General Chemistry II - Assignment 1
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General Chemistry II - Assignment 1

Course Number: CHEM UG07120_02, Summer 2008

College/University: Logan University

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General Chemistry II - Assignment 1 Due at 11:00am on Tuesday, July 15, 2008 [102%] Deriving Gas Law Formulas is for 1 point(s) [102%] The Kinetic Molecular Theory of Gases is for 1 point(s) [103%] Avogadro's Law is for 1 point(s) [102%] Boyle's Law is for 1 point(s) [100%] Charles's Law is for 1 point(s) [99%] Partial Pressure and the Ideal Gas Law is for 1 point(s) [107%] The Ideal Gas Law is for 1 point(s)...

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Chemistry General II - Assignment 1 Due at 11:00am on Tuesday, July 15, 2008 [102%] Deriving Gas Law Formulas is for 1 point(s) [102%] The Kinetic Molecular Theory of Gases is for 1 point(s) [103%] Avogadro's Law is for 1 point(s) [102%] Boyle's Law is for 1 point(s) [100%] Charles's Law is for 1 point(s) [99%] Partial Pressure and the Ideal Gas Law is for 1 point(s) [107%] The Ideal Gas Law is for 1 point(s) [89%] Biochemical Applications of the Gas Laws is for 1 point(s) [106%] Changes in Temperature is for 1 point(s) [106%] Changes in Volume is for 1 point(s) [104%] Gas Laws is for 1 point(s) [109%] Ideal vs. Real Gases is for 1 point(s) [105%] The van der Waals Equation is for 1 point(s) [106%] The Behavior of Gas Molecules is for 1 point(s) [103%] Gas Pressure is for 1 point(s) [110%] Partial Pressure is for 1 point(s) [100%] Problem 9.29: Key Concept Problem is for 1 point(s) [100%] Problem 9.80: Section Problem is for 1 point(s) [100%] Problem 9.81: Section Problem is for 1 point(s) [105%] Properties of Liquids is for 1 point(s) [105%] Vapor Pressure and Phase Changes is for 1 point(s) [100%] Problem 10.59: Section Problem is for 1 point(s) [100%] Problem 10.66: Section Problem is for 1 point(s) [100%] Problem 9.34: Key Concept Problem is for 1 point(s) [100%] Problem 9.27: Key Concept Problem is for 1 point(s) [88%] Problem 9.97: Chapter Problem is for 1 point(s) [100%] Problem 9.39: Section Problem is for 1 point(s) [100%] Problem 9.50: Section Problem is for 1 point(s) [100%] Problem 9.52: Section Problem is for 1 point(s) [100%] Problem 9.53: Section Problem is for 1 point(s) Print View ] General Chemistry II Assignment 1 Due at 11:00am on Tuesday, July 15, 2008 View Grading Details Hide Grading Details Number of answer attempts per question is: 5 You gain credit for: correctly answering a question in a Part, correctly answering a question in a Hint, or not opening a Hint (2% bonus). You lose credit for: exhausting all attempts or requesting the answer to a question in a Part or Hint, incorrectly answering a question in a Part, or opening a Hint (3% deduction). Late submissions: receive no credit. Hints are helpful clues or simpler questions that guide you to the answer. Hints are not available for all questions. There is no penalty for leaving questions in Hints unanswered. Grading of Incorrect Answers For Multiple-Choice or True/False questions, you lose 100%/(# of options - 1) credit per incorrect answer. For any other question, you lose 3% credit per incorrect answer. Deriving Gas Law Formulas Learning Goal: To understand how to determine the appropriate formula for a given gas law problem. When you are unsure of which formula to use to solve a gas law problem, it is often helpful to make a chart of initial and final values of pressure , volume , number of moles , and temperature Initial Final , as shown in the table. This type of chart will help you to determine which quantities are changing, and which quantities remain the same. For example, use the formula for the product of pressure and volume at two different points in time when pressure and volume are changing but the number of moles of gas and the temperature are constant. Here is how this formula is derived from the ideal gas law, 1. If and are constant, the entire quantity is constant. at any point in time will be the same value as at any other point in time. . : is constant. 2. Therefore, 4. Therefore, 3. It follows that the quantity Now imagine that your chart looks like this: Initial 2 1.7 Final 2 2.5 0.45 273 It is clear that , , and 1. 2. 3. 4. . is constant. . and together, we need to do a little bit of algebra: . 0.45 ? and are constant. In order to put all constants are the quantities that are changing, while 5. Therefore the quantity 6. at any point in time will be the same value as at any other point in time. This is the best formula to use when only volume and temperature are changing. Now to solve for the missing value, simply substitute numbers into this formula: You are confronted with the following question: One mole of an ideal gas is sealed in a 22.4- container at a pressure of 1 and a temperature of 273 . The temperature is then increased to 305 not expand. What will the new pressure be? Part A The most appropriate formula for solving this problem includes which variables? Hint A.1 How to approach the problem Hint not displayed Part A.2 Make a chart Part not displayed Enter the required variables, separated by commas. For example: P,V,T. ANSWER: P,T Part B You have determined in Part A that and are the only variables needed in the formula. Which statement is true? Hint B.1 How to approach the problem Hint not displayed ANSWER: is constant. , but the container does is constant. Since the quantity Part C Solve the problem stated in the problem introduction by finding the new pressure. ANSWER: = 1.12 The numbers of moles and liters were given, but they are not necessary for solving the problem. is constant, it follows that . The Kinetic Molecular Theory of Gases Learning Goal: To understand some aspects of molecular motion in the gas phase. The kinetic molecular theory of gases explains how gas molecules behave in terms of motion, speed, and energy. One important aspect of this theory deals with the relationship between temperature and the average speed of the gas molecules. Increasing the temperature of a gas sample increases the average kinetic energy of the molecules. The kinetic energy of a molecule determines its speed. It is important to realize that not all molecules in a sample will have the same kinetic energy, which is why we refer to the average kinetic energy and the average speed. The speed of a particle with average kinetic energy is called the root mean square (rms) speed, . The rms speed may be expressed by the following equation: where is the ideal gas constant, kilograms per mole. is the absolute temperature, and is the molar mass of the substance in The constant motion of gas molecules causes diffusion and effusion. Diffusion is the gradual mixing of two substances due to the movement of their particles. Effusion is the gradual escape of gas molecules through microscopic holes in their container. Part A Which of the following statements are true? Hint A.1 The relationship among temperature, average kinetic energy, and speed Hint not displayed Check all that apply. ANSWER: The average kinetic energy of gas molecules increases with temperature. There are gas molecules that move faster than the average. There are gas molecules that move slower than the average. All the gas molecules in a sample have the same kinetic energy. The average speed of gas molecules increases with temperature. Part B What are each of the following observations an example of? Hint B.1 How to approach the problem Hint not displayed Drag the appropriate items to their respective bins. ANSWER: View Part C Which statements about a sample of gas containing molecules of different masses are true? Hint C.1 How to approach the problem Hint not displayed Check all that apply. ANSWER: The rms speed depends only on temperature, and so all types of particles in the sample have the same rms speed. More-massive gas molecules in the sample have higher rms speed than less-massive ones. More-massive gas molecules in the sample have lower rms speed than less-massive ones. The average kinetic energy depends only on the temperature. However, a heavier molecule will move more slowly than a lighter molecule if both have the same kinetic energy. Avogadro's Law If one had a box that was filled to the brim with oranges, it would make intuitive sense that if we had twice as many oranges, the box would need to be twice as big. But, is the same thing true if one is concerned with gas particles and not oranges? The answer is Yes. More than a century ago Italian chemist Amadeo Avagadro postulated that , the volume of a gas and , the number of gas particles are directly proportional, when the temperature and pressure are held constant. Part A Avogadro's Law states that at a given temperature and pressure, what quantity is constant? Part A.1 Direct relationship Part not displayed ANSWER: Since and are directly proportional, the quantity temperature. Part B will always have the same value at a given pressure and If it takes three "breaths" to blow up a balloon to 1.2 , and each breath supplies the balloon with 0.060 moles of exhaled air, how many moles of air are in a 3.0 balloon? Hint B.1 How to approach the problem Hint not displayed Part B.2 Calculate n 1 Part not displayed Express your answer in moles using two significant figures. ANSWER: 0.45 Boyle's Law In the mid 1600s, the Irish chemist Robert Boyle carried out experments that determined the quantitative relationship between the pressure and volume of a gas. His data showed that for gas at a constant temperature, pressure and volume are inversely proportional. Part A According to Boyle's law, for a fixed quantity of gas at a given temperature, what quantity relating pressure volume is constant? Hint A.1 Inverse proportionality Hint not displayed Express your answer as a mathematical expression using the variables ANSWER: and . and Application of Boyle's law A 12-liter tank contains helium gas pressurized to 160 Part B . )? What size tank would be needed to contain this same amount of helium at atomspheric pressure (1 Hint B.1 Inverse proportionality Hint not displayed ANSWER: Part C 1920 How many 3-liter balloons could the 12-L helium tank fill? Keep in mind that an "exhausted" helium tank is not empty. In other words, once the gas inside the tank reaches atmospheric pressure, it will no longer be able to fill balloons. Part C.1 Determine the volume of "unused" helium Part not displayed Part C.2 Determine the volume of "usable" helium Part not displayed ANSWER: 636 balloons Charles's Law Experiments carried out by French chemists Jacques Alexandre Csar Charles and Joseph-Louis Gay-Lussac and British physicist Lord Kelvin determined a quantitative relationship between the volume and temperature of a gas. Their data showed that for a container of gas held at constant pressure, the volume and temperature are directly proportional. For example, if you inflate a balloon outdoors with cold air on a cold day and then take it inside, it will expand, and it might even burst. This happens because as the temperature of the air inside the balloon increases, the volume of the balloon increases as well. What is less obvious is the quantitative relation: If the pressure is held constant, then when the temperature is doubled, the volume is doubled as well. Part A According to Charles's law, for a fixed quantity of gas at constant pressure, which of the given quantities is constant? Hint A.1 Direct proportionality Hint not displayed ANSWER: Part B A balloon was filled to a volume of 2.50 when the temperature was the temperature dropped to . Hint B.1 Direct proportionality Hint not displayed ANSWER: Part C Refrigerators are usually kept at about , while room temperature is about . If you were to take an "empty" sealed 2-liter soda bottle at room temperature and place it in the fridge, would you expect it to contract to onefourth its original volume? ANSWER: Yes, because 5 is one-fourth of 20. No, because there is no gas inside the bottle. No, because Celsius is not an absolute temperature scale. Note that and . 2.34 . What would the volume become if So the absolute temperature has not decreased by very much at all, certainly not by a factor of four. Therefore, the change in size of the plastic bottle would not be very noticeable. Partial Pressure and the Ideal Gas Law The ideal gas law, is independent of the kind of gas. In other words, the pressure exerted by a given number of ideal gas particles is the same whether the sample consists of all one type of particle or a mixture of different kinds of particles. Therefore, the pressure exerted by a mixture of gases can be expressed as follows: A partial pressure is the pressure exerted by just one type of gas in a mixture. A partial pressure is calculated using only the number of moles of that particular gas, instead of the total number of moles: , , , etc. The sum of the partial pressures is equal to the total pressure in the mixture: Air is about 78.0% nitrogen molecules and 21.0% oxygen molecules. Several other gases make up the remaining 1% of air molecules. Part A What is the partial pressure of nitrogen in air at atmospheric pressure (1 atm)? Assume ideal behavior. Hint A.1 How to approach the problem The number of molecules of a gas is directly related to the number of moles of that gas. The number of moles of a gas is directly related to its partial pressure (because volume and temperature must be the same for all the gases in a mixture). Therefore, if 78.0% of the air molecules are nitrogen, then 78.0% of the moles are nitrogen, and also 78.0% of the total pressure is due to the presence of nitrogen. Express your answer numerically in atmospheres. ANSWER: 0.780 Part B An "empty" container is not really empty if it contains air. How may moles of nitrogen are in an "empty" two-liter cola bottle at atmospheric pressure and room temperature ( Hint B.1 How to approach the problem )? Assume ideal behavior. Hint not displayed Express your answer numerically to three significant figures. ANSWER: 6.3810-2 moles of nitrogen Part C What is the partial pressure of oxygen in air at atmospheric pressure (1 atm)? Assume ideal behavior. Hint C.1 How to approach the problem Hint not displayed Express your answer numerically, in atmospheres. ANSWER: 0.21 Part D An "empty" container is not really empty if it contains air. How may moles of oxygen are in an "empty" two-liter cola bottle at atmospheric pressure (1 ) and room temperature ( )? Assume ideal behavior. Hint D.1 How to approach the problem Hint not displayed Express your anwer numerically to three significant figures. ANSWER: 1.7210-2 moles of oxygen The Ideal Gas Law The ideal gas law describes the relationship among the pressure temperature , volume , number of moles , and absolute of an ideal gas. Here is the relationship expressed mathematically: where is a proportionality constant. The units of are determined by the units of pressure and volume used in the value is . equation. When Part A is used for pressure and for volume, the appropriate How many air molecules are in a temperature of 20.0 , and ideal behavior. room? Assume atmospheric pressure of 1.00 , a room Volume conversion:There are 28.2 liters in one cubic foot. Hint A.1 How to approach the problem Hint not displayed Part A.2 Calculate the volume in liters Part not displayed Part A.3 ANSWER: Calculate the number of moles of air Part not displayed 1.021027 molecules Biochemical Applications of the Gas Laws The various gas laws can be used to describe air, which is a mixture of gases. In some cases, these laws have direct application to the air that we breathe. Part A How long does it take a person at rest to breathe one mole of air if the person breathes 78.0 measured at 25 and 755 ? Part A.1 Determine the volume of air Part not displayed Part A.2 Identify how to find the time from the volume and respiration rate Part not displayed Express your answer numerically in seconds. of air that is ANSWER: Part B 315 Exercise greatly increases respiration rate and would reduce the time needed to breathe one mole of air. Typically, when a person coughs, he or she first inhales about 2.40 of air at 1.00 and 25 . The epiglottis and the vocal cords then shut, trapping the air in the lungs, where it is warmed to 37 and compressed to a volume of about 1.70 by the action of the diaphragm and chest muscles. The sudden opening of the epiglottis and vocal cords releases this air explosively. Just prior to this release, what is the approximate pressure of the gas inside the lungs? Part B.1 Determine the equation for final pressure Part not displayed Part B.2 Convert the temperatures to kelvins Part not displayed Express your answer numerically in atmospheres. ANSWER: 1.47 Part C Helium-oxygen mixtures are used by divers to avoid the bends and are used in medicine to treat some respiratory ailments. What percent (by moles) of is present in a helium-oxygen mixture having a density of 0.528 at 25 and 721 ? Hint C.1 How to approach the problem First, use the density and the ideal gas law to calculate the molar mass of the mixture. At this point, the problem contains two unknown quantities: (1) the mole percent of of . Thus, two equations are needed to relate the quantities. and (2) the mole percent Set up the equation for the molar mass of the mixture in terms of the percentage of each gas. Then, set up a second equation that relates the percentage of equations, solve for the percentage of . to the percentage of in the mixture. Using these two Part C.2 Determine the average molar mass of the gas mixture What is the molar mass of this oxygen-helium mixture? Part C.2.a Determine the equation that relates density to molar mass Part not displayed Part C.2.b Convert the pressure to atmospheres Part not displayed Part C.2.c Convert the temperature to kelvins Part not displayed Express your answer numerically in grams per mole. ANSWER: = 13.6 Part C.3 Complete the equation that relates percentage to molar mass If and are the percentages (by mole) of be expressed as and , respectively, then the molar mass of the mixture, , can What values should be used for and , respectively? Enter the molar masses of helium and oxygen in grams per mole using four significant figures separated by a comma. ANSWER: , = 4.003,32.00 With the values of the molar masses substituted, the general equation becomes You can also plug in the value of the molar mass for the mixture. Part C.4 Find the percentage of helium and oxygen If is the percentage (by mole) of and is the percentage (by mole) of two-gas mixture? Hint C.4.a How to determine the total percentage Hint not displayed Express your answer numerically as a percentage. ANSWER: = 100 By isolating the equation becomes , what is the sum of and for this Now, you can substitute this expression of in the equation for the molar mass of the mixture. Part C.5 Identify the final form of the equation to calculate the percent He Based on the following two simultaneous equations: 1. 2. when Equation 2 is substituted into Equation 1, what is the correct form of the resulting equation? ANSWER: With the average molar mass of the mixture included the equation becomes All that remains is to solve the equation for the value of . Express your answer numerically as a percentage. ANSWER: 65.7 In addition to relating gas densities to frequently measured gas parameters, the ideal gas equation can also relate these parameters to the molar mass of a gas. As such, it provides a method to determine molar masses of unknown gases. Changes in Temperature The ideal gas law ( amount . When and ) describes the relationship among pressure , volume , temperature , and molar are fixed, the equation can be rearranged to take the following form where is a constant: or When and are fixed, the expression becomes or Part A The pressure inside a hydrogen-filled container was 2.10 was heated to 91 ? Hint A.1 How to approach the problem Hint not displayed Part A.2 Convert the initial temperature to kelvins Part not displayed Part A.3 Convert the final temperature to kelvins Part not displayed Express your answer numerically in atmospheres. ANSWER: = 2.60 The pressure of a fixed amount of gas at constant volume is directly proportional to its temperature: the higher the temperature, the greater the pressure. That is why warnings on many household containers read: "Contents under pressure. Do not place near sources of heat." If such a container is overheated, it might explode. Part B At standard temperature and pressure (0 and 1.00 ), 1.00 of an ideal gas occupies a volume of 22.4 . ? at 21 . What would the pressure be if the container What volume would the same amount of gas occupy at the same pressure and 55 Hint B.1 How to approach the problem Hint not displayed Part B.2 Part B.3 Convert the initial temperature to kelvins Part not displayed Convert the final temperature to kelvins Part not displayed Express your answer numerically in liters. ANSWER: = 26.9 When the amount of gas and the pressure are fixed, the volume is directly proportional to the temperature in kelvins. When the temperature increases, the gas expands. When the gas is cooled, its volume decreases. Changes in Volume The ideal gas law ( ) describes the relationship among pressure , volume , temperature , and molar amount . When some of these variables are constant, the ideal gas law can be rearranged in different ways to take the following forms where is a constant: Name Boyle's law Charles's law Expression Constant and and Avogadro's law and Part A A certain amount of chlorine gas was placed inside a cylinder with a movable piston at one end. The initial volume was 3.00 and the initial pressure of chlorine was 1.45 . The piston was pushed down to change the volume to 1.00 . Calculate the final pressure of the gas if the temperature and number of moles of chlorine remain constant.. Part A.1 Identify the law that applies to the problem Part not displayed Hint A.2 How to set up the equation Hint not displayed Express your answer numerically in atmospheres. ANSWER: = 4.35 Boyle's law describes the inverse relationship between volume and pressure of an ideal gas. When the temperature and amount of a gas remain fixed, an increase in volume results in a lower pressure, and a decrease in volume results in a higher pressure. Part B In an air-conditioned room at 19.0 , a spherical balloon had the diameter of 50.0 . When taken outside on a hot summer day, the balloon expanded to 51.0 in diameter. What was the temperature outside? Assume that the balloon is a perfect sphere and that the pressure and number of moles of air molecules remains the same. Hint B.1 How to approach the problem Hint not displayed Part B.2 Part B.3 Part Part B.4 B.5 Calculate the initial volume of the balloon Part not displayed Calculate the final volume of the balloon Part not displayed Convert the initial temperature to kelvins Part not displayed Calculate the final temperature in kelvins Part not displayed Express your answer numerically in degrees Celsius. ANSWER: outside temperature = 37.3 Since, at a constant pressure, the volume of an ideal gas is directly proportional to its temperature, the plot of volume versus temperature is a straight line. When temperature is plotted on the Celsius scale, the straight line can be extrapolated to the point where the volume is zero and the temperature is a negative volume, Part C A cylinder with a movable piston contains 2.00 of helium, , at room temperature. More helium was added to the cylinder and the volume was adjusted so that the gas pressure remained the same. How many grams of helium were added to the cylinder if the volume was changed from 2.00 to 3.30 ? (The temperature was held constant.) Hint C.1 How to approach the problem Hint not displayed Part C.2 Calculate the initial number of moles Part not displayed Part C.3 Calculate the final number of moles Part not displayed Express your answer numerically in grams. ANSWER: mass of helium added = 1.30 According to Avogadro's law, the volume of an ideal gas is directly proportional to its molar amount. Put another way, the law also says that equal volumes of different gases at the same temperature and pressure contain the same molar amounts. At 0 molar volume). and 1 (standard conditions), 1 of any gas occupies a volume of 22.4 (standard 273 273 . Because matter cannot have must be the lowest possible temperature, absolute zero. Gas Laws This figure movable piston. Inside the container is a ideal gas at 1.00 all parts of this problem A, B, and C. Part A shows a container that is sealed at the top by a , 20.0 , and 1.00 . This information will apply to What will the pressure inside the container become if the piston is moved to the 1.80 mark while the temperature of the gas is kept constant? Part A.1 Determine the relationship between variables Part not displayed ANSWER: Part B The gas sample has now returned to its original state of 1.00 , 20.0 and 1.00 . What will the pressure = 0.556 become if the temperature of the gas is raised to 200.0 and the piston is not allowed to move? Part B.1 Determine the relationship between variables Part not displayed ANSWER: Part C The gas described in parts A and B has a mass of 1.66 grams. The sample is most likely which monoatomic gas? Hint C.1 How to approach the problem Hint not displayed Part C.2 Determine the number of moles Part not displayed Part C.3 Determine molar mass Part not displayed Type the elemental symbol of the gas below. ANSWER: symbol = Ar = 1.61 Ideal vs. Real Gases The ideal gas law describes the relationship among the volume of an ideal gas ( ), its pressure ( ), its absolute temperature ( ), and number of moles ( ): Under standard conditions, the ideal gas law does a good job of approximating these properites for any gas. However, the ideal gas law does not account for all the properties of real gases such as intermolecular attraction and molecular volume, which become more pronouced at low temperatures and high pressures. The van der Waals equation corrects for these factors with the constants and , which are unique to each substance: The gas constant Part A is equal to 0.08206 . A 3.00- flask is filled with gaseous ammonia, . The gas pressure measured at 15.0 ideal-gas behavior, how many grams of ammonia are in the flask? Hint A.1 How to approach the problem Hint not displayed Part A.2 Convert temperature to kelvins Part not displayed Part A.3 Calculate the number of moles Part not displayed Part A.4 Calculate the molar mass of ammonia Part not displayed Express your answer numerically in grams. ANSWER: mass of = 2.48 is 1.15 . Assuming An ideal gas concept assumes that the volume of gas particles is negligible compared to the total volume the gas occupies and there are no attractive forces between individual gas particles. All real gases deviate slightly from the ideal behavior. Under most conditions the difference is very small and we can still use the ideal gas law. But at higher pressures the deviation from the ideal behavior becomes greater, and we need to take both the size of individual gas particles and the attractive forces between individual gas particles into account. Ideal versus real behavior for gases In the following part you can see how the behavior of real gases deviates from the ideal behavior. You will calculate the pressure values for a gas using the ideal gas law and also the van der Waals equation. Take note of how they differ. Part B If 1.00 of argon is placed in a 0.500- container at 23.0 , what is the difference between the ideal pressure (as predicted by the ideal gas law) and the real pressure (as predicted by the van der Waals equation)? For argon, and . Hint B.1 How to approach the problem Hint not displayed Part B.2 Part B.3 Part B.4 Convert temperature to kelvins Part not displayed Calculate the pressure using the ideal gas law Part not displayed Calculate the pressure using the van der Waals equation Part not displayed Enter the pressure difference in atmospheres using two significant figures. ANSWER: = 2.0 At higher pressures, gas particles are much closer together and the attractive forces between them become more important. The attraction draws the molecules together slightly. The result is a decrease in the actual pressure of most real gases when compared with ideal gases (at pressures up to 300 ). The van der Waals Equation Learning Goal: To be able to predict and calculate properties of real gases. Kinetic molecular theory makes certain assumptions about gases that are, in fact, not true for real gases. Therefore, the measured properties of a gas are often slightly different from the values predicted by the ideal gas law. The van der Waals equation is a more exact way of calculating properties of real gases. The formula includes two constants, and , that are unique for each gas. The van der Waals equation is the volume, the temperature, and the gas constant. , where is the pressure, the number of moles of gas, All real gases possess intermolecular forces that can slightly decrease the observed pressure. In the van der Waals equation, the variable is adjusted to , where is the attractive force between molecules. The volume of a gas is defined as the space in which the gas molecules can move. When using the ideal gas law we make the assumption that this space is equal to the volume of the container. However, the gas molecules themselves take up some space in the container. So in the van der Waals equation, the variable the container minus the space taken up by the molecules, Part A In general, which of the following gases would you expect to behave the most ideally? Hint A.1 How to approach the problem Hint not displayed Hint A.2 Relative sizes Hint not displayed Hint A.3 Relative intermolecular attraction Hint not displayed is adjusted to be the volume of , where is the volume of a mole of molecules. ANSWER: Part B In general, which of the following gases would you expect to behave the least ideally? Hint B.1 How to approach the problem Hint not displayed Hint B.2 Relative sizes Hint not displayed Hint B.3 ANSWER: Relative intermolecular attraction Hint not displayed Part C 15.0 moles of gas are in a 8.00 tank at 23.2 . Calculate the difference in pressure between methane and an and ideal gas under these conditions. The van der Waals constants for methane are . Hint C.1 How to approach the problem Hint not displayed Part C.2 Find the pressure of an ideal gas Part not displayed Part C.3 ANSWER: Determine the pressure of methane Part not displayed pressure difference = 4.09 atm The answer to this question represents the error incurred by using the ideal gas law for real gases. This error is very tiny at low pressure and high temperature, but becomes quite significant for compressed gases (high pressure). The Behavior of Gas Molecules Learning Goal: To calculate the average speed and relative rates of effusion of gas molecules. In a given sample of gas, the particles move at varying speeds. The average speed of particles in a gas sample, , is given by the formula where is the Kelvin temperature, is the molar mass in , and is the gas constant. Effusion is the escape of gas molecules through a tiny hole into a vacuum. The rate of effusion of a gas is directly related to the average speed of the gas molecules, so it's inversely proportional to the square root of its mass. In other words, the lighter the molecules of the gas, the more rapidly it effuses. Mathematically, this can be expressed as The relative rate of effusion can be expressed in terms of molecular masses and as Part A Calculate the average speed of an oxygen gas molecule, Hint A.1 How to approach the problem Part A.2 Calculate the molar mass of O2 Part not displayed Express your answer numerically in meters per second. ANSWER: 488 The average speed of an oxygen gas molecule, 488 (which equals 1090 ), is faster than the speed of an airplane! However, an oxygen molecule cannot travel from Boston to Miami in one hour. A gas particle travels only a very short distance before it collides with another particle and bounces off in a different direction. For an oxygen molecule at 25 Part B Rank the following gases in order of decreasing rate of effusion. Hint B.1 How to approach the problem Hint not displayed Part B.2 Determine the relationship between effusion rate and molecular mass Part not displayed Hint B.3 How to calculate the molecular mass Hint not displayed Part B.4 Part B.5 Part B.6 Calculate the molecular mass of propane Part not displayed Calculate the molecular mass of hydrogen sulfide Part not displayed Find the molecular masses of helium, argon, and hydrogen Part not displayed Rank from the highest to lowest effusion rate. To rank items as equivalent, overlap them. ANSWER: and 1 , an average distance between collisions is about . , at 33.0 . Hint not displayed View The lighter the molecules of a gas, the greater the average speed of the molecules, and the more rapidly the gas effuses. Part C How much faster do ammonia ( ) molecules effuse than carbon monoxide ( Hint C.1 How to approach the problem Hint not displayed Hint C.2 How to calculate the molecular mass Hint not displayed Part C.3 Calculate the molecular mass of carbon monoxide Part not displayed Part C.4 Calculate the molecular mass of ammonia Part not displayed Enter the ratio of the rates of effusion. Express your answer numerically using three significant figures. ANSWER: 1.28 = Carbon monoxide molecules are times heavier than ammonia molecules, so ammonia molecules will effuse ) molecules? times faster than carbon monoxide molecules. Different rates of effusion can be helpful in isoloating one kind of gas molecule from a mixture. For example, if a mixture of gases is separated from a vacuum by a membrane, the lighter molecules effuse into the new chamber more quickly. So the gas sample that fills the new chamber will have a higher percentage of the lighter gas than the previous chamber did. Repeating the process several times would result in a nearly pure sample of the lighter gas. Gas Pressure Learning Goal: To understand several methods of measuring gas pressure. Gases exert a measurable pressure ( ) on the walls of their container. The SI unit for pressure is the pascal ( Other pressure units frequently used in chemistry are millimeters of mercury ( ) and atmospheres ( can convert between units of pressure using conversion factors such as those indicated below. ). ). You Part A On a rainy day, a barometer reads 743 . Convert this value to atmospheres. Hint A.1 How to manipulate the conversion factor Hint not displayed Express your answer numerically in atmospheres. ANSWER: 0.978 The standard atmospheric pressure at a sea level is 1 (760 ). Higher pressure usually means sunny and dry weather, and lower pressure often brings clouds and precipitation. Part B A closed container is filled with oxygen. The pressure in the container is 305 millimeters of mercury? Hint B.1 How to approach the problem Hint not displayed Part B.2 Convert the pressure to pascals Part not displayed Express the pressure numerically in millimeters of mercury. ANSWER: 2290 Part C What is the pressure (in atmospheres) of the gas inside the container connected to an open-end, mercury-filled . What is the pressure in manometer as shown in the picture? Hint C.1 How to approach the problem Hint not displayed The atmospheric pressure is 0.95 . Part C.2 Convert the difference in mercury levels from centimeters to millimeters Part not displayed Part C.3 Convert the pressure difference to atmospheres Part not displayed Express your answer numerically in atmospheres. ANSWER: gas pressure = 1.19 Mercury is used in barometers and manometers because it has the highest density among liquids at room temperature (13.6 ). It is possible to make a manometer filled with any liquid, and it would work, but the tube would need to be much longer. When the density of a liquid is low, relatively small changes in pressure would result in large changes in the level of the liquid. Partial Pressure Dalton's law states that the total pressure, , of a mixture of gases in a container equals the sum of the pressures of each individual gas: The partial pressure of the first component, pressure of the mixture: , is equal to the mole fraction of this component, , times the total The mole fraction, , represents the concentration of the component in the gas mixture, so Part A Three gases (8.00 of methane, , 18.0 of ethane, , and an unknown amount of propane, ) were added to the same 10.0- container. At 23.0 partial pressure of each gas in the container. Hint A.1 How to approach the problem , the total pressure in the container is 4.90 . Calculate the Hint not displayed Part A.2 Calculate the partial pressure of methane Part not displayed Part A.3 Calculate the partial pressure of ethane Part not displayed Express the pressure values numerically in atmospheres, separated by commas. Enter the partial pressure of methane first, then ethane, then propane. ANSWER: 1.21,1.45,2.24 Part B A gaseous mixture of and contains 31.8 nitrogen by mass. What is the partial pressure of oxygen in the mixture if the total pressure is 445 ? Hint B.1 How to approach the problem Hint not displayed Part B.2 Part B.3 Part B.4 Calculate the number of moles of nitrogen Part not displayed Calculate the number of moles of oxygen Part not displayed Calculate the mole fraction of oxygen Part not displayed Express you answer numerically in millimeters of mercury. ANSWER: = 290 If we now wanted to find the partial pressure of nitrogen in the mixture, we could calculate the mole fraction of nitrogen and multiply it by the total pressure of the mixture. Or we could take a shortcut and simply subtract the partial pressure of oxygen from the total pressure, because the pressure of the mixture is the sum of the partial pressures of its components. Problem 9.29: Key Concept Problem Assume that you have a sample of gas at 350 in a sealed container, as represented in Part A Which of the drawings (a)-(d) represents the gas after the temperature is lowered from 350 ANSWER: Drawing (a). Drawing (b). Drawing (c). Drawing (d). to 150 ? Problem 9.80: Section Problem Magnesium metal reacts with aqueous to yield gas: and a pressure of 747 . The gas that forms is found to have a volume of 3.557 at 25 Part A Assuming that the gas is saturated with water vapor at a partial pressure of 23.8 pressure (in millimeters of mercury) of the ANSWER: = 723 Part B How many grams of magnesium metal were used in the reaction? ANSWER: = 3.36 ? , what is the partial Problem 9.81: Section Problem Chlorine gas was first prepared in 1774 by the oxidation of with : Assume that the gas produced is saturated with water vapor at a partial pressure of 28.7 of 0.597 at 27 Part A and 755 pressure. and that it has a volume What is the mole fraction of ANSWER: = 0.962 Part B in the gas? How many grams of were used in the experiment, assuming complete reaction? ANSWER: = 2.71 Properties of Liquids Learning Goal: To understand the relation between intermolecular forces and the observable properties of liquids. Viscosity, surface tension, boiling point, and vapor pressure are properties of liquids that are affected by the intermolecular forces within them. Viscosity is a liquid's resistance to flowing. Examples of viscous liquids are molasses, olive oil, and maple syrup. Surface tension is a phenomenon that allows insects to walk on water and causes water to "bead" on windows. Surface tension is defined as the energy required to increase the surface area of a given amount of liquid by a given unit of area. Boiling point is the temperature at which a liquid becomes a gas. It is also defined as the temperature at which the vapor pressure reaches atmospheric pressure. Vapor pressure is the pressure exerted by the vapor above a liquid. Liquids that readily evaporate have relatively high vapor pressures. Part A Classify each property as associated with a liquid that has strong or weak intermolecular forces. Hint A.1 Intermolecular forces and properties of liquids Hint not displayed Drag the appropriate items to their respective bins. ANSWER: View Part B Arrange the liquids bromine ( ), formaldehyde ( surface tension, and boiling point. Hint B.1 How to approach the problem ), and water ( ) in order of decreasing viscosity, Hint not displayed Hint B.2 Types of intermolecular forces Hint not displayed Part B.3 Part B.4 Part B.5 Identify substances that exhibit hydrogen bonding Part not displayed Identify substances that exhibit dipole-dipole forces Part not displayed Identify substances that exhibit dispersion forces Part not displayed Rank from highest to lowest viscosity, surface tension, and boiling point. To rank items as equivalent, overlap them. ANSWER: View Vapor Pressure and Phase Changes All liquids evaporate to a certain extent. The pressure exerted by the gas phase in equilibrium with the liquid is called vapor pressure, . The vapor pressure of a particular substance is determined by the strength of the intermolecular forces. But for any given substance, the vapor pressure only changes with temperature. The Clausius-Clapeyron equation expresses the relationship between vapor pressure and temperature: where and are the vapor pressures that correspond to temperatures is the gas constant. and , respectively, and Part A Consider the following two substances and their vapor pressures at 298 Substance Vapor pressure ( A 275 . ) B 459 Based on this information, compare the characteristics of the two substances. Part A.1 Describe the relationship between vapor pressure and boiling point Part not displayed Part A.2 Describe the relationship between vapor pressure and enthalpy of vaporization Part not displayed Part A.3 Describe the relationship between vapor pressure and intermolecular forces Part not displayed Part A.4 Describe the relationship between vapor pressure and phase. Part not displayed Drag each item to the appropriate bin. ANSWER: View Part B The vapor pressure of dichloromethane, , at 0 is 134 . The normal boiling point of dichloromethane is 40. . Calculate its molar heat of vaporization. Hint B.1 How to approach the problem Hint not displayed Part B.2 Part B.3 Part B.4 Part B.5 Determine the value of P 2 Part not displayed Convert T 1 to kelvins Part not displayed Convert T 2 to kelvins Part not displayed Calculate the heat of vaporization Part not displayed Express your answer numerically in kilojoules per mole. ANSWER: = 30.9 Problem 10.59: Section Problem The vapor pressure of is 100 at 5.4 , and the normal boiling point is 56.8 . Part A What is for in ? ANSWER: = 30.2 Problem 10.66: Section Problem Part A Choose any two temperatures and corresponding vapor pressures in the table given, and use those values to calculate ( ) ( for dichloromethane (in ) ). 263 80.1 273 133.6 283 213.3 293 329.6 303 495.4 313 724.4 ANSWER: = 30.1 Problem 9.34: Key Concept Problem Effusion of a 1:1 mixture of two gases through a small pinhole produces the results shown below. Part A Which gas molecules - yellow or blue - have a higher average speed? ANSWER: Yellow molecules have a higher average speed. Blue molecules have a higher average speed. Part B If the yellow molecules have a molecular mass of 50 Express your answer using two significant figures. ANSWER: = 72 , what is the molecular mass of the blue molecules? Problem 9.27: Key Concept Problem The following drawing represents a container holding a mixture of four gases, red, blue, green, and black. Part A If the total pressure inside the container is 470 ANSWER: = 134 Part B ANSWER: Part C ANSWER: Part D ANSWER: = 33.6 = 201 = 101 , what is the partial pressure of each individual component? Problem 9.97: Chapter Problem Part A What would the atmospheric pressure be (in millimeters of mercury) if our atmosphere were composed of pure gas? Express your answer using four significant figures. ANSWER: = 1155 Problem 9.39: Section Problem Carry out the following conversions. Part A 303 to ANSWER: Part B 0.300 to 303 = 40.4 ANSWER: Part C 0.0264 ANSWER: 0.300 = 228 to 0.0264 = 3.52 Problem 9.50: Section Problem Oxygen gas is commonly sold in 48.0 Part A steel containers at a pressure of 162 . What volume (in liters) would the gas occupy at a pressure of 1.20 ANSWER: = 6480 Part B if its temperature remained unchanged? What volume (in liters) would the gas occupy if its temperature was raised from 18.0 162 ? ANSWER: = 51.6 to 40.0 at constant Problem 9.52: Section Problem Part A If 17.0 of ANSWER: gas has a volume of 0.10 4 = 7.1110 at 295 , what is its pressure in millimeters of mercury? Problem 9.53: Section Problem Part A If 2.8 of gas has a volume of 0.50 and a pressure of 6.6 Express your answer using two significant figures. ANSWER: = 400 , what is its Kelvin temperature?

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Logan University - CHEM - UG07120_02
General Chemistry II - Assignment 2 Due at 11:00am on Tuesday, July 22, 2008 View Grading Details [102%] The Equilibrium Constant Expression is for 1 point(s) [99%] Analysis of Equilibria is for 1 point(s) [108%] Applying Le Chtelier's Principle is f
Logan University - CHEM - UG07120_02
Logan University - CHEM - UG07120_02
Logan University - CHEM - UG07120_02
Logan University - CHEM - UG07120_02
Logan University - CHEM - UG07120_02
Logan University - CHEM - UG07120_02
Logan University - CHEM - UG07120_02
[ Print View ] General Chemistry II Assignment 3 Due at 11:00am on Thursday, July 31, 2008 View Grading Details [104%] Definitions of Acids and Bases is for 1 point(s) [102%] Conjugate Pairs is for 1 point(s) [103%] Sulfuric Acid is for 1 point(s) [1
Logan University - CHEM - UG07120_02
Logan University - CHEM - UG07120_02
[ Print View ] General Chemistry II Assignment 5 Due at 11:00am on Tuesday, August 12, 2008 View Grading DetailsCommon Ion Effect on SolubilityLead thiocyanate, Part A , has a of .Calculate the molar solubility of lead thiocyanate in pure water.
Logan University - CHEM - UG07120_02
[ Print View ] General Chemistry II Assignment 4 Due at 11:00am on Wednesday, August 6, 2008 View Grading DetailsAdding a Strong Acid to a BufferLearning Goal: To understand how buffers use their reserves of conjugate acid and conjugate base to am
Logan University - CHEM - UG07120_02
Logan University - CHEM - UG07120_02
Logan University - CHEM - UG07120_02
Logan University - CHEM - UG07120_02
Logan University - CHEM - UG07120_02
Logan University - CHEM - UG07120_02
Logan University - CHEM - UG07120_02
Logan University - CHEM - UG07120_02
Logan University - CHEM - UG07120_02
Logan University - CHEM - UG07120_02
Logan University - CHEM - UG07120_02
Logan University - CHEM - UG07120_02
Logan University - CHEM - UG07120_02
Logan University - CHEM - UG07120_02
Logan University - CHEM - UG07120_02
Logan University - CHEM - UG07120_02
Logan University - CHEM - UG07120_02
Logan University - CHEM - UG07120_02
Logan University - CHEM - UG07120_02
Logan University - CHEM - UG07120_02
Logan University - CHEM - UG07120_02
Logan University - CHEM - UG07120_02
Logan University - CHEM - UG07120_02
Logan University - CHEM - UG07120_02
Logan University - CHEM - UG07120_02
Logan University - CHEM - UG07120_02
Logan University - CHEM - UG07120_02
Logan University - CHEM - UG07120_02
Logan University - CHEM - UG07120_02
Logan University - CHEM - UG07120_02
Logan University - CHEM - UG07120_02
Logan University - CHEM - UG07120_02
Logan University - CHEM - UG07120_02
Logan University - CHEM - UG07120_02
Logan University - CHEM - UG07120_02
Logan University - CHEM - UG07120_02
Logan University - CHEM - UG07120_02
Logan University - CHEM - UG07120_02
Logan University - CHEM - UG07120_02
Logan University - CHEM - UG07120_02
Logan University - CHEM - UG07120_02
Logan University - CHEM - UG07120_02
Logan University - CHEM - UG07120_02
Logan University - CHEM - UG07120_02
Logan University - CHEM - UG07120_02
Logan University - CHEM - UG07120_02
Logan University - CHEM - UG07120_02
Logan University - CHEM - UG07120_02
Logan University - CHEM - UG07120_02
Logan University - CHEM - UG07120_02