POGIL EXERCISE 18 All You Need to Know About Gas Laws

POGIL 18 Page 1 of 11 POGIL EXERCISE 18 All You Need to Know About Gas Laws Each member should assume his or her role at this time. The new manager takes charge of the POGIL folder and hands out the GRF and RRF to the appropriate members. The new recorder should record the names of the group members on the new GRF. Table 1. Group Member Role Assignments GROUP TYPE -> GROUPS OF THREE GROUPS OF FOUR MEMBER NO. -> 1 2 3 1 2 3 4 Manager + + Reporter + + Recorder + + Reflector + + Technician Encourager + + SFUC * + + * OBSERVATION IA. As you know, a gas has no shape or volume of its own. The shape of the gas depends on the container. The condition where the container is open to the outside environment is known as an open system and the system where the container is sealed is called a closed system. The volume of the gas in the container depends on two environmental conditions: the absolute temperature (K) and the pressure (P). The absolute temperature is an important variable because the kinetic energy (energy of motion) of the molecules increases proportionately with the absolute temperature. As the molecules move faster, they take up more room in the container and thus the volume will tend to increase in an open system which causes some gas to leave the container. When gas leaves the container it means that the amount of gas in the container decreases. On the other hand, in a closed system, the gas cannot escape and so the movement of the molecules causes the molecules to exert more pressure on the sides of the container. 1. Suppose you have 10 L of a gas in an open system. If the temperature were to decrease around the container, would the amount of the gas in the container be higher or lower than the starting conditions? Explain your answer. 5 MIN

POGIL 18 Page 2 of 11 2. Suppose you have 10 L of a gas in an open system. If the temperature were to increase around the container, would the volume of the gas in the container be higher or lower than the starting conditions? Explain your answer. Please discuss your answers to Items #1 and #2 with your instructor before continuing. OBSERVATION II. The overall conclusion to be gathered from Observation I is that the difference between an open and closed system is that in an open system, the amount (i.e. number of moles) of gas can vary and, in a closed system, the amount (i.e. number of moles) of gas is a constant. Thus, there are two basic gas laws to allow quantification of components of either the open or closed system. The open system can be handled only by the ideal gas law. Although the closed system can also be handled by the ideal gas law, it is far more convenient to use the combined gas law to calculate components in a closed system. 3. Supposed you needed 100 L of hydrogen gas at 25 C and 1 atmosphere (atm) to fill a small balloon. You know you can generate hydrogen by reacting zinc with an acid. You need to know how much zinc and acid you need to generate 100 L of hydrogen. Which of the two basic laws would you use to calculate the amounts of reactants you need? What is your rational for this answer? 4. Suppose you added the 100 L of hydrogen to the balloon and it rose to an altitude of 20,000 ft. where the pressure was 0.8 atm and the temperature was 5 C and you wanted to know what the volume of the balloon would be under those conditions. Which of the two basic laws would you use to calculate the volume? What is your rational for this answer? 10 MIN Please discuss your answers to Items #6 and #7 with your instructor before continuing.

POGIL 18 Page 3 of 11 OBSERVATION III. Let us now consider the quantitative relationship between pressure (P), volume (V) and temperature in Kelvin (T) in a closed system. These relationships are always used when a constant (but not necessarily known) amount of gas at a certain volume, temperature and pressure is changed to a different volume, temperature and or pressure. The combined gas law is described by Equation 1 below: EQ1: VsPs = VfPf OR VsPs = VfPf The subscript, s, in EQ1 stands for starting conditions and the subscript, f, stands for the final conditions. Just to emphasize the point: All the parameters can change, but in a closed system, the amount (i.e. the number of moles) of the gas does not. Let s take this equation for a ride: A container of nitrogen (or any other gas) with a volume of 1.58 L at a pressure of 1.21 atm and a temperature of 12.2 C was heated to 32.3 C and the pressure was increased to 1.54 atm. What is the new volume in the container? Looks kind of complicated doesn t it? OK, let s take it one step at a time. a. Because we are changing a gas from one set of conditions to another we know that we are dealing with a closed system; therefore, we must use the combined gas law (EQ1) to solve the problem: VsPs = VfPf OR VsPs = VfPf b. We make out a variable table and fill in the values given in the problem: VsPs = VfPf OR VsPs = VfPf Vs = 1.58 L_ Vf =?? Ps = 1.21 atm Pf = _1.54 atm = 12.2 C = _32.3 C c. After filling out the variable table we learn that we are solving Equation 1 for the volume after the conditions were changed (Vf). So we solve the equation for that variable before we do anything else. VsPs = VfPf; Vf = VsPs 20 MIN Pf

POGIL 18 Page 4 of 11 d. We notice that the given temperature is in centigrade; however, we know that the temperature must be in Kelvin work in any gas law. So we convert the temperatures to their kelvin equivalents using the following equivalence equation: K = 273 + C. K = 273 + 12.2 = 285.2 and K = 273 + 32.3 = 305.3 and plug the values into our variable table: Vf = VsPs Pf Vs = 1.58 L_ Vf =?? Ps = 1.21 atm Pf = _1.54 atm = 285.2 K = _305.3 K e. Using the second form of EQ1, we solve the expression algebraically, then plug and chug. VsPs = VfPf OR VsPs = VfPf Vf = VsPs = (305.3 K)(1.58 L)(1.21 atm) Pf (285.2 K)(1.54 atm) Vs = 1.58 L_ Vf =?? Ps = 1.21 atm Pf = _1.54 atm = 285.2 K = _305.3 K = 1.3289 L = 1.33 L Solve the following problems by using the above example and process 5. Exactly 141 ml of oxygen at 721 torr and 135 K was subjected to 801 mm Hg pressure with a resulting volume of 0.152 L. What was the temperature after the change of conditions? a. Fill in all the known values in the spaces provided to the right. b. Solve Equation 4 for the unknown parameter, plug in the values, and solve the problem. Remember to include units in the calculations. Vs = Vf = Ps = Pf = = = +35 MIN

POGIL 18 Page 5 of 11 6. Helium, 879 ml, under 5.51 atm and 22.1 C was increased in volume to 1.05 L and a temperature of 101 F. What was the pressure in the container after the change? a. Fill in all the known values in the spaces provided to the right. b. Solve Equation 4 for the unknown parameter, plug in the values, and solve the problem. Remember to include units in the calculations. Vs = Vf = Ps = Pf = = = OBSERVATION IV. The name for EQ1, the combined gas law, is derived from the fact that not all the relationships incorporated into the equation were known at the same time. The first special case was described by Boyle and became known as Boyle s Law. Through experimentation Boyle determined that the volume of a gas at constant temperature is inversely proportional to the pressure; i.e., if you increase the pressure and bleed off the excess kinetic energy to keep the temperature constant, the volume will decrease proportionally. The mathematical derivation of the special case described by Boyle s law is easily accomplished using the combined gas law: EQ3: VsPs = VfPf Since the temperatures are the same, they cancel out and you have expression for Boyle s law (Equation 2). EQ4: VsPs = VfPf BOYLES LAW. Note: It is important for the student to know the definition of Boyle s law and that it is a special case of a closed system. It is not necessary for the student to remember the formula since all Boyle s Law problems can be worked with the combined gas Law. +35 MIN

POGIL 18 Page 6 of 11 Let s look at a problem that utilizes this concept. 7. What would be the resulting pressure if you have 44.5 L of a gas at 1.22 atm and you expanded the volume to 178 L? a. Fill in the data table at right: b. Which of the four parameters is missing? c. Solve Equation 1 for this parameter and then substitute the known values. Remember to include the units as well as the numerical value in the calculations. Variable Table Vs = Vf = Ps = Pf = = = Present your answers to the instructor for validation before continuing with processing. OBSERVATION IV: The next special case of the combined gas law describes the relationship in a closed system with a constant amount of gas between temperature and volume at CONSTANT PRESSURE. 1. This direct proportional relationship of volume to temperature at constant pressure is known as Charles s Law. Charles s law can also be derived from the combined gas law: EQ5: VsPs = VfPf leads to EQ6: VS = VF. OR EQ3: VSTF = VFTS CHARLES S LAW TS TF This relationship is true only if T is expressed as the absolute temperature (K). In fact, in all gas laws where temperature is included as operator, the temperature must be expressed as the absolute temperature (K). +40 MIN

POGIL 18 Page 7 of 11 Note: It is important for the student to know the definition of Charles s law and that it is a special case of a closed system. It is not necessary for the student to remember the formula since all Charles s Law problems can be worked with the combined gas Law. Let s now learn to use this relationship solve by solving a problem: 8. What would be the resulting temperature if you had 6.11 L of a gas at 21.2 C and you expanded the volume to 25.90 L? Variable Table OBSERVATION V. The final special case defines the relationship between temperature and pressure at CONSTANT VOLUME in a closed container with a constant amount of gas. This direct proportional relationship of pressure to temperature at CONSTANT VOLUME is known as Gay-Lussac s Law. Since the amount of gas is constant, Gay- Lussac s law allows us to calculate the temperature or pressure at any condition. Thus, this law too can be derived directly from the combined gas law: EQ7: VsPs = VfPf leads to EQ8: PS = PF. TS TF OR PS*TF = PF*TS GAY-LUSSAC S LAW 45 MIN

POGIL 18 Page 8 of 11 9. What would be the resulting temperature if you had a gas at 1.00 atm and 37 C and you increased the pressure to 10 atm and kept the volume the same? Variable Table OBSERVATION VI. There are many situations in our environment where there are mixtures of gases air for example. Dry air at sea level is composed 78% nitrogen, 21% oxygen, 0.9% argon and 0.04% carbon dioxide. One might ask the question as to the pressure exerted by each of these gases. Since temperature and volume of the mixture are constants, this can also be considered a closed system question. Dalton studied this situation and in the end proposed his Law of Partial Pressure that says the total pressure of a gaseous mixture is equal to the sum of the pressure exerted by each gas in the mixture (EQ10). EQ9: Ttot = Pa + Pb + Pc. So if the atmospheric pressure is equal to one atm, it means that the pressure component of nitrogen (partial pressure of nitrogen) is equal to 0.78 atm (1 atm x 78/100). Let s use this new law to solve some problems 10. A gas mixture contains each of the following gases with their partial pressures in parenthesis: N2 (285 torr), O2 (116 torr), He (267 torr). What is the total pressure of this mixture? List of Knowns: 50 MIN

POGIL 18 Page 9 of 11 11. A heliox (a commercial mixture of helium and oxygen) deep-sea diving mixture delivers a partial pressure of oxygen of 0.30 atm when the total pressure of the mixture is 11.0 atm. What is the partial pressure of helium in the mixture? List of Knowns: 55 MIN OBSERVATION VI. Let us now consider the open system problems. As stated in Observation I, these types of problems can only be solved using the Ideal Gas Law. The primary variable to concentrate on when dealing with these problems is the number of moles of the gas (the amount of gas) which must be known or calculated to work the problem. The Ideal Gas Law defines the relationship between the absolute temperature (T), the pressure (P) which must be expressed in atmospheres, the volume (V) always expressed in liters and the number of moles (n) of the gas present. This relationship is seen in Equation 6. EQ17: PV = nrt R = 8.21 x 10-2 L atm/nk IDEAL GAS LAW R is one of the marvels of the universal constants of nature and is the reason why for our problems the volume, temperature and pressure must be converted to liters, Kelvin, and atmospheres, respectively. The ideal gas law can be rearranged to give yet another definition of a mole: EQ18: n = PV/RT The ideal gas law can also be used to calculate volume of one mole of any gas under standard temperature and pressure (STP). STP is 0 C and 1 atm. PV = nrt; V = nrt = (1 n)( 8.21 x 10-2 L atm/nk)(2.73 x 10 2 K) = 22.4 L P 1 atm

POGIL 18 Page 10 of 11 Supposed you needed 100 L of hydrogen gas at 27 C and 1 atmosphere (atm) to fill a small balloon. You know you can generate hydrogen by reacting zinc with and hydrochloric acid. You need to know the minimum amount zinc and acid required to generate 100 L of hydrogen. a. First you have to recognize that this is a stoichiometric problem that must be solved from knowing how many moles of molecular hydrogen are in 100 L of the gas at 27 C and 1 atm. Since it involves moles, you must use the Ideal Gas Law, rearranging to solve for n and then plug and chug. PV = nrt; n = PV. = (1 atm)(100 L). = 4.06 n H2 RT ( 8.21 x 10-2 L atm/nk)(3.00 x 10 2 K) b. Write a balanced molecular equation for the reaction: Zn + 2 HCl ZnCl2 + H2 c. We convert n hydrogen to moles Zn using the conversion factor from the balanced equation: 4.06 n H2 x n Zn = 4.06 n Zn n H2 d. Finally we calculate the grams of Zn we need: 4.06 n Zn x 65.39 g = 16.484 g n Zn e. For HCl we first find n HCl using the conversion factor of the balanced equation 4.06 n H2 x 2n HCl = 8.12 n HCl n H2 f. Finally, since HCl comes as a 12.0 M solution, we have to calculate the minimum volume of concentrated HCl required: n = M x L; L = n/m = 8.12 n HCl = 0.677 L 12.0 M HCl 12.0 n/l Let s work a problem using the Ideal Gas Law: 60 MIN

POGIL 18 Page 11 of 11 12. What volume of hydrogen at 25 C and 740 mmhg can be produced from 100 g of magnesium with excess sulfuric acid? List of Knowns: Present your results to the instructor for validation. Please complete the matching exercise in Table 4. Table 4. Matching Conditions with Gas Laws Condition 13. Relates volume and temperature when pressure is constant. 14. Relates partial pressures of a gaseous mixture to total pressure. 15. The product of the volume and pressure is a constant amount of gas. 16. Relates volume, temperature and pressure in a closed system. 17. Relates pressure and volume at constant temperature 18. VsPs = VfPf Gas Law a. Boyle s Law. b. Charles s Law c. Gay-Lussac s Law d. Dalton s Law of Partial Pressure e. Combined Gas Law +65 MIN 19. VsPs = VfPf 20. T = PV/nR Present your results to the instructor for validation. EXERCISE END. Managers should collect the GRF and RRF, staple them together, and place in the back of the left pocket of the folder. The folder should be closed and left on the table. NOTIFY INSTRUCTOR WHEN FINISHED. Wait for further instructions.