{"id":787,"date":"2018-03-20T15:57:31","date_gmt":"2018-03-20T15:57:31","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/suny-orgbiochemistry\/?post_type=chapter&p=787"},"modified":"2018-08-08T13:35:55","modified_gmt":"2018-08-08T13:35:55","slug":"8-4-gas-laws","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/suny-orgbiochemistry\/chapter\/8-4-gas-laws\/","title":{"raw":"8.4 Gas Laws","rendered":"8.4 Gas Laws"},"content":{"raw":"
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Learning Objectives<\/h3>\r\n
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  1. Predict the properties of gases using the gas laws.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\nExperience has shown that several properties of a gas can be related to each other under certain conditions. The properties are pressure (<\/span>P<\/em>), volume (<\/span>V<\/em>), temperature (<\/span>T<\/em>, in kelvins), and amount of material expressed in moles (<\/span>n<\/em>). What we find is that a sample of gas cannot have any random values for these properties. Instead, only certain values, dictated by some simple mathematical relationships, will occur.<\/span>\r\n\r\n<\/div>\r\n<\/div>\r\n
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    Boyle\u2019s Law<\/h2>\r\n

    The first simple relationship, referred to as a gas law<\/span><\/strong><\/span>, is between the pressure of a gas and its volume. If the amount of gas in a sample and its temperature are kept constant, then as the pressure of a gas is increased, the volume of the gas decreases proportionately. Mathematically, this is written as<\/p>\r\n [latex]P\\,\u221d\\,\\frac{1}{V}[\/latex] <\/span>\r\n

    where the \u201c\u221d\u201d symbol means \u201cis proportional to.\u201d This is one form of Boyle\u2019s law<\/span><\/strong><\/span>, which relates the pressure of a gas to its volume.<\/p>\r\n

    A more useful form of Boyle\u2019s law involves a change in conditions of a gas. For a given amount of gas at a constant temperature, if we know the initial pressure and volume of a gas sample and the pressure or volume changes, we can calculate what the new volume or pressure will be. That form of Boyle\u2019s law is written<\/p>\r\nP<\/em>i<\/sub>V<\/em>i<\/sub> = P<\/em>f<\/sub>V<\/em>f<\/sub><\/span><\/span>\r\n

    where the subscript i<\/em> refers to initial conditions and the subscript f<\/em> refers to final conditions.<\/p>\r\n

    To use P<\/em>i<\/sub>V<\/em>i<\/sub> = P<\/em>f<\/sub>V<\/em>f<\/sub>, you need to know any three of the variables so that you can algebraically calculate the fourth variable. Also, the pressure quantities must have the same units, as must the two volume quantities. If the two similar variables don\u2019t have the same variables, one value must be converted to the other value\u2019s unit.<\/p>\r\n\r\n

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    Example 4<\/h3>\r\n

    What happens to the volume of a gas if its pressure is increased? Assume all other conditions remain the same.<\/p>\r\n

    Solution<\/p>\r\n

    If the pressure of a gas is increased, the volume decreases in response.<\/p>\r\n\r\n<\/div>\r\n

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    Skill-Building Exercise<\/h3>\r\n
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    1. \r\n
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      What happens to the pressure of a gas if its volume is increased? Assume all other conditions remain the same.<\/p>\r\n\r\n<\/div><\/li>\r\n<\/ol>\r\n<\/div>\r\n<\/div>\r\n

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      Example 5<\/h3>\r\n

      If a sample of gas has an initial pressure of 1.56 atm and an initial volume of 7.02 L, what is the final volume if the pressure is reduced to 0.987 atm? Assume that the amount and the temperature of the gas remain constant.<\/p>\r\n

      Solution<\/p>\r\n

      The key in problems like this is to be able to identify which quantities represent which variables from the relevant equation. The way the question is worded, you should be able to tell that 1.56 atm is P<\/em>i<\/sub>, 7.02 L is V<\/em>i<\/sub>, and 0.987 atm is P<\/em>f<\/sub>. What we are looking for is the final volume\u2014V<\/em>f<\/sub>. Therefore, substituting these values into P<\/em>i<\/sub>V<\/em>i<\/sub> = P<\/em>f<\/sub>V<\/em>f<\/sub>:<\/p>\r\n(1.56 atm)(7.02 L) = (0.987 atm) \u00d7 V<\/em>f<\/sub><\/span><\/span>\r\n

      The expression has atmospheres on both sides of the equation, so they cancel algebraically:<\/p>\r\n(1.56)(7.02 L) = (0.987) \u00d7 V<\/em>f<\/sub><\/span><\/span>\r\n

      Now we divide both sides of the expression by 0.987 to isolate V<\/em>f<\/sub>, the quantity we are seeking:<\/p>\r\n [latex]\\frac{(1.56)(7.02\\,L)}{0.987}\\,=\\,V_f[\/latex]<\/span>\r\n

      Performing the multiplication and division, we get the value of V<\/em>f<\/sub>, which is 11.1 L. The volume increases. This should make sense because the pressure decreases, so pressure and volume are inversely related.<\/p>\r\n\r\n<\/div>\r\n

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      Skill-Building Exercise<\/h3>\r\n
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        If a sample of gas has an initial pressure of 3.66 atm and an initial volume of 11.8 L, what is the final pressure if the volume is reduced to 5.09 L? Assume that the amount and the temperature of the gas remain constant.<\/p>\r\n\r\n<\/div><\/li>\r\n<\/ol>\r\n<\/div>\r\n<\/div>\r\n

        If the units of similar quantities are not the same, one of them must be converted to the other quantity\u2019s units for the calculation to work out properly. It does not matter which quantity is converted to a different unit; the only thing that matters is that the conversion and subsequent algebra are performed properly. The following example illustrates this process.<\/p>\r\n\r\n

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        Example 6<\/h3>\r\n

        If a sample of gas has an initial pressure of 1.56 atm and an initial volume of 7.02 L, what is the final volume if the pressure is changed to 1,775 torr? Does the answer make sense? Assume that the amount and the temperature of the gas remain constant.<\/p>\r\n

        Solution<\/p>\r\n

        This example is similar to Example 5, except now the final pressure is expressed in torr. For the math to work out properly, one of the pressure values must be converted to the other unit. Let us change the initial pressure to torr:<\/p>\r\n [latex]1.56\\,\\rlap{\\text{-------}}atm\\,\u00d7\\,\\frac{760\\,torr}{1\\,\\rlap{\\text{------}}atm}\\,=\\,1,190\\,torr[\/latex] <\/span>\r\n

        Now we can use Boyle\u2019s law:<\/p>\r\n(1,190 torr)(7.02 L) = (1,775 torr) \u00d7 V<\/em>f<\/sub><\/span><\/span>\r\n

        Torr cancels algebraically from both sides of the equation, leaving<\/p>\r\n(1,190)(7.02 L) = (1,775) \u00d7 V<\/em>f<\/sub><\/span><\/span>\r\n

        Now we divide both sides of the equation by 1,775 to isolate V<\/em>f<\/sub> on one side. Solving for the final volume,<\/p>\r\n [latex]V_f\\,=\\,\\frac{(1,190)(7.02\\,L)}{1,775}\\,=\\,4.71\\,L[\/latex] <\/span>\r\n

        Because the pressure increases, it makes sense that the volume decreases.<\/p>\r\n\r\n<\/div>\r\n

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        Note<\/h3>\r\n

        The answer for the final volume is essentially the same if we converted the 1,775 torr to atmospheres: 1,775\u00a0torr\u00d7[latex]\\frac{1\\,atm}{760\\,torr}[\/latex]=2.336\u00a0atm. <\/span> Using Boyle\u2019s law: (1.56 atm)(7.02 L) = (2.335 atm) \u00d7 V<\/em>f<\/sub>; [latex]V_f\\,=\\,\\frac{(1.56\\,atm)(7.02\\,L)}{2.336\\,atm}\\,=\\,4.69\\,L[\/latex].<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n

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        Skill-Building Exercise<\/h3>\r\n
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          If a sample of gas has an initial pressure of 375 torr and an initial volume of 7.02 L, what is the final pressure if the volume is changed to 4,577 mL? Does the answer make sense? Assume that amount and the temperature of the gas remain constant.<\/p>\r\n\r\n<\/div><\/li>\r\n<\/ol>\r\n<\/div>\r\n<\/div>\r\n

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          To Your Health: Breathing<\/h3>\r\n

          Breathing certainly is a major contribution to your health! Without breathing, we could not survive. Curiously, the act of breathing itself is little more than an application of Boyle\u2019s law.<\/p>\r\n

          The lungs are a series of ever-narrowing tubes that end in a myriad of tiny sacs called alveoli. It is in the alveoli that oxygen from the air transfers to the bloodstream and carbon dioxide from the bloodstream transfers to the lungs for exhalation. For air to move in and out of the lungs, the pressure inside the lungs must change, forcing the lungs to change volume\u2014just as predicted by Boyle\u2019s law.<\/p>\r\n

          The pressure change is caused by the diaphragm, a muscle that covers the bottom of the lungs. When the diaphragm moves down, it expands the size of our lungs. When this happens, the air pressure inside our lungs decreases slightly. This causes new air to rush in, and we inhale. The pressure decrease is slight\u2014only 3 torr, or about 0.4% of an atmosphere. We inhale only 0.5\u20131.0 L of air per normal breath.<\/p>\r\n

          Exhaling air requires that we relax the diaphragm, which pushes against the lungs and slightly decreases the volume of the lungs. This slightly increases the pressure of the air in the lungs, and air is forced out; we exhale. Only 1\u20132 torr of extra pressure is needed to exhale. So with every breath, our own bodies are performing an experimental test of Boyle\u2019s law.<\/p>\r\n\r\n<\/div>\r\nCharles\u2019s Law<\/span>\r\n\r\n<\/div>\r\n<\/div>\r\n

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          Another simple gas law relates the volume of a gas to its temperature. Experiments indicate that as the temperature of a gas sample is increased, its volume increases as long as the pressure and the amount of gas remain constant. The way to write this mathematically is<\/p>\r\nV \u221d T<\/span><\/span>\r\n

          At this point, the concept of temperature must be clarified. Although the Kelvin scale is the preferred temperature scale, the Celsius scale is also a common temperature scale used in science. The Celsius scale is based on the melting and boiling points of water and is actually the common temperature scale used by most countries around the world (except for the United States, which still uses the Fahrenheit scale). The value of a Celsius temperature is directly related to its Kelvin value by a simple expression:<\/p>\r\nKelvin temperature = Celsius temperature + 273<\/span><\/span>\r\n

          Thus, it is easy to convert from one temperature scale to another.<\/p>\r\n\r\n

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          Note<\/h3>\r\n

          The Kelvin scale is sometimes referred to as the absolute scale because the zero point on the Kelvin scale is at absolute zero, the coldest possible temperature. On the other temperature scales, absolute zero is \u2212260\u00b0C or \u2212459\u00b0F.<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n

          The expression relating a gas volume to its temperature begs the following question: to which temperature scale is the volume of a gas related? The answer is that gas volumes are directly related to the Kelvin temperature<\/em>. Therefore, the temperature of a gas sample should always be expressed in (or converted to) a Kelvin temperature.<\/p>\r\n\r\n

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          Example 7<\/h3>\r\n

          What happens to the volume of a gas if its temperature is decreased? Assume that all other conditions remain constant.<\/p>\r\n

          Solution<\/p>\r\n

          If the temperature of a gas sample is decreased, the volume decreases as well.<\/p>\r\n\r\n<\/div>\r\n

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          Skill-Building Exercise<\/h3>\r\n
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            What happens to the temperature of a gas if its volume is increased? Assume that all other conditions remain constant.<\/p>\r\n\r\n<\/div><\/li>\r\n<\/ol>\r\n<\/div>\r\n<\/div>\r\n

            As with Boyle\u2019s law, the relationship between volume and temperature can be expressed in terms of initial and final values of volume and temperature, as follows:<\/p>\r\n [latex]\\frac{V_i}{T_i}\\,=\\,\\frac{V_f}{T_f}[\/latex] <\/span>\r\n

            where V<\/em>i<\/sub> and T<\/em>i<\/sub> are the initial volume and temperature, and V<\/em>f<\/sub> and T<\/em>f<\/sub> are the final volume and temperature. This is Charles\u2019s law<\/span><\/strong><\/span>. The restriction on its use is that the pressure of the gas and the amount of gas must remain constant. (Charles\u2019s law is sometimes referred to as Gay-Lussac\u2019s law, after the scientist who promoted Charles\u2019s work.)<\/p>\r\n\r\n

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            Example 8<\/h3>\r\n

            A gas sample at 20\u00b0C has an initial volume of 20.0 L. What is its volume if the temperature is changed to 60\u00b0C? Does the answer make sense? Assume that the pressure and the amount of the gas remain constant.<\/p>\r\n

            Solution<\/p>\r\n

            Although the temperatures are given in degrees Celsius, we must convert them to the kelvins before we can use Charles\u2019s law. Thus,<\/p>\r\n20\u00b0C + 273 = 293 K = T<\/em>i<\/sub><\/span><\/span>\r\n60\u00b0C + 273 = 333 K = T<\/em>f<\/sub><\/span><\/span>\r\n

            Now we can substitute these values into Charles\u2019s law, along with the initial volume of 20.0 L:<\/p>\r\n [latex]\\frac{20.0\\,L}{293\\,K}\\,=\\,\\frac{V_f}{333K}[\/latex] <\/span>\r\n

            Multiplying the 333 K to the other side of the equation, we see that our temperature units will cancel:<\/p>\r\n [latex]\\frac{(333\\,\\rlap{\\text{---}}K)(20.0\\,L)}{293\\,\\rlap{\\text{---}}K}\\,=\\,V_f[\/latex] <\/span>\r\n

            Solving for the final volume, V<\/em>f<\/sub> = 22.7 L. So, as the temperature is increased, the volume increases. This makes sense because volume is directly proportional to the absolute temperature (as long as the pressure and the amount of the remain constant).<\/p>\r\n\r\n<\/div>\r\n

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            Skill-Building Exercise<\/h3>\r\n
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              A gas sample at 35\u00b0C has an initial volume of 5.06 L. What is its volume if the temperature is changed to \u221235\u00b0C? Does the answer make sense? Assume that the pressure and the amount of the gas remain constant.<\/p>\r\n\r\n<\/div><\/li>\r\n<\/ol>\r\n<\/div>\r\n \r\n\r\n<\/div>\r\n<\/div>\r\n

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              Combined Gas Law<\/h2>\r\n

              Other gas laws can be constructed, but we will focus on only two more. The combined gas law<\/span><\/strong><\/span>\u00a0brings Boyle\u2019s and Charles\u2019s laws together to relate pressure, volume, and temperature changes of a gas sample:<\/p>\r\n [latex]\\frac{P_i\\,V_i}{T_i}\\,=\\,\\frac{P_f\\,V_f}{T_f}[\/latex] <\/span>\r\n

              To apply this gas law, the amount of gas should remain constant. As with the other gas laws, the temperature must be expressed in kelvins, and the units on the similar quantities should be the same. Because of the dependence on three quantities at the same time, it is difficult to tell in advance what will happen to one property of a gas sample as two other properties change. The best way to know is to work it out mathematically.<\/p>\r\n\r\n

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              Example 9<\/h3>\r\n

              A sample of gas has P<\/em>i<\/sub> = 1.50 atm, V<\/em>i<\/sub> = 10.5 L, and T<\/em>i<\/sub> = 300 K. What is the final volume if P<\/em>f<\/sub> = 0.750 atm and T<\/em>f<\/sub> = 350 K?<\/p>\r\n

              Solution<\/p>\r\n

              Using the combined gas law, substitute for five of the quantities:<\/p>\r\n [latex]\\frac{(1.50\\,atm)(10.5\\,L)}{300\\,K}\\,=\\,\\frac{(0.750\\,atm)(V_f)}{350\\,K}[\/latex] <\/span>\r\n

              We algebraically rearrange this expression to isolate V<\/em>f<\/sub> on one side of the equation:<\/p>\r\n [latex]V_f\\,=\\,\\frac{(1.50\\,\\rlap{\\text{------}}atm)(10.5\\,L)(350\\,\\rlap{\\text{---}}K)}{(300\\,\\rlap{\\text{---}}K)(0.750\\,\\rlap{\\text{------}}atm)}\\,=\\,24.5\\,L[\/latex] <\/span>\r\n

              Note how all the units cancel except the unit for volume.<\/p>\r\n\r\n<\/div>\r\n

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              Skill-Building Exercise<\/h3>\r\n
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                A sample of gas has P<\/em>i<\/sub> = 0.768 atm, V<\/em>i<\/sub> = 10.5 L, and T<\/em>i<\/sub> = 300 K. What is the final pressure if V<\/em>f<\/sub> = 7.85 L and T<\/em>f<\/sub> = 250 K?<\/p>\r\n\r\n<\/div><\/li>\r\n<\/ol>\r\n<\/div>\r\n<\/div>\r\n

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                Example 10<\/h3>\r\n

                A balloon containing a sample of gas has a temperature of 22\u00b0C and a pressure of 1.09 atm in an airport in Cleveland. The balloon has a volume of 1,070 mL. The balloon is transported by plane to Denver, where the temperature is 11\u00b0C and the pressure is 655 torr. What is the new volume of the balloon?<\/p>\r\n

                Solution<\/p>\r\n

                The first task is to convert all quantities to the proper and consistent units. The temperatures must be expressed in kelvins, and the pressure units are different so one of the quantities must be converted. Let us convert the atmospheres to torr:<\/p>\r\n22\u00b0C + 273 = 295 K = T<\/em>i<\/sub><\/span><\/span>\r\n11\u00b0C + 273 = 284 K = T<\/em>f<\/sub><\/span><\/span>\r\n [latex]1.09\\,\\rlap{\\text{------}}atm\\,\u00d7\\,\\frac{760\\,torr}{1\\,\\rlap{\\text{------}}atm}\\,=\\,828\\,torr\\,=\\,P_i[\/latex]<\/span>\r\n

                Now we can substitute the quantities into the combined has law:<\/p>\r\n [latex]\\frac{(828\\,torr)(1,070\\,mL)}{295\\,K}\\,=\\,\\frac{(655\\,torr)\\,\u00d7\\,V_f}{284\\,K}[\/latex] <\/span>\r\n

                To solve for V<\/em>f<\/sub>, we multiply the 284 K in the denominator of the right side into the numerator on the left, and we divide 655 torr in the numerator of the right side into the denominator on the left:<\/p>\r\n [latex]\\frac{(828\\,\\rlap{\\text{------}}torr)(1,070\\,mL)(284\\,\\rlap{\\text{---}}K)}{(295\\,\\rlap{\\text{---}}K)(655\\,\\rlap{\\text{------}}torr)}\\,=\\,V_f[\/latex] <\/span>\r\n

                Notice that torr and kelvins cancel, as they are found in both the numerator and denominator. The only unit that remains is milliliters, which is a unit of volume. So V<\/em>f<\/sub> = 1,300 mL. The overall change is that the volume of the balloon has increased by 230 mL.<\/p>\r\n\r\n<\/div>\r\n

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                Skill-Building Exercise<\/h3>\r\n
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                  A balloon used to lift weather instruments into the atmosphere contains gas having a volume of 1,150 L on the ground, where the pressure is 0.977 atm and the temperature is 18\u00b0C. Aloft, this gas has a pressure of 6.88 torr and a temperature of \u221215\u00b0C. What is the new volume of the gas?<\/p>\r\n\r\n<\/div><\/li>\r\n<\/ol>\r\n<\/div>\r\n \r\n\r\n<\/div>\r\n<\/div>\r\n

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                  The Ideal Gas Law<\/h2>\r\n

                  So far, the gas laws we have used have focused on changing one or more properties of the gas, such as its volume, pressure, or temperature. There is one gas law that relates all the independent properties of a gas under any particular condition, rather than a change in conditions. This gas law is called the ideal gas law<\/span><\/strong><\/span>. The formula of this law is as follows:<\/p>\r\nPV<\/em> = nRT<\/em><\/span><\/span>\r\n

                  In this equation, P<\/em> is pressure, V<\/em> is volume, n<\/em> is amount of moles, and T<\/em> is temperature. R<\/em> is called the ideal gas law constant<\/span><\/strong><\/span>\u00a0and is a proportionality constant that relates the values of pressure, volume, amount, and temperature of a gas sample. The variables in this equation do not have the subscripts i<\/em> and f<\/em> to indicate an initial condition and a final condition. The ideal gas law relates the four independent properties of a gas under any<\/em> conditions.<\/p>\r\n

                  The value of R<\/em> depends on what units are used to express the other quantities. If volume is expressed in liters and pressure in atmospheres, then the proper value of R<\/em> is as follows:<\/p>\r\n [latex]R\\,=\\,0.08205\\,\\frac{L\u22c5atm}{mol\u22c5K}[\/latex]<\/span>\r\n

                  This may seem like a strange unit, but that is what is required for the units to work out algebraically.<\/p>\r\n\r\n

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                  Example 11<\/h3>\r\n

                  What is the volume in liters of 1.45 mol of N2<\/sub> gas at 298 K and 3.995 atm?<\/p>\r\n

                  Solution<\/p>\r\n

                  Using the ideal gas law where P<\/em> = 3.995 atm, n<\/em> = 1.45, and T<\/em> = 298,<\/p>\r\n [latex](3.995\\,atm)\\,\u00d7\\,V\\,=\\,(1.45\\,mol)\\,(0.08205\\,\\frac{L\u22c5atm}{mol\u22c5K})\\,(298\\,K)[\/latex] <\/span>\r\n

                  On the right side, the moles and kelvins cancel. Also, because atmospheres appear in the numerator on both sides of the equation, they also cancel. The only remaining unit is liters, a unit of volume. So<\/p>\r\n3.995 \u00d7 V<\/em> = (1.45)(0.08205)(298) L<\/span><\/span>\r\n

                  Dividing both sides of the equation by 3.995 and evaluating, we get V<\/em> = 8.87 L. Note that the conditions of the gas are not changing. Rather, the ideal gas law allows us to determine what the fourth property of a gas (here, volume) must<\/em> be if three other properties (here, amount, pressure, and temperature) are known.<\/p>\r\n\r\n<\/div>\r\n

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                  Skill-Building Exercise<\/h3>\r\n
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                  1. \r\n
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                    What is the pressure of a sample of CO2<\/sub> gas if 0.557 mol is held in a 20.0 L container at 451 K?<\/p>\r\n\r\n<\/div><\/li>\r\n<\/ol>\r\n<\/div>\r\n<\/div>\r\n

                    For convenience, scientists have selected 273 K (0\u00b0C) and 1.00 atm pressure as a set of standard conditions for gases. This combination of conditions is called standard temperature and pressure (STP)<\/span><\/strong><\/span>. Under these conditions, 1 mol of any gas has about the same volume. We can use the ideal gas law to determine the volume of 1 mol of gas at STP:<\/span><\/p>\r\n\r\n<\/div>\r\n [latex](1.00\\,atm)\\,\u00d7\\,V\\,=\\,(1.00\\,mol)\\,(0.08205\\,\\frac{L\u22c5atm}{mol\u22c5K})\\,(273\\,K)[\/latex]<\/span>\r\n

                    This volume is 22.4 L. Because this volume is independent of the identity of a gas, the idea that 1 mol of gas has a volume of 22.4 L at STP makes a convenient conversion factor:<\/p>\r\n1 mol gas = 22.4 L (at STP)<\/span><\/span>\r\n

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                    Example 12<\/h3>\r\n

                    Cyclopropane (C3<\/sub>H6<\/sub>) is a gas that formerly was used as an anesthetic. How many moles of gas are there in a 100.0 L sample if the gas is at STP?<\/p>\r\n

                    Solution<\/p>\r\n

                    We can set up a simple, one-step conversion that relates moles and liters:<\/p>\r\n [latex]100.0\\,\\rlap{\\text{---}}L\\,C_3H_6\\,\u00d7\\,\\frac{1\\,mol}{22.4\\,\\rlap{\\text{---}}L}\\,=\\,4.46\\,mol\\,C_3H_6[\/latex] <\/span>\r\n

                    There are almost 4.5 mol of gas in 100.0 L.<\/p>\r\n\r\n<\/div>\r\n

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                    Note<\/h3>\r\n

                    Because of its flammability, cyclopropane is no longer used as an anesthetic gas.<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n

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                    Skill-Building Exercise<\/h3>\r\n
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                    1. \r\n
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                      Freon is a trade name for a series of fluorine- and chlorine-containing gases that formerly were used in refrigeration systems. What volume does 8.75 mol of Freon have at STP?<\/p>\r\n\r\n<\/div><\/li>\r\n<\/ol>\r\n<\/div>\r\n<\/div>\r\n

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                      \r\n

                      Note<\/h3>\r\n

                      Many gases known as Freon are no longer used because their presence in the atmosphere destroys the ozone layer, which protects us from ultraviolet light from the sun.<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n

                      \r\n
                      \r\n

                      Career Focus: Respiratory Therapist<\/h3>\r\n

                      Certain diseases\u2014such as emphysema, lung cancer, and severe asthma\u2014primarily affect the lungs. Respiratory therapists help patients with breathing-related problems. They can evaluate, help diagnose, and treat breathing disorders and even help provide emergency assistance in acute illness where breathing is compromised.<\/p>\r\n

                      Most respiratory therapists must complete at least two years of college and earn an associate\u2019s degree, although therapists can assume more responsibility if they have a college degree. Therapists must also pass state or national certification exams. Once certified, respiratory therapists can work in hospitals, doctor\u2019s offices, nursing homes, or patient\u2019s homes. Therapists work with equipment such as oxygen tanks and respirators, may sometimes dispense medication to aid in breathing, perform tests, and educate patients in breathing exercises and other therapy.<\/p>\r\n

                      Because respiratory therapists work directly with patients, the ability to work well with others is a must for this career. It is an important job because it deals with one of the most crucial functions of the body.<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n

                      \r\n
                      \r\n
                      \r\n

                      Concept Review Exercises<\/h3>\r\n
                        \r\n \t
                      1. \r\n
                        \r\n

                        What properties do the gas laws help us predict?<\/p>\r\n\r\n<\/div><\/li>\r\n \t

                      2. \r\n
                        \r\n

                        What makes the ideal gas law different from the other gas laws?<\/p>\r\n\r\n<\/div><\/li>\r\n<\/ol>\r\n<\/div>\r\n

                        \r\n

                        Answers<\/h3>\r\n

                        [reveal-answer q=\"141044\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"141044\"]<\/p>\r\n\r\n

                          \r\n \t
                        1. Gas laws relate four properties: pressure, volume, temperature, and number of moles.<\/li>\r\n \t
                        2. The ideal gas law does not require that the properties of a gas change.[\/hidden-answer]<\/li>\r\n<\/ol>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n
                          \r\n
                          \r\n

                          Key Takeaway<\/h3>\r\n
                            \r\n \t
                          • The physical properties of gases are predictable using mathematical formulas known as gas laws.<\/li>\r\n<\/ul>\r\n<\/div>\r\n<\/div>\r\n
                            \r\n
                            \r\n
                            \r\n

                            Exercises<\/h3>\r\n
                            \r\n
                              \r\n \t
                            1. \r\n
                              \r\n

                              What conditions of a gas sample should remain constant for Boyle\u2019s law to be used?<\/p>\r\n\r\n<\/div><\/li>\r\n \t

                            2. \r\n
                              \r\n

                              What conditions of a gas sample should remain constant for Charles\u2019s law to be used?<\/p>\r\n\r\n<\/div><\/li>\r\n \t

                            3. \r\n
                              \r\n

                              Does the identity of a gas matter when using Boyle\u2019s law? Why or why not?<\/p>\r\n\r\n<\/div><\/li>\r\n \t

                            4. \r\n
                              \r\n

                              Does the identity of a gas matter when using Charles\u2019s law? Why or why not?<\/p>\r\n\r\n<\/div><\/li>\r\n \t

                            5. \r\n
                              \r\n

                              A sample of nitrogen gas is confined to a balloon that has a volume of 1.88 L and a pressure of 1.334 atm. What will be the volume of the balloon if the pressure is changed to 0.662 atm? Assume that the temperature and the amount of the gas remain constant.<\/p>\r\n\r\n<\/div><\/li>\r\n \t

                            6. \r\n
                              \r\n

                              A sample of helium gas in a piston has a volume of 86.4 mL under a pressure of 447 torr. What will be the volume of the helium if the pressure on the piston is increased to 1,240 torr? Assume that the temperature and the amount of the gas remain constant.<\/p>\r\n\r\n<\/div><\/li>\r\n \t

                            7. \r\n
                              \r\n

                              If a gas has an initial pressure of 24,650 Pa and an initial volume of 376 mL, what is the final volume if the pressure of the gas is changed to 775 torr? Assume that the amount and the temperature of the gas remain constant.<\/p>\r\n\r\n<\/div><\/li>\r\n \t

                            8. \r\n
                              \r\n

                              A gas sample has an initial volume of 0.9550 L and an initial pressure of 564.5 torr. What would the final pressure of the gas be if the volume is changed to 587.0 mL? Assume that the amount and the temperature of the gas remain constant.<\/p>\r\n\r\n<\/div><\/li>\r\n \t

                            9. \r\n
                              \r\n

                              A person draws a normal breath of about 1.00 L. If the initial temperature of the air is 18\u00b0C and the air warms to 37\u00b0C, what is the new volume of the air? Assume that the pressure and amount of the gas remain constant.<\/p>\r\n\r\n<\/div><\/li>\r\n \t

                            10. \r\n
                              \r\n

                              A person draws a normal breath of about 1.00 L. If the initial temperature of the air is \u221210\u00b0C and the air warms to 37\u00b0C, what is the new volume of the air? Assume that the pressure and the amount of the gas remain constant.<\/p>\r\n\r\n<\/div><\/li>\r\n \t

                            11. \r\n
                              \r\n

                              An air\/gas vapor mix in an automobile cylinder has an initial temperature of 450 K and a volume of 12.7 cm3<\/sup>. The gas mix is heated to 565\u00b0C. If pressure and amount are held constant, what is the final volume of the gas in cubic centimeters?<\/p>\r\n\r\n<\/div><\/li>\r\n \t

                            12. \r\n
                              \r\n

                              Given the following conditions for a gas: V<\/em>i<\/sub> = 0.665 L, T<\/em>i<\/sub> = 23.6\u00b0C, V<\/em>f<\/sub> = 1.034 L. What is T<\/em>f<\/sub> in degrees Celsius and kelvins?<\/p>\r\n\r\n<\/div><\/li>\r\n \t

                            13. \r\n
                              \r\n

                              Assuming the amount remains the same, what must be the final volume of a gas that has an initial volume of 387 mL, an initial pressure of 456 torr, an initial temperature of 65.0\u00b0C, a final pressure of 1.00 atm, and a final temperature of 300 K?<\/p>\r\n\r\n<\/div><\/li>\r\n \t

                            14. \r\n
                              \r\n

                              When the nozzle of a spray can is depressed, 0.15 mL of gas expands to 0.44 mL, and its pressure drops from 788 torr to 1.00 atm. If the initial temperature of the gas is 22.0\u00b0C, what is the final temperature of the gas?<\/p>\r\n\r\n<\/div><\/li>\r\n \t

                            15. \r\n
                              \r\n

                              Use the ideal gas law to show that 1 mol of a gas at STP has a volume of about 22.4 L.<\/p>\r\n\r\n<\/div><\/li>\r\n \t

                            16. \r\n
                              \r\n

                              Use a standard conversion factor to determine a value of the ideal gas law constant R<\/em> that has units of L\u00b7torr\/mol\u00b7K.<\/p>\r\n\r\n<\/div><\/li>\r\n \t

                            17. \r\n
                              \r\n

                              How many moles of gas are there in a 27.6 L sample at 298 K and a pressure of 1.44 atm?<\/p>\r\n\r\n<\/div><\/li>\r\n \t

                            18. \r\n
                              \r\n

                              How many moles of gas are there in a 0.066 L sample at 298 K and a pressure of 0.154 atm?<\/p>\r\n\r\n<\/div><\/li>\r\n \t

                            19. \r\n
                              \r\n

                              A 0.334 mol sample of carbon dioxide gas is confined to a volume of 20.0 L and has a pressure of 0.555 atm. What is the temperature of the carbon dioxide in kelvins and degrees Celsius?<\/p>\r\n\r\n<\/div><\/li>\r\n \t

                            20. \r\n
                              \r\n

                              What must V<\/em> be for a gas sample if n<\/em> = 4.55 mol, P<\/em> = 7.32 atm, and T<\/em> = 285 K?<\/p>\r\n\r\n<\/div><\/li>\r\n \t

                            21. \r\n
                              \r\n

                              What is the pressure of 0.0456 mol of Ne gas contained in a 7.50 L volume at 29\u00b0C?<\/p>\r\n\r\n<\/div><\/li>\r\n \t

                            22. \r\n
                              \r\n

                              What is the pressure of 1.00 mol of Ar gas that has a volume of 843.0 mL and a temperature of \u221286.0\u00b0C?<\/p>\r\n\r\n<\/div><\/li>\r\n<\/ol>\r\n<\/div>\r\n

                              \r\n

                              Answers<\/h3>\r\n[reveal-answer q=\"546073\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"546073\"]\r\n\r\n1. temperature and amount of the gas\r\n\r\n3. The identity does not matter because the variables of Boyle\u2019s law do not identify the gas.\r\n\r\n5. 3.89 L\r\n\r\n7. 92.1 mL\r\n\r\n9. 1.07 L\r\n\r\n11. 23.7 cm3<\/sup>\r\n\r\n13. 206 mL\r\n\r\n15. The ideal gas law confirms that 22.4 L equals 1 mol.\r\n\r\n17. 1.63 mol\r\n\r\n19. 405 K; 132\u00b0C\r\n\r\n21. 0.151 atm \u00a0[\/hidden-answer]\r\n
                              <\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>","rendered":"
                              \n
                              \n
                              \n

                              Learning Objectives<\/h3>\n
                              \n
                              \n
                              \n
                                \n
                              1. Predict the properties of gases using the gas laws.<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n

                                Experience has shown that several properties of a gas can be related to each other under certain conditions. The properties are pressure (<\/span>P<\/em>), volume (<\/span>V<\/em>), temperature (<\/span>T<\/em>, in kelvins), and amount of material expressed in moles (<\/span>n<\/em>). What we find is that a sample of gas cannot have any random values for these properties. Instead, only certain values, dictated by some simple mathematical relationships, will occur.<\/span><\/p>\n<\/div>\n<\/div>\n

                                \n
                                \n
                                \n

                                Boyle\u2019s Law<\/h2>\n

                                The first simple relationship, referred to as a gas law<\/span><\/strong><\/span>, is between the pressure of a gas and its volume. If the amount of gas in a sample and its temperature are kept constant, then as the pressure of a gas is increased, the volume of the gas decreases proportionately. Mathematically, this is written as<\/p>\n

                                [latex]P\\,\u221d\\,\\frac{1}{V}[\/latex] <\/span><\/p>\n

                                where the \u201c\u221d\u201d symbol means \u201cis proportional to.\u201d This is one form of Boyle\u2019s law<\/span><\/strong><\/span>, which relates the pressure of a gas to its volume.<\/p>\n

                                A more useful form of Boyle\u2019s law involves a change in conditions of a gas. For a given amount of gas at a constant temperature, if we know the initial pressure and volume of a gas sample and the pressure or volume changes, we can calculate what the new volume or pressure will be. That form of Boyle\u2019s law is written<\/p>\n

                                P<\/em>i<\/sub>V<\/em>i<\/sub> = P<\/em>f<\/sub>V<\/em>f<\/sub><\/span><\/span><\/p>\n

                                where the subscript i<\/em> refers to initial conditions and the subscript f<\/em> refers to final conditions.<\/p>\n

                                To use P<\/em>i<\/sub>V<\/em>i<\/sub> = P<\/em>f<\/sub>V<\/em>f<\/sub>, you need to know any three of the variables so that you can algebraically calculate the fourth variable. Also, the pressure quantities must have the same units, as must the two volume quantities. If the two similar variables don\u2019t have the same variables, one value must be converted to the other value\u2019s unit.<\/p>\n

                                \n

                                Example 4<\/h3>\n

                                What happens to the volume of a gas if its pressure is increased? Assume all other conditions remain the same.<\/p>\n

                                Solution<\/p>\n

                                If the pressure of a gas is increased, the volume decreases in response.<\/p>\n<\/div>\n

                                \n
                                \n

                                Skill-Building Exercise<\/h3>\n
                                  \n
                                1. \n
                                  \n

                                  What happens to the pressure of a gas if its volume is increased? Assume all other conditions remain the same.<\/p>\n<\/div>\n<\/li>\n<\/ol>\n<\/div>\n<\/div>\n

                                  \n

                                  Example 5<\/h3>\n

                                  If a sample of gas has an initial pressure of 1.56 atm and an initial volume of 7.02 L, what is the final volume if the pressure is reduced to 0.987 atm? Assume that the amount and the temperature of the gas remain constant.<\/p>\n

                                  Solution<\/p>\n

                                  The key in problems like this is to be able to identify which quantities represent which variables from the relevant equation. The way the question is worded, you should be able to tell that 1.56 atm is P<\/em>i<\/sub>, 7.02 L is V<\/em>i<\/sub>, and 0.987 atm is P<\/em>f<\/sub>. What we are looking for is the final volume\u2014V<\/em>f<\/sub>. Therefore, substituting these values into P<\/em>i<\/sub>V<\/em>i<\/sub> = P<\/em>f<\/sub>V<\/em>f<\/sub>:<\/p>\n

                                  (1.56 atm)(7.02 L) = (0.987 atm) \u00d7 V<\/em>f<\/sub><\/span><\/span><\/p>\n

                                  The expression has atmospheres on both sides of the equation, so they cancel algebraically:<\/p>\n

                                  (1.56)(7.02 L) = (0.987) \u00d7 V<\/em>f<\/sub><\/span><\/span><\/p>\n

                                  Now we divide both sides of the expression by 0.987 to isolate V<\/em>f<\/sub>, the quantity we are seeking:<\/p>\n

                                  [latex]\\frac{(1.56)(7.02\\,L)}{0.987}\\,=\\,V_f[\/latex]<\/span><\/p>\n

                                  Performing the multiplication and division, we get the value of V<\/em>f<\/sub>, which is 11.1 L. The volume increases. This should make sense because the pressure decreases, so pressure and volume are inversely related.<\/p>\n<\/div>\n

                                  \n
                                  \n

                                  Skill-Building Exercise<\/h3>\n
                                    \n
                                  1. \n
                                    \n

                                    If a sample of gas has an initial pressure of 3.66 atm and an initial volume of 11.8 L, what is the final pressure if the volume is reduced to 5.09 L? Assume that the amount and the temperature of the gas remain constant.<\/p>\n<\/div>\n<\/li>\n<\/ol>\n<\/div>\n<\/div>\n

                                    If the units of similar quantities are not the same, one of them must be converted to the other quantity\u2019s units for the calculation to work out properly. It does not matter which quantity is converted to a different unit; the only thing that matters is that the conversion and subsequent algebra are performed properly. The following example illustrates this process.<\/p>\n

                                    \n

                                    Example 6<\/h3>\n

                                    If a sample of gas has an initial pressure of 1.56 atm and an initial volume of 7.02 L, what is the final volume if the pressure is changed to 1,775 torr? Does the answer make sense? Assume that the amount and the temperature of the gas remain constant.<\/p>\n

                                    Solution<\/p>\n

                                    This example is similar to Example 5, except now the final pressure is expressed in torr. For the math to work out properly, one of the pressure values must be converted to the other unit. Let us change the initial pressure to torr:<\/p>\n

                                    [latex]1.56\\,\\rlap{\\text{-------}}atm\\,\u00d7\\,\\frac{760\\,torr}{1\\,\\rlap{\\text{------}}atm}\\,=\\,1,190\\,torr[\/latex] <\/span><\/p>\n

                                    Now we can use Boyle\u2019s law:<\/p>\n

                                    (1,190 torr)(7.02 L) = (1,775 torr) \u00d7 V<\/em>f<\/sub><\/span><\/span><\/p>\n

                                    Torr cancels algebraically from both sides of the equation, leaving<\/p>\n

                                    (1,190)(7.02 L) = (1,775) \u00d7 V<\/em>f<\/sub><\/span><\/span><\/p>\n

                                    Now we divide both sides of the equation by 1,775 to isolate V<\/em>f<\/sub> on one side. Solving for the final volume,<\/p>\n

                                    [latex]V_f\\,=\\,\\frac{(1,190)(7.02\\,L)}{1,775}\\,=\\,4.71\\,L[\/latex] <\/span><\/p>\n

                                    Because the pressure increases, it makes sense that the volume decreases.<\/p>\n<\/div>\n

                                    \n
                                    \n

                                    Note<\/h3>\n

                                    The answer for the final volume is essentially the same if we converted the 1,775 torr to atmospheres: 1,775\u00a0torr\u00d7[latex]\\frac{1\\,atm}{760\\,torr}[\/latex]=2.336\u00a0atm. <\/span> Using Boyle\u2019s law: (1.56 atm)(7.02 L) = (2.335 atm) \u00d7 V<\/em>f<\/sub>; [latex]V_f\\,=\\,\\frac{(1.56\\,atm)(7.02\\,L)}{2.336\\,atm}\\,=\\,4.69\\,L[\/latex].<\/p>\n<\/div>\n<\/div>\n

                                    \n
                                    \n

                                    Skill-Building Exercise<\/h3>\n
                                      \n
                                    1. \n
                                      \n

                                      If a sample of gas has an initial pressure of 375 torr and an initial volume of 7.02 L, what is the final pressure if the volume is changed to 4,577 mL? Does the answer make sense? Assume that amount and the temperature of the gas remain constant.<\/p>\n<\/div>\n<\/li>\n<\/ol>\n<\/div>\n<\/div>\n

                                      \n
                                      \n

                                      To Your Health: Breathing<\/h3>\n

                                      Breathing certainly is a major contribution to your health! Without breathing, we could not survive. Curiously, the act of breathing itself is little more than an application of Boyle\u2019s law.<\/p>\n

                                      The lungs are a series of ever-narrowing tubes that end in a myriad of tiny sacs called alveoli. It is in the alveoli that oxygen from the air transfers to the bloodstream and carbon dioxide from the bloodstream transfers to the lungs for exhalation. For air to move in and out of the lungs, the pressure inside the lungs must change, forcing the lungs to change volume\u2014just as predicted by Boyle\u2019s law.<\/p>\n

                                      The pressure change is caused by the diaphragm, a muscle that covers the bottom of the lungs. When the diaphragm moves down, it expands the size of our lungs. When this happens, the air pressure inside our lungs decreases slightly. This causes new air to rush in, and we inhale. The pressure decrease is slight\u2014only 3 torr, or about 0.4% of an atmosphere. We inhale only 0.5\u20131.0 L of air per normal breath.<\/p>\n

                                      Exhaling air requires that we relax the diaphragm, which pushes against the lungs and slightly decreases the volume of the lungs. This slightly increases the pressure of the air in the lungs, and air is forced out; we exhale. Only 1\u20132 torr of extra pressure is needed to exhale. So with every breath, our own bodies are performing an experimental test of Boyle\u2019s law.<\/p>\n<\/div>\n

                                      Charles\u2019s Law<\/span><\/p>\n<\/div>\n<\/div>\n

                                      \n

                                      Another simple gas law relates the volume of a gas to its temperature. Experiments indicate that as the temperature of a gas sample is increased, its volume increases as long as the pressure and the amount of gas remain constant. The way to write this mathematically is<\/p>\n

                                      V \u221d T<\/span><\/span><\/p>\n

                                      At this point, the concept of temperature must be clarified. Although the Kelvin scale is the preferred temperature scale, the Celsius scale is also a common temperature scale used in science. The Celsius scale is based on the melting and boiling points of water and is actually the common temperature scale used by most countries around the world (except for the United States, which still uses the Fahrenheit scale). The value of a Celsius temperature is directly related to its Kelvin value by a simple expression:<\/p>\n

                                      Kelvin temperature = Celsius temperature + 273<\/span><\/span><\/p>\n

                                      Thus, it is easy to convert from one temperature scale to another.<\/p>\n

                                      \n
                                      \n

                                      Note<\/h3>\n

                                      The Kelvin scale is sometimes referred to as the absolute scale because the zero point on the Kelvin scale is at absolute zero, the coldest possible temperature. On the other temperature scales, absolute zero is \u2212260\u00b0C or \u2212459\u00b0F.<\/p>\n<\/div>\n<\/div>\n

                                      The expression relating a gas volume to its temperature begs the following question: to which temperature scale is the volume of a gas related? The answer is that gas volumes are directly related to the Kelvin temperature<\/em>. Therefore, the temperature of a gas sample should always be expressed in (or converted to) a Kelvin temperature.<\/p>\n

                                      \n

                                      Example 7<\/h3>\n

                                      What happens to the volume of a gas if its temperature is decreased? Assume that all other conditions remain constant.<\/p>\n

                                      Solution<\/p>\n

                                      If the temperature of a gas sample is decreased, the volume decreases as well.<\/p>\n<\/div>\n

                                      \n
                                      \n

                                      Skill-Building Exercise<\/h3>\n
                                        \n
                                      1. \n
                                        \n

                                        What happens to the temperature of a gas if its volume is increased? Assume that all other conditions remain constant.<\/p>\n<\/div>\n<\/li>\n<\/ol>\n<\/div>\n<\/div>\n

                                        As with Boyle\u2019s law, the relationship between volume and temperature can be expressed in terms of initial and final values of volume and temperature, as follows:<\/p>\n

                                        [latex]\\frac{V_i}{T_i}\\,=\\,\\frac{V_f}{T_f}[\/latex] <\/span><\/p>\n

                                        where V<\/em>i<\/sub> and T<\/em>i<\/sub> are the initial volume and temperature, and V<\/em>f<\/sub> and T<\/em>f<\/sub> are the final volume and temperature. This is Charles\u2019s law<\/span><\/strong><\/span>. The restriction on its use is that the pressure of the gas and the amount of gas must remain constant. (Charles\u2019s law is sometimes referred to as Gay-Lussac\u2019s law, after the scientist who promoted Charles\u2019s work.)<\/p>\n

                                        \n

                                        Example 8<\/h3>\n

                                        A gas sample at 20\u00b0C has an initial volume of 20.0 L. What is its volume if the temperature is changed to 60\u00b0C? Does the answer make sense? Assume that the pressure and the amount of the gas remain constant.<\/p>\n

                                        Solution<\/p>\n

                                        Although the temperatures are given in degrees Celsius, we must convert them to the kelvins before we can use Charles\u2019s law. Thus,<\/p>\n

                                        20\u00b0C + 273 = 293 K = T<\/em>i<\/sub><\/span><\/span>
                                        \n60\u00b0C + 273 = 333 K = T<\/em>f<\/sub><\/span><\/span><\/p>\n

                                        Now we can substitute these values into Charles\u2019s law, along with the initial volume of 20.0 L:<\/p>\n

                                        [latex]\\frac{20.0\\,L}{293\\,K}\\,=\\,\\frac{V_f}{333K}[\/latex] <\/span><\/p>\n

                                        Multiplying the 333 K to the other side of the equation, we see that our temperature units will cancel:<\/p>\n

                                        [latex]\\frac{(333\\,\\rlap{\\text{---}}K)(20.0\\,L)}{293\\,\\rlap{\\text{---}}K}\\,=\\,V_f[\/latex] <\/span><\/p>\n

                                        Solving for the final volume, V<\/em>f<\/sub> = 22.7 L. So, as the temperature is increased, the volume increases. This makes sense because volume is directly proportional to the absolute temperature (as long as the pressure and the amount of the remain constant).<\/p>\n<\/div>\n

                                        \n
                                        \n

                                        Skill-Building Exercise<\/h3>\n
                                          \n
                                        1. \n
                                          \n

                                          A gas sample at 35\u00b0C has an initial volume of 5.06 L. What is its volume if the temperature is changed to \u221235\u00b0C? Does the answer make sense? Assume that the pressure and the amount of the gas remain constant.<\/p>\n<\/div>\n<\/li>\n<\/ol>\n<\/div>\n

                                           <\/p>\n<\/div>\n<\/div>\n

                                          \n

                                          Combined Gas Law<\/h2>\n

                                          Other gas laws can be constructed, but we will focus on only two more. The combined gas law<\/span><\/strong><\/span>\u00a0brings Boyle\u2019s and Charles\u2019s laws together to relate pressure, volume, and temperature changes of a gas sample:<\/p>\n

                                          [latex]\\frac{P_i\\,V_i}{T_i}\\,=\\,\\frac{P_f\\,V_f}{T_f}[\/latex] <\/span><\/p>\n

                                          To apply this gas law, the amount of gas should remain constant. As with the other gas laws, the temperature must be expressed in kelvins, and the units on the similar quantities should be the same. Because of the dependence on three quantities at the same time, it is difficult to tell in advance what will happen to one property of a gas sample as two other properties change. The best way to know is to work it out mathematically.<\/p>\n

                                          \n

                                          Example 9<\/h3>\n

                                          A sample of gas has P<\/em>i<\/sub> = 1.50 atm, V<\/em>i<\/sub> = 10.5 L, and T<\/em>i<\/sub> = 300 K. What is the final volume if P<\/em>f<\/sub> = 0.750 atm and T<\/em>f<\/sub> = 350 K?<\/p>\n

                                          Solution<\/p>\n

                                          Using the combined gas law, substitute for five of the quantities:<\/p>\n

                                          [latex]\\frac{(1.50\\,atm)(10.5\\,L)}{300\\,K}\\,=\\,\\frac{(0.750\\,atm)(V_f)}{350\\,K}[\/latex] <\/span><\/p>\n

                                          We algebraically rearrange this expression to isolate V<\/em>f<\/sub> on one side of the equation:<\/p>\n

                                          [latex]V_f\\,=\\,\\frac{(1.50\\,\\rlap{\\text{------}}atm)(10.5\\,L)(350\\,\\rlap{\\text{---}}K)}{(300\\,\\rlap{\\text{---}}K)(0.750\\,\\rlap{\\text{------}}atm)}\\,=\\,24.5\\,L[\/latex] <\/span><\/p>\n

                                          Note how all the units cancel except the unit for volume.<\/p>\n<\/div>\n

                                          \n
                                          \n

                                          Skill-Building Exercise<\/h3>\n
                                            \n
                                          1. \n
                                            \n

                                            A sample of gas has P<\/em>i<\/sub> = 0.768 atm, V<\/em>i<\/sub> = 10.5 L, and T<\/em>i<\/sub> = 300 K. What is the final pressure if V<\/em>f<\/sub> = 7.85 L and T<\/em>f<\/sub> = 250 K?<\/p>\n<\/div>\n<\/li>\n<\/ol>\n<\/div>\n<\/div>\n

                                            \n

                                            Example 10<\/h3>\n

                                            A balloon containing a sample of gas has a temperature of 22\u00b0C and a pressure of 1.09 atm in an airport in Cleveland. The balloon has a volume of 1,070 mL. The balloon is transported by plane to Denver, where the temperature is 11\u00b0C and the pressure is 655 torr. What is the new volume of the balloon?<\/p>\n

                                            Solution<\/p>\n

                                            The first task is to convert all quantities to the proper and consistent units. The temperatures must be expressed in kelvins, and the pressure units are different so one of the quantities must be converted. Let us convert the atmospheres to torr:<\/p>\n

                                            22\u00b0C + 273 = 295 K = T<\/em>i<\/sub><\/span><\/span>
                                            \n11\u00b0C + 273 = 284 K = T<\/em>f<\/sub><\/span><\/span>
                                            \n [latex]1.09\\,\\rlap{\\text{------}}atm\\,\u00d7\\,\\frac{760\\,torr}{1\\,\\rlap{\\text{------}}atm}\\,=\\,828\\,torr\\,=\\,P_i[\/latex]<\/span><\/p>\n

                                            Now we can substitute the quantities into the combined has law:<\/p>\n

                                            [latex]\\frac{(828\\,torr)(1,070\\,mL)}{295\\,K}\\,=\\,\\frac{(655\\,torr)\\,\u00d7\\,V_f}{284\\,K}[\/latex] <\/span><\/p>\n

                                            To solve for V<\/em>f<\/sub>, we multiply the 284 K in the denominator of the right side into the numerator on the left, and we divide 655 torr in the numerator of the right side into the denominator on the left:<\/p>\n

                                            [latex]\\frac{(828\\,\\rlap{\\text{------}}torr)(1,070\\,mL)(284\\,\\rlap{\\text{---}}K)}{(295\\,\\rlap{\\text{---}}K)(655\\,\\rlap{\\text{------}}torr)}\\,=\\,V_f[\/latex] <\/span><\/p>\n

                                            Notice that torr and kelvins cancel, as they are found in both the numerator and denominator. The only unit that remains is milliliters, which is a unit of volume. So V<\/em>f<\/sub> = 1,300 mL. The overall change is that the volume of the balloon has increased by 230 mL.<\/p>\n<\/div>\n

                                            \n
                                            \n

                                            Skill-Building Exercise<\/h3>\n
                                              \n
                                            1. \n
                                              \n

                                              A balloon used to lift weather instruments into the atmosphere contains gas having a volume of 1,150 L on the ground, where the pressure is 0.977 atm and the temperature is 18\u00b0C. Aloft, this gas has a pressure of 6.88 torr and a temperature of \u221215\u00b0C. What is the new volume of the gas?<\/p>\n<\/div>\n<\/li>\n<\/ol>\n<\/div>\n

                                               <\/p>\n<\/div>\n<\/div>\n

                                              \n

                                              The Ideal Gas Law<\/h2>\n

                                              So far, the gas laws we have used have focused on changing one or more properties of the gas, such as its volume, pressure, or temperature. There is one gas law that relates all the independent properties of a gas under any particular condition, rather than a change in conditions. This gas law is called the ideal gas law<\/span><\/strong><\/span>. The formula of this law is as follows:<\/p>\n

                                              PV<\/em> = nRT<\/em><\/span><\/span><\/p>\n

                                              In this equation, P<\/em> is pressure, V<\/em> is volume, n<\/em> is amount of moles, and T<\/em> is temperature. R<\/em> is called the ideal gas law constant<\/span><\/strong><\/span>\u00a0and is a proportionality constant that relates the values of pressure, volume, amount, and temperature of a gas sample. The variables in this equation do not have the subscripts i<\/em> and f<\/em> to indicate an initial condition and a final condition. The ideal gas law relates the four independent properties of a gas under any<\/em> conditions.<\/p>\n

                                              The value of R<\/em> depends on what units are used to express the other quantities. If volume is expressed in liters and pressure in atmospheres, then the proper value of R<\/em> is as follows:<\/p>\n

                                              [latex]R\\,=\\,0.08205\\,\\frac{L\u22c5atm}{mol\u22c5K}[\/latex]<\/span><\/p>\n

                                              This may seem like a strange unit, but that is what is required for the units to work out algebraically.<\/p>\n

                                              \n

                                              Example 11<\/h3>\n

                                              What is the volume in liters of 1.45 mol of N2<\/sub> gas at 298 K and 3.995 atm?<\/p>\n

                                              Solution<\/p>\n

                                              Using the ideal gas law where P<\/em> = 3.995 atm, n<\/em> = 1.45, and T<\/em> = 298,<\/p>\n

                                              [latex](3.995\\,atm)\\,\u00d7\\,V\\,=\\,(1.45\\,mol)\\,(0.08205\\,\\frac{L\u22c5atm}{mol\u22c5K})\\,(298\\,K)[\/latex] <\/span><\/p>\n

                                              On the right side, the moles and kelvins cancel. Also, because atmospheres appear in the numerator on both sides of the equation, they also cancel. The only remaining unit is liters, a unit of volume. So<\/p>\n

                                              3.995 \u00d7 V<\/em> = (1.45)(0.08205)(298) L<\/span><\/span><\/p>\n

                                              Dividing both sides of the equation by 3.995 and evaluating, we get V<\/em> = 8.87 L. Note that the conditions of the gas are not changing. Rather, the ideal gas law allows us to determine what the fourth property of a gas (here, volume) must<\/em> be if three other properties (here, amount, pressure, and temperature) are known.<\/p>\n<\/div>\n

                                              \n
                                              \n
                                              \n

                                              Skill-Building Exercise<\/h3>\n
                                                \n
                                              1. \n
                                                \n

                                                What is the pressure of a sample of CO2<\/sub> gas if 0.557 mol is held in a 20.0 L container at 451 K?<\/p>\n<\/div>\n<\/li>\n<\/ol>\n<\/div>\n<\/div>\n

                                                For convenience, scientists have selected 273 K (0\u00b0C) and 1.00 atm pressure as a set of standard conditions for gases. This combination of conditions is called standard temperature and pressure (STP)<\/span><\/strong><\/span>. Under these conditions, 1 mol of any gas has about the same volume. We can use the ideal gas law to determine the volume of 1 mol of gas at STP:<\/span><\/p>\n<\/div>\n

                                                [latex](1.00\\,atm)\\,\u00d7\\,V\\,=\\,(1.00\\,mol)\\,(0.08205\\,\\frac{L\u22c5atm}{mol\u22c5K})\\,(273\\,K)[\/latex]<\/span><\/p>\n

                                                This volume is 22.4 L. Because this volume is independent of the identity of a gas, the idea that 1 mol of gas has a volume of 22.4 L at STP makes a convenient conversion factor:<\/p>\n

                                                1 mol gas = 22.4 L (at STP)<\/span><\/span><\/p>\n

                                                \n

                                                Example 12<\/h3>\n

                                                Cyclopropane (C3<\/sub>H6<\/sub>) is a gas that formerly was used as an anesthetic. How many moles of gas are there in a 100.0 L sample if the gas is at STP?<\/p>\n

                                                Solution<\/p>\n

                                                We can set up a simple, one-step conversion that relates moles and liters:<\/p>\n

                                                [latex]100.0\\,\\rlap{\\text{---}}L\\,C_3H_6\\,\u00d7\\,\\frac{1\\,mol}{22.4\\,\\rlap{\\text{---}}L}\\,=\\,4.46\\,mol\\,C_3H_6[\/latex] <\/span><\/p>\n

                                                There are almost 4.5 mol of gas in 100.0 L.<\/p>\n<\/div>\n

                                                \n
                                                \n

                                                Note<\/h3>\n

                                                Because of its flammability, cyclopropane is no longer used as an anesthetic gas.<\/p>\n<\/div>\n<\/div>\n

                                                \n
                                                \n

                                                Skill-Building Exercise<\/h3>\n
                                                  \n
                                                1. \n
                                                  \n

                                                  Freon is a trade name for a series of fluorine- and chlorine-containing gases that formerly were used in refrigeration systems. What volume does 8.75 mol of Freon have at STP?<\/p>\n<\/div>\n<\/li>\n<\/ol>\n<\/div>\n<\/div>\n

                                                  \n
                                                  \n

                                                  Note<\/h3>\n

                                                  Many gases known as Freon are no longer used because their presence in the atmosphere destroys the ozone layer, which protects us from ultraviolet light from the sun.<\/p>\n<\/div>\n<\/div>\n

                                                  \n
                                                  \n

                                                  Career Focus: Respiratory Therapist<\/h3>\n

                                                  Certain diseases\u2014such as emphysema, lung cancer, and severe asthma\u2014primarily affect the lungs. Respiratory therapists help patients with breathing-related problems. They can evaluate, help diagnose, and treat breathing disorders and even help provide emergency assistance in acute illness where breathing is compromised.<\/p>\n

                                                  Most respiratory therapists must complete at least two years of college and earn an associate\u2019s degree, although therapists can assume more responsibility if they have a college degree. Therapists must also pass state or national certification exams. Once certified, respiratory therapists can work in hospitals, doctor\u2019s offices, nursing homes, or patient\u2019s homes. Therapists work with equipment such as oxygen tanks and respirators, may sometimes dispense medication to aid in breathing, perform tests, and educate patients in breathing exercises and other therapy.<\/p>\n

                                                  Because respiratory therapists work directly with patients, the ability to work well with others is a must for this career. It is an important job because it deals with one of the most crucial functions of the body.<\/p>\n<\/div>\n<\/div>\n

                                                  \n
                                                  \n
                                                  \n

                                                  Concept Review Exercises<\/h3>\n
                                                    \n
                                                  1. \n
                                                    \n

                                                    What properties do the gas laws help us predict?<\/p>\n<\/div>\n<\/li>\n

                                                  2. \n
                                                    \n

                                                    What makes the ideal gas law different from the other gas laws?<\/p>\n<\/div>\n<\/li>\n<\/ol>\n<\/div>\n

                                                    \n

                                                    Answers<\/h3>\n

                                                    \n

                                                    Show Answer<\/span><\/p>\n
                                                    \n
                                                      \n
                                                    1. Gas laws relate four properties: pressure, volume, temperature, and number of moles.<\/li>\n
                                                    2. The ideal gas law does not require that the properties of a gas change.<\/div>\n<\/div>\n<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n
                                                      \n
                                                      \n

                                                      Key Takeaway<\/h3>\n