
SKILLBUILDER 11.3 Charles’s Law
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Introductory Chemistry, Third EditionBy Nivaldo J. Tro
You are given an initial pressure, temperature, and volume as well as a final temperature and volume. You are asked to find the final pressure.
EXAMPLE 11.4 The Combined Gas Law
Given:P1 = 735 mm Hg t1 = 34 °C t2 = 85 °CV1 = 158 mL V2 = 108 mL
Find: P2
A sample of gas has an initial volume of 158 mL at a pressure of 735 mm Hg and a temperature of 34°C. If the gas is compressed to a volume of 108 mL and heated to a temperature of 85°C, what is its final pressure in millimeters of mercury?
This problem requires the use of thecombined gas law.
Draw a solution map beginning with thegiven quantities. The combined gas lawshows the relationship necessary to get to thefind quantity.
Equation:
Solution Map:
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Introductory Chemistry, Third EditionBy Nivaldo J. Tro
Solve the equation for the quantity you aretrying to find (P2).
Before you substitute in the numerical values, you must convert the temperatures tokelvins.
Once you have converted the temperatureto kelvins, substitute into the equation tofind P2 .
EXAMPLE 11.4 The Combined Gas Law
Continued
Solution:
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Introductory Chemistry, Third EditionBy Nivaldo J. Tro
EXAMPLE 11.4 The Combined Gas Law
Continued
FOR MORE PRACTICE Example 11.15; Problems 55, 56, 57, 58, 59, 60.
A balloon has a volume of 3.7 L at a pressure of 1.1 atm and a temperature of 30 °C. If the balloon is submerged in water to a depth where the pressure is 4.7 atm and the temperature is 15 °C, what will its volume be (assume that any changes in pressure caused by the skin of the balloon are negligible)?
SKILLBUILDER 11.4 The Combined Gas Law
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Introductory Chemistry, Third EditionBy Nivaldo J. Tro
You are given an initial volume, an initial number of moles, and a final volume. You are (essentially) asked to find the final number of moles.
This problem requires the use of Avogadro’slaw
EXAMPLE 11.5 Avogadro’s Law
A 4.8-L sample of helium gas contains 0.22 mol of helium. How many additional moles of helium gas must be added to the sample to obtain a volume of 6.4 L? Assume constant temperature and pressure.
Draw a solution map beginning with the given quantities. Avogadro’s law shows the relationship necessary to get to the find quantity.
Given:V1 = 4.8 L n1 = 0.22 mol V2 = 6.4 L
Find: n2
Equation:
Solution Map:
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Introductory Chemistry, Third EditionBy Nivaldo J. Tro
EXAMPLE 11.5 Avogadro’s Law
Continued
FOR MORE PRACTICE Problems 49, 50, 51, 52.
A chemical reaction occurring in a cylinder equipped with a moveable piston produces 0.58 mol of a gaseous product. If the cylinder contained 0.11 mol of gas before the reaction and had an initial volume of 2.1 L, what was its volume after the reaction?
SKILLBUILDER 11.5 Avogadro’s Law
Solve the equation for the quantity you aretrying to find (n2)and substitute the appropriate quantities to compute n2.
Since the balloon already contains 0.22 mol,subtract this quantity from the final numberof moles to determine how much must beadded
Solution:
mol to add = 0.29 – 0.22 = 0.07 mol
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Introductory Chemistry, Third EditionBy Nivaldo J. Tro
You are given the number of moles, the pressure, and the temperature of a gas sample. You are asked to find the volume.
This problem requires the use of the idealgas law.
EXAMPLE 11.6 The Ideal Gas Law
Given:n = 0.845 molP = 1.37 atm T = 315 K
Find: VEquation: PV = nRT
Calculate the volume occupied by 0.845 mol of nitrogen gas at a pressure of 1.37 atm and a temperature of 315 K.
Draw a solution map beginning with thegiven quantities. The ideal gas law shows therelationship necessary to get to the findquantity.Solve the equation for the quantity you aretrying to find (V) and substitute the appropriate quantities to compute V.
Solution Map:
Solution:
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Introductory Chemistry, Third EditionBy Nivaldo J. Tro
EXAMPLE 11.6 The Ideal Gas Law
Continued
FOR MORE PRACTICE Example 11.16; Problems 63, 64, 65, 66.
An 8.5-L tire is filled with 0.55 mol of gas at a temperature of 305 K. What is the pressure of the gas in the tire?
SKILLBUILDER 11.6 The Ideal Gas Law
Copyright ©2009 by Pearson Education, Inc.Upper Saddle River, New Jersey 07458
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Introductory Chemistry, Third EditionBy Nivaldo J. Tro
You are given the pressure, the volume, andthe temperature of a gas sample. You are asked to find the number of moles.
This problem requires the use of the ideal gaslaw.
EXAMPLE 11.7 The Ideal Gas Law Requiring Unit Conversion
Given:P = 24.2 psiV = 3.2 L t = 25 °C
Find: nEquation: PV = nRT
Calculate the number of moles of gas in a basketball inflated to a total pressure of 24.2 psi with a volume of 3.2 L at25 °C.
Draw a solution map beginning with thegiven quantities. The ideal gas law shows therelationship necessary to get to the findquantity.
Solution Map:
Copyright ©2009 by Pearson Education, Inc.Upper Saddle River, New Jersey 07458
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Introductory Chemistry, Third EditionBy Nivaldo J. Tro
Solve the equation for the quantity you aretrying to find (n).
Before substituting into the equation, youmust convert P and t into the correct units.(Since 1.6462 is an intermediate answer,mark the least significant digit, but don’tround until the end.)
Finally, substitute into the equation to compute n.
EXAMPLE 11.7 The Ideal Gas Law Requiring Unit Conversion
Continued
How much volume does 0.556 mol of gas occupy when its pressure is 715 mm Hg and its temperature is 58 °C?
SKILLBUILDER 11.7 The Ideal Gas Law Requiring Unit Conversion
Solution:
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Introductory Chemistry, Third EditionBy Nivaldo J. Tro
EXAMPLE 11.7 The Ideal Gas Law Requiring Unit Conversion
Continued
FOR MORE PRACTICE Problems 67, 68, 71, 72.
Find the pressure in millimeters of mercury of a 0.133-g sample of helium gas at 32 °C and contained in a 648-mL container.
SKILLBUILDER PLUS
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Introductory Chemistry, Third EditionBy Nivaldo J. Tro
You are given the mass, the volume, the temperature, and the pressure of a gas sample. You are asked to find the molar mass of the gas.
This problem requires the use of the ideal gaslaw and the definition of molar mass.
EXAMPLE 11.8 Molar Mass Using the Ideal Gas Law and Mass Measurement
A sample of gas has a mass of 0.311 g. Its volume is 0.225 L at a temperature of 55 °C and a pressure of 886 mm Hg. Find its molar mass.
In the first part of the solution map, use theideal gas law to find the number of moles ofgas from the other given quantities. In thesecond part, use the number of moles fromthe first part, as well as the given mass, tofind the molar mass.
Given:m = 0.311 gV = 0.225 L t = 55 °C P = 886 mm Hg
Find: molar mass (g/mol)Equation:
Solution Map:
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Introductory Chemistry, Third EditionBy Nivaldo J. Tro
First, solve the ideal gas law for n.
Before substituting into the equation, youmust convert the pressure to atm and temperature to K.
Now, substitute into the equation to computen, the number of moles.
Finally, use the number of moles just foundand the given mass (m) to find the molarmass.
EXAMPLE 11.8 Molar Mass Using the Ideal Gas Law and Mass Measurement
Continued
Solution:
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Introductory Chemistry, Third EditionBy Nivaldo J. Tro
EXAMPLE 11.8 Molar Mass Using the Ideal Gas Law and Mass Measurement
Continued
FOR MORE PRACTICE Problems 73, 74, 75, 76.
A sample of gas has a mass of 827 mg. Its volume is 0.270 L at a temperature of 88 °C and a pressure of 975 mm Hg. Find its molar mass.
SKILLBUILDER 11.8 Molar Mass Using the Ideal Gas Law and Mass Measurement
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Introductory Chemistry, Third EditionBy Nivaldo J. Tro
You are given the total pressure of a gas mixture and the partial pressures of two (of its three) components. You are asked to find the partial pressure of the third component.
This problem requires the use of Dalton’slaw of partial pressures.
EXAMPLE 11.9 Total Pressure and Partial Pressure
Given:Ptot = 558 mm HgPHe = 341 mm HgPNe = 112 mm Hg
Find: PAr
Equation: Ptot = Pa + Pb + Pc + . . .
A mixture of helium, neon, and argon has a total pressure of 558 mm Hg. If the partial pressure of helium is 341 mm Hg and the partial pressure of neon is 112 mm Hg, what is the partial pressure of argon?
To solve this problem, simply solve Dalton’slaw for the partial pressure of argon and substitute the correct values to compute it.
Solution:Ptot = PHe + PNe + PAr
PAr = Ptot + PHe + PNe
= 558 mm Hg – 341 mm Hg – 112 mm Hg= 105 mm Hg
A sample of hydrogen gas is mixed with water vapor. The mixture has a total pressure of 745 torr, and the water vapor has a partial pressure of 24 torr. What is the partial pressure of the hydrogen gas?
SKILLBUILDER 11.9 Total Pressure and Partial Pressure
FOR MORE PRACTICE Example 11.17; Problems 77, 78, 79, 80.
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Introductory Chemistry, Third EditionBy Nivaldo J. Tro
You are given the percent oxygen in the mixture and the total pressure. You are asked to find the partial pressure of oxygen.
You will need the equation that relates partial pressure to total pressure.
EXAMPLE 11.10 Partial Pressure, Total Pressure, and Percent Composition
Calculate the partial pressure of oxygen that a diver breathes with a heliox mixture containing 2.0% oxygen at adepth of 100 m where the total pressure is 10.0 atm.
Calculate the fractional composition of O2 by dividing the percent composition by 100.Calculate the partial pressure of O2 by multiplying the fractional composition by the total pressure.
FOR MORE PRACTICE Problems 83, 84, 85, 86.
Solution:
Given:O2 percent = 2.0%Ptot = 10.0 atm
Find: Equation:Partial pressure of component = Fractional composition of component × Total pressure
What must the total pressure be for a diver breathing heliox with an oxygen composition of 5.0% to breathe = 0.21 atm?
SKILLBUILDER 11.10 Partial Pressure, Total Pressure, and Percent Composition
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Introductory Chemistry, Third EditionBy Nivaldo J. Tro
You are given the mass of a reactant in achemical reaction. You are asked to findthe volume of a gaseous product at agiven pressure and temperature.
You will need the molar mass of KClO3 and the stoichiometric relationship between KClO3 and O2 (from the balanced chemical equation). You will also need the ideal gas law.
EXAMPLE 11.11 Gases in Chemical Reactions
Given: 294 g KClO3
P = 755 mm Hg (of oxygen gas)T = 305 K
Find: Volume of O2 in litersEquations and Conversion Factors:
1 mol KClO3 = 122.6 g2 mol KClO3 = 3 mol O2
PV = nRT
The solution map has two parts. In the first part, convert from g KClO3 to mol KClO3
and then to mol O2 In the second part, use mol O2 as n in the ideal gas law to find the volume of O2 .
Solution Map:
How many liters of oxygen gas form when 294 g of KClO3 completely react in the following reaction (which is used in the ignition of fireworks)?
Assume that the oxygen gas is collected at P = 755 mm Hg and T = 305 K.
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Introductory Chemistry, Third EditionBy Nivaldo J. Tro
Begin by converting mass KClO3 to molKClO3 and then to mol O2.
Then solve the ideal gas equation for V.
Before substituting the values into this equation, you must convert the pressure to atm.
Finally, substitute the given quantities along with the number of moles just calculated to compute the volume.
EXAMPLE 11.11 Gases in Chemical Reactions
Continued
FOR MORE PRACTICE Problems 93, 94, 95, 96, 97, 98.
Solution:
In the following reaction, 4.58 L of O2 were formed at 745 mm Hg and 308 K. How many grams of Ag2O must have decomposed?
SKILLBUILDER 11.11 Gases in Chemical Reactions
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Introductory Chemistry, Third EditionBy Nivaldo J. Tro
You are given the volume of a reactant at STP and asked to find the mass of the product formed.You will need the molar volume at STP, the stoichiometric relationship between H2 and H2O (from the balanced chemical equation), and the molar mass of H2O.
EXAMPLE 11.12 Using Molar Volume in Calculations
Given: 1.24 L H2
Find: g H2OConversion Factors:
1 mol = 22.4 L (at STP)2 mol H2 = 2 mol H2O18.02 g H2O = 1 mol H2O
The solution map has two parts. In the first part, convert from g KClO3 to mol KClO3
and then to mol O2 In the second part, use mol O2 as n in the ideal gas law to find the volume of O2 .
Solution Map:
How many grams of water form when 1.24 L of H2 gas at STP completely reacts with O2?
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Introductory Chemistry, Third EditionBy Nivaldo J. Tro
Begin with the volume of H2 and follow the solution map to arrive at mass H2O in grams.
EXAMPLE 11.12 Using Molar Volume in Calculations
Continued
FOR MORE PRACTICE Problems 99, 100, 101, 102.
Solution:
How many liters of oxygen (at STP) are required to form 10.5 g of H2O?
SKILLBUILDER 11.12 Using Molar Volume in Calculations
Copyright ©2009 by Pearson Education, Inc.Upper Saddle River, New Jersey 07458
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Introductory Chemistry, Third EditionBy Nivaldo J. Tro
EXAMPLE 11.13 Pressure Unit Conversion
Convert 18.4 in. Hg to torr.
Given: 18.4 in. Hg
Find: torr
Conversion Factors: 1 atm = 29.92 in. Hg760 torr = 1 atm
Solution Map:
Solution:
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Introductory Chemistry, Third EditionBy Nivaldo J. Tro
EXAMPLE 11.14 Simple Gas Laws
A gas has a volume of 5.7 L at a pressure of 3.2 atm. What is its volume at 4.7 atm? (Assume constant temperature.)Given:
P1 = 3.2 atmV1 = 5.7 LP2 = 4.7 atm
Find: V2
Equation: P1V1 = P2V2
Solution Map:
Solution:
Copyright ©2009 by Pearson Education, Inc.Upper Saddle River, New Jersey 07458
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Introductory Chemistry, Third EditionBy Nivaldo J. Tro
EXAMPLE 11.15 The Combined Gas Law
A sample of gas has an initial volume of 2.4 L at a pressure of 855 mm Hg and a temperature of 298 K. If the gas is heated to a temperature of 387 K and expanded to a volume of 4.1 L, what is its final pressure in millimeters of mercury?Given: P1 = 855 mm Hg
V1 = 2.4 LT1 = 298 KV2 = 4.1 LT2 = 387 K
Find: P2
Equation:
Solution Map:
Copyright ©2009 by Pearson Education, Inc.Upper Saddle River, New Jersey 07458
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Introductory Chemistry, Third EditionBy Nivaldo J. Tro
EXAMPLE 11.15 The Combined Gas Law
Continued
Solution:
Copyright ©2009 by Pearson Education, Inc.Upper Saddle River, New Jersey 07458
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Introductory Chemistry, Third EditionBy Nivaldo J. Tro
EXAMPLE 11.16 The Ideal Gas Law
Calculate the pressure exerted by 1.2 mol of gas in a volume of 28.2 L and at a temperature of 334 K.Given:
n = 1.2 molV = 28.2 LT = 334 K
Find: P
Equation: PV = nRT
Solution Map:
Solution:
Copyright ©2009 by Pearson Education, Inc.Upper Saddle River, New Jersey 07458
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Introductory Chemistry, Third EditionBy Nivaldo J. Tro
EXAMPLE 11.17 Total Pressure and Partial Pressure
A mixture of three gases has the following partial pressures.
What is the total pressure of the mixture?
Given:
Find: Ptot
Equation: Ptot = Pa + Pb + Pc + . . .
Solution:
= 289 mm Hg + 342 mm Hg + 122 mm H = 289 mm Hg + 342 mm Hg + 122 mm Hg