11. Gases - Chapter Overview PDF

11: GASES  CHAPTER OVERVIEW 11: Gases Of the three basic phases of matter—solids, liquids, and gases—only one of them has predictable physical properties: gases. In fact, the study of the properties of gases was the beginning of the development of modern chemistry from its alchemical roots. The in...

11: GASES  CHAPTER OVERVIEW 11: Gases Of the three basic phases of matter—solids, liquids, and gases—only one of them has predictable physical properties: gases. In fact, the study of the properties of gases was the beginning of the development of modern chemistry from its alchemical roots. The interesting thing about some of these properties is that they are independent of the identity of the gas. That is, it doesn’t matter if the gas is helium gas, oxygen gas, or sulfur vapors; some of their behavior is predictable and very similar. In this chapter, we will review some of the common behaviors of gases. Gases have no definite shape or volume; they tend to fill whatever container they are in. They can compress and expand, sometimes to a great extent. Gases have extremely low densities, a one-thousandth or less of the density of a liquid or solid. Combinations of gases tend to mix together spontaneously—that is, they form solutions. Air, for example, is a solution of mostly nitrogen and oxygen. Any understanding of the properties of gases must be able to explain these characteristics. 11.1: Extra-Long Straws 11.2: Kinetic Molecular Theory- A Model for Gases 11.3: Pressure - The Result of Constant Molecular Collisions 11.4: Boyle’s Law - Pressure and Volume 11.5: Charles’s Law- Volume and Temperature 11.6: The Combined Gas Law- Pressure, Volume, and Temperature 11.7: Avogadro’s Law- Volume and Moles 11.8: The Ideal Gas Law- Pressure, Volume, Temperature, and Moles 11.9: Mixtures of Gases - Why Deep-Sea Divers Breathe a Mixture of Helium and Oxygen 11.11: Gay-Lussac's Law- Temperature and Pressure 11: Gases is shared under a CK-12 license and was authored, remixed, and/or curated by Marisa Alviar-Agnew & Henry Agnew. 1 11.1: EXTRA-LONG STRAWS A drinking straw is a tube for transferring a beverage from its container to the mouth of the drinker and is typically a thin tube of plastic (such as polypropylene and polystyrene) or other material. Many people believe that when they drink a liquid they are sucking the liquid up, however the liquid is really being pushed up. A straw works because when you suck the air out of the straw, it creates a vacuum. This causes a decrease in air pressure on the inside of the straw. Since the atmospheric pressure is greater on the outside of the straw, liquid is forced into and up the straw and into your mouth (Figure 11.1.1 ). Figure 11.1.1: As you suck the air out of the straw, it creates a low pressure zone inside of it. With a low pressure zone, there is nothing pushing down on the juice, so it moves upward easily. (CC BY-NC 4.0; Ümit Kaya)  HOW LONG OF A STRAW IS POSSIBLE? With the straw just sitting in the glass, the pressure on the surface of the tea is the same all over, including on the little bit of surface inside the straw. When you suck the air out of the straw, you decrease the pressure inside the straw, allowing the higher pressure on the rest of the surface to push the tea up the straw and into your mouth. Because it is really the atmosphere that is doing the pushing, the atmospheric pressure limits how high water will go up a straw. If you formed a perfect vacuum within the straw, the pressure outside of the straw at sea level would be enough to push water to a total height of about 10.3 m. A 10.3-m column of water exerts the same pressure—101,325 N/m2 or 14.7 lb/in2 (psi)—as do the gas molecules in our atmosphere. At sea level, the air pressure is enough to support a column of water about thirty feet high. This means that even if you could suck all the air out of a forty-foot straw, the water would not rise more than thirty feet. CONTRIBUTIONS & ATTRIBUTIONS Lisa Peck's Conceptual Physics Class 11.1: Extra-Long Straws is shared under a CC BY-SA 3.0 license and was authored, remixed, and/or curated by Marisa Alviar-Agnew & Henry Agnew. 11.1.1 https://chem.libretexts.org/@go/page/47523 11.2: Kinetic Molecular Theory- A Model for Gases  Learning Objectives State the major concepts behind the kinetic theory of gases. Relate the general properties of gases to the kinetic theory. Gases were among the first substances studied in terms of the modern scientific method, which was developed in the 1600s. It did not take long to recognize that gases all shared certain physical behaviors, suggesting that all gases could be described by one all- encompassing theory. Today, that theory is the kinetic theory of gases. It is based on the following statements: 1. Gases consist of tiny particles of matter that are in constant motion. 2. Gas particles are constantly colliding with each other and the walls of a container. These collisions are elastic; that is, there is no net loss of energy from the collisions. 3. Gas particles are separated by large distances, with the size of a gas particle tiny compared to the distances that separate them. 4. There are no interactive forces (i.e., attraction or repulsion) between the particles of a gas. 5. The average speed of gas particles is dependent on the temperature of the gas. Figure 11.2.1 shows a representation of how we mentally picture the gas phase. Figure 11.2.1 : The Kinetic Theory of Gases. The kinetic theory of gases describes this state of matter as composed of tiny particles in constant motion with a lot of distance between the particles. This model of gases explains some of the physical properties of gases. Because most of a gas is empty space, a gas has a low density and can expand or contract under the appropriate influence. The fact that gas particles are in constant motion means that two or more gases will always mix, as the particles from the individual gases move and collide with each other. An ideal gas is a gas that exactly follows the statements of the kinetic theory. Unfortunately, real gases are not ideal. Many gases deviate slightly from agreeing perfectly with the kinetic theory of gases. However, most gases adhere to the statements so well that the kinetic theory of gases is well accepted by the scientific community. The physical behavior of gases is explained by the kinetic theory of gases. An ideal gas adheres exactly to the kinetic theory of gases. 11.2: Kinetic Molecular Theory- A Model for Gases is shared under a CC BY-NC-SA 3.0 license and was authored, remixed, and/or curated by Marisa Alviar-Agnew & Henry Agnew. 11.2.1 https://chem.libretexts.org/@go/page/47525 11.3: Pressure - The Result of Constant Molecular Collisions  Learning Objectives Define pressure. Learn the units of pressure and how to convert between them. The kinetic theory of gases indicates that gas particles are always in motion and are colliding with other particles and the walls of the container holding them. Although collisions with container walls are elastic (i.e., there is no net energy gain or loss because of the collision), a gas particle does exert a force on the wall during the collision. The accumulation of all these forces distributed over the area of the walls of the container causes something we call pressure. Pressure (P ) is defined as the force of all the gas particle/wall collisions divided by the area of the wall: force pressure = area All gases exert pressure; it is one of the fundamental measurable quantities of this phase of matter. Even our atmosphere exerts pressure—in this case, the gas is being “held in” by the earth’s gravity, rather than the gas being in a container. The pressure of the atmosphere is about 14.7 pounds of force for every square inch of surface area: 14.7 lb/in2. Pressure has a variety of units. The formal, SI-approved unit of pressure is the pascal (Pa), which is defined as 1 N/m2 (one newton of force over an area of one square meter). However, this is usually too small in magnitude to be useful. A common unit of pressure is the atmosphere (atm), which was originally defined as the average atmospheric pressure at sea level. However, “average atmospheric pressure at sea level” is difficult to pinpoint because of atmospheric pressure variations. A more reliable and common unit is millimeters of mercury (mmHg), which is the amount of pressure exerted by a column of mercury exactly 1 mm high. An equivalent unit is the torr, which equals 1 mmHg. (The torr is named after Evangelista Torricelli, a seventeenth-century Italian scientist who invented the mercury barometer.) With these definitions of pressure, the atmosphere unit is redefined: 1 atm is defined as exactly 760 mmHg, or 760 torr. We thus have the following equivalents: 1 atm=760 mmHg=760 torr We can use these equivalents as with any equivalence—to perform conversions from one unit to another. Relating these to the formal SI unit of pressure, 1 atm = 101,325 Pa.  Example 11.3.1: Pressure Conversion How many atmospheres are there in 595 torr? Solution Solutions to Example 11.3.1 Steps for Problem Solving Unit Conversion Identify the "given” information and what the problem is asking you Given: 595 torr to "find." Find: ? atm List other known quantities. 1 atm = 760 torr Prepare a concept map. 1 atm Cancel units and calculate. 595 torr × = 0.783 atm 760 torr 595 torr is less than 760 torr so the final answer should be less than 1 Think about your result. atm. 11.3.1 https://chem.libretexts.org/@go/page/47527   Exercise 11.3.1 How many atmospheres are there in 1,022 torr? Answer 1.345 atm  Example 11.3.2: Mars The atmosphere on Mars is largely CO2 at a pressure of 6.01 mmHg. What is this pressure in atmospheres? Solution Solutions to Example 11.3.2 Steps for Problem Solving Unit Conversion Identify the "given” information and what the problem is asking you Given: 6.01mmHg to "find." Find: ? atm List other known quantities. 1 atm = 760 mmHg Prepare a concept map. 1 atm −3 Cancel units and calculate. 6.01 mmHg × = 0.00791 atm = 7.91 × 10 atm 760 mmHg 6.01 is a very small number relative to 760 mmHg, just like the value Think about your result. in atmospheres.  Exercise 11.3.2 Atmospheric pressure is low in the eye of a hurricane. In a 1979 hurricane in the Pacific Ocean, a pressure of 0.859 atm was reported inside the eye. What is this pressure in torr? Answer 652 torr Summary Pressure is a force exerted over an area. Pressure has several common units that can be converted. 11.3: Pressure - The Result of Constant Molecular Collisions is shared under a CK-12 license and was authored, remixed, and/or curated by Marisa Alviar-Agnew & Henry Agnew. 11.3.2 https://chem.libretexts.org/@go/page/47527 11.4: Boyle’s Law - Pressure and Volume  Learning Objectives Learn what is meant by the term gas laws. Learn and apply Boyle’s Law. When seventeenth-century scientists began studying the physical properties of gases, they noticed some simple relationships between some of the measurable properties of the gas. Take pressure (P) and volume (V), for example. Scientists noted that for a given amount of a gas (usually expressed in units of moles [n]), if the temperature (T) of the gas was kept constant, pressure and volume were related: as one increases, the other decreases. As one decreases, the other increases. This means that pressure and volume are inversely related. There is more to it, however: pressure and volume of a given amount of gas at constant temperature are numerically related. If you take the pressure value and multiply it by the volume value, the product is a constant for a given amount of gas at a constant temperature: P × V = constant at constant n and T If either volume or pressure changes while amount and temperature stay the same, then the other property must change so that the product of the two properties still equals that same constant. That is, if the original conditions are labeled P and V and the new 1 1 conditions are labeled P and V , we have 2 2 P1 V1 = constant = P2 V2 where the properties are assumed to be multiplied together. Leaving out the middle part, we have simply P1 V1 = P2 V2 at constant n and T This equation is an example of a gas law. A gas law is a simple mathematical formula that allows you to model, or predict, the behavior of a gas. This particular gas law is called Boyle's Law, after the English scientist Robert Boyle, who first announced it in 1662. Figure 11.4.1 shows two representations of how Boyle’s Law works. Figure 11.4.1 : Boyle’s Law. A piston having a certain pressure and volume (left piston) will have half the volume when its pressure is twice as much (right piston). One can also plot P versus V for a given amount of gas at a certain temperature; such a plot will look like the graph on the right. Boyle’s Law is an example of a second type of mathematical problem we see in chemistry—one based on a mathematical formula. Tactics for working with mathematical formulas are different from tactics for working with conversion factors. First, most of the questions you will have to answer using formulas are word-type questions, so the first step is to identify what quantities are known and assign them to variables. Second, in most formulas, some mathematical rearrangements (i.e., algebra) must be performed to solve for an unknown variable. The rule is that to find the value of the unknown variable, you must mathematically isolate the unknown variable by itself and in the numerator of one side of the equation. Finally, units must be consistent. For example, in Boyle’s Law there are two pressure variables; they must have the same unit. There are also two volume variables; they also must have the same unit. In most cases, it won’t matter what the unit is, but the unit must be the same on both sides of the equation. 11.4.1 https://chem.libretexts.org/@go/page/47530   Example 11.4.1 A sample of gas has an initial pressure of 2.44 atm and an initial volume of 4.01 L. Its pressure changes to 1.93 atm. What is the new volume if temperature and amount are kept constant? Solution Solutions to Example 11.8.1 Steps for Problem Solving Given: P1 = 2.44 atm and V1 = 4.01 L Identify the "given" information and what the problem is asking you P2 = 1.93 atm to "find." Find: V2 = ? L List other known quantities. none First, rearrange the equation algebraically to solve for V. 2 Plan the problem. P1 × V1 V2 = P2 Now substitute the known quantities into the equation and solve. Cancel units and calculate. 2.44 atm × 4.01 L V2 = = 5.07 L 1.93 atm We know that pressure and volume are inversely related; as one decreases, the other increases. Pressure is decreasing (from 2.44 atm Think about your result. to 1.93 atm), so volume should be increasing to compensate, and it is (from 4.01 L to 5.07 L). So the answer makes sense based on Boyle’s Law.  Exercise 11.4.1 If P1 = 334 torr, V1 = 37.8 mL, and P2 = 102 torr, what is V2? Answer 124 mL As mentioned, you can use any units for pressure and volume, but both pressures must be expressed in the same units, and both volumes must be expressed in the same units.  Example 11.4.2: A sample of gas has an initial pressure of 722 torr and an initial volume of 88.8 mL. Its volume changes to 0.663 L. What is the new pressure? Solution Solutions to Example 11.8.2 Steps for Problem Solving Given: P1 = 722 torr and V1 = 88.8 mL Identify the "given" information and what the problem is asking you V2 = 0.633 L to "find." Find: P2 = ? torr List other known quantities. 1 L = 1000 mL to have the same units for volume. 1. Perform the conversion of the second volume unit from L to mL. 2. Rearrange the equation algebraically to solve for P. 2 Plan the problem. P1 × V1 P2 = V2 11.4.2 https://chem.libretexts.org/@go/page/47530  Steps for Problem Solving 1. 1000 ml 0.663 L × = 663 ml 1 L Cancel units and calculate. 2. Substitute the known quantities into the equation and solve. 722 torr × 88.8 mL P2 = = 96.7 torr 663 mL When the volume increased, the pressure decreased, which is as Think about your result. expected for Boyle’s Law.  Exercise 11.4.2 If V1 = 456 mL, P1 = 308 torr, and P2 = 1.55 atm, what is V2? Answer 119 mL Summary The behavior of gases can be modeled with gas laws. Boyle’s Law relates the pressure and volume of a gas at constant temperature and amount. 11.4: Boyle’s Law - Pressure and Volume is shared under a CK-12 license and was authored, remixed, and/or curated by Marisa Alviar-Agnew & Henry Agnew. 11.4.3 https://chem.libretexts.org/@go/page/47530 11.5: Charles’s Law- Volume and Temperature  Learning Objectives Learn and apply Charles's Law. Everybody enjoys the smell and taste of freshly-baked bread. It is light and fluffy as a result of the action of yeast on sugar. The yeast converts the sugar to carbon dioxide, which at high temperatures causes the dough to expand. The end result is an enjoyable treat, especially when covered with melted butter. Charles's Law French physicist Jacques Charles (1746-1823) studied the effect of temperature on the volume of a gas at constant pressure. Charles's Law states that the volume of a given mass of gas varies directly with the absolute temperature of the gas when pressure is kept constant. The absolute temperature is temperature measured with the Kelvin scale. The Kelvin scale must be used because zero on the Kelvin scale corresponds to a complete stop of molecular motion. Figure 11.5.1 : As a container of confined gas is heated, its molecules increase in kinetic energy and push the movable piston outward, resulting in an increase in volume. Mathematically, the direct relationship of Charles's Law can be represented by the following equation: V =k T As with Boyle's Law, k is constant only for a given gas sample. The table below shows temperature and volume data for a set amount of gas at a constant pressure. The third column is the constant for this particular data set and is always equal to the volume divided by the Kelvin temperature. Table 11.5.1 : Temperature-Volume Data V mL Temperature (K) Volume (mL) = k ( ) T K 50 20 0.40 100 40 0.40 150 60 0.40 200 80 0.40 300 120 0.40 500 200 0.40 1000 400 0.40 When this data is graphed, the result is a straight line, indicative of a direct relationship, shown in the figure below. 11.5.1 https://chem.libretexts.org/@go/page/47533  Figure 11.5.2 :The volume of a gas increases as the Kelvin temperature increases. Notice that the line goes exactly toward the origin, meaning that as the absolute temperature of the gas approaches zero, its volume approaches zero. However, when a gas is brought to extremely cold temperatures, its molecules would eventually condense into the liquid state before reaching absolute zero. The temperature at which this change into the liquid state occurs varies for different gases. Charles's Law can also be used to compare changing conditions for a gas. Now we use V and T to stand for the initial volume 1 1 and temperature of a gas, while V and T stand for the final volume and temperature. The mathematical relationship of Charles's 2 2 Law becomes: V1 V2 = T1 T2 This equation can be used to calculate any one of the four quantities if the other three are known. The direct relationship will only hold if the temperatures are expressed in Kelvin. Temperatures in Celsius will not work. Recall the relationship that C + 273. o K=  Example 11.5.1: A balloon is filled to a volume of 2.20 L at a temperature of 22 o C. The balloon is then heated to a temperature of 71 o C. Find the new volume of the balloon. Solution Solutions to Example 11.5.1 Steps for Problem Solving Given: V1 = 2.20 L and Identify the "given" information and what the problem is asking you o T1 = 22 C = 295 K to "find." o T2 = 71 C = 344 K Find: V2 = ? L List other known quantities. The temperatures have first been converted to Kelvin. First, rearrange the equation algebraically to solve for V. 2 Plan the problem. V1 × T2 V2 = T1 Now substitute the known quantities into the equation and solve. Cancel units and calculate. 2.20 L × 344 K V2 = = 2.57 L 295 K The volume increases as the temperature increases. The result has Think about your result. three significant figures. 11.5.2 https://chem.libretexts.org/@go/page/47533   Exercise 11.5.1 If V1 = 3.77 L and T1 = 255 K, what is V2 if T2 = 123 K? Answer 1.82 L  Example 11.5.2: A sample of a gas has an initial volume of 34.8 L and an initial temperature of −67°C. What must be the temperature of the gas for its volume to be 25.0 L? Solution Solutions to Example 11.5.2 Steps for Problem Solving Given: Identify the "given" information and what the problem is asking you Given:T1 = -27oC and V1 = 34.8 L to "find." V2 = 25.0 L Find: T2 = ? K List other known quantities. K = -27oC + 273 1. Convert the initial temperature to Kelvin 2. Rearrange the equation algebraically to solve for T.2 Plan the problem. V2 × T1 T2 = V1 1. −67°C + 273 = 206 K 2. Substitute the known quantities into the equation and solve. Cancel units and calculate. 25.0 L × 206 K T2 = = 148 K 34.8 L This is also equal to −125°C. As temperature decreases, volume Think about your result. decreases—which it does in this example.  Exercise 11.5.2 If V1 = 623 mL, T1 = 255°C, and V2 = 277 mL, what is T2? Answer 235 K, or −38°C Summary Charles’s Law relates the volume and temperature of a gas at constant pressure and amount. 11.5: Charles’s Law- Volume and Temperature is shared under a CK-12 license and was authored, remixed, and/or curated by Marisa Alviar- Agnew & Henry Agnew. 11.5.3 https://chem.libretexts.org/@go/page/47533 11.6: The Combined Gas Law- Pressure, Volume, and Temperature  Learning Objectives Learn and apply the Combined Gas Law. One thing we notice about all the gas laws is that, collectively, volume and pressure are always in the numerator, and temperature is always in the denominator. This suggests that we can propose a gas law that combines pressure, volume, and temperature. This gas law is known as the Combined Gas Law, and its mathematical form is P1 V1 P2 V2 = at constant n T1 T2 This allows us to follow changes in all three major properties of a gas. Again, the usual warnings apply about how to solve for an unknown algebraically (isolate it on one side of the equation in the numerator), units (they must be the same for the two similar variables of each type), and units of temperature must be in Kelvin.  Example 11.6.1: A sample of gas at an initial volume of 8.33 L, an initial pressure of 1.82 atm, and an initial temperature of 286 K simultaneously changes its temperature to 355 K and its volume to 5.72 L. What is the final pressure of the gas? Solution Solutions to Example 11.4.1 Steps for Problem Solving Given: Identify the "given" information and what the problem is asking you V1 = 8.33 L, P1 = 1.82 atm, and T1 = 286 K to "find." V2 = 5.72 L and T2 = 355 K Find: P2 = ? atm List other known quantities. none First, rearrange the equation algebraically to solve for V. 2 Plan the problem. P1 V1 T2 P2 = T1 V2 Now substitute the known quantities into the equation and solve. Calculate. (1.82 atm)(8.33 L )(355 K ) P2 = = 3.22atm (286 K )(5.72 L ) Ultimately, the pressure increased, which would have been difficult to Think about your result. predict because two properties of the gas were changing.  Exercise 11.6.1 If P1 = 662 torr, V1 = 46.7 mL, T1 = 266 K, P2 = 409 torr, and T2 = 371 K, what is V2? Answer 105 mL As with other gas laws, if you need to determine the value of a variable in the denominator of the combined gas law, you can either cross-multiply all the terms or just take the reciprocal of the combined gas law. Remember, the variable you are solving for must be in the numerator and all by itself on one side of the equation. 11.6.1 https://chem.libretexts.org/@go/page/47535 Summary The Combined Gas Law relates pressure, volume, and temperature of a gas. 11.6: The Combined Gas Law- Pressure, Volume, and Temperature is shared under a CK-12 license and was authored, remixed, and/or curated by Marisa Alviar-Agnew & Henry Agnew. 11.6.2 https://chem.libretexts.org/@go/page/47535 11.7: Avogadro’s Law- Volume and Moles A flat tire is not very useful. It does not cushion the rim of the wheel and creates a very uncomfortable ride. When air is added to the tire, the pressure increases as more molecules of gas are forced into the rigid tire. How much air should be put into a tire depends on the pressure rating for that tire. Too little pressure and the tire will not hold its shape. Too much pressure and the tire could burst. Avogadro's Law You have learned about Avogadro's hypothesis: equal volumes of any gas at the same temperature and pressure contain the same number of molecules. It follows that the volume of a gas is directly proportional to the number of moles of gas present in the sample. Avogadro's Law states that the volume of a gas is directly proportional to the number of moles (or number of particles) of gas when the temperature and pressure are held constant. The mathematical expression of Avogadro's Law is: V = k×n or V1 V2 = n1 n2 where n is the number of moles of gas and k is a constant. Avogadro's Law is in evidence whenever you blow up a balloon. The volume of the balloon increases as you add moles of gas to the balloon by blowing it up. If the container holding the gas is rigid rather than flexible, pressure can be substituted for volume in Avogadro's Law. Adding gas to a rigid container makes the pressure increase.  Example 11.7.1 A balloon has been filled to a volume of 1.90 L with 0.0920 mol of helium gas. If 0.0210 mol of additional helium is added to the balloon while the temperature and pressure are held constant, what is the new volume of the balloon? Solution Solutions to Example 11.11.1 Steps for Problem Solving Given: Identify the "given" information and what the problem is asking you V1 = 1.90 L to "find." n1 = 0.0920 mol Find: V 2 =? L Note that the final number of moles has to be calculated by adding the List other known quantities. original number of moles to the moles of added helium. n2 = 0.0920 + 0.0210 = 0.1130 mol First, rearrange the equation algebraically to solve for V. 2 Plan the problem. V1 × n2 V2 = n1 Now substitute the known quantities into the equation and solve. Calculate. 1.90 L × 0.1130 mol V2 = = 2.33 L 0.0920 mol Since a relatively small amount of additional helium was added to the Think about your result. balloon, its volume increases slightly. 11.7.1 https://chem.libretexts.org/@go/page/47537   Exercise 11.7.1 A 12.8 L volume of gas contains.000498 moles of oxygen gas. At constant temperature and pressure, what volume does.0000136 moles of the gas fill? Answer 0.350 L Summary Calculations for relationships between volume and number of moles of a gas can be performed using Avogadro's Law. 11.7: Avogadro’s Law- Volume and Moles is shared under a CK-12 license and was authored, remixed, and/or curated by Marisa Alviar-Agnew & Henry Agnew. 11.7.2 https://chem.libretexts.org/@go/page/47537 11.8: The Ideal Gas Law- Pressure, Volume, Temperature, and Moles  Learning Objectives Explain the Ideal Gas Law. There are a number of chemical reactions that require ammonia. In order to carry out the reaction efficiently, we need to know how much ammonia we have for stoichiometric purposes. Using gas laws, we can determine the number of moles present in the tank if we know the volume, temperature, and pressure of the system. Ideal Gas Law The Combined Gas Law shows that the pressure of a gas is inversely proportional to volume and directly proportional to temperature. Avogadro's Law shows that volume or pressure is directly proportional to the number of moles of gas. Putting these together leaves us with the following equation: P1 × V1 P2 × V2 = T1 × n1 T2 × n2 (P × V ) As with the other gas laws, we can also say that is equal to a constant. The constant can be evaluated provided that the (T × n) gas being described is considered to be ideal. The Ideal Gas Law is a single equation which relates the pressure, volume, temperature, and number of moles of an ideal gas. If we substitute in the variable R for the constant, the equation becomes: P ×V =R T ×n The Ideal Gas Law is conveniently rearranged to look this way, with the multiplication signs omitted: P V = nRT The variable R in the equation is called the ideal gas constant. Evaluating the Ideal Gas Constant The value of R , the ideal gas constant, depends on the units chosen for pressure, temperature, and volume in the ideal gas equation. It is necessary to use Kelvin for the temperature and it is conventional to use the SI unit of liters for the volume. However, pressure is commonly measured in one of three units: kPa , atm, or mm Hg. Therefore, R can have three different values. We will demonstrate how R is calculated when the pressure is measured in kPa. The volume of 1.00 mol of any gas at STP (Standard temperature, 273.15 K and pressure, 1 atm) is measured to be 22.414 L. We can substitute 101.325 kPa for pressure, 22.414 L for volume, and 273.15 K for temperature into the ideal gas equation and solve for R. PV R = nT 101.325 kPa × 22.414 L = 1.000 mol × 273.15 K = 8.314 kPa ⋅ L/K ⋅ mol This is the value of R that is to be used in the ideal gas equation when the pressure is given in kPa. The table below shows a summary of this and the other possible values of R. It is important to choose the correct value of R to use for a given problem. Table 11.8.1 : Values of the Ideal Gas Constant Unit of P Unit of V Unit of n Unit of T Value and Unit of R kPa L mol K 8.314 J/K ⋅ mol atm L mol K 0.08206 L ⋅ atm/K ⋅ mol 11.8.1 https://chem.libretexts.org/@go/page/47539  Unit of P Unit of V Unit of n Unit of T Value and Unit of R mm Hg L mol K 62.36 L ⋅ mm Hg/K ⋅ mol Notice that the unit for R when the pressure is in kPa has been changed to J/K ⋅ mol. A kilopascal multiplied by a liter is equal to the SI unit for energy, a joule (J).  Example 11.8.1 Oxygen Gas What volume is occupied by 3.76 g of oxygen gas at a pressure of 88.4 kPa and a temperature of 19 o C ? Assume the oxygen is ideal. Solution Solutions to Example 11.5.1 Steps for Problem Solving Given: P = 88.4 kPa Identify the "given" information and what the problem is asking you o T = 19 C = 292 K to "find." Mass O = 3.76 g 2 Find: V = ? L O = 32.00 g/mol List other known quantities. 2 R = 8.314 J/K ⋅ mol 1. First, determine the number of moles of O2 from the given mass and the molar mass. Plan the problem. 2. Then, rearrange the equation algebraically to solve for V nRT V = P 1. 1 mol O 2 3.76 g × = 0.1175 mol O 2 32.00 g O 2 Calculate. 2. Now substitute the known quantities into the equation and solve. 0.1175 mol × 8.314 J/K ⋅ mol × 292 K nRT V = = = 3.23 L O 2 P 88.4 kPa The number of moles of oxygen is far less than one mole, so the volume should be fairly small compared to molar volume (22.4 L/mol) since the pressure and temperature are reasonably close Think about your result. to standard. The result has three significant figures because of the values for T and P. Since a joule (J) = kPa ⋅ L , the units cancel out correctly, leaving a volume in liters.  Example 11.8.2: Argon Gas A 4.22 mol sample of Ar has a pressure of 1.21 atm and a temperature of 34°C. What is its volume? Solution Solutions to Example 11.5.2 Steps for Problem Solving 11.8.2 https://chem.libretexts.org/@go/page/47539  Steps for Problem Solving Given: n = 4.22 mol Identify the "given" information and what the problem is asking you P = 1.21 atm to "find." T = 34°C Find: V = ? L List other known quantities. none 1. The first step is to convert temperature to Kelvin. 2. Then, rearrange the equation algebraically to solve for V Plan the problem. nRT V = P 1. 34 + 273 = 307 K 2. Now substitute the known quantities into the equation and solve. L. atm (4.22 mol )(0.08205 )(307 K ) Calculate. mol. K V = 1.21 atm = 87.9 L The number of moles of Ar is large so the expected volume should Think about your result. also be large.  Exercise 11.8.1 A 0.0997 mol sample of O2 has a pressure of 0.692 atm and a temperature of 333 K. What is its volume? Answer 3.94 L  Exercise 11.8.2 For a 0.00554 mol sample of H2, P = 23.44 torr and T = 557 K. What is its volume? Answer 8.21 L Summary The Ideal Gas Law is a single equation which relates the pressure, volume, temperature, and number of moles of an ideal gas. 11.8: The Ideal Gas Law- Pressure, Volume, Temperature, and Moles is shared under a CK-12 license and was authored, remixed, and/or curated by Marisa Alviar-Agnew & Henry Agnew. 11.8.3 https://chem.libretexts.org/@go/page/47539 11.9: Mixtures of Gases - Why Deep-Sea Divers Breathe a Mixture of Helium and Oxygen  Learning Objectives Explain Dalton's Law of Partial Pressures. The atmosphere of Venus is markedly different from that of Earth. The gases in the Venusian atmosphere are 96.5% carbon dioxide and 3% nitrogen. The atmospheric pressure on Venus is roughly 92 times that of Earth, so the amount of nitrogen on Venus would contribute a pressure well over 2700 mm Hg. And there is no oxygen present, so we couldn't breathe there. Not that we would want to go to Venus, as the surface temperature is usually over 460 C. o Dalton's Law of Partial Pressures Gas pressure results from collisions between gas particles and the inside walls of their container. If more gas is added to a rigid container, the gas pressure increases. The identities of the two gases do not matter. John Dalton, the English chemist who proposed the atomic theory, also studied mixtures of gases. He found that each gas in a mixture exerts a pressure independently of every other gas in the mixture. For example, our atmosphere is composed of about 78% nitrogen and 21% oxygen, with smaller amounts of several other gases making up the rest. Since nitrogen makes up 78% of the gas particles in a given sample of air, it exerts 78% of the pressure. If the overall atmospheric pressure is 1.00 atm, then the pressure of just the nitrogen in the air is 0.78 atm. The pressure of the oxygen in the air is 0.21 atm. The partial pressure of a gas is the contribution that gas makes to the total pressure when the gas is part of a mixture. The partial pressure of nitrogen is represented by P. Dalton's Law of Partial Pressures states that the total pressure of a mixture of gases is N2 equal to the sum of all of the partial pressures of the component gases. Dalton's Law can be expressed with the following equation: Ptotal = P1 + P2 + P3 + ⋯ The figure below shows two gases that are in separate, equal-sized containers at the same temperature and pressure. Each exerts a different pressure, P and P , reflective of the number of particles in the container. On the right, the two gases are combined into 1 2 the same container, with no volume change. The total pressure of the gas mixture is equal to the sum of the individual pressures. If P = 300 mm Hg and P = 500 mm Hg , then P 1 2 = 800 mm Hg. total Figure 11.9.1 : Dalton's Law states that the pressure of a gas mixture is equal to the partial pressures of the combining gases. Collecting Gases Over Water You need to do a lab experiment where hydrogen gas is generated. In order to calculate the yield of gas, you have to know the pressure inside the tube where the gas is collected. But how can you get a barometer in there? Very simple: you don't. All you need 11.9.1 https://chem.libretexts.org/@go/page/47541 is the atmospheric pressure in the room. As the gas pushes out the water, it is pushing against the atmosphere, so the pressure inside is equal to the pressure outside. Gas Collection by Water Displacement Gases that are produced in laboratory experiments are often collected by a technique called water displacement (Figure 11.9.2). A bottle is filled with water and placed upside-down in a pan of water. The reaction flask is fitted with rubber tubing, which is then fed under the bottle of water. As the gas is produced in the reaction flask, it exits through the rubber tubing and displaces the water in the bottle. When the bottle is full of the gas, it can be sealed with a lid. Figure 11.9.2 : A gas produced in a chemical reaction can be collected by water displacement. Because the gas is collected over water, it is not pure, but is mixed with vapor from the evaporation of the water. Dalton's Law can be used to calculate the amount of the desired gas by subtracting the contribution of the water vapor. PTotal = Pg + PH O 2 where P is the pressure of the desired gas, which can be solved for: g Pg = PT otal − PH O 2 In order to solve a problem, it is necessary to know the vapor pressure of water at the temperature of the reaction (see table below). The sample problem illustrates the use of Dalton's Law when a gas is collected over water. Table 11.9.1 : Vapor Pressure of Water (mm Hg) at Selected Temperatures ( o C) 0 5 10 15 20 25 30 35 40 45 50 55 60 4.58 6.54 9.21 12.79 17.54 23.76 31.82 42.18 55.32 71.88 92.51 118.04 149.38  Example 14.14.1 A certain experiment generates 2.58 L of hydrogen gas, which is collected over water. The temperature is 20 C o and the atmospheric pressure is 98.60 kPa. Find the volume that the dry hydrogen would occupy at STP. Solution Step 1: List the known quantities and plan the problem. Known VTotal = 2.58 L o T = 20 C = 293 K PTotal = 98.60 kPa = 739.7 mm Hg Unknown V at STP =? L H2 The atmospheric pressure is converted from kPa to mm Hg in order to match units with the table. The sum of the pressures of the hydrogen and the water vapor is equal to the atmospheric pressure. The pressure of the hydrogen is found by subtraction. Then, the volume of the gas at STP can be calculated by using the combined gas law. Step 2: Solve. PH = PTotal − PH O 2 2 = 739, 7 mm Hg − 17.54 mm Hg = 722.2 mm Hg 11.9.2 https://chem.libretexts.org/@go/page/47541  Now the combined gas law is used, solving for V , the volume of hydrogen at STP. 2 P1 × V1 × T2 V2 = P2 × T1 722.2 mm Hg × 2.58 L × 273 K = 760 mm Hg × 293 K = 2.28 L H 2 Step 3: Think about your result. If the hydrogen gas were to be collected at STP and without the presence of the water vapor, its volume would be 2.28 L. This is less than the actual collected volume because some of that is water vapor. The conversion using STP is useful for stoichiometry purposes. Summary Dalton's Law of Partial Pressures states that the total pressure in a system is equal to the sum of the partial pressures of the gases present. The vapor pressure due to water in a sample can be corrected for, in order to get the true value for the pressure of the gas. 11.9: Mixtures of Gases - Why Deep-Sea Divers Breathe a Mixture of Helium and Oxygen is shared under a CK-12 license and was authored, remixed, and/or curated by Marisa Alviar-Agnew & Henry Agnew. 11.9.3 https://chem.libretexts.org/@go/page/47541 11.11: Gay-Lussac's Law- Temperature and Pressure  Learning Objectives Explain Gay-Lussac's Law. Propane tanks are widely used with barbeque grills. But it's not fun to find out halfway through grilling that you have run out of gas. You can buy gauges that measure the pressure inside the tank to see how much is left. The gauge measures pressure and will register a higher pressure on a hot day than it will on a cold day. So you need to take the air temperature into account when you decide whether or not to refill the tank before your next cook-out. Gay-Lussac's Law When the temperature of a sample of gas in a rigid container is increased, the pressure of the gas increases as well. The increase in kinetic energy results in the molecules of gas striking the walls of the container with more force, resulting in a greater pressure. The French chemist Joseph Gay-Lussac (1778-1850) discovered the relationship between the pressure of a gas and its absolute temperature. Gay-Lussac's Law states that the pressure of a given mass of gas varies directly with the absolute temperature of the gas, when the volume is kept constant. Gay-Lussac's Law is very similar to Charles's Law, with the only difference being the type of container. Whereas the container in a Charles's Law experiment is flexible, it is rigid in a Gay-Lussac's Law experiment. Figure 11.11.1: Joseph Gay-Lussac. The mathematical expressions for Gay-Lussac's Law are likewise similar to those of Charles's Law: P P1 P2 and = T T1 T2 A graph of pressure vs. temperature also illustrates a direct relationship. As a gas is cooled at constant volume, its pressure continually decreases until the gas condenses to a liquid.  Example 11.11.1 The gas in an aerosol can is under a pressure of 3.00 atm at a temperature of 25 C. It is dangerous to dispose of an aerosol can o by incineration. What would the pressure in the aerosol can be at a temperature of 845 C? o Solution Solutions to Example 11.10.1 Steps for Problem Solving Given: P1 = 3.00 atm Identify the "given" information and what the problem is asking you o T1 = 25 C = 298 K to "find." o T2 = 845 C = 1118 K Find: P 2 =? atm List other known quantities. The temperatures have first been converted to Kelvin. First, rearrange the equation algebraically to solve for P. 2 Plan the problem. P1 × T2 P2 = T1 11.11.1 https://chem.libretexts.org/@go/page/52227  Steps for Problem Solving Now substitute the known quantities into the equation and solve. Calculate. 3.00 atm × 1118 K P2 = = 11.3 atm 298 K The pressure increases dramatically due to a large increase in Think about your result. temperature. Summary Pressure and temperature at constant volume are directly proportional. 11.11: Gay-Lussac's Law- Temperature and Pressure is shared under a CK-12 license and was authored, remixed, and/or curated by Marisa Alviar-Agnew & Henry Agnew. 11.11.2 https://chem.libretexts.org/@go/page/52227