Science 3

Percent by volume of solute = volume of solute × 100% volume of solution Sample Problem: If you are asked to prepare 500 mL of pineapple-orange juice whose concentration is 60% (v/v) orange juice and 40% (v/v) pineapple juice, how are you going to mix it with the right combination? Explanation: Remember that 60% (v/v) orange juice is the same as 60 mL of orange juice in 100 mL solution. The same goes for 40% (v/v) pineapple juice which is 40 mL of pineapple juice in 100 mL solution. So, in a 500 mL pineapple-orange juice, the volume would be: 60 mL orange juice × 500 mL = 300 mL of orange juice 100 mL juice solution 40 mL pineapple juice × 500mL = 200 mL of pineapple juice 100 mL juice solution Therefore, in order to get the right combination, you just have to mix 300 mL of orange juice to 200 mL pineapple juice to make a pineapple-orange juice, the total volume of which is 500 mL.3. Percent mass by volume (% m/v) The same is true for percent by weight-volume. It is expressed the same as percent by mass but the denominator is changed to volume instead, as given by the equation: Percent by weight - volume of solute = mass of solute × 100% volume of solution Sample Problem: You are asked to add 30 mg salt (NaCl) in 20 mL solution to a favorite adobo dish. This concentration is just enough to give your adobo a salty taste. Express this in percent concentration. (Note: 1 mg = 0.001 g) Explanation: 0.03 g NaCl × 100 = 0.15% NaCl solution 20 mL solution - 13 -

What you will do Practice Exercise 3.11. An alcoholic drink claims 12% alcohol by volume. Calculate the volume in mL of alcohol present in a 0.8 L wine.2. How much sugar (C12H22O11) in grams must be added to 450 mL water to make a 20% sugar solution?3. How much bagoong (fish paste) in grams will you add to 1.5 kg mixed vegetables to be able to come up with a delicious pinakbet with percent by mass concentration of 30%? (1 kg = 1000 g) Key to answers on page 26.B. Mole fraction Another method of expressing concentration is by ratio solution. A 1:10 salt solutionis 1 g of salt in 10 mL solution. The first number in the ratio indicates the number of gramsof solute and the second number gives the number of milliliters of solution. One way ofexpressing ratio solutions is by Mole fraction. It is the ratio of the individual component ofthe solution to the total component of the solution.Mole fraction of component A (XA ) = sum of moles of A moles of all componentsSince it is a ratio of two similar quantities, the mole fraction has no units.Sample Problem: Calculate the mole fraction of solute and solvent in a 15% aqueous sugar solution(C12H22O11). Note: An aqueous solution is a solution where water is the solvent.Explanation: Always assume the mass of the solution as 100 g. Therefore, in a 15% aqueoussugar solution (C12H22O11), there are 15 g of sugar (solute) to be dissolved in 85 g water(solvent). And the number of moles is computed as: - 14 -

15 g C12H22O11 × 1mol C12H22O11 = 0.044 mol C12H22O11 342 g C12H22O1185 g H2 O × 1mol C12H22O11 = 4.72 mol H2O 18 g H2OThen, the mole fraction of solute and solvent in the solution is computed as:Mole fraction of C12H22O11 = 0.044 = 0.01 0.044 + 4.72Mole fraction of H2O = 4.72 = 0.99 0.044 + 4.72 What you will do Practice Exercise 3.2Calculate the mole fraction of solute and solvent in a 30% aqueous sugar solution(C12H22O11). Note: An aqueous solution is a solution where water is the solvent. Key to answers on page 26.C. Molarity (M) Molar solutions are used most frequently by chemists. A molar solution is defined asone that contains 1 mole of solute per liter of solution. Molarity = moles of solute liters of solution Sample Problem: For you to better understand molarity, let us use the same problem in ratio solutions but let us change the concentration. You need to prepare 1 L (1 L = 1000 mL) of “arnibal” (it is a brown sugar solution) for your sago-gulaman juice, the concentration of which is 2 M. How much sugar will you need to add to water? - 15 -

Explanation: The problem calls for the preparation of 1 L of 2 M sugar solution, C6H12O6(molecular weight is 180 g/mol ). Recall that molarity means moles per liter. So, 1 L x 2 M C6H12O6 = 2 moles C6H12O6Then, since 1 mol weighs 180 g, 2 moles C6H12O6 x 180 g C6H12O6 = 360 g C6H12O6 1mole C6H12O6Therefore, we take 360 g C6H12O6, dissolve it in water, and dilute to a total of 1 L. What you will do Practice Exercise 3.3A juice advertised as \"low sodium\" contains 12 mg Na+ (molecular weight is 23 g/mol) ineach 8-oz serving (8 oz = 250 mL). What is the concentration of sodium in the beverage interms of molarity? Key to answers on page 26.D. Molality (m) Molality is the number of moles of solute dissolved in 1 kg (1000 g) of solvent: thatis, Molality = moles of solute mass of solvent (kg) Sample Problem: How will you express 30% aqueous sugar solution (C12H22O11) in molal concentration? Explanation: Always assume the mass of the solution as 100 g. Therefore, in a 30% aqueous - 16 -

sugar solution (C12H22O11), there are 30 g of sugar (solute) to be dissolved in 70 gwater (solvent). So:30 g C12H22O11 × 1mol C12H22O11 × 1000 g H2O = 1.25 m C12H22O11 70 g H2O 342 g C12H22O11 1kg H2O Molarity is most often used over molality because it is generally easier to measurethe volume of the solution using precisely calibrated volumetric flask than to weigh thesolvent. However, when accuracy of the experiment is at stake, molality is preferredover molarity because the volume of the solution typically increases with increasingtemperature. For example, a solution at 1.0 M at 25oC may become 0.95 M at 45oCbecause of the increase in volume. What you will do Practice Exercise 3.4How will you express 60% aqueous sugar solution (C12H22O11) in terms of molality? Key to answers on page 26.Dilution Solutions Sometimes, it is often necessary to prepare a weaker (dilute) solution from a stronger(concentrated) solution. To do so, we must add water to the stronger solution. But howmuch water must be added? To answer this question, we use the relationship: Initial concentration x Initial volume = Final concentration x Final volume C1V1 = C2V2where the initial concentration and initial volume are the strength and amount ofconcentrated solution to be used; and the final concentration and final volume are thestrength and amount of diluted solution to be used. Sample Problem: At dinner, the visitors found the orange juice prepared by your brother too sweet. Your brother said he just mixed 70 g of powdered orange juice in 500 mL of water. When you read the label, the suggested preparation was 60 g of powdered juice for 500 mL of water. The best thing to do is to dilute the orange juice prepared by adding water. - 17 -

How much water will you add? Explanation: Using the above formula for diluting solutions, the amount of water to be added can be computed as: 60 g x (x mL of 60 g powdered orange) = 70 g x 500 mL x = 70 g x 500 mL 60g x = 583 mL Therefore, in order to have the right orangy taste as prescribed by the manufacturer, you must add 83 mL of water (583 – 500 = 83) to your brother’s prepared juice. What you will do Practice Exercise 3.5Let us make a scenario.You were unable to copy your mom’s specialty “hinilabos na talangka” (salted crablets) andyour guests found it too salty. You prepared it by adding 10% salt (NaCl) solution to yourcrablets before steaming. How much water are you going to add if your mom said that sheprepared her specialty by mixing 20 mL of 0.5% salt (NaCl) solution instead? Key to answers on page 26. What you will do Self-Test 3.1Let us try to check how much you have learned from this lesson by answering the followingquestions.Multiple Choice. Choose the letter of the best answer. Write the chosen letter on aseparate sheet of paper. - 18 -

1. Which of the following is the most concentrated solution?a. 2 M NaCl solution c. 2 M Na2SO4b. 3 m NaCl solution d. 2 g of NaCl in 100 mL solution2. A solution labeled \"5% NaCl\" is assumed to contain 5 g NaCl per __________.a. 100 mL water c. 100 mL solutionb. 100 mL solvent d. none of the above3. How much salt is present in a 5 M salt solution (Molecular weight = 58 g/mol)?a. 0.09 g c. 290 gb. 11.6 g d. none of the above4. A juice claims to contain 180 mg of Na+ (molecular weight is 23 g/mol) in each 8-ozserving (8 oz = 250 mL). What is the concentration of sodium in terms of ppm?a. 0.72 ppm c. 72 ppmb. 7.2 ppm d. 720 ppm5. The active ingredient in one brand of antacid is magnesium oxide, MgO (MolecularWeight = 40.31 g / mol). If it is dissolved in 250 mL water at a ratio of 1:10, how manygrams of MgO must be dissolved in the solution?a. 25 g c. 2500 gb. 250 g d. none of the above Key to answers on page 26.Lesson 4. Why Do You Need to Study Solutions? Water is usually referred to as the universal solvent. The electrical charges in watermolecules help dissolve different kinds of substances. Solutions form when the force ofattraction between the solute and the solvent is greater than the force of attraction betweenthe particles in the solute. This type of solutions where water is the solvent is calledaqueous solutions. One example of an important process is chemical weathering.Chemical weathering begins to take place when carbon dioxide in the air dissolves inrainwater. A solution called carbonic acid is formed. The process is then completed as theacidic water seeps into rocks and dissolves underground limestone deposits. Sometimes,the dissolving of soluble minerals in rocks can even lead to the formation of caves. If one takes a moment to consider aqueous solutions, one quickly observes that theyexhibit many interesting properties. For example, the tap water in your kitchen sink doesnot freeze at exactly 0ºC. This is because tap water is not pure water; it contains dissolvedsolutes. Some tap water, commonly known as hard water, contains mineral solutes such ascalcium carbonate, magnesium sulfate, calcium chloride, and iron sulfate. - 19 -

Did you know? The amount of oxygen required to oxidize a given quantity of organic material is known as the biochemical oxygen demand (BOD). This is one indicator of water pollution. Interestingly, some microorganisms, the anaerobic bacteria, can “eat” certain highly toxic substances and convert it to less toxic ones. The reduced solubility of oxygen in hot water has a direct bearing on thermalpollution—that is, the heating of the environment (usually waterways) to temperatures thatare harmful to its living inhabitants. Fish, like all other cold-blooded animals, have muchmore difficulty coping with rapid temperature fluctuation in the environment than we humansdo. The increase in water temperature accelerates their rate of metabolism, which generallydoubles with each 10oC rise. The results can be drastic because the rate of oxygendemand for fishes does not support the oxygen supply available because of its lowersolubility in heated water. Thus, effective ways of cooling power plants with minimaldamage to biological environment are being sought.Environmental Awareness We often don’t think about what we throw away, or how our household wastes canaffect groundwater, lakes, rivers, and coastlines. Thus, we should be more vigilant toparticipate in recycling programs. A substance is biodegradable if it can be broken downby microorganisms. Some organic compounds are non-biodegradable, presumablybecause their structures are such that microorganisms cannot use them for food, orbecause they are toxic and can kill the microorganisms. Have you heard of the term pH? I am sure you have heard of the term pH-balancedin a TV commercial. Have you ever wondered what it means? pH indicates whether a solution is acidic or basic. If the pH is less than 7, thesolution is considered acidic. If the pH is greater than 7, the solution is basic. At pH 7, thesolution is said to be neutral. Solutions in the body have to maintain a specific pH range. For example, in order tohave a healthy hair, the pH must be maintained at pH between 4 and 7. This means thatour hair is slightly acidic, and for it to remain healthy, it must be exposed to shampoo orconditioners of same pH range. - 20 -

Table 4.1 pH of Body FluidsBody Fluids pHPlasma 7.4Interstitial fluid 7.4Cerebrospinal fluid 7.3Saliva 5.8-7.1Urine 4.6-8.0Sweat / perspiration 4.0-6.8Gastric juice 1.6-1.8 Could you justify the pH of the body fluids listed in table 4.1? What should you do tomaintain a healthy pH range of each body fluids? What you will do Self-Test 4.1Let us try to check how much you have learned from the lesson by answering the followingquestions.Fill in the blanks with the correct answer._______________1. A substance that can be broken down by microorganisms_______________2. It begins to take place when carbon dioxide in the air dissolves in rainwater forming carbonic acid._______________3. A microorganism which determines the biochemical oxygen demand (BOD) number in water pollution_______________4. It refers to the heating of the environment (usually waterways) to temperatures that are harmful to its living inhabitants._______________5. It is the term used when tap water contains mineral solutes such as calcium carbonate, magnesium sulfate, calcium chloride, and iron sulfate. Did you encounter any problem? Well, compare your answers with the answer keyand see for yourself the items you missed. Good luck! Key to answers on page 26. - 21 -

Let’s SummarizeA. Summary of Key EquationsPercent by mass of solute (%m/m) = mass of solute × 100% mass of solutionPercent by volume of solute (%v/v) = volume of solute × 100% volume of solutionPercent by weight - volume of solute (%m/v) = mass of solute × 100% volume of solutionMole fraction of component A (XA ) = moles of A sum of moles of all componentsMolarity (M) = moles of solute liters of solutionMolality (m) = moles of solute mass of solvent (kg) Dilution Solutions: C1V1 = C2V2B. Summary of Facts and Concepts 1. Solutions are homogenous mixtures of two or more substances, which may either be solids, liquids or gases. They are clear, have a variable composition, do not settle, can be separated by physical means and can be separated by filtration. 2. Solutions can also be classified as diluted or concentrated according to relative amount of components present. 3. Suspension is defined as fine particles of solid in a liquid and because it is transparent, it does not settle out and cannot be separated by filtration. 4. Solutions are composed of solutes, the substance dissolved and the solvent, the liquid material in which the solute has dissolved. 5. The dissolving process involves physical changes (phase changes) and both absorption and evolution of energy. If more energy was absorbed than evolved, the process is endothermic. If more energy was evolved than absorbed, the dissolution process is exothermic. 6. Solubility is the maximum amount of solute that will dissolve in a solvent at a specific temperature. The solubility of a solute is affected by temperature, pressure, and nature of the solvent. The rate of dissolution is affected by surface area, rate of stirring and temperature. - 22 -

7. Increasing temperature usually increases the solubility of solid and liquid substances, and decreases the solubility of gases in water.8. The greater the pressure, the greater the solubility of a gas in liquid.9. A saturated solution is when there is no more solute that will dissolve in the solvent. An unsaturated solution contains less solute than it has the capacity to dissolve. A supersaturated solution contains more solute than is present in a saturated solution.10. The concentration of a solution may be expressed in terms of percentage (by weight, by volume, or by weight-volume), ratio (mole fraction), molarity and molality. The choice of units depends on the purpose of the measurement.11. Numerical values of pH indicate whether a solution is acidic or basic.PosttestI. Multiple Choice. Choose the letter of the best answer. Write the chosen letter on aseparate sheet of paper.1. A solution has all the following properties EXCEPTa. clear c. cannot pass through filter paperb. homogenous d. can be separated by physical means2. A solution that contains the maximum amount of dissolved solute is:a. saturated c. concentratedb. unsaturated d. supersaturated3. When there is less solute in relation to the solvent, the solution is said to be:a. dilute c. supersaturatedb. unsaturated d. none of the above4. Water in a creek near a busy highway was analyzed for salt content and found to contain0.0276 g NaCl per 100 mL solution. What is the concentration of the NaCl expressed inppm?a. 0.276 c. 27.6b. 2.76 d. 2765. All of the following affects the solubility of a solid in a liquid EXCEPTa. stirring c. surface areab. pressure d. temperature6. When alcohol is mixed with water, the mixture is said to be _________ to one another.a. soluble c. insolubleb. miscible d. immiscible - 23 -

7. Liquids that DO NOT dissolve freely in one another in any proportion are known as:a. solutions c. homogenous mixturesb. emulsions d. heterogenous mixtures8. All of the following are biodegradable substances EXCEPTa. glass c. animal fecesb. tissue paper d. fruit peelings9. A \"10% NaCl solution\" contains 10 g NaCl per __________.a. 100 mL water c. 100 mL solutionb. 100 mL solvent d. 1000 mL solution10. What kind of solution contains less solute than it has the capacity to dissolve?a. diluted c. unsaturatedb. saturated d. supersaturated11. Which factor will increase the amount of dissolved oxygen in a body of water? a. onset of cooler season b. presence of more fishes c. contamination of body of water d. increased volume of oxygen in the atmosphere12. The dissolution of sodium hydroxide pellets in water is an exothermic process. Which can be done to increase the solubility of the sodium hydroxide pellets in 100 mL of water? a. increase pressure of the solution b. decrease the pressure of the solution c. increase temperature of the solution d. decrease temperature of the solution13. Sterling silver contains 95% silver. If a necklace made of sterling silver weighs 15 g,what is the mass of silver dissolved in the necklace?a. 1.6 g c. 9.0 gb. 6.2 g d. 14.25 g14. The pH of gastric juice is 1.6-1.8. Which of the following is the reason for its acidic nature? a. It kills most of the bacteria taken in along with food. b. It triggers secretion of acidic fluid containing bicarbonate ions. c. It facilitates build up of proteins from an inactive to active form. d. It inhibits the secretion of pancreas of an alkaline fluid containing bicarbonate ions.15. Which of the following triggers Thermal Pollution?a. increase in fishes c. increase in carbon dioxide supplyb. increase in temperature d. all of the above - 24 -

II. True or False. Write the word TRUE if the statement is correct and FALSE if it isincorrect.________ 1. Solutions are heterogenous mixtures.________ 2. Solubility of a solid in liquid increases with increasing temperature.________ 3. Water can dissolve anything that is why it is called the universal solvent.________ 4. Solubility of a gas in liquid increase with increasing pressure.________ 5. Solutions expressed in molarity and molarity are the same when water is the solvent. Key to answers on page 27. Key to AnswersPretestI. Multiple Choice 6. a II. True or False1. a 7. a 1. True2. a 8. c 2. True3. a 9. a 3. False4. d 10. b 4. False5. c 5. FalseLesson 1 - 25 -Self-Test 1.11. c2. b3. d4. c5. dLesson 2Self-Test 2.11. d2. d3. d4. a5. a

Lesson 3Practice Exercise 3.1 1. 96 mL alcohol 2. 31 g sugar 3. 300 g bagoongPractice Exercise 3.2 Mole fraction of C12H22O11 is 0.02 and water is 0.98.Practice Exercise 3.3 0.002 M Na+Practice Exercise 3.4 4.39 mPractice Exercise 3.5 19 mL waterSelf-Test 3.1 1. b 2. c 3. a 4. d 5. aLesson 4Self-Test 4.1 1. Biodegradable 2. Chemical weathering 3. Anaerobic Bacteria 4. Thermal pollution 5. Hard water - 26 -

PosttestI. Multiple Choice 6. b 11. a II. True or False1. c 7. d 12. c 1. False2. a 8. a 13. d 2. False3. a 9. c 14. a 3. False4. d 10. c 15. b 4. True5. b 5. TrueReferencesAraneta, F.L. & Catris, L.V. (2002). The world of chemistry: Exploring the natural world series). Manila: SIBS Publishing House Inc.Chang, R.N. (2004). Chemistry. (8th ed.) USA: McGraw-Hill, Inc.Green, J. & Damji, S. (2001). Chemistry. (2nd ed.) Australia: IBID Press, Victoria.Magno, M.C., Tan, M.C. & Punzalan, A.G. (2000). Chemistry. (3rd ed.) Manila: DIWA Learning Systems, Inc.Moore, J.W., Stanitski, C.L. & Wood J.L. (1998). The chemical world: Concepts and applications. (2nd ed.) USA: Harcourt Brace & Company.Petrucci, R.H. & Harwood, W.S. (1998). General chemistry: Principles and modern applications. (7th ed.) New Jersey: Prentice-Hall International.Regalado, E.C. (2000). Chemistry. Philippines: St. Augustine Publications, Inc.Sackheim, G.I. & Lehman, D.D. (1990). Chemistry for the health sciences. (6th ed.) New York: Macmillan Publishing Company. - 27 -

Module 8 Behavior of Gases What this module is about This module will require a lot of your imagination because many gases are colorless,unlike solids and liquids that are visible. For instance, you need to visualize the molecularmotion of particles in the gas phase of matter as if there’s a group of people in chaos insidea burning house. The kinetic molecular theory of gases is an attempt to explain the behavior of theparticles of gases such as its rate of motion, the distances between its particles and theforce that exists between the gas molecules. It highlights the differences in the property ofthe three (3) phases of matter namely: solid, liquid and gas. The fourth phase of matter,plasma, is not considered in the discussion because it exists only in extreme temperatureand pressure. The conditions considered in this module are those that exist within therange of normal temperature and pressure. The module contains the following lessons:  Lesson 1 – Kinetic Molecular Theory of Gases  Lesson 2 – Diffusion and Effusion Rates of Gases  Lesson 3 – Practical Applications of Behavior of Gases What you are expected to learn After going through this module, you should be able to: 1. state and illustrate the kinetic molecular theory of gas; 2. differentiate diffusion and effusion of gases; 3. state the relation of rate of diffusion and effusion to the molecular mass of gas substance; and 4. cite some practical applications of the behavior of gases.

How to learn from this module Take enough time to read and study each lesson in this module. It is a must that youread it with understanding and with the help of a dictionary, if necessary. Perform theactivities diligently as these will deepen your conceptual understanding of the lessons. Just be honest in evaluating yourself in the self-test portion of this module. Try hardto answer the test without looking back at the lecture. Go over the lesson again if you fail toget at least 70% of the items in the test.What to do before (Pretest)Multiple Choice: Choose the letter of the best answer. Write the chosen letter on aseparate sheet of paper.1. Which of the following best describes the molecules of a gas? a. All the molecules are connected by bending and stretching. b. The molecules are not connected and are moving very rapidly. c. The molecules are arranged in strings and are moving very rapidly. d. The molecules are not connected and are moving slowly by bending and stretching.2. All of these are properties of gases EXCEPT a. can conduct electricity b. can be compressed easily c. can move from one place to another randomly d. can expand when the temperature is increased by increasing the distances between molecules.3. This scientist formulated the law of diffusion and effusion of gases. Who was he?a. John Dalton c. Thomas Grahamb. Robert Boyle d. Jacque Charles4. Which among these gases contributes primarily to air pollution?a. H2 c. CO2b. CO d. NO25. Which among these substances will diffuse fastest?a. HCl c. CO2b. NH3 d. NO2 -2-

6. A gas confined in a closed container is heated. What will happen to the gas? a. The gas molecules move faster. b. The gas molecules settle to the bottom of the container. c. The gas molecules collide with each other less frequently. d. The gas molecules collide with the wall of the container less frequently.7. All of the following proves that gas diffuses EXCEPT a. Toxic chemicals in cigarette smoke spread in a car. b. The sweet fragrance of perfume spreads in a room. c. The aroma of the coffee is confined in one corner of the room. d. The CFC from aerosol spray reaches the upper atmosphere and destroys the ozone layer.8. As gas molecules collide, which will not happen?a. They stick together. c. Their kinetic energies are conserved.b. They gain their kinetic energies. d. They will explode.9. Which of the following gases is a product of incomplete combustion in automobileengine?a. CH4 c. CO2b. CO d. NO210. The following statements illustrate the kinetic theory of gases EXCEPT a. The gas molecules move slowly in all direction. b. The distances between gas molecules are negligible. c. The gas molecules do not interact except during collisions. d. The gas molecules often collide with each other or with their container. Key to answers on page 16.Lesson 1. Kinetic Molecular Theory of Gases The word “gas” seems to mean different things tovarious people. Some think that it refers to the ‘gas’ that weuse for cooking and heating. Others think that ‘gas’ is thefuel for automobiles and industrial machines. So, what is meant by gas? Gas is found everywhere.It enters our body everyday every time we inhale andexhale. Do you know that almost 16 kg of gases get into ourlungs everyday? Every time we use perfume spray oraerosol spray, we encounter gases. Inflating a balloon andopening a refrigerator are examples of gas in pressure. -3-

What you will do plastic ice bagActivity 1.1 Properties of GasesMaterials: empty soft drinks can rubber plastic ice bag band rubber band alcohol burner soft drinksProcedure: (Refer to the diagram for the assembly.) can1. Put the plastic ice bag on the top edge of the empty soft drinks can.2. Tie the plastic with rubber band. See to it that no air escapes the system.3. Heat the bottom part of the can. Observe what happens. ______________________________________________________________________4. Remove the burner and let it cool. What happens to the plastic ice bag? ______________________________________________________________________ Key to answers on page 17. You’re right in your observations! Why? This activity illustrates that gases exhibit expansibility and compressibility.Expansibility is the property of gases to increase the spaces between its molecules if thetemperature is increased. Compressibility is its ability to compress if the temperature isdecreased. Another property of gases is its ability to diffuse or intermingle with the molecules ofother substances. This property is called diffusibility. When your mother is cooking“bagoong”, what do you observe? You can smell the cooked bagoong with sauted garlic ata distance, right? This is because the molecules of the sauted garlic in bagoong mingledwith the molecules of the air. This phenomenon can be further explained by the kinetictheory of gases.What is the Kinetic Theory of Gases? In the 18th century, a Flemish scientist and Physicist named Jan Baptista VanHelmont was the first to use the word “gas”. He derived the term from the Greek word“chaos”, which means formless mass. He felt that gases were a disordered, formless typeof matter and can be found anywhere. -4-

The properties and characteristics of gases can clearly be explained by a series ofstatements called the Kinetic Theory of Gases. This theory states that: 1. gases consist of independent molecules at normal temperature. The distance between gas molecules are so great compared to the size of the molecule itself. 2. gas molecules do not interact except during collisions. The force of interaction between molecules is almost negligible. 3. gas molecules move rapidly, randomly and constantly in any direction. 4. gas molecules often collide with each other or with their container. The collision of gas molecules is completely translational. 5. When the temperature increases, gas molecules move faster. When the temperature decreases, the gas molecules move more slowly. The kinetic energy of the molecules of gas is directly proportional to its absolute temperature. The fourth property of gases is its ability to move rapidly and randomly because it haslow density. The movement of gases is dependent to its molecular mass. Helium, He2 gashaving a molecular mass of 8 g/mol (2 x 4 g/mol) travel faster than a chlorine gas, Cl2, witha molecular mass of 70 g/mol (2 x 35 g/mol). This is why helium gas is used to inflateballoons. What you will do Self-Test 1.1 Kinetic Theory of GasesFill in the blanks to complete the statements of the Kinetic Theory of Gases.1. Gases consist of ____________molecules.2. At normal atmospheric pressure and standard temperature, the distances between gas molecules are ___________.3. Gas molecules do not interact except during _____________.4. The ___________ between gas molecules is almost negligible.5. Gas molecules move rapidly, ______________, and constantly in any direction.6. Gas molecules are always in ____________.7. Gas molecules often collide with other gas molecules or with their container. The collisions are completely _______________.8. When the temperature ________, gas molecules move faster. When the temperature _______, gas molecules move more slowly.9. The average _______________ of gas molecules is directly proportional to the absolute temperature of the gas.10. The ability of gases to intermingle with the molecules of other substances is called _________________. Key to answers on page 17. -5-

Lesson 2. Diffusion and Effusion of Gases If someone opens a bottle of perfume at one end of a room, a person at the other endof the room soon smells the scent of the perfume. Why? The vapors from the perfume mixwith the air inside the room and spread out evenly and slowly throughout the room. This isa direct demonstration of the random motion of vapor molecules of the perfume, which isprovided by diffusion. So what is diffusion? Diffusion is the gradual mixing of themolecules of one gas with the molecules of another gas because of their kinetic energies. The diffusion process takes a relatively long period of time to complete. It will takesome time before a person at the other end of the room can smell the perfume scent. Whyis that so? You see, the molecules of a perfume experience numerous collisions with theother air molecules in the room while moving from one end of the room to the other. Diffusion of gases always happens gradually, and not instantly as molecular speedsseem to suggest. Furthermore, since the average speed of light gas molecules is greaterthan the average speed of heavier gas molecules, a lighter gas will diffuse through a certainspace faster than a heavier one. How do gas molecules pass through a small tiny openingor pinhole of a container? If a gas container has a tiny opening, gas molecules inside the container graduallyescape from the container. Since gas molecules are in constant motion, they randomlystrike the tiny opening and pass through the pinholes. This behavior of gas is calledeffusion. Effusion can be utilized to determine the leakage of a certain gas container like theL.P.G. tank that we use in cooking. In 1829, a Scottish chemist named Thomas Graham experimented with the diffusionand effusion of gases. He discovered that gas molecules with high velocity diffuse or effusefaster than molecules with low velocity. But what determines the velocity of gas molecules? Graham also discovered that when the temperature and pressure of gases are equal,their velocities are determined by the mass of their particles. He proposed the Graham’slaw of diffusion /effusion as follows: Under conditions of equal temperature and pressure, the rates of effusion or diffusionof two gases are inversely proportional to the square roots of their molar massesRA = MB Where: RA = rate of effusion/diffusion of gas ARB MA RB = rate of effusion/diffusion of gas B MA = molar mass of gas A MB = molar mass of gas B -6-

In fact, according to the Kinetic Molecular Theory of gases, the temperature of a gasdirectly affects the kinetic energy of its molecules. At constant temperature and pressure, the kinetic energies of gases are the sameand from the definition of kinetic energy we have, KEA = KEB ½ MARA2 = ½ MB RB2canceling the common ½ on both sides of the equations and regrouping thevariables; RA2 = MB RB2 MASimplifying,  2 R = MB A MA R BGetting the square root of both sides of the equations: R A = MB RB MA Put simply, Graham’s law of diffusion/effusion assumes that gas particles with alower molar mass (lighter one) diffuse or effuse faster than those having high molar mass(heavier gas).Application of Graham’s Law of Diffusion/Effusion of Gases: At 250C, the average speed of oxygen molecule is 482 m/s. Find the average speedof a hydrogen molecule at the same temperature. Which gas will diffuse faster using therelationship of relative rates of diffusion of gases to their molar masses? We know that H2has 2g/mole and O2 has 32g/mole. Since H2 has a lighter molar mass, does it mean thatH2 will diffuse faster?RH2 = M O2RO2 MH2 -7-

RH2 = 32 g/mole = 3.984RO2 2.016 g/moleRH2 = 4 (H2 diffuses 4 times as fast as O2 )RO2Solving for the rate of diffusion of H2 if rate of diffusion of O2 is 482 m/s;RH2 = 3.984 × 482m/s = 1.92 × 103 m/s Furthermore, based on the definition of rate of motion of gas particles (rate ofeffusion/diffusion) and the distance traveled by gas molecules per unit time (R=d/t),Graham’s law of diffusion/effusion can be simplified as:RA = MBRB MAdA / t A = MBdB /tB MAIf the distance traveled by the 2 gases are the same; dA = dB1 / t A = MB1/ tB MAFurthermoretB = MBtA MAWhere: tB = time it takes for gas A to diffuse/ effuse the same distance as gas B tA = time it takes for gas B to diffuse/ effuse the same distance as gas A Do you know that you can identify the molar mass of an unknown gas by using therelationship of time of effusion / diffusion of gases to their molar masses? Suppose that an experiment on effusion is performed to identify the molar mass of anunknown gas. It requires 45 seconds for a certain amount of an unknown gas to passthrough a pinhole into a vacuum under the same condition. On the other hand, it only takes26 seconds for the same amount of Argon (Ar) to effuse. With the given values, find themolar mass of the unknown gas. -8-

Use the relation of time of effusion and molar mass of gases:tu = Mut Ar MArMu = (45)2MAr (26)2Mu = 2.996MArM u = MAr × 2.996 = 39.95 g/mole × 2.996Molar mass of unknown gas = M u = 119.67 g / moleWhat you will doActivity 2.1 Problem Solving in Graham’s law1. Helium gas (a gas that enables the balloon to float in air) effuses 2.0 times faster than another gas. What is the molar mass of the other gas?2. Which of the following is most likely to be the unknown gas in problem no. 1?a. CH4 c. CO2b. NO2 d. NH4 Key to answers on page 17.Lesson 3. Practical Application of the Behavior of Gases Learning the behavior of gases based from its properties such as diffusibility,compressibility, expansibility and low density, this lesson enables us to see the applicationof the said behavior in our daily life. -9-

Do you know that altering the surface of the earth by covering it with concrete,asphalt and metal structures causes tremendous change in our atmosphere? Radiantenergy of the sun in the form of heat reaches the earth’s surface and is reflected back bythe concrete and asphalt covering of the earth. The reflected heat is absorbed by the airparticles causing them to become more mobile thus circulating hot air in our land. This is proven by one of our postulates in kinetic molecular theory of gases whichstates that absolute temperature of gas substances is directly proportional to its averagetranslational kinetic energy. Likewise, the increase in the kinetic energy of gas particlesincreases the diffusion rate of even the polluted gases in the air. Figure 3.1 When a fuel is burned in the cylinder of a machine, it produces the gaseous product of combustion like CO2, H2O and sometimes CO and unburned fuel, if the combustion is incomplete.What you will doActivity 3.1 Paper Turbine (Expansion and Convection of Air)Materials: Paper or Cardboard Match Small Burner Stick Scissor ThreadProcedure:1. Cut a circle from a piece of cardboard. Cut it further Figure Aspirally. (Figure A)2. Tie a thread to the center of the spiral and suspend it Figure B above the flame of a burner. (Be careful not to burn it).(Figure B)3. What happened to the spiral board?______________________________________________________________________4. Lift the spiral board up and down over the flame.a. What do you notice? ________________________________________________b. Why? ____________________________________________________________ - 10 -

Through this experiment we can create a simple turbine. We all know that a “steamturbine” operates with the use of vapor from boiling water, but this time we will make thepaper spiral spin with the use of convection current of the hot air coming from the flame.The lighted lamp produces heat, which heats up the particles of the air around it. The airparticles that gets heated expands, therefore decreasing its density. In a mixture of warmand cold air, the warm air will rise forcing the cold air downward. Then the convectioncurrent in air occurs, which makes the spiral spin. In cooking, particularly in tenderizing meat faster, a pressure cooker is used. Theincrease in pressure is due to the air trapped in the cooking device through theaccumulation of steam. This delays the boiling point of the liquid in the pressure cooker,which in turn allows it to boil at a temperature higher than 100oC. Thus, the food absorbsmore heat causing it to become soft quickly. Figure 3.2 Pressure cooker increases the boiling point of the water solution in it by building up the pressure of the air trapped from the evaporated steam in the cooking device. Another technology that uses the principle of air convection is the exhaust fan. Anexhaust fan should be placed at the higher portions of the wall in order to maximize itspurpose. It aims to release hot air from a kitchen for proper ventilation. From the propertiesof gases, light dense gas like hot air floats and more dense gas like cold air sinks. As hottergases float in the air, the fan causes the colder air to circulate in the room by sucking hot airout of the room. The inflation of the tire with air keeps its shaperound. Air in the tires allows the vehicle to make a turnalong a sharp curve without being deformed. Itcushions the ride of the passengers, especially whenthe car is passing through rough roads or pavements. Figure 3.3 The pressure of the air inside the tire causes it to be always in good shape for a comfortable ride.- 11 -

The differences in the densities of gases like Figure 3.4those between air and helium enable the production A weather balloon in flightof a weather balloon. Helium, which is less densethan air, is commonly used to inflate weatherballoons. These types of balloons are used mainly inthe measurement and evaluation of upperatmospheric conditions. Weather information may begathered during the vertical ascent of the balloonthrough the atmosphere or during its motions once ithas reached a maximum altitude, which ispredetermined. Atmospheric pressure, temperature,and humidity information may be sent by radio from aballoon. Monitoring of its movement providesinformation about winds at its flight level. Gases are also involved in atmospheric pollution. Numerous invisible cycles of naturehappen in our surrounding everyday. Majority of these cycles involve gaseous substances.They even change our climate naturally. A tremendous change in our climate occurredsince the start of the industrial revolution in the 18th century. The changing agricultural andindustrial practices in the industrial revolution began altering our climate and environment. Before the industrial revolution, human activity releases very few gases into theatmosphere, but now because of population growth, fossil fuel burning and deforestation,we are affecting the mixture of gases in the atmosphere. Carbon dioxide (CO2) is releasedto the atmosphere when solid waste like garbage, fossil fuel such as oil, natural gas andcoal are burned. Don’t you know that every 1-gram of fossil fuel burned releases about 3 grams ofCO2 into the atmosphere? Plants in photosynthesis use parts of this CO2, but deforestationlessens the accumulation of CO2. Human activity has altered the chemical composition of the atmosphere through thebuild up of greenhouse gases primarily carbon dioxide (CO2), methane (CH4) and nitrousoxide. These gases together with water vapor create the natural greenhouse effect. Theability of these gases to absorb infrared radiation (heat) warms the earth. In this way, CO2and H2O act as an insulating blanket to prevent heat from escaping into outer space, whichis often referred to as the greenhouse effect.- 12 -

(a) (b)Exhaust gas of automobile contributes Deforestation and burning of forests lessen the users of polluted gas, CO2 to air pollutionFigure 5. Causes of air pollution What you will do Self-Test 3.1From the behavior and properties of gases, identify what principle is involved in eachsituation: 1. People are warned not to dispose empty aerosols or spray cans in an incinerator or flame because these might explode. 2. The volume of the floating balloon in the air increases as it ascends upward. 3. A basketball player trained at sea level experiences hardship of breathing during the first few hours of playing in an elevated place like Baguio. 4. Food is cooked in a shorter time using the pressure cooker. 5. It is advisable not to increase the tire pressure to its full limit when driving during summertime. Key to answers on page 17. - 13 -

Let‘s SummarizeYou have learned that: 1. According to Jan Baptiste Van Helmont, gases are a disordered, formless type of matter. 2. The kinetic molecular theory (KMT) of gases describes the behavior of molecules of gases. Through the KMT of gases, we can explain logically what happens at the molecular level that results to the changes that we observe in the macroscopic level. 3. The Kinetic Molecular Theory of gases is based on the following assumptions:  gas molecules are separated by distances bigger than their own dimensions, in which we can consider the total volume of the gas molecules to be negligible,  they are in constant motion and they frequently collide with one another and with the walls of the container without losing their kinetic energy,  the molecules neither attract nor repel one another. 4. The forces of attraction between liquid and solid molecules are much greater than those between gas molecules. As a result, the molecules of a liquid are very close to each other giving the liquid a definite volume but no specific shape. 5. The particles of a solid are close together and are held in a fixed position resulting to the definite volume of solid. 6. Diffusion is the gradual mixing of two gases due to the spontaneous random movement of gas particles while effusion is the behavior of gas particles to pass through a very small opening of its container. 7. Graham’s law of diffusion/effusion states that at a constant temperature and pressure, the rates of effusion or diffusion of two gases are inversely proportional to the square roots of their molar masses. - 14 -

PosttestMultiple Choice. Choose the letter of the best answer. Write the chosen letter on aseparate sheet of paper.1. A gas confined in a closed container is cooled. Which of the following does NOT happen to the gas? a. The gas molecules move slower. b. The gas molecules collide with the wall of the container less frequently. c. The gas molecules settle to the bottom of the container. d. The gas molecules collide with each other less frequently.2. As gas molecules collide, what happens?a. They stick together. c. Their kinetic energies are conserved.b. They lose their kinetic energies. d. They will explode.3. The gradual mixing of two gases due to the spontaneous, random motion of the gasparticles is calleda. effusion c. diffusionb. cohesion d. viscosity4. Which property shows that gas molecules are always moving?a. Have a high density c. Have a high rate of diffusion /effusionb. Easily compressible d. Have a definite shape and volume5. Which of the following proves that gas diffuses? a. Toxic chemicals in cigarette smoke spread in a car. b. The sweet fragrance of perfume spreads in a room. c. The CFC from aerosol spray reaches the upper atmosphere and destroys the ozone layer. d. All of the above6. Which statement about gas is NOT true? a. It is incompressible. b. It expands when heated. c. It has lesser density than solid and liquid. d. It has no definite shape but has definite volume.7. When air is added into an automobile tire, which of the following does NOT happen? a. The gas density increases. b. The gas molecules move faster. c. The gas molecules collide more frequently. d. The space between the molecules decreases. - 15 -

8. The diffusion process of a gas takes a relatively long period of time to complete because ____________________. a. gas molecules are very near each other. b. gas molecules are crowded because of high density. c. gas molecules have a relatively low average kinetic energy. d. gas molecules experience numerous collisions while diffusing.9. According to Graham’s law of diffusion, the rates of diffusion of gases are ___________ to the square roots of their molar masses . a. not directly related b. directly proportional c. inversely proportional d. related squarely proportional10. The average kinetic energy of gas molecules is measured by their ___________.a. pressure c. volumeb. temperature d. densityII. From the behavior and properties of gases, identify what principle is involved in each situation:1. It is advisable not to remove the tab of a can of a warm soft drink after shaking it vigorously.2. Fully inflated balloons should not be exposed to extreme heat like sunlight because it might explode.3. The freshness of flowers reaches our nose.4. Helium gas is used to inflate balloons.5. The light of a candle will be extinguished when it is placed under an inverted jar or bottle. Key to answers on page 18.Key to AnswersPretest 6. a 7. c 1. c 8. a 2. a 9. b 3. c 10. a 4. b 5. b - 16 -

Lesson 1Activity 1.13. The plastic blew-up.4. The plastic bag returned to its original size and shape.Self-Test 1.11. independent 6. motion2. great or large 7. elastic3. collision 8. rises ; lowers4. force of attraction 9. kinetic energy ; directly5. randomly 10. diffusibilityLesson 2Activity 2.11. MHC = 4 g /mole RHe = 2RU MU = ?RHe 2 = MU RU MHe  2MU = MHe 2R = 4 g/mole × 4 U R UMU = 16 g/mole2. (a) CH4Lesson 3Self-Test 3.11. expansibility2. expansibility3. compressibility4. expansibility5. expansibility - 17 -

Posttest 6. d II. 7. b 1. expansibility I. 8. d 2. expansibility 1. c 9. c 3. diffusibility 2. c 10. b 4. low-density 3. c 5. compressibilty 4. c 5. dReferencesBooks:Araneta, F.L., Catris, L.V. & Deauna, M.C. (2002). The world of chemistry III. (2nd ed.) Quezon City: SIBS Publishing House, Inc.Chang, R. (2005). Chemistry. (8th ed.) New York: Mc Graw-Hill Companies.Magno, M.C., Tan, M.C. & Punzalan, A.G. (2000). Science and technology for a better life (3rd ed.) Manila: Diwa Scholastic Press Inc.Masterton, W.L. & Hurley, C.N. (1993). Chemistry principles and reactions. (2nd ed.) USA: Saunders College Publishing.Regalado, E.C. (2001). Chemistry. Manila: St. Augustine Publications,Inc.Sherman, S.J. & Sherman, A. (1999). Essential concepts of chemistry. New York: Houghton Muffin Company. - 18 -

Module 9 Gas Laws What this module is about Every time we breathe through our lungs, pump air into a tire, blow up soap bubbles,or use a spray can, we are depending on gases to work in a predictable way. Have youever wondered why gases act the way they do? This module will introduce you to the properties of gases and the four variables weare concerned with when dealing with gases. These variables are temperature (T),pressure (P), volume (V), and the quantity of a gas in moles (n). Equations that express therelationships between these four variables are called gas laws. These concepts as well asthe analytical skills needed to understand them will be developed as you try the activitiesand read the discussions presented in the four lessons of this module.  Lesson 1 – Boyle’s Law  Lesson 2 – Charles’ Law  Lesson 3 – Combined Gas Law and Gay-Lussac’s Law  Lesson 4 – Ideal Gas Law What you are expected to learnAfter going through this module, you should be able to: 1. state the meaning associated with the characteristic properties of gases: volume, pressure, temperature and amount; 2. recognize the following symbols used to describe gases: n, P, R, T, V and STP; 3. explain the interrelationships of pressure, volume and temperature of gases; 4. solve problems involving changes in the condition of the gas using the equations for Boyle’s Law, Charles’ Law, Combined Gas Law, Gay-Lussac’s Law and Ideal Gas Law; 5. describe the properties of an ideal gas and recognize the conditions at which gases behave ideally; and 6. describe how gas density is related to pressure, temperature and number of moles by applying the ideal gas law.

How to learn from this module1. Familiarize yourself with the objectives of the module. Take the Pretest.2. Read the Introduction to the lessons, beginning with Lesson 1. Think of the skills that will help you understand the lesson. If necessary, briefly review the concepts that you must know beforehand3. Read and understand the Discussion. From time to time, ask yourself if you are able to acquire the concepts and skills listed in the objectives.4. Perform the Activities. Record your observations and answer the guide questions.5. The self-tests are provided to give ample practice in problem solving.6. Do the Self-test to assess your understanding of the topics covered.7. Repeat Steps 2 to 5 for the remaining lessons. Refresh your memory by reading the Summary of all the lessons.8. Take the Posttest to see how well you understood the concepts in the module.What to do before (Pretest)Multiple Choice. Choose the letter of the best answer. Write the chosen letter on aseparate sheet of paper.1. Who was the English scientist who made accurate observations on how pressure andvolume are related?a. Boyle c. Combineb. Charles d. Gay-Lussac2. When pressure on a gas goes down what happens to its volume?a. rises c. stays the sameb. goes down d. rises, then falls3. In the equation for Boyle’s Law, P1 stands for:a. new pressure c. difference in pressureb. original pressure d. standard pressure, 1 atm4. As the volume of a gas goes up, what happens to the temperature of the gas?a. goes up c. stays the sameb. goes down d. goes down, then up5. The temperature and volume of a gas are directly related. This is a statement of:a. Boyle's Law c. Combined Gas Lawb. Charles' Law d. the Ideal Gas Law -2-

6. Which equation gives the correct relationship between volume and temperature of agas?a. T1 V1 = T2 V2 c. T2 V1 = T1 V2b. T1 V2 = T2 V1 d. T2 V2 = T1 V1.7. What is the standard temperature in Kelvin?a. 0 Kelvin c. 273 Kelvinb. 25 Kelvin d. 373 Kelvin8. The greater the ________ of a gas at a given temperature, the smaller its volume.a. mass c. pressureb. diffusion d. expansion9. In the ideal gas equation, the universal gas constant R has the value of _______L . atm/ mol . K.a. 0.008206 c. 0.820600b. 0.082060 d. 8.20600010. The molar volume of a gas is the volume occupied by 1 mole of a gas at:a. 1oC and 1 atm c. 1 K and 0 atmb. 273oC and 0 atm d. 273 K and 1 atm11. Gases having molecules that behave as predicted by the gas laws are called:a. ideal gases c. rare gasesb. noble gases d. real gases12. A toy balloon left under the sun at noontime is likely to expand and burst. This illustrates:a. Boyles’ Law c. Gay-Lussac’s Lawb. Charles’ Law d. Avogradro’s Law13. At standard temperature and pressure, what is the volume occupied by a mole of anygas?a. 0.224 L c. 22.4 Lb. 2.24 mL d. 22.4 mL14. What happens to the density of a gas when the temperature is increased at constantpressure?a. decreases c. remains the sameb. increases d. cannot be determined15. “The density of a gas decreases with increasing temperature at constant pressure.” This statement: a. is true b. is false c. does not apply to ideal gases d. can neither be scientifically proven nor disproved -3-

16. To expand a 15 L sample of gas to 20 L, it is necessary to:a. increase the pressure c. increase the temperatureb. decrease the number of moles d. decrease the temperature17. A sample of carbon dioxide occupies 3 L at 35ºC and 1 atm. What will happen to itsvolume if its condition is changed to 48ºC and 1.5 atm?a. The volume will decrease. c. The volume will remain the same.b. The volume will increase. d. The volume cannot be determined.18. The J-shaped glass tube containing mercury and a sample of trapped gas was used by________ to study the effect of pressure on volume.a. Boyle c. Avogadrob. Charles d. Gay-Lussac19. Gases behave non-ideally under which of the following conditions? a. at high pressure and low temperature b. at low pressure and low temperature c. near the boiling point of water d. ideal gases can never exhibit non-ideal behavior20. A graph showing the variation of volume (vertical axis) with pressure at constant temperature would look like: a. b. c. d.Error! Key to answers on page 20.Lesson 1. Boyle’s LawIntroduction Air is all around us. We breathe in the air so that our body can receive adequatesupply of oxygen gas. Our lungs expand as they fill with air and take in oxygen, and relaxas they release carbon dioxide. Plants in turn, use up the carbon dioxide during the processof photosynthesis to manufacture sugars. Life as we know it would not have been possiblewithout the life-sustaining gases found in the atmosphere Like breathing, many other human activities involve gases. When air is pumped into -4-

a bicycle or automobile tire, a mixture of gases is compressed into a small volume. Heliumgas make toy balloons float. Gas used to fill rubber lifeboats and vests exerts pressure onits containers, giving them rigidity and shape. For centuries now, scientists are curious about how gases behave. Investigations onthe behavior of gases mainly concern the relationship among the four important propertiesof gases: volume, pressure, temperature and amount in moles. This lesson introduces therelationship between volume and pressure at constant temperature, which is also known asBoyle’s Law. What you will do Activity 1.1Guess the words described below. Clues are given in the jumbled letters in the balloons.1. It is an instrument consisting of a mercury- 2 1 5 filled tube inverted in a dish of mercury. MEPR Torricelli invented it and used it to measure SPER RME ERTE the pressure of a gas. USE RABT AUT2. This is the most easily measured gas R OE property defined as the force exerted upon a 4 unit area of a surface. 3 EU VML3. It is a more fundamental temperature scale LVN O than Celsius or Fahrenheit EK4. Related to mass and density, it is the total I space occupied by an object.5. This is what you measure when you want to know the degree of hotness or coldness of an object. It is defined as the average kinetic energy possessed by a sample of matter. Key to answers on page 21. What you will do Activity 1.2Fill in the table with commonly used units for the temperature, pressure and volume of agas. -5-

PHYSICAL QUANTITIES UNITSTemperature Key to answers on page 21.PressureVolumeDiscussion One of the first scientists to study the behavior of gases was Robert Boyle. In 1661, he made a device using a J-shaped glass tube containing mercury and a sample of trapped gas similar to the figure below. He observed that the volume of the trapped gas decreased in proportion to the pressure exerted by the addition of more mercury. Figure 1.1 Robert Boyle http://www.woodrow.org/teachers/chemistry/ institutes/1992/BOYLE.GIF Figure 1.2 Boyle’s J-shaped device -6-

Checkpoint What is the relationship between the different units of pressure? Answer: Measurements of pressure can be easily interconverted from one unit to another. Just use the following relationships: 1 atmosphere = 760 mm of Hg = 760 torricellis = 101,325 Pascals. Boyle’s Law states that if the temperature is heldconstant, the volume of a given amount of gas is inverselyproportional to its pressure, V α 1/P. Mathematically, Boyle’sLaw is expressed as: P1V1 = P2V2. The subscript of 1 refers tothe original conditions while 2 refers to the new conditions. Thefigure on the right shows what happens to the volume of asample of gas when pressure is increased while maintaining thetemperature. Note the inverse relationship of pressure andvolume. Figure 1.3 Illustration of Boyle’s Law Checkpoint Explain in terms of Boyle’s law what happens when you alternately squeeze and release a hollow rubber ball. Answer: When you squeeze a hollow rubber ball, the volume decreases and the pressure within the ball increases. When the squeezing ceases, the volume increases and the pressure decreases within the ball. Are you ready to experience Boyle’s Law in action? Try these activities on your ownor together with a friend. -7-

What you will do Activity 1.3 In this activity, you will demonstrate Boyle’s Law using simple materials. You willneed several small marshmallows and a plastic syringe with a diameter large enough to fitthe marshmallows. You will also need the plastic cap but not the needle of the syringe forthis. Remove the plunger of the syringe and put the marshmallows inside. Return theplunger allowing only a small space for the marshmallows. Place the cap tightly (you maywant to use wax to seal it). Slowly pull the plunger away and see how the marshmallowsmagically expand! They will return to the original size if you release the plunger. Can youexplain these observations in terms of Boyle’s Law? Key to answers on page 21. Try these Self-Test questions to check how well you understood Lesson 1. What you will do Self-Test 1.1Directions: Read each item carefully and supply the required information.1. A certain model of a car has gas-filled shock absorbers to make the car run smoother and less “bumpy.” Describe the gases inside the shock absorbers when the car is full of passengers compared to when the car is empty.2. Which graph demonstrates Boyle’s law? Vertical axis is V and horizontal axis is P a. b. c. d.3. If the pressure on a gas is decreased by one-half, what will happen to its volume?4. A 40 L balloon is filled with gas at 4 atm. What will be its new volume at standard pressure of 1 atm?5. A gas at 30.0°C occupies 500 mL at a pressure of 1.00 atm. What will be its volume at a pressure of 2.50 atm? Key to answers on page 21. -8-

Lesson 2. Charles’ LawIntroduction In this lesson, we will investigate Charles' Law, which relates changes in thetemperature of a confined gas kept at a constant pressure to the volume of the gas. Youwill be introduced to another equation that determines the variation of gas volume withchange in temperature.Discussion Jacques Charles was a French chemist famousfor his experiments in ballooning. Instead of hot air, heused hydrogen gas to fill balloons that could stay afloatlonger and travel farther. Figure 2.1 Jacques Charles web.fccj.org/~ethall/ gaslaw/gaslaw.htm Charles’ Law states that for a given amount ofgas at constant pressure, the volume is directlyproportional to the temperature in Kelvin, V α T.Charles Law is expressed in equation form as: T1V2 =T2V1. The subscript of 1 refers to the originalconditions while 2 refers to the new conditions.Temperatures should be expressed in Kelvin beforesubstituting the values in the equation. The figure onthe right shows what happens to the volume of asample of gas when temperature is increased atconstant pressure. Note the direct relationship oftemperature and volume. Figure 2.2 Illustration of Charles’ Law-9-

Checkpoint What happens when a 450 mL sample of a gas kept at constant pressure is cooled from 60ºC to 20ºC? Answer: It decreases in volume. Identify the given data and the unknown in the problem. To apply Charles Law, convert first the temperature to Kelvin scale by adding 273. Rearrange the equation to isolate V2, arriving at V2 = T2V1 / T1. If we substitute the given values, we get 396 mL as a result. Are you ready to experience Charles’ Law in action? Try these activities on your ownor together with a friend. What you will do Activity 2.1For this activity, you will need 2 small basins and 3 toy balloons inflated to approximately thesame size. Fill the first basin with warm water and the second with cold water. Immersethe two balloons in the separate basins. Leave the third balloon untouched. 1. Compare the balloons after 1 and 3 minutes. Which balloon expanded and which one shrunk? 2. Remove the balloons from the basin and leave at room temperature for 10 minutes. What did you observe? Key to answers on page 21. What you will do Activity 2.2For this activity, you will need to take a trip to the supermarket (or your mother’s kitchen) toobtain any previously unopened jar of preserved food (examples are pickles, sandwichspread, and “Gerber” baby food) with a metal lid indicating a \"freshness button.\" 1. Open the metal lid slowly and listen to the popping sound. What happens to the “freshness button”? What caused this sound? 2. If you put back the cover, will the button return to the “down” position? How was the button fixed in place during packing of the jar’s contents? Key to answers on page 21. - 10 -

Try these Self-Test questions to check how well you understood Lesson 2. What you will do Self-Test 2.1Directions: Read each item carefully and supply the required information.1. For Charles’ Law to apply, the gas must be kept at constant _________.2. Which graph demonstrates Charles’ law? Vertical axis is V and horizontal axis is T. a. b. c. d. Key to answers on page 22. What you will do Self-Test 2.2Directions: Read each item carefully and supply the required information.1. A balloon was inflated to a volume of 2.5 L at 11 am when the temperature is 30°C. At 9 pm, the temperature fell to 10°C. What will be the volume of the balloon if the pressure remains constant?2. A sample of 50.0 L of nitrogen at 20°C is compressed to 5.0 L. What must the new temperature (in Kelvin) be to maintain constant pressure?3. Compute the decrease in temperature when 2.0 L at 280 K is compressed to 1.5 L. Key to answers on page 22. - 11 -

Lesson 3. Combined Gas Law and Gay-Lussac’s LawIntroduction The volume of a gas is greatly affected by changes in pressure and temperature,hence temperature and pressure at the time of measurement must always be specified. Inthis lesson, you will learn more about the Combined Gas Law, which connects the variablespressure, temperature and volume of gas. Also, you will learn about Gay-Lussac's Law,which describes how changing the temperature of a gas that is kept at constant volumeaffects the pressure of the gas.Discussion Because pressure and temperature will change from day to day and from location tolocation, it is common to use more than one of the gas laws to determine the resultingvolume of the gas. If we combine the relationships expressing Boyle’s Law, V α 1 / P, andCharles’ Law, V α T / P, we obtain the relationship: V α T / P. The formula for the CombinedGas Law can be expressed as: P1V1T2 = P2V2T1 , where the subscript of 1 refers to theoriginal conditions while 2 refers to the new conditions. Checkpoint A 350 mL sample of argon gas is collected at 295 K and 99.3 kPa. What volume would this gas occupy at STP? STP means standard temperature and pressure (1 atm and 273 K). Answer: 317.5 mL. First, analyze the given data and the unknown. We manipulate the equation for Combined Gas Law to isolate the unknown variable V2: V2 = V1T2P1 / P2T1. Now, substituting the values into the equation gives us the new volume, 317.5 mL. Gay-Lussac studied many reactions involving gases andgeneralized that at constant temperature and pressure, thevolumes of gases can be expressed as a ratio of small wholenumbers. For example, at 1 atm and 100ºC, one volume ofoxygen gas combines with two volumes of hydrogen gas to givetwo volumes of water vapor, also at 1 atm and 100ºC. Thisexample illustrates the Law of Combining Volumes. Figure 3.1 Gay-Lussac http://www.ac-nancy-metz.fr/pres- etab/lapicque/Opinfo00/Genin/GayLussac.jpg - 12 -

Figure 3.2 Gay-Lussac’s Law Gay-Lussac’s Law states that if the temperature is held constant, the volume of agiven amount of gas is inversely proportional to its pressure. This relationship is expressedby the equation: P1T1 = P2T2. What you will do Activity 3.1 Blow up a small balloon with as much air as possible without bursting it. Tie a secureknot to ensure that no air escapes. Leave the balloon at room temperature for an hour orso. Then place the balloon in hot water and check the tension of the balloon's skin forseveral minutes. Make a scale from 1-5 (1 = very low tension and 5 = very high tension) torate the tension of the balloon's skin. Can you explain your results? Key to answers on page 22. Try these Self-Test questions to check how well you understood Lesson 3. What you will do Self-Test 3.1Directions: Read the item carefully and supply the required information. A 5000 mL container is filled with helium gas to a pressure of 3.0 atm at 250ºC. Approximately how many toy balloons at STP can be filled by helium from this container, assuming each balloon can contain 1 L? Recall that STP means standard temperature and pressure (1 atm and 273 K). Key to answers on page 22. - 13 -

What you will do Self-Test 3.2Directions: Read each item carefully and supply the required information.1. A bicycle tire was inflated to a pressure of 3.74 atm during early morning when the temperature was 15ºC. At noontime, the temperature rose to 35ºC. What was the resulting pressure in the tire (assuming that its volume did not change)?2. Helium-filled balloons can be used to carry weather instruments high into the atmosphere. Before launching, a certain balloon has a volume of 1.0 x 106 L at 22.5 C and 754 mm Hg. What will happen to the balloon when it reaches the height of 30 km, where the pressure is 76.0 mm Hg and the temperature is 240 K? Key to answers on page 22.Lesson 4. Ideal Gas LawIntroduction Certain hair spray products are packaged in aerosol cans. Serious accidents couldoccur if you throw the empty cans into the fire because the pressurized gas could explode.These cans usually have a printed warning such as “Do not puncture or incinerate” butusers seldom read what is on the label. A gas that behaves exactly as described by the gas laws is called an ideal gas.Many gases, especially at high pressure or low temperatures do not behave quite ideally,hence they are called real gases. This lesson applies the gas laws covered in Lessons 1and 2 to derive the Ideal Gas Law and analyze how it governs ideal gas behavior.Discussion If we combine the relationships expressing Boyle’s Law, V α 1 / P,Charles’ Law, VαT - 14 -

and the proportionality Vα nwhere (n stands for the number of moles of gas), we obtain the relationship: V α nT / P.By introducing a constant, this relationship can be expressed as the equation V = RnT/P,and further simplified to PV = nRTwhere R is called the universal gas constant.CheckpointIf you plug in the standard units of pressure, volume, amountand temperature of gas in the equation, what is the resulting unitof R?Answer: The unit of R is liter atmosphere/mole Kelvin. Thevalue of R if atmosphere is used as the unit of pressure is0.08206 L . atm/mol K. If kilopascal is used as the unit ofpressure, R becomes 8.314 L kPa/mol K. Notice that if you know the values of any three characteristic properties of gas, youcan easily calculate the fourth property. Remember that temperatures should be expressedin Kelvin before substituting the values in the equation.CheckpointWhat is the volume of 1 mole of an ideal gas at STP?Answer: 22.4 L. Substitute the values for STP (1 atm and 273K) and see for yourself. 22.4 L is the volume of a mole of anygas at standard conditions. This is called the molar volume ofan ideal gas. A chemist named Amadeo Avogadro studied the properties of many different gases.In 1811, he concluded that equal volumes of all gases, when measured under the sameconditions of temperature and pressure, always contain the same number of molecules.This relationship, V α n, is known as the Avogadro’s Law, which you encountered at thebeginning of the lesson. It is very useful because it tells us that one mole of an ideal gas (itdoesn’t matter which type of gas!) at STP will always occupy 22.4 L. - 15 -

Checkpoint Is the following statement true or false? “The density of a gas varies with temperature and pressure.” Answer: True. Notice that the ideal gas equation can be rearranged as: n/V = RT/P. n is number of moles, obtained by dividing mass in grams by the molar mass, M. Substituting m/M. gives us the equation m/V = PM/ RT. Obviously, m/V is mass divided by volume, also known as density. Thus, density is directly proportional to P and inversely proportional to T. As implied in the equation above, the density of the gas is also proportional to itsmolar mass. Thus, He gas (molar mass = 4 g/mol) is less dense than O2 gas (molar mass =32 g/mol). Here are some useful tips: When three of the variables P, V, n, and T are known andyou are looking for the value of the fourth variable, use the ideal gas equation. When youhave only one gas sample in one set of conditions and you are asked to determine thechange in one variable under a new set of conditions, just use the Combined Gas Law.What you will doActivity 4.1Complete the following table:Data on a sample of oxygen gas.n? T V? ? 0.082060.00625 1.0 atm 293 K mol Key to answers on page 22.Try these Self-test questions to check how well you understood Lesson 4. - 16 -

What you will doSelf-Test 4.1Directions: The following list contains the key terms discussed from Lesson 1 to Lesson 4.Briefly explain the concepts associated with each term.1. pressure, P ____________________________________2. volume, V ____________________________________3. temperature, T ____________________________________4. Boyle’s Law ____________________________________5. Charles Law ____________________________________6. Combined Gas Law ____________________________________7. Gay-Lussac’s Law ____________________________________8. standard temperature and pressure, STP _____________________9. gas density ____________________________________10. ideal gas constant, R ____________________________________ Key to answers on page 22. Let’s Summarize1. Gases exhibit definite behaviors that may be identified and analyzed.2. Standard Temperature and Pressure, STP, refers to accepted experimental conditions for the study of gases, 1 atm and 273 K.3. Boyle’s Law demonstrates that at constant temperature, gas volume varies inversely with pressure. P1V1 = P2V24. Charles’ Law predicts that at constant pressure, gas volume varies directly with temperature changes. V1T2 = V2T15. Temperature, pressure and volume are all interrelated and several changes may occur simultaneously and can be determined using the Combined Gas Law. P1V1 = P2 V2 T1 T26. Gay-Lussac’s Law shows that at constant volume, the pressure of a gas is proportional to its temperature. - 17 -