{"id":284,"date":"2016-01-11T19:59:53","date_gmt":"2016-01-11T19:59:53","guid":{"rendered":"https:\/\/opentextbc.ca\/introductorychemistryclone\/chapter\/the-ideal-gas-law-and-some-applications-2\/"},"modified":"2020-05-07T20:18:31","modified_gmt":"2020-05-07T20:18:31","slug":"the-ideal-gas-law-and-some-applications","status":"publish","type":"chapter","link":"https:\/\/opentextbc.ca\/introductorychemistryclone\/chapter\/the-ideal-gas-law-and-some-applications\/","title":{"raw":"The Ideal Gas Law and Some Applications","rendered":"The Ideal Gas Law and Some Applications"},"content":{"raw":"[latexpage]\r\n
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Learning Objectives<\/p>\r\n\r\n<\/header>\r\n

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    \r\n \t
  1. Learn the ideal gas law.<\/li>\r\n \t
  2. Apply the ideal gas law to any set of conditions of a gas.<\/li>\r\n \t
  3. Apply the ideal gas law to molar volumes, density, and stoichiometry problems.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<\/div>\r\nSo far, the gas laws we have considered have all required that the gas change its conditions; then we predict a resulting change in one of its properties. Are there any gas laws that relate the physical properties of a gas at any given time?\r\n\r\nConsider a further extension of the combined gas law to include n<\/i>. By analogy to Avogadro\u2019s law, n<\/i> is positioned in the denominator of the fraction, opposite the volume. So:\r\n\r\n\\[\\dfrac{PV}{nT}=\\text{constant}\\]\r\n\r\nBecause pressure, volume, temperature, and amount are the only four independent physical properties of a gas, the constant in the above equation is truly a constant; indeed, because we do not need to specify the identity of a gas to apply the gas laws, this constant is the same for all gases. We define this constant with the symbol R<\/i>, so the previous equation is written as:\r\n\r\n\\[\\dfrac{PV}{nT}=R\\]\r\n\r\nwhich is usually rearranged as:\r\n\r\n\\[PV=nRT\\]\r\n\r\nThis equation is called the [pb_glossary id=\"1578\"]ideal gas law[\/pb_glossary]. It relates the four independent properties of a gas at any time. The constant R<\/i> is called the ideal gas law constant. Its value depends on the units used to express pressure and volume. Table 6.1 \"Values of the Ideal Gas Law Constant R<\/i>\u201d<\/a> lists the numerical values of R<\/i>.\r\n\r\n\r\n\r\n\r\n\r\n\r\n
    Table 6.1<\/span> Values of the Ideal Gas Law Constant R<\/i><\/caption>\r\n
    Numerical Value<\/th>\r\nUnits<\/th>\r\n<\/tr>\r\n<\/thead>\r\n
    0.08205<\/td>\r\n\\(\\dfrac{\\text{L}\\cdot \\text{atm}}{\\text{mol}\\cdot \\text{K}}\\)<\/td>\r\n<\/tr>\r\n
    62.36<\/td>\r\n\\(\\dfrac{\\text{L}\\cdot \\text{torr}}{\\text{mol}\\cdot \\text{K}}=\\dfrac{\\text{L}\\cdot\\text{mmHg}}{\\text{mol}\\cdot \\text{K}}\\)<\/td>\r\n<\/tr>\r\n
    8.314<\/td>\r\n\\(\\dfrac{\\text{J}}{\\text{mol}\\cdot \\text{K}}\\)<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nThe ideal gas law is used like any other gas law, with attention paid to the units and making sure that temperature is expressed in kelvins. However, the ideal gas law does not require a change in the conditions of a gas sample<\/em>. The ideal gas law implies that if you know any three of the physical properties of a gas, you can calculate the fourth property.\r\n
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    Example 6.9<\/p>\r\n\r\n<\/header>\r\n

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    Problem<\/h1>\r\nA 4.22 mol sample of Ar has a pressure of 1.21 atm and a temperature of 34\u00b0C. What is its volume?\r\n

    Solution<\/h2>\r\nThe first step is to convert temperature to kelvins:\r\n

    34 + 273 = 307 K<\/p>\r\nNow we can substitute the conditions into the ideal gas law:\r\n\r\n\\[(1.21\\text{ atm})(V)=(4.22\\text{ mol})\\left(0.08205\\cdot \\dfrac{\\text{L}\\cdot\\text{atm}}{\\text{mol}\\cdot \\text{K}}\\right)(307 \\text{ K})\\]\r\n\r\nThe atm unit is in the numerator of both sides, so it cancels. On the right side of the equation, the mol and K units appear in the numerator and the denominator, so they cancel as well. The only unit remaining is L, which is the unit of volume we are looking for. We isolate the volume variable by dividing both sides of the equation by 1.21:\r\n\r\n\\[V=\\dfrac{(4.22)(0.08205)(307)}{1.21}\\text{ L}\\]\r\n\r\nThen solving for volume, we get:\r\n

    V<\/i> = 87.9 L<\/p>\r\n\r\n

    Test Yourself<\/h1>\r\nA 0.0997 mol sample of O2<\/sub> has a pressure of 0.692 atm and a temperature of 333 K. What is its volume?\r\n

    Answer<\/h2>\r\n3.94 L\r\n\r\n<\/div>\r\n<\/div>\r\n
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    Example 6.10<\/p>\r\n\r\n<\/header>\r\n

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    Problem<\/h1>\r\nAt a given temperature, 0.00332 g of Hg in the gas phase has a pressure of 0.00120 mmHg and a volume of 435 L. What is its temperature?\r\n

    Solution<\/h2>\r\nWe are not given the number of moles of Hg directly, but we are given a mass. We can use the molar mass of Hg to convert to the number of moles.\r\n\r\n\\[0.00332\\text{ \\cancel{g Hg}}\\times \\dfrac{1\\text{ mol Hg}}{200.59\\text{ \\cancel{g Hg}}}=0.0000165\\text{ mol}=1.65\\times 10^{-5}\\text{ mol}\\]\r\n\r\nPressure is given in units of millimetres of mercury. We can either convert this to atmospheres or use the value of the ideal gas constant that includes the mmHg unit. We will take the second option. Substituting into the ideal gas law:\r\n\r\n\\[(0.00120\\text{ mmHg})(435\\text{ L})=(1.65\\times 10^{-5}\\text{ mol})\\left(62.36\\cdot \\dfrac{\\text{L}\\cdot \\text{mmHg}}{\\text{mol}\\cdot\\text{K}}\\right)T\\]\r\n\r\nThe mmHg, L, and mol units cancel, leaving the K unit, the unit of temperature. Isolating T<\/em> all by itself on one side, we get:\r\n\r\n\\[T=\\dfrac{(0.00120)(435)}{(1.65\\times 10^{-5})(62.36)}\\text{ K}\\]\r\n\r\nThen solving for K, we get:\r\n

    T<\/i> = 507 K<\/p>\r\n\r\n

    Test Yourself<\/h1>\r\nFor a 0.00554 mol sample of H2<\/sub>, P<\/i> = 23.44 torr and T<\/i> = 557 K. What is its volume?\r\n

    Answer<\/h2>\r\n8.21 L\r\n\r\n<\/div>\r\n<\/div>\r\nThe ideal gas law can also be used in stoichiometry problems.\r\n
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    Example 6.11<\/p>\r\n\r\n<\/header>\r\n

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    Problem<\/h1>\r\nWhat volume of H2<\/sub> is produced at 299 K and 1.07 atm when 55.8 g of Zn metal react with excess HCl?\r\n

    Zn(s) + 2HCl(aq) \u2192 ZnCl2<\/sub>(aq) + H2<\/sub>(g)<\/p>\r\n\r\n

    Solution<\/h2>\r\nHere we have a stoichiometry problem where we need to find the number of moles of H2<\/sub> produced. Then we can use the ideal gas law, with the given temperature and pressure, to determine the volume of gas produced. First, the number of moles of H2<\/sub> is calculated:\r\n\r\n\\[55.8\\text{ \\cancel{g Zn}}\\times \\dfrac{1\\text{ \\cancel{mol Zn}}}{65.41\\text{ \\cancel{g Zn}}}\\times \\dfrac{1\\text{ mol }\\ce{H2}}{1\\text{ \\cancel{mol Zn}}}=0.853\\text{ mol }\\ce{H2}\\]\r\n\r\nNow that we know the number of moles of gas, we can use the ideal gas law to determine the volume, given the other conditions:\r\n\r\n\\[(1.07\\text{ atm})V=(0.853\\text{ mol})\\left(0.08205\\cdot \\dfrac{\\text{L}\\cdot\\text{atm}}{\\text{mol}\\cdot\\text{K}}\\right)(299\\text{ K})\\]\r\n\r\nAll the units cancel except for L, for volume, which means:\r\n

    V<\/i> = 19.6 L<\/p>\r\n\r\n

    Test Yourself<\/h1>\r\nWhat pressure of HCl is generated if 3.44 g of Cl2<\/sub> are reacted in 4.55 L at 455 K?\r\n

    H2<\/sub>(g) + Cl2<\/sub>(g) \u2192 2HCl(g)<\/p>\r\n\r\n

    Answer<\/h2>\r\n0.796 atm\r\n\r\n<\/div>\r\n<\/div>\r\nIt should be obvious by now that some physical properties of gases depend strongly on the conditions. What we need is a set of standard conditions so that properties of gases can be properly compared to each other. [pb_glossary id=\"1579\"]Standard temperature and pressure (STP)[\/pb_glossary]\u00a0is defined as exactly 100 kPa of pressure (0.986 atm) and 273 K (0\u00b0C). For simplicity, we will use 1 atm as standard pressure. Defining STP allows us to compare more directly the properties of gases that differ from each other.\r\n\r\nOne property shared among gases is a molar volume. The [pb_glossary id=\"1580\"]molar volume[\/pb_glossary]\u00a0is the volume of 1 mol of a gas. At STP, the molar volume of a gas can be easily determined by using the ideal gas law:\r\n\r\n\\[(1\\text{ atm})V=(1\\text{ mol})\\left(0.08205\\cdot \\dfrac{\\text{L}\\cdot\\text{atm}}{\\text{mol}\\cdot\\text{K}}\\right)(273\\text{ K})\\]\r\n\r\nAll the units cancel except for L, the unit of volume. So:\r\n

    V<\/i> = 22.4 L<\/p>\r\nNote that we have not specified the identity of the gas; we have specified only that the pressure is 1 atm and the temperature is 273 K. This makes for a very useful approximation: any gas at STP has a volume of 22.4 L per mole of gas<\/i>; that is, the molar volume at STP is 22.4 L\/mol (Figure 6.3 \"Molar Volume\"<\/a>). This molar volume makes a useful conversion factor in stoichiometry problems if the conditions are at STP. If the conditions are not at STP, a molar volume of 22.4 L\/mol is not applicable. However, if the conditions are not at STP, the combined gas law can be used to calculate the volume of the gas at STP; then the 22.4 L\/mol molar volume can be used.\r\n\r\n[caption id=\"attachment_274\" align=\"aligncenter\" width=\"300\"]\"A<\/a> Figure 6.3 \"Molar Volume.\" A mole of gas at STP occupies 22.4 L, the volume of a cube that is 28.2 cm on a side.[\/caption]\r\n\r\n

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    Example 6.12<\/p>\r\n\r\n<\/header>\r\n

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    Problem<\/h1>\r\nHow many moles of Ar are present in 38.7 L at STP?\r\n

    Solution<\/h2>\r\nWe can use the molar volume, 22.4 L\/mol, as a conversion factor, but we need to reverse the fraction so that the L units cancel and mol units are introduced. It is a one-step conversion:\r\n\r\n\\[38.7\\text{ \\cancel{L}}\\times \\dfrac{1\\text{ mol}}{22.4\\text{ \\cancel{L}}}=1.73\\text{ mol}\\]\r\n

    Test Yourself<\/h1>\r\nWhat volume does 4.87 mol of Kr have at STP?\r\n

    Answer<\/h2>\r\n109 L\r\n\r\n<\/div>\r\n<\/div>\r\n
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    Example 6.13<\/p>\r\n\r\n<\/header>\r\n

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    Problem<\/h1>\r\nWhat volume of H2<\/sub> is produced at STP when 55.8 g of Zn metal react with excess HCl?\r\n

    Zn(s) + 2HCl(aq) \u2192 ZnCl2<\/sub>(aq) + H2<\/sub>(g)<\/p>\r\n\r\n

    Solution<\/h2>\r\nThis is a stoichiometry problem with a twist: we need to use the molar volume of a gas at STP to determine the final answer. The first part of the calculation is the same as in a previous example:\r\n\r\n\\[55.8\\text{ \\cancel{g Zn}}\\times\\dfrac{1\\text{ \\cancel{mol Zn}}}{65.41\\text{ \\cancel{g Zn}}}\\times \\dfrac{1\\text{ mol }\\ce{H2}}{1\\text{ \\cancel{mol Zn}}}=0.853\\text{ mol }\\ce{H2}\\]\r\n\r\nNow we can use the molar volume, 22.4 L\/mol, because the gas is at STP:\r\n\r\n\\[0.853\\cancel{\\text{ mol }\\ce{H2}}\\times \\dfrac{22.4\\text{ L}}{1\\cancel{\\text{ mol }\\ce{H2}}}=19.1\\text{ L }\\ce{H2}\\]\r\n\r\nAlternatively, we could have applied the molar volume as a third conversion factor in the original stoichiometry calculation.\r\n

    Test Yourself<\/h1>\r\nWhat volume of HCl is generated if 3.44 g of Cl2<\/sub> are reacted at STP?\r\n

    H2<\/sub>(g) + Cl2<\/sub>(g) \u2192 2HCl(g)<\/p>\r\n\r\n

    Answer<\/h2>\r\n2.17 L\r\n\r\n<\/div>\r\n<\/div>\r\nThe ideal gas law can also be used to determine the densities of gases. Recall that density is defined as the mass of a substance divided by its volume:\r\n\r\n\\[d=\\dfrac{m}{V}\\]\r\n\r\nAssume that you have exactly 1 mol of a gas. If you know the identity of the gas, you can determine the molar mass of the substance. Using the ideal gas law, you can also determine the volume of that mole of gas, using whatever the temperature and pressure conditions are. Then you can calculate the density of the gas by using:\r\n\r\n\\[\\text{density}=\\dfrac{\\text{molar mass}}{\\text{molar volume}}\\]\r\n
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    Example 6.14<\/p>\r\n\r\n<\/header>\r\n

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    Problem<\/h1>\r\nWhat is the density of N2<\/sub> at 25\u00b0C and 0.955 atm?\r\n

    Solution<\/h2>\r\nFirst, we must convert the temperature into kelvins:\r\n

    25 + 273 = 298 K<\/p>\r\nIf we assume exactly 1 mol of N2<\/sub>, then we know its mass: 28.0 g. Using the ideal gas law, we can calculate the volume:\r\n\r\n\\[(0.955\\text{ atm})V=(1\\text{ mol})\\left(0.08205\\cdot \\dfrac{\\text{L}\\cdot\\text{atm}}{\\text{mol}\\cdot\\text{K}}\\right)(298\\text{ K})\\]\r\n\r\nAll the units cancel except for L, the unit of volume. So:\r\n

    V<\/i> = 25.6 L<\/p>\r\nKnowing the molar mass and the molar volume, we can determine the density of N2<\/sub> under these conditions:\r\n\r\n\\[d=\\dfrac{28.0\\text{ g}}{25.6\\text{ L}}=1.09\\text{ g\/L}\\]\r\n

    Test Yourself<\/h1>\r\nWhat is the density of CO2<\/sub> at a pressure of 0.0079 atm and 227 K? (These are the approximate atmospheric conditions on Mars.)\r\n

    Answer<\/h2>\r\n0.019 g\/L\r\n\r\n<\/div>\r\n<\/div>\r\n
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    Chemistry Is Everywhere: Breathing<\/h1>\r\nBreathing (more properly called respiration<\/i>) is the process by which we draw air into our lungs so that our bodies can take up oxygen from the air. Let us apply the gas laws to breathing.\r\n\r\nStart by considering pressure. We draw air into our lungs because the diaphragm, a muscle underneath the lungs, moves down to reduce pressure in the lungs, causing external air to rush in to fill the lower-pressure volume. We expel air by the diaphragm pushing against the lungs, increasing pressure inside the lungs and forcing the high-pressure air out. What are the pressure changes involved? A quarter of an atmosphere? A tenth of an atmosphere? Actually, under normal conditions, a pressure difference of\u00a0only 1 or 2 torr\u00a0makes us breathe in and out.\r\n\r\n[caption id=\"attachment_282\" align=\"aligncenter\" width=\"300\"]\"Breathing<\/a> Figure 6.4 \"Breathing Mechanics.\" Breathing involves pressure differences between the inside of the lungs and the air outside. The pressure differences are only a few torr.[\/caption]\r\n\r\nA normal breath is about 0.50 L. If room temperature is about 22\u00b0C, then the air has a temperature of about 295 K. With normal pressure being 1.0 atm, how many moles of air do we take in for every breath? The ideal gas law gives us an answer:\r\n\r\n\\[(1.0\\text{ atm})(0.50\\text{ L})=n\\left(0.08205\\cdot \\dfrac{\\text{L}\\cdot\\text{atm}}{\\text{mol}\\cdot \\text{K}}\\right)(295\\text{ K})\\]\r\n\r\nSolving for the number of moles, we get:\r\n

    n<\/i> = 0.021 mol air<\/p>\r\nThis ends up being about 0.6 g of air per breath \u2014 not much, but enough to keep us alive.\r\n\r\n<\/div>\r\n

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    Key Takeaways<\/p>\r\n\r\n<\/header>\r\n

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