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Assignment 5.1 Gas Laws Topic Materials: Boyle’s Law 2 https://www.youtube.com/watch?v=NB1aCBId6qA Chemistry: Charles’ Law (Gas Laws) with 2 examples https://www.youtube.com/watch? v=7ZpuMBkf1Ss Dalton’s Law and Partial Pressures https://www.youtube.com/watch?v=jbmOcH1fFKs Respiration Partial Pressure Changes https://www.youtube.com/watch?v=332FPuvNXZU Gases Intro https://phet.colorado.edu/sims/html/gases-intro/latest/gases-intro_en.html Chemistry: An Introduction to General, Organic, and Biological Chemistry Review Chapter 8; sections 8.2, 8.3, and 8.7 in Chemistry: An Introduction to General, Organic, and Biological Chemistry . http://www.gcumedia.com/digital-resources/pearson/2017/chemistry_an-introduction-to-general- organic-and-biological-chemistry_13e.php Boyle’s Law Go to http://phet.colorado.edu/simulations/ , find the simulation links in the Chemistry simulations, and open the Gases Intro Simulation . Open the simulation by clicking the “play” button and then select Laws . Click on the Width. This will cause a ruler to appear. The ruler’s units are in nanometers (nm) but we are going to use the ruler to give us an estimated measurement of volume. You will use the ruler to measure the width of the box. We will then change the units of measurement to liters . For example: initially the box should have a width of 5.5 nm which will be recorded in your data table as 5.5 L (liters). When you are asked to change/measure the volume of the box, use the ruler to do so. First, you need to add a gas to your container. Click on the handle of the pump and add ONE PUMPFUL of gas to your container. Locate the Particles section on the right and click on the “+” sign. How many gas particles did you add to your container? 50 gas particles. What type of gas did you add? Heavy gas particles. Describe the motion of the particles: The particles are moving at random motion and at different speeds. The particles are colliding both with the walls and each other. Boyle’s Law looks at the relationship between volume and pressure when there is a constant temperature. You must set your container to constant temperature. Click on the Temperature button in the Hold Constant section on the upper right corner. This will cause the temperature to automatically adjust to whatever the initial value is set at.

Set your temperature to constant. What is the temperature of your box? 300 Kelvin. You are going to adjust the volume of the container by clicking on the handle on the left side of the container and dragging it to various widths. Change the gas to 100 molecules of the HEAVY species by manually setting this in the right box. According to the Kinetic Molecular Theory, what action causes pressure on the inside of the container? The Kinetic Molecular Theory states that the pressure on the inside of the container is the total force exerted by the gas particles on the wall upon collision. Hypothesize: If you will make the container smaller, how will this affect the answer to the previous question? If the container were to become smaller, the gas particles will be colliding with the wall much more frequently than before. Thus, this leads to an increase in pressure. If you make it larger? Conversely, if the container were to become larger, the gas particles will be colliding with the wall less frequently than before. Thus, this leads to a decrease in pressure. Fill in the following chart by selecting various Volumes . Measure the volume of the container using the ruler. Calculate the values as indicated in the other columns. Trials Volume (V) Pressure (P) Calculate k 1 = ( PxV ) Calculate k 2 = P V Trial 1 15.0 L 7.8 atm 117 L·atm 0.62 atm/L 2 12.5 L 9.3 atm 116.25 L·atm 0.744 atm/L 3 10.0 L 11.7 atm 117 L·atm 1.17 atm/L 4 7.5 L 15.7 atm 117.75 L·atm 2.09 atm/L 5 6.3 L 18.6 atm 117.18 L·atm 2.95 atm/L 6 5.0 L 23.4 atm 117 L·atm 4.68 atm/L Which variable did you control (independent)? The independent variable was the volume since it was the one being changed.

Show the k -value calculation: Point #1: k 1 = P 1 V 1 = (7.8 atm)(15.0 L) = 117 L·atm Point #2: k 1 = P 2 V 2 = (11.7 atm)(10.0 L) = 117 L·atm Write an equation for Boyles’s Law : P 1 V 1 = P 2 V 2 where P 1 is the initial pressure, V 1 is the initial volume, P 2 is the final pressure, and V 2 is the final volume. We can use this formula to predict the pressure (P 2 ) or volume (V 2 ) of any gas. Use this formula to complete the following calculations. Show your work. 1) If a gas has a volume of 1.25 L and a pressure of 1.75 atm, what will the pressure be if the volume is changed to 3.15 L? P 1 = 1.75 atm, V 1 = 1.25 L, V 2 = 3.15 L P 1 V 1 = P 2 V 2 (1.75 atm)(1.25 L) = P 2 (3.15 L) P 2 = (1.75 atm)(1.25 L)/(3.15 L) P 2 = 0.69 atm The pressure will be 0.69 atm if the volume is changed to 3.15 L. 2) If a gas has a volume of 3.67 L and a pressure of 790 mm Hg, what will the pressure be if the volume is compressed to 2.12 L? What is the pressure in atmospheres (atm)? Convert pressure units. P 1 = 790 mm Hg, V 1 = 3.67 L, V 2 = 2.12 L P 1 V 1 = P 2 V 2 (790 mm Hg)(3.67 L) = P 2 (2.12 L) P 2 = (790 mm Hg)(3.67 L)/(2.12 L) P 2 = 1368 mm Hg P 2 = 1368 mm Hg x (1 atm/760 mm Hg) = 1.80 atm The pressure will be 1.80 atm after the change in volume. 3) A container has a volume of 5.85 L and a pressure of 4.25 atm. What will the volume be if the container’s pressure is changed to 2.75 atm?

P 1 = 4.25 atm, V 1 = 5.85 L, P 2 = 2.75 atm P 1 V 1 = P 2 V 2 (4.25 atm)(5.85 L) = (2.75 atm)V 2 V 2 = (4.25 atm)(5.85 L) /(2.75 atm) V 2 = 9.04 L The volume will become 9.04 L if the container’s pressure is changed to 2.75 atm. 4) A container has a volume of 2.79 L and a pressure of 5.97 atm. If the pressure changes to 1460 mm Hg, what is the container’s new volume? Convert pressure units. P 1 = 5.97 atm, V 1 = 2.79 L P 2 = 1460 mm Hg x (1 atm/760 mm Hg) = 1.92 atm P 1 V 1 = P 2 V 2 (5.97 atm)(2.79 L) = (1.92 atm)V 2 V 2 = (4.25 atm)(5.85 L) /(2.75 atm) V 2 = 12.9 L The container’s new volume will be 12.9 L. 5) Using the information collected in the Boyle’s Law simulation, what would be the number of moles (n) present at each volume? The number of moles of gas particles can be calculated using the Ideal Gas Law: PV = nRT where P is pressure (in atm); V is volume (in L); n is the number of gas particles (in moles); T is the temperature (in K); and R is gas constant (0.0821 L atm / mol K). a. Trial 1 15.0 L n = 4.75 mol b. Trial 2 12.5 L n = 4.72 mol c. Trial 3 10.0 L n = 4.75 mol d. Trial 4 7.5 L n = 4.78 mol e. Trial 5 6.3 L n = 4.76 mol f. Trial 6 5.0 L n = 4.75 mol 6) Does the value of n change at different volumes? Why or why not? The number of moles is constant at different volumes. This is due to the fact that PV or k 1 is constant as well as RT for all the trials. This would lead to a constant number of moles. This just shows that change in the volume of the container will not change the number of particles in the container.

Trials Temperature (T) Volume (V) Calculate k 1 = ( VxT ) Calculate k 2 = V T Trial 1 400 K 13.3 L 5320 K·L 0.033 L/K 2 350 K 11.7 L 4095 K·L 0.033 L/K 3 300 K 10.0 L 3000 K·L 0.033 L/K 4 250 K 8.3 L 2075 K·L 0.033 L/K 5 200 K 6.7 L 1340 K·L 0.034 L/K 6 150 K 5.0 L 750 K·L 0.033 L/K Which variable did you control (independent)? The independent variable was the temperature since it was the one being changed. Which variable is the dependent variable? The dependent variable was the volume since it was the one being measured after changing the independent variable (i.e., temperature). Graph Volume vs. Temperature in the following graph. Use proper scaling. Label the graph appropriately. Graph the line of best fit. In the graph above, the scale used is 1 unit: 100 K for the x-axis and 1 unit:2.5 L for the y-axis. Volume (L) 25 20 15 10 5 Temperature (K) 600 500 400 300 200 100 0

Looking at your data and graph, describe the relationship between temperature and volume. Based on the graph, the temperature and volume are directly proportional. This means that as temperature increases, volume also increases; conversely, as temperature decreases, volume also decreases. As the temperature gets colder and approaches 0 Kelvin, what happens to the volume of the gas? As the temperature gets colder and approaches 0 Kelvin, the volume also approaches 0 liters. Which value remains consistent in the data table? k 1 or k 2 This k-value is constant; the ratio between volume and temperature of any point on the graph will be the same. Pick any two points from the graph or table: Point #1 Point #2 V 1 = 10.0 L V 2 = 5.0 L T 1 = 300 K T 2 = 150 K Do the calculation: V 1 T 1 = k V 2 T 2 = k (10.0 L)/(300 K) = 0.033 L/K (5.0 L)/(150 K) = 0.033 L/K Since both fractions equal “k”, then we can conclude that: V 1 T 1 = V 2 T 2 This is Charles’s Law. We can use this formula to predict the volume (V 2 ) or temperature (T 2 ) of any gas. Use this formula to complete the following calculations. 8) If a gas has a volume of 1.25 L at a temperature of 300 K, what will the volume change to if the container is cooled to 200 K? V 1 = 1.25 L, T 1 = 300 K, T 2 = 200 K V 1 /T 1 = V 2 /T 2 (1.25 L)/(300 K) = V 2 /(200 K)

Dalton’s Law of Partial Pressure After reviewing the Dalton’s Law and Partial Pressures resource listed in the study materials complete the following questions. 12) If a scuba tank at 675 mmHg is made up of a mixture of 79.0% nitrogen and 21.0% oxygen what is the partial pressure of each the nitrogen and the oxygen in the mixture? To determine the partial pressure of the two gases, we will simply multiply their corresponding percentage composition to the total pressure. Total Pressure, P T = 675 mm Hg P N2 = (79.0%)(675 mm Hg) = 533.25 mm Hg P O2 = (21.0%)(675 mm Hg) = 141.75 mm Hg The partial pressure of nitrogen and oxygen in the scuba tank are 533.25 mm Hg and 141.75 mm Hg, respectively. 13) Explain how Dalton’s Law of Partial Pressure relates to arterial blood gases. The Dalton’s Law of Partial Pressure governs how the transport of gas molecules or diffusion in the blood vessels. Gas molecules move based on the partial pressure gradient between two parts of the respiration. That is, gas molecules will move from a segment with higher partial pressure to a segment with lower partial pressure. For example, the partial pressure of Oxygen and Carbon dioxide in the alveolus are 104 and 40, respectively. On the other hand, the partial pressure of Oxygen and Carbon dioxide in the artery are 40 and 45, respectively. Because the partial pressure of Oxygen is lower in the artery, the Oxygen molecules will diffuse from the alveolus to the artery. Similarly, since the partial pressure of Carbon dioxide is lower in the alveolus, the Carbon Dioxide will diffuse from the artery to the alveolus. Once the partial pressures of the a gas is equal on both sides of the gradient, there will be no more net movement of gas molecules.