1 atm 760 mmhg =.454 atm (3 points/6) 14.7 psi =.816 atm (3 points/9)

Transcription

1 Chapter 5 Homework acket 1. Gases generally have a) low density b) high density c) closely packed particles d) no increase in volume when temperature is increased e) no decrease in volume when pressure is increased Your brain should contain an instantly accessible visual of particles being far apart in gases and much closer together in liquids and solids this visual should also come with the instant recognition that this depends on interparticle forces and that you are aware of the properties of particles that impact interparticle forces. Answer A (1 point1/1). ressure is a) defined as the mass that an object exerts when at rest b) measured in Newtons c) defined as the number of moles of substance divided by the mass of the substance d) defined as the force per unit area e) measured in grams ressure is defined as force per unit of area you don t need to memorize this it just makes sense that pressure is a push that is spread out over a certain area. This makes it easy to remember what the SI units are as well unit of force is Newtons, unit of area is meters squared (N/m ). This unit is also called a pascal. Along with this knowledge, you should also have immediately available to you in your brain that we mainly use atmospheres in our calculations and 1 atm is about 100,000 a (or N/m ), and 760 torr (or mmhg) Answer D (1 point/) 3. The SI unit of pressure is the a) ampere b) kilojoule c) newton d) gram e) pascal See discussion for answer questio. Answer E (1 point/3) 4. Convert the following into atmospheres which represents the greatest pressure? As with any conversion, start by writing out what you have been given, (already done in these problems), and ask what it is you need to get rid of, placing the units for this on the opposite side of the dividing line. Then ask what units you are trying to create, placing these units on the same side of the dividing line. Then add in equivalent values for the units written down. Do this as many times as you need to to get from the given value to what you want. Then perform the multiplications and divisions necessary to complete the calculation. a) 345 mmhg x 1 atm 760 mmhg.454 atm (3 points/6) b) 1.0 psi x 1 atm 14.7 psi.816 atm (3 points/9)

2 Chapter Error! Unknown document property name.: Error! Unknown document property name. 95 c) a x 1 atm 10135a.5771 atm (3 points/1) d) 16.9 in Hg x 5.4mm x 1 atm 1in 760 mm.565 atm (3 points/15) 5. A glass column is filled with mercury and inverted in a pool of mercury. The mercury column stabilizes at a height of 75 mm above the pool of mercury. What is the pressure of the atmosphere in atmospheres? This is a simple conversion. Be aware that this is the classic method of making a barometer to measure atmospheric pressure. 75 mm Hg x 1 atm 760 mm.989 atm (3 points/18) 6. The local weather forecaster reports that the current barometric pressure is 8. inches of mercury. What is the current pressure in atmospheres? Write down the given value and create the conversion factor by getting rid of inches of Hg and creating atmospheres, knowing that 1 atm is equivalent to 9.9 inches of mercury. Check your answer by knowing that the given value, 8. in Hg, is only a little less tha9.9 atm of mercury, which represents 1 atm. 8. in Hg 1 atm.943 atm (3 points/1) 9.9 in Hg 7. A physics experiment is conducted at a pressure of 14.0 ka. What is this pressure in mmhg? Write down the given value and create the conversion factor by getting rid of ka and creating mm Hg, knowing that both ka and 760 mm Hg are equivalent to 1 atm. Check your answer by knowing that the given value, 14 ka, is about.15 atm, and your answer of 105 mm Hg, is also about 105/760 mm, or about.15 atm. 760 mm Hg 14.0 ka 105 mm Hg (3 points/4) ka 8. The air pressure in the inner tube of a tire on a typical racing bike is held at a pressure of about 117 psi. Convert this pressure to atm. Write down the given value and create the conversion factor by getting rid of psi and creating atm, knowing. Check your answer by knowing that the given value, is between 5 and 10 times the value of 14.7, meaning that the answer should be between 5 and 10 atm. 117 psi 1 atm psi 9. Boyle's law states that: 7.96 atm (3 points/7) a) Equal amounts of gases occupy the same volume at constant temperature and pressure. b) The volume of a fixed amount of gas is inversely proportional to its pressure at constant temperature. c) The volume of a fixed amount of gas is directly proportional to its temperature in Kelvin at constant pressure.

3 Chapter Error! Unknown document property name.: Error! Unknown document property name. 96 d) The total pressure of a mixture of gases is the simple sum of the partial pressure of all of the gaseous compounds. e) The rates of effusion of gases are inversely proportional to the square roots of their molar masses. You should be able to remind yourself within a few seconds that the main laws relate everything else to volume, Avogadro relates n to V; Boyle relates to V; Charles relates T to V ABC-nT the beginning letters of both names and variables are alphabetical. Once you remember that, you intuitively know that increasing pressure decreases volume so the relationship is inverse. Be aware, all of the statement listed are true statuements-only (b) describes the asked for relationship Answer B (1 point/8) 10. Avogadro's law states that: a) Equal amounts of gases occupy the same volume at constant temperature and pressure. b) The volume of a fixed amount of gas is inversely proportional to its pressure at constant temperature. c) The volume of a fixed amount of gas is directly proportional to its temperature in Kelvin at constant pressure. d) The total pressure of a mixture of gases is the simple sum of the partial pressure of all of the gaseous compounds. e) The rates of effusion of gases are inversely proportional to the square roots of their molar masses. See discussion for question 9. Only answer (a) relates moles to volume and so is Avogadro s law. Answer A (1 point/9) 11. Charles's law states that: a) Equal amounts of gases occupy the same volume at constant temperature and pressure. b) The volume of a fixed amount of gas is inversely proportional to its pressure at constant temperature. c) The volume of a fixed amount of gas is directly proportional to its temperature in Kelvin at constant pressure. d) The total pressure of a mixture of gases is the simple sum of the partial pressure of all of the gaseous compounds. e) The rates of effusion of gases are inversely proportional to the square roots of their molar masses. See discussion for question 9. Only answer (c) relates temperature to volume and so is Charles Law. Answer C (1 point/30) 1. Consider a sample of helium gas in a container fitted with a piston, as pictured below. The piston is frictionless, but has a mass of 10.0 kg. How many of the following processes will cause the piston to move away from the base and decrease the pressure of the gas? Assume ideal behavior 10.0 kg I. heating the helium II. removing some of the helium from the container III. turning the container on its side IV. decreasing the pressure outside the container

4 Chapter Error! Unknown document property name.: Error! Unknown document property name. 97 a) 0 b) 1 c) d) 3 e) 4 I. Increasing the temperature will increase the volume and so cause the piston to move away from the base. However, after the volume finishes increasing, as the piston will no longer be moving, the forces on both sides of the piston will be balanced, so the pressure inside will balance the weight of the piston and plus that of the atmosphere, the same as it was doing before the heating and so the pressure will not have changed as a result of heating it will remain constant. So process I will not both cause the piston to move away from the base and decrease pressure. II. Removing some of the helium will decrease the number of particles and so, decrease the volume. This will cause the piston to move toward the base, not away from it. Additionally, after the helium is removed, as the piston will no longer be moving, the forces on both sides of the piston will be balanced, so the pressure inside will balance the weight of the piston and plus that of the atmosphere, the same as it was doing before the heating and so the pressure will not have changed as a result of loss of particles it will remain constant. III. Turning the container on its side will result in a decrease in pressure from the outside as the pressure inside will no longer be pushing against the force due to gravity acting on the piston. Because the pressure will momentarily be greater inside than outside the piston will move away from the base. After it stops, the pressures inside and out will be equal, but now the pressure insides is only the same as the atmospheric pressure on the outside, which is less than the atmospheric pressure plus pushing against the force due to gravity on the piston so, both conditions are satisfied. IV. Decreasing the pressure outside the container will accomplish the same thing as in process III, except that now it will be the atmospheric pressure component of the force that is decreasing rather than the force due to gravity on the piston. So, two processes Answer C (1 point/31) 13. A gas sample is held at constant pressure. The gas occupies 3.6 L of volume when the temperature is 1.6 C. Determine the temperature at which the volume of the gas is 3.45 L. This problem presents a comparison of conditions situation. Either write out 1 V and drop out constant values, or be able to visualize this relationship in your mind, cancel out constant values in your mind and write the remaining relationship. In this problem, pressure is stated to be constant and you are not given information about moles so you can assume it is also constant. You are asked for so isolate this and plug in the values. Write the isolation so you can see what the ratio of the changed variables does to the needed variable in this case, the ratio of V to is slightly less tha so you know that has to be slightly less than this helps you check you answer and also to find answers quickly on multiple choices without needing a calculator. Because the temperature we are multiplying the volume ratio by is the absolute (K) temperature, we must use the K scale. 1 V V T V T 3.45 L L 94.6K 81 K (3 points/34) 14. You have 30.4 g of O gas in a container with twice the volume as one with CO gas. The pressure and temperature of both containers are the same. Calculate the mass of carbon dioxide gas you have in the container. This is an comparison of conditions problem embedded in a mole to mass conversion problem. g (mol)(mm) so you need moles of CO to calculate the number of grams write this down.

5 Chapter Error! Unknown document property name.: Error! Unknown document property name. 98 You recognize that there are two different sets of conditions and that and T are constant so you can set the ratio of V/n for carbon dioxide equal to V/n for oxygen and isolate moles of carbon dioxide (you do not have to write out the whole relationship but do so if it guides you through the problem). You are not told the volumes of the gases but you are given the ratio (1 L CO for L of O ) so you can use this ratio. Also see that as V of the CO container is half the volume of the O container, it must have half the moles. Once you isolate n CO, plug in the values and solve. (4 points/38) gco (mol CO )(mm CO ) (.475mol)( ) 0.9 g 1 V V CO V V O CO n n CO n CO n 30.4g 1 O O V O 3.0g / mol.475 mol 15. Gaseous chlorine is held in two separate containers at identical temperature and pressure. The volume of container 1 is 1.30 L, and it contains 6.70 mol of the gas. The volume of container is.05 L. How many moles of the gas are in container? Because we are given information that describes more than one set of gas law data this is a comparison of conditions problem. and T are constant so you can set the ratio of / for the first situation equal to V / for the second situation and isolate moles of gas for the second situation (you do not have to write out the whole relationship but do so if it guides you through the problem). lug in the values and solve. Notice that the ratio of volumes is almost :1 and that the number of moles in the second situation is almost twice that of the first situation. (3 points/41) 1 V V n.05 L 6.70 mol 10.6 mol L 16. A balloon has a volume of.36 liters at 4.0 C. The balloon is heated to 48.0 C. Calculate the new volume of the balloon. Because we are given information that describes more than one set of gas law data this is a comparison of conditions problem. Either write out 1 V and drop out constant values, or be able to visualize this relationship in your mind, cancel out constant values in your mind and write the remaining relationship. In this problem you are not given information about or n so you can assume they are constant. You are asked for V so isolate this and plug in the values. Be careful don t think that the temperature has doubled here so that you can just double the volume you have to use the K scale. Write the isolation so you can see what the ratio of the changed variables does to the needed variable in this case, the ratio of to is only slightly greater tha:1 so the V will not be that much greater than. (3 points/44) 1 V V V V 31 K.36 L.55 L 1 97 K 17. Consider a sample of gas in a container on a comfortable spring day. The Celsius temperature suddenly doubles, and you transfer the gas to a container with twice the volume of the first container. If the original pressure was 1 atm, what is a good estimate for the new pressure? This problem is a little awkward because you are probably not familiar with a Celsius temperature on a comfortable spring day. This would probably be considered to be about room temperature which is usually stated to be 5 0 C. Doubling this would be 50 0 C but remember, we need to use K. Because we are given information that describes more than one set of gas law data this is a comparison of conditions problem. We have been given initial and final information for T and V, and initial information on, and are asked to find final. As you are given no information on n you can assume this is constant.

6 Chapter Error! Unknown document property name.: Error! Unknown document property name. 99 Either write out 1 V and drop out constant values, or be able to visualize this relationship in your mind, cancel out constant values in your mind and write the remaining relationship. Isolate, plug in the values given and solve (you are not given values for volume but a ratio, which works as well). Write the isolation so you can see what the ratio of the changed variables does to the needed variable in this case, the increase in temperature will only increase the pressure slightly, while the doubling of the volume would halve the pressure. Combined, the final pressure should be a little more than half of the original pressure. (3 points/47) 1 V V 1 1 V T V 1 33 K 1 98 K (1atm) 6.5atm 18. Body temperature is about 309 K. On a cold day, what volume of air at 71 K must a person with a lung capacity of 1.5 L breathe in to fill the lungs? Because we are given information that describes more than one set of gas law data this is a comparison of conditions problem. Even if you cannot see through the problem initially, write out the relationship for this type of problem and see what you can eliminate. You are not given information about pressure () or number of moles (n) so these can drop out. Looking more carefully at what you are given, you are asked to say what happens to a volume of air at 71K (the initial state) is breathed into a body at 309 K to fill a volume of 1.5L (the final state). You can now see that you need and have been given,, and V. Isolate in your relationship, plug in the values and solve. Write the isolation so you can see what the ratio of the changed variables does to the needed variable in this case, the original slightly colder temperature would mean the original volume is slightly lower than the final volume. (3 points/50) 1 V V V V 71 K 1.5 L 1.3L 309 K 19. You have a certain mass of helium gas (He) in a rigid steel container. You add the same mass of neon gas (Ne) to this container. Which of the following best describes what happens? Assume the temperature is constant. a) The pressure in the container doubles. b) The pressure in the container increases but does not double. c) The pressure in the container more than doubles. d) The volume of the container doubles. e) The volume of the container more than doubles. If the container is rigid its volume will not change so you can immediately eliminate (d) and (e). T also remains constant so the only ratios you need to be concerned about are n and. If you have a certain mass of He, this will be equivalent to a certain number of moles. If you add the same mass of neon, this will be a lesser number of moles because neon has more mass, so the pressure will increase, but not by double. Answer B (1 point/51) If you need to, write out the initial vs final condition relationship, cancel constant values, and isolate as this is the variable they are really concerned about. This will show that you need to consider what is happening to the number of moles with the addition of neon and may lead you to the answer if your

7 Chapter Error! Unknown document property name.: Error! Unknown document property name. 100 initial reasoning can t get you there. 1 V 1 0. You are holding two balloons, an orange balloon and a blue balloon. The orange balloon is filled with neon (Ne) gas and the blue balloon is filled with argon (Ar) gas. The orange balloon has twice the volume of the blue balloon. Which of the following best represents the mass ratio of Ne:Ar in the balloons? a) 1:1 b) 1: c) :1 d) 1:3 e) 3:1 You are given no information regarding T so you can assume it is equal for both balloons. Also, as neither balloon is stated to by changing in shape or size, the pressure inside the balloon must be the same as the pressure outside the balloon (atmospheric pressure). Therefore, the difference in volume is due to the difference in number of particles. If this is the case, the orange balloon filled with Ne has to have twice the number of particles lets say moles of gas compared to the number of particles in the blue balloon lets say 1mole. As the atomic weight of argon is twice that of neon, the mass ratio of each has to be 1:1. Answer A (1 point/5) If you need to, write out the initial vs final condition relationship and cancel constant values. In this case, we are concerned about mass and mole ratios, so we will want to visualize what will show us the mole ratio our relationship should tell us that the ratio of volume of neon to volume of argon will equal the ratio of moles of argon to neon. Therefore, as neon has twice the volume it should have twice the number of moles. In order for this to happen, the number of grams of neon would have to be the same as the number of grams of argon. V neon 1 n neon V argon n argon V argon V neon n argon n neon 1. You are holding four identical balloons each containing 10.0 g of a different gas. The balloon containing which gas is the largest balloon? a) H b) He c) Ne d) O e) All have the same volume. Because each balloon contains a different substance, the 10.0 g of each will represent a different number of moles. We are asked for which will have the greatest volume. As temperature and pressure are not given we can assume they are constant. Therefore, the largest balloon will also contain the gas of the greatest number of moles. As you are give0.0 g of each, the one with greatest number of moles with be the one with the lowest molar mass. Quick inspection reveals this to be H. You should probably be able to reason this out without a visual representation of the gas law in front of you trying to use the initial vs final condition relationship in this case would probably increase rather than decrease your time spent on this problem. Answer A (1 point/53) Use the following to answer questions -4: Consider three 1-L flasks at ST. Flask A contains NH 3 gas, flask B contains NO gas, and flask C contains N gas.

8 Chapter Error! Unknown document property name.: Error! Unknown document property name Which contains the largest number of molecules? a) Flask A b) Flask B c) Flask C d) All are the same. e) More information is need to answer this. At standard temperature and pressure, with a 1-L flask, all gases must have the same number of moles. Therefore, all flasks will have the same number of molecules. Answer D (1 point/54) 3. In which flask are the molecules least polar and therefore most ideal in behavior? a) Flask A b) Flask B c) Flask C d) All are the same. e) More information is needed to answer this. We have made a big deal about interparticle forces and the effect they have on physical processes. We have said the ideal gases have minimal interparticle forces so there is little deviation except at extreme pressures and extreme low temperatures. These substances also have physical properties reflecting minimal interparticle forces, like low melting and boiling points, and small molar mass. On the other hand, substances that have strong interparticle forces (hydrogen bonding, greater molar mass) deviate more from ideal gas behavior and also have physical properties reflecting greater interparticle forces like higher melting and boiling points. In this problem, NH 3 is highly polar, NO would be of intermediate polarity, and N is non-polar. Therefore, the substance most ideal in behavior is in flask C. Answer C (1 point/55) 4. In which flask do the molecules have the highest average velocity? a) Flask A b) Flask B c) Flask C d) All are the same. e) More information is needed to answer this. According to kinetic molecular theory and our understanding of kinetic energy, the average velocity is inversely proportional to the square root of the molar mass. Therefore, the particles with the highest average velocity will have the smallest molar mass, in this case, NH 3, so flask A. Answer A (1 point/56) Use the following to answer questions 5-6: You have two samples of the same gas in the same size container, with the same pressure. The gas in the first container has a Kelvin temperature four times that of the gas in the other container. 5. The ratio of the number of moles of gas in the first container compared to that in the second is a) 1:1 b) 4:1 c) 1:4 d) :1 e) 1: You are told that volume and pressure are constant but that temperature is different and you want to know how the temperature relates to the ratio of moles of particles given these circumstances. Because we are given information that describes more than one set of gas law data this is a comparison of conditions problem. If you need to write our the initial vs final condition relationship and eliminate V and leaving 1/ 1/. Then, instead of isolating a specific variable, you are asked about the ratio of numbers of moles of particles. Rearrange this formula so the ratio of to is in terms of the ratio of temperatures. Then plug the ratio of temperatures in to see what the ratio of moles should be. (1 point/57) 1 V or 1: 4 Answer C 4

9 Chapter Error! Unknown document property name.: Error! Unknown document property name The ratio of the average velocity of particles in the first container compared to that in the second is a) 1:1 b) 4:1 c) 1:4 d) :1 e) 1: Because we are given information that describes more than one set of gas law data this is a comparison of conditions problem. If the Kelvin temperature in the second container is four times that of the first container, its average kinetic energy is also four times that of particles in the second container. As the KE energy is related to the square of the velocity, if the kinetic energy 4is times as great, the velocity must be twice as great to accomplish this. Answer D (1 point/58) Use the following to answer questions 7-8: Three 1.00-L flasks at 5 C and 75 torr contain the gases CH 4 (flask A), CO (flask B), and C H 6 (flask C). 7. In which flask is there mol of gas? a) Flask A b) Flask B c) Flask C d) all e) none All three flasks have the same volume, temperature and pressure, so they should contain the same number of moles. You do not need to perform any calculations or really even need to visualize the initial vs final conditions relationship to see this. However, you are given the choice of all or none containing.39 moles so you do have to perform a calculation to determine what the number of moles is that is common to all three flasks. This now becomes a problem of the second type of gas law problem where you are given 3 of the 4 values and simply asked to find the missing variable. Isolate n in the ideal gas law, plug in the values and solve. n V RT 75torr ( 760torr )(1.00 L) (.0806 Latm )(98 K).039 mol Answer D (1 point/59) 8. In which single flask do the molecules have the greatest mass, the greatest average velocity, and the highest kinetic energy? a) Flask A b) Flask B c) Flask C d) All are the same. e) No one flask has all these. Because mass and average velocity are inversely related, and because all of the gases are at the same temperature and so have the same kinetic energy, non of the gases can simultaneously have the greatest mass, the greatest average velocity and the highest kinetic energy. Answer E (1 point/60) 9. A gas sample is heated from -0.0 C to 57.0 C and the volume is increased from.00 L to 4.50 L. If the initial pressure is atm, what is the final pressure? Because we are given information that describes more than one set of gas law data this is a comparison of conditions problem. Either write out 1 V and drop out constant values, or be able to visualize this relationship in your mind, cancel out constant values in your mind and write the remaining relationship. In this problem you are given information about changes in temperature and volume, you are given an initial pressure, and you are asked for a final temperature. You are not given information about numbers of moles so you can assume this is constant. Isolate and plug in the values. Write the isolation so you can see what the ratio of the changed variables does to the needed variable in this case final temperature is greater than the initial temperature so this would increase the pressure, while final volume is greater than the initial volume,

10 Chapter Error! Unknown document property name.: Error! Unknown document property name. 103 which would cause a decrease in pressure. In this case the volume increase is great enough that it wins. The pressure decreases. (3 points/63) 1 V T V 1 (330K)(.00L).109 atm.063 atm (53 K)(4.50 L) 30. A sample of oxygen gas has a volume of 3.5 L at 7 C and torr. How many oxygen molecules does it contain? This is a solving for a missing variable problem embedded in mole to particle conversion problem. You have been asked for number of particles and you know this is the product of the number of moles and Avogadro s number. Write this relationship down. Then, you have been given a single set of gas law data including V, T and and know you can use this to find moles. Substitute the equivalent gas law expression isolated for moles into the original equation, plug in values and solve. Recognize that you will have to convert T to kelvin and torr to atm. Try to combine calculation is this fashion whenever possible. (3 points/66) 800torr ( 760torr/atm )(3.5L) O molecules (mol O )( AN ) V RT ( AN ) (.0806 Latm )(300K) (6.0x103 ) 9.06x10 molecules 31. You fill a balloon with.50 moles of gas at 4 C at a pressure of 1.78 atm. What is the volume of the balloon? This is a simple solving for a missing variable problem. You have been asked for V and have been given n, T and. lug in the values and solve. Make sure you convert T to Kelvin (3 points/69) V nrt Latm (.50 mol)(.0806 )(97K) 34. L 1.78 atm 3. A sample of helium gas occupies 19.5 L at 3 C and atm. What volume will it occupy at 40 C and 1.0 atm? Because we are given information that describes more than one set of gas law data this is a comparison of conditions problem. Either write out 1 V and drop out constant values, or be able to visualize this relationship in your mind, cancel out constant values in your mind and write the remaining relationship. In this problem you are given information about changes in temperature and pressure, you are given an initial volume, and you are asked for a final volume. You are not given information about numbers of moles so you can assume this is constant. Isolate V and plug in the values. Write the isolation so you can see what the ratio of the changed variables does to the needed variable in this case final temperature is greater than the initial temperature so this would increase the volume, while the final pressure is greater than the initial pressure, which would cause a decrease in volume. In this case the pressure increase is great enough that it wins. The volume decreases. (3 points/7) 1 V V T 1 V 1 (313K)(.956atm) 19.5L 16.4L (96K)(1.0atm) 33. A 4.8-L sample of carbon monoxide is collected at 55 C and atm. What volume will the gas occupy at 1.05 atm and 5 C? Because we are given information that describes more than one set of gas law data this is a comparison of conditions problem. Either write out 1 V and drop out constant values, or be able to visualize this relationship in your mind, cancel out constant values in your mind and write the remaining relationship. In this problem you are given information about changes in temperature and pressure, you are given an initial volume, and you are asked for a final volume. You are not given information about numbers of moles so you can assume this is constant. Isolate V and plug in the values. Write the isolation so you can see what the ratio of the changed variables does to the needed variable in this case final temperature is less than the initial temperature so this would decrease the volume, while the

11 Chapter Error! Unknown document property name.: Error! Unknown document property name. 104 final pressure is greater than the initial pressure, which would also cause a decrease in volume therefore the volume decreases. (3 points/75) 1 V V 1 V 1 (98K)(.833atm) 4.8L 3.47L (38K)(1.05atm) 34. Mercury vapor contains Hg atoms. What is the volume of 01 g of mercury vapor at 8 K and 0.13 atm? This is a gram to mole conversion problem embedded in a solving for a missing variable problem. As you are asked for volume write out the ideal gas law isolating for V. You don t have moles but you know molg/mm so substitute this in for moles. lug values in and solve. (3 points/78) V nrt V nrt g ( mm ) RT ( 01g Latm )(.0806 )(8 K) 00.59g/mol 317 L.13 atm 35. What volume is occupied by 0.7 g of methane (CH 4 ) at 7 C and 1.64 atm? This is a gram to mole conversion problem embedded in a solving for a missing variable problem. As you are asked for volume write out the ideal gas law isolating for V. You don t have moles but you know molg/mm so substitute this in for moles. lug values in and solve. (3 points/81) V nrt g ( mm ) RT ( 0.7 g Latm )(.0806 )(300 K) g/mol 19.4 L 1.64 atm 36. An automobile tire is filled with air at a pressure of 5.9 lb/i at 5 C. A cold front moves through and the temperature drops to 5 C. Assuming no change in volume, what is the new tire pressure? Because we are given information that describes more than one set of gas law data this is a comparison of conditions problem. Either write out 1 V and drop out constant values, or be able to visualize this relationship in your mind, cancel out constant values in your mind and write the remaining relationship. In this problem you are given information about change in temperature, you are given an initial pressure, and you are asked for a final pressure. You are not given information about numbers of moles so you can assume this is constant, and you are told volume is constant, so these drop out. Isolate and plug in the values you are not asked to so there is no need to convert pressure to atmospheres in initial vs final condition problems it is the ratios of variables that matter, not the units (in contrast to solving for missing variable problems). Write the isolation so you can see what the ratio of the changed variables does to the needed variable in this case final temperature is less than the initial temperature so this would decrease the pressure. (3 points/84) 1 78K 5.9 psi 4. psi 1 98 K 37. Which conditions of, T, and n, respectively, are most ideal? a) high, high T, high n b) low, low T, low n c) high, low T, high n d) low, high T, high n e) low, high T, low n The best way to think about this is to recognize that any condition that lessens the chance of interparticle forces of attraction and lessens the volume the particles take up will lead to more ideal behavior. Low pressure leads to particles being farther apart, high temperature leads to particles being farther apart, and low n leads to low volume of particles. Answer E (1 point/85) 38. A 9.70-g piece of solid CO (dry ice) is allowed to sublime in a balloon. The final volume of the balloon is 1.00 L at 98 K. What is the pressure of the gas?

12 Chapter Error! Unknown document property name.: Error! Unknown document property name. 105 This may seem like a comparison of conditions type of problem but in fact there is really only one set of gas law data we are only asking about the state of a system at one point in time. This is a gram to mole conversion problem embedded in a solving for a missing variable problem. As you are asked for pressure, write out the ideal gas law isolating for. You don t have moles but you know molg/mm so substitute this in for moles. lug values in and solve. Make sure you understand that 9.70 g of solid CO is still 9.70 g of gaseous CO, and so, the same number of moles, just in a different form. (3 points/88) nrt V ( g mm )RT V ( 9.70 g g/mol Latm )(.0806 )(98 K) 5.39atm 1.00 L 39. A sample of 35.1 g of methane gas has a volume of 4.45 L at a pressure of.70 atm. Calculate the temperature. This is a gram to mole conversion problem embedded in a solving for a missing variable problem. As you are asked for temperature, write out the ideal gas law isolating for T. You don t have moles but you know molg/mm so substitute this in for moles. lug values in and solve. (3 points/91) T V nr (.70atm)(4.45L) 35.1 g ( 16.04g/mol )(.0806 Latm ) 66.9 K 40. A 30.1-g sample of Ne gas exerts a certain pressure in a container of fixed volume. What mass of Ar is required to exert half the pressure at the same conditions of volume and temperature? Because we are given information that describes more than one set of gas law data this is a comparison of conditions problem. Further, it is embedded in a gram to mole conversion problem. You are asked for number of grams (mass of Ar) so start with this. g Ar (mol)(mm). As you don t have mol, you will have to find it through the initial vs final gas problem strategy. As always, write out the intial vs final relationship, using subscripts for argon and neon rather than 1 and. V and T drop out because they are constant. From this point, as it is n Ar that you want simply isolate this. You can find n Ne because you know the number of grams of neon. And, you know that whatever the of neon is, of Ar is half of that with initial vs final problems the ratio is all that matters. Substitute this values in and solve for n Ar. (4 points/95) g Ar (mol Ar)(mm Ar) (.745 mol)(39.95g / mol) 9.8 g Ar V Ar n Ar T Ar Ne V Ne n Ne T Ne n Ar Ar n Ne 1 Ne g Ne mm Ne (.5) 30.1g 0.18g / mol.745mol 41. A sample of gas is in a 50.0-mL container at a pressure of 645 torr and a temperature of 5 C. The entire sample is heated to a temperature of 35 C and transferred to a new container whose volume is 77. ml. The pressure of the gas in the second container is about: Because we are given information that describes more than one set of gas law data this is a comparison of conditions problem. Either write out 1 V and drop out constant values, or be able to visualize this relationship in your mind, cancel out constant values in your mind and write the remaining relationship. In this problem you are given information about change in volume and temperature, you are given an initial pressure, and you are asked for a final pressure. You are not given information about numbers of moles so you can assume this is constant so this drops out. Isolate and plug in the values you are not asked to so there is no need to convert pressure to atmospheres in initial vs final condition problems it is the ratios of variables that matter, not the units (in contrast to solving for missing variable problems). Write

13 Chapter Error! Unknown document property name.: Error! Unknown document property name. 106 the isolation so you can see what the ratio of the changed variables does to the needed variable in this case final temperature is greater than the initial temperature so this would increase the pressure. The final volume is greater than the initial volume so this would decrease the pressure. The effect of the increase in volume appears to be greater overall the pressure decreases (3 points/98) 1 V T V 1 (308K)(.0500L) 645torr 43 torr (98K)(.077L) 4. Given a cylinder of fixed volume filled with 1 mol of argon gas, which of the following is correct? (Assume all gases obey the ideal gas law.) a) If the temperature of the cylinder is changed from 5 C to 50 C, the pressure inside the cylinder will double. b) If a second mole of argon is added to the cylinder, the ratio T/ would remain constant. c) A cylinder of identical volume filled with the same pressure of helium must contain more atoms of gas because He has a smaller atomic radius than argon. d) Two of the above. e) None of the above. (a) false--the pressure will increase if T goes from 5C to 50C but it will not double it will only double if the absolute T (K) doubles; (b) false doubling the number of particles will double the number of collisions with the wall of the container so will double. However, there is no reason for us to expect the addition of the particles to increase the average kinetic energy of the particles, so the temperature would also not increase. So, the T/ ratio will not remain constant. (c) false at constant temperature and volume, the pressure depends on the number of particles, not their size. We are given no information regarding temperature so we assume it is the same for both circumstances. Therefore, with the same pressure, the number of particles must be the same. Answer E (1 point/99) 43. For an ideal gas, which pairs of variables are inversely proportional to each other (if all other factors remain constant)? 1. and T. T and n 3. n and 4. V and T a) 1 and only b) 3 and 4 only c) only d) 1 and 3 only e) 1, 3, and 4 only For 1: think the greater the T the greater the direct relationship (Amonton s Law); For : think the greater the n (numbers of particles) the less the T because the energy is spread out over more particles inverse relationship; For 3: think the greater then number of particles, n, the greater the direct relationship; For 4: think the greater the T, the greater the V direct relationship (Charles s Law). So only Answer C (1 point/100) 44. For a gas, which two variables are directly proportional to each other (if all other conditions remain constant)? 1. and V. V and n 3. V and T

14 Chapter Error! Unknown document property name.: Error! Unknown document property name. 107 a) 1 only b) only c) 3 only d) 1 and only e) and 3 only For 1: think the greater the the less the V greater pressure compresses inverse relationship (Boyle s Law); For : think the greater the number of particles, n, the greater the V direct relationship (Avogadro s Law); For 3: think the greater the T the greater the V direct relationship (Charles s Law) So, and 3 only Answer E (1 point/101) 45. The temperature of a specific amount of gas in a sealed container changes from 0.0 C to 40.0 C. If the volume remains constant, the pressure will change from 755 mmhg to Because we are given information that describes more than one set of gas law data this is a comparison of conditions problem. Either write out 1 V and drop out constant values, or be able to visualize this relationship in your mind, cancel out constant values in your mind and write the remaining relationship. In this problem you are given information about change in temperature, you are given an initial pressure, and you are asked for a final pressure. You are not given information about numbers of moles so you can assume this is constant, and you are told volume is constant, so these drop out. Isolate and plug in the values you are not asked to so there is no need to convert pressure to atmospheres in initial vs final condition problems it is the ratios of variables that matter, not the units (in contrast to solving for missing variable problems). Write the isolation so you can see what the ratio of the changed variables does to the needed variable in this case final temperature is greater than the initial temperature so this would increase the pressure. (1 point/10) 1 V 1 313K 755mmHg 807 mmhg 1 93K Use the following to answer questions 46-51: Four identical 1.0-L flasks contain the gases He, Cl, CH 4, and NH 3, each at 0 C and 1 atm pressure. Because we are given information that describes more than one set of gas law data this is a comparison of conditions problem. 46. Which gas has the highest density? a) He b) Cl c) CH 4 d) NH 3 e) all gases the same Density is g/ml. Because all four gases are at the same, T and V, they must all have the same number of particles or moles per ml this means, the one with the greatest density will be the one with the greatest mm. You should be able to quickly see that this must be chlorine gas at a mm of g/mol. Answer B (1 point/103)

15 Chapter Error! Unknown document property name.: Error! Unknown document property name For which gas do the molecules have the highest average velocity? a) He b) Cl c) CH 4 d) NH 3 e) all gases the same As all of the gases are at the same temperature, which means they all have the same average kinetic energy, and as under those circumstances velocity is inversely proportional to mass, the gas with the highest velocity will be the gas with the lowest molar mass, which would be He. Answer A (1 point/104) 48. Which gas sample has the greatest number of molecules? a) He b) Cl c) CH 4 d) NH 3 e) all gases the same Because all four gases are at the same, T and V, they must all have the same number of moles of particles which means they also must have the same number of particles. Answer E (1 point/105) 49. For which gas are the molecules diatomic? a) He b) Cl c) CH 4 d) NH 3 e) all gases the same As diatomic means two of the same type of atom, this would be Cl. Answer B (1 point/106) 50. For which gas are the collisions elastic? a) He b) Cl c) CH 4 d) NH 3 e) all gases the same At standard temperature and pressure, these gasses would all behave very closely to ideally so they would all have elastic collisions none would be under the influence of interparticle forces. Answer E (1 point/107) 51. For which gas do the molecules have the smallest average kinetic energy? a) He b) Cl c) CH 4 d) NH 3 e) all gases the same Because all of the gases are at the same T all of them have the same KE. Answer E (1 point/108) Use the following to answer questions 5-54: A plastic bag is weighed and then filled successively with two gases, X and Y. The following data are gathered: Temperature: 0.0 C (73 K) ressure: 1.00 atmosphere Mass of empty bag: 0.77 g Mass of bag filled with gas X: 4.97 g Mass of 1.1 liters of air at conditions given: 1.30 g Volume of bag: 1.1 liter Molar volume at ST:.4 liters

16 Chapter Error! Unknown document property name.: Error! Unknown document property name. 109 This problem is not entirely clear in what it is asking. By saying that they are successively adding gas X and Y, they literally mean, first they are adding gas X to the empty bag; then they are emptying the bag and adding gas Y to the empty bag. Given that the volume and pressure are the same, they are comparing the features of gas X and Y in the first scenario, and then starting over, comparing the features of gas X and Y at the higher temperature, still at the same pressure and temperature. 5. The mass of 1.1 liters of gas Y is found to be 6.3 g. The density of gas Y is Density is g/ml (or g/l is often used for gases as you are given a number of grams and L there is no reason not to use g/l in this case). (3 points/111) d g L 6.3g 1.1 L 5.56 g / L 53. The molar mass of gas Y is You know that the ideal gas law can be used to find molar mass according to the moles pee under dirt version of the formula. Write this relationship down. You just calculated density and have been given T and so plug in the values and solve you don t need to do anything else. Notice that the units all cancel to g/mol. (3 points/114) mm drt (6.3 g Latm )(.0806 )(73K) L 15 g / mol 1.00 atm 54. The bag is emptied and refilled, successively, with gases X and Y, this time at 1 atm pressure and a temperature 30 C higher. Assume that the volume of the bag is the same as before. Which one of the following statements is wrong? a) The full bag contains fewer molecules of each gas than it did at 0.0 C. b) The ratio of the density of gas Y to the density of gas X is the same as at 0.0 C. c) The molar masses of the two gases are the same as they were at 0.0 C. d) The mass of each gas filling the bag is now 303/73 times the mass held at 0.0 C. e) The average velocity of the molecules of gas X at 30 C is higher than it was at 0.0 C. (a) True With the increase in temperature, it should make sense that in order to maintain the same volume (1.1L) and pressure (1.00atm), the number of particles must decrease. (b) True Although there will be fewer particles overall, the proportions of the decrease in particles should be the same for both gases so the ratio of the densities should be the same as it was at 0 C.; (c) True as the substances X and Y are the same, their molar masses must remain the same; (d) False we have already said that the number of particles of both gases will be decreased, so, the overall mass must decrease if the mass was 303/73 times the mass in the first scenario the mass would increase; (e) True Increase in temperature means an increase in KE, and this is accomplished by an increase in average velocity of the particles. Answer D (1 point/115) 55. Argon has a density of 1.78 g/l at ST. How many of the following gases have a density at ST greater than that of argon? Cl He NH 3 NO a) 0 b) 1 c) d) 3 e) 4 As density is proportional to molar mass, and, conditions are the same for all listed gases (ST) any of the substances with a greater molar mass than argon will have a greater density than argon. The molar mass of argon is g/mol. Quickly estimate the molar masses of the others: Cl : x 3570 >40; He: 4 < 40; NH 3 : 14+3<40; NO : 14+3>40 so of the gases will have a density greater than that of Argon. Answer C (1 point/116)

17 Chapter Error! Unknown document property name.: Error! Unknown document property name Which of the following is the best qualitative graph of versus molar mass of a 1-g sample of different gases at constant volume and temperature? a) b) c) d) e) none of these As mm grt V, mm and are inversely related this is represented by a hyperbolic relationship, which is none of the graphs shown. Recognize that given a 1 gram sample of each gas, what you are really doing by increasing the molar mass is decreasing the number of moles of gas present, which is leading to decreased pressure. Remember, if the number of moles of particles is constant, it does not matter what the molar mass is. When asked to describe or interpret graphical information representing these relationships the best thing to do is write the relationship down so you can see how the variables being asked about relate (that is, directly or inversely). Answer E (1 point/117) 57. It is found that 50. ml of a gas at ST has a mass of g. What is the molar mass? This is molar mass problem embedded in a solving for single variable problem. Write down the molar mass relationship, plug in the given values and solve. (3 points/10) mm grt V Latm (.677g)(.0806 )(73K) 60.7g / mol (.50L)(1atm) 58. Given reactionh 3 (g) + 3Cl (g) N (g) + 6HCl(g), you react 5.0 L of NH 3 with 5.0 L of Cl measured at the same conditions in a closed container. Calculate the ratio of pressures in the container ( final / initial ). This is a stoichiometry problem embedded in a comparison of conditions problem. The key phrase is that the same volume (5.0L) of each gas reacting is measured at the same conditions. Stoichiometrically this means you must realize that the number of moles of each is also identical. You are not given than number of moles but you know it is the same. You are asked to find out what the ratio of pressures is before and after the reaction ( final : initial ). You can use the comparison of conditions relationship to help you. You know that T and V are constant so they drop out of the relationship. Therefore, 1. As you want the ratio of : 1, you can rearrange this to reflect that. 1. This tell you that you need to find the ratio of moles of gas present before and after the reaction. But you haven t been given number of moles. As the ratio is what s important, under circumstances where you don t know what two values are specifically, but you know they are the same, you are free to choose whatever values are most

18 Chapter Error! Unknown document property name.: Error! Unknown document property name. 111 advantageous to you and work the problem with those values. Inspecting the chemical equation, you see that no matter what value you choose for moles of each reactant, there will be a limiting reagent. As you know the numbers of moles of each are the same, I would choose 3 moles for each. This means that you would start with 6 moles of gas. In that case, you would consume all of the Cl, and leave one mole of NH 3 behind. The reaction would generate one mole of N gas and 6 moles of HCl gas. For a total of 7 moles. But, you still have one mole of reactant left for a total of 8 moles after the reaction is over a ratio of 8:6, or 1.33 The following is probably what I would show as work for this problem on a test: (3 points/13) Start with 3 mol each of NH 3 and Cl, which will form 1 mol N and 6 mol HCl. 1 mol NH 3 will be left over giving n final 8 mol, n initial 6 mol 1 V 1 or final initial n final n initial Given reaction N + 3H NH 3, you mix 1 mol each of nitrogen and hydrogen gases under the same conditions in a container fitted with a piston. Calculate the ratio of volumes of the container (V final /V initial ). This is a stoichiometry problem embedded in a comparison of conditions problem. You are asked to find out what the ratio of volumes is before and after the reaction (V final:v initial). You can use the comparison of conditions relationship to help you. You know that and T are constant so they drop out of the relationship. Therefore, V. As you want the ratio of V :, you can rearrange this to reflect that. V, so it doesn t matter that you don t know the specific volumes, their ratio will be the same as the ratio of moles, and you can figure this out. If you start with one mole of each gas, there will be moles of reactants ( moles). For every 1 mole of H consumed, only 1/3 of a mole of N will be used so the entire mole of H will be used but /3 mole of N will be left. For every one mole of N used, moles of NH 3 will be formed. As only 1/3 mole of N was used, only /3 of a mole of NH 3 will form. /3 mole of product plus /3 mole of leftover reactant equals 4/3 mole of gas left in the container ( 4/3 mole). Therefore, : (4/3)/() /3.67. (3 points/16) The following is probably what I would show as work for this problem on a test: Starting with 1 mol of each reactant, n initial mol. Only 1/3 mol of N will react leaving /3 mol N after the reaction. Only /3 mol NH 3 will form. n final 4 / 3 mol1.33 mol 1 V As and T are constant, V or V final V initial n final n initial You carry out the reaction represented by the following balanced equation: N (g) + 3H (g) NH 3 (g) You add an equal number of moles of nitrogen and hydrogen gases in a balloon. The volume of the balloon is 1.00 L before any reaction occurs. Determine the volume of the balloon after the reaction is complete. Assume constant temperature. This is a stoichiometry problem embedded in a comparison of conditions problem. You are given a set of conditions before a reaction occurs and are asked to determine conditions after the reaction occurs. You have been told that T is constant. You are not given any pressure information so you must also