{"id":294,"date":"2016-01-11T19:59:54","date_gmt":"2016-01-11T19:59:54","guid":{"rendered":"https:\/\/opentextbc.ca\/introductorychemistryclone\/chapter\/gas-mixtures-2\/"},"modified":"2020-05-07T21:46:43","modified_gmt":"2020-05-07T21:46:43","slug":"gas-mixtures","status":"publish","type":"chapter","link":"https:\/\/opentextbc.ca\/introductorychemistryclone\/chapter\/gas-mixtures\/","title":{"raw":"Gas Mixtures","rendered":"Gas Mixtures"},"content":{"raw":"[latexpage]\r\n
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Learning Objectives<\/p>\r\n\r\n<\/header>\r\n

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  1. Learn Dalton\u2019s law of partial pressures.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<\/div>\r\nOne of the properties of gases is that they mix with each other. When they do so, they become a solution \u2014 a homogeneous mixture. Some of the properties of gas mixtures are easy to determine if we know the composition of the gases in the mix.\r\n\r\nIn gas mixtures, each component in the gas phase can be treated separately. Each component of the mixture shares the same temperature and volume. (Remember that gases expand to fill the volume of their container; gases in a mixture do that as well.) However, each gas has its own pressure. The [pb_glossary id=\"1608\"]partial pressure[\/pb_glossary]\u00a0of a gas, P<\/i>i<\/sub>, is the pressure that an individual gas in a mixture has. Partial pressures are expressed in torr, millimetres of mercury, or atmospheres like any other gas pressure; however, we use the term pressure<\/i> when talking about pure gases and the term partial pressure<\/i> when we are talking about the individual gas components in a mixture.\r\n\r\n[pb_glossary id=\"1610\"]Dalton\u2019s law of partial pressures[\/pb_glossary]\u00a0states that the total pressure of a gas mixture, P<\/i>tot<\/sub>, is equal to the sum of the partial pressures of the components, P<\/i>i<\/sub>:\r\n\r\n\\[P_{\\text{tot}}=P_1+P_2+P_3+ \\ldots = \\sum_{\\#\\text{ of gases}} P_{\\text{i}}\\]\r\n\r\nAlthough this may seem to be a trivial law, it reinforces the idea that gases behave independently of each other.\r\n
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    Example 6.15<\/p>\r\n\r\n<\/header>\r\n

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    Problem<\/h1>\r\nA mixture of H2<\/sub> at 2.33 atm and N2<\/sub> at 0.77 atm is in a container. What is the total pressure in the container?\r\n

    Solution<\/h2>\r\nDalton\u2019s law of partial pressures states that the total pressure is equal to the sum of the partial pressures. We simply add the two pressures together:\r\n

    P<\/i>tot<\/sub> = 2.33 atm + 0.77 atm = 3.10 atm<\/p>\r\n\r\n

    Test Yourself<\/h1>\r\nAir can be thought of as a mixture of N2<\/sub> and O2<\/sub>. In 760 torr of air, the partial pressure of N2<\/sub> is 608 torr. What is the partial pressure of O2<\/sub>?\r\n

    Answer<\/h2>\r\n152 torr\r\n\r\n<\/div>\r\n<\/div>\r\n
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    Example 6.16<\/p>\r\n\r\n<\/header>\r\n

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    Problem<\/h1>\r\nA 2.00 L container with 2.50 atm of H2<\/sub> is connected to a 5.00 L container with 1.90 atm of O2<\/sub> inside. The containers are opened, and the gases mix. What is the final pressure inside the containers?\r\n

    Solution<\/h2>\r\nBecause gases act independently of each other, we can determine the resulting final pressures using Boyle\u2019s law and then add the two resulting pressures together to get the final pressure. The total final volume is 2.00 L + 5.00 L = 7.00 L. First, we use Boyle\u2019s law to determine the final pressure of H2<\/sub>:\r\n\r\n\\[(2.50\\text{ atm})(2.00\\text{ L})=P_2(7.00\\text{ L})\\]\r\n\r\nSolving for P<\/i>2<\/sub>, we get:\r\n\r\n\\[P_2=0.714\\text{ atm}=\\text{partial pressure of }\\ce{H2}\\]\r\n\r\nNow we do that same thing for the O2<\/sub>:\r\n\r\n\\[\\begin{array}{rrcll}\r\n(1.90\\text{ atm})(5.00\\text{ L})&=&P_2(7.00\\text{ L})&& \\\\ \\\\\r\nP_2&=&1.36\\text{ atm}&=&\\text{partial pressure of }\\ce{O2}\r\n\\end{array}\\]\r\n\r\nThe total pressure is the sum of the two resulting partial pressures:\r\n\r\n\\[P_{\\text{tot}}=0.714\\text{ atm}+1.36\\text{ atm}=2.07\\text{ atm}\\]\r\n

    Test Yourself<\/h1>\r\nIf 0.75 atm of He in a 2.00 L container is connected to a 3.00 L container with 0.35 atm of Ne and the connection between the containers is opened, what is the resulting total pressure?\r\n

    Answer<\/h2>\r\n0.51 atm\r\n\r\n<\/div>\r\n<\/div>\r\nOne of the reasons we have to deal with Dalton\u2019s law of partial pressures is that gases are frequently collected by bubbling through water. As we will see in Chapter 10 \"Solids and Liquids,\"<\/a>\u00a0liquids are constantly evaporating into a vapour until the vapour achieves a partial pressure characteristic of the substance and the temperature. This partial pressure is called a [pb_glossary id=\"1611\"]vapour pressure[\/pb_glossary]. Table 6.2 \"Vapour Pressure of Water versus Temperature\"<\/a> lists the vapour pressures of H2<\/sub>O versus temperature. Note that if a substance is normally a gas under a given set of conditions, the term partial pressure<\/i> is used; the term vapour pressure<\/i> is reserved for the partial pressure of a vapour when the liquid is the normal phase under a given set of conditions.\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n
    Table 6.2 Vapour Pressure of Water versus Temperature<\/caption>\r\n
    Temperature (\u00b0C)<\/th>\r\nVapour Pressure (torr)<\/th>\r\n<\/tr>\r\n<\/thead>\r\n
    5<\/td>\r\n6.54<\/td>\r\n<\/tr>\r\n
    10<\/td>\r\n9.21<\/td>\r\n<\/tr>\r\n
    15<\/td>\r\n12.79<\/td>\r\n<\/tr>\r\n
    20<\/td>\r\n17.54<\/td>\r\n<\/tr>\r\n
    21<\/td>\r\n18.66<\/td>\r\n<\/tr>\r\n
    22<\/td>\r\n19.84<\/td>\r\n<\/tr>\r\n
    23<\/td>\r\n21.08<\/td>\r\n<\/tr>\r\n
    24<\/td>\r\n22.39<\/td>\r\n<\/tr>\r\n
    25<\/td>\r\n23.77<\/td>\r\n<\/tr>\r\n
    30<\/td>\r\n31.84<\/td>\r\n<\/tr>\r\n
    35<\/td>\r\n42.20<\/td>\r\n<\/tr>\r\n
    40<\/td>\r\n55.36<\/td>\r\n<\/tr>\r\n
    50<\/td>\r\n92.59<\/td>\r\n<\/tr>\r\n
    60<\/td>\r\n149.5<\/td>\r\n<\/tr>\r\n
    70<\/td>\r\n233.8<\/td>\r\n<\/tr>\r\n
    80<\/td>\r\n355.3<\/td>\r\n<\/tr>\r\n
    90<\/td>\r\n525.9<\/td>\r\n<\/tr>\r\n
    100<\/td>\r\n760.0<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nAny time a gas is collected over water, the total pressure is equal to the partial pressure of the gas plus<\/em> the vapour pressure of water. This means that the amount of gas collected will be less than the total pressure suggests.\r\n
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    Example 6.17<\/p>\r\n\r\n<\/header>\r\n

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    Problem<\/h1>\r\nHydrogen gas is generated by the reaction of nitric acid and elemental iron. The gas is collected in an inverted 2.00 L container immersed in a pool of water at 22\u00b0C. At the end of the collection, the partial pressure inside the container is 733 torr. How many moles of H2<\/sub> gas were generated?\r\n

    Solution<\/h2>\r\nWe need to take into account that the total pressure includes the vapour pressure of water. According to Table 6.2 \"Vapour Pressure of Water versus Temperature,\"<\/a> the vapour pressure of water at 22\u00b0C is 19.84 torr. According to Dalton\u2019s law of partial pressures, the total pressure equals the sum of the pressures of the individual gases, so:\r\n

    733 torr = PH2<\/sub> + PH2<\/sub>O = PH2<\/sub> + 19.84 torr<\/p>\r\nWe solve by subtracting:\r\n

    PH2<\/sub> = 713 torr<\/p>\r\nNow we can use the ideal gas law to determine the number of moles (remembering to convert the temperature to kelvins, making it 295 K):\r\n\r\n\\[(713\\text{ torr})(2.00\\text{ L})=n\\left(62.36\\cdot \\dfrac{\\text{L}\\cdot\\text{atm}}{\\text{mol}\\cdot\\text{K}}\\right)(295\\text{ K})\\]\r\n\r\nAll the units cancel except for mol, which is what we are looking for. So:\r\n

    n<\/i> = 0.0775 mol H2<\/sub> collected<\/p>\r\n\r\n

    Test Yourself<\/h1>\r\nCO2<\/sub>, generated by the decomposition of CaCO3<\/sub>, is collected in a 3.50 L container over water. If the temperature is 50\u00b0C and the total pressure inside the container is 833 torr, how many moles of CO2<\/sub> were generated?\r\n

    Answer<\/h2>\r\n0.129 mol\r\n\r\n<\/div>\r\n<\/div>\r\nFinally, we introduce a new unit that can be useful, especially for gases. The [pb_glossary id=\"1612\"]mole fraction[\/pb_glossary], \u03c7i<\/sub>, is the ratio of the number of moles of component i<\/i> in a mixture divided by the total number of moles in the sample:\r\n\r\n\\[\\chi_{\\text{i}}=\\dfrac{\\text{moles of component }i}{\\text{total number of moles}}\\]\r\n\r\n(\u03c7 is the lowercase Greek letter chi<\/i>.) Note that mole fraction is not<\/em> a percentage; its values range from 0 to 1. For example, consider the combination of 4.00 g of He and 5.0 g of Ne. Converting both to moles, we get:\r\n

    \\(4.00\\text{ \\cancel{g He}} \\times \\dfrac{1\\text{ mol He}}{4.00\\text{ \\cancel{g He}}}=1.00\\text{ mol He and 5.0 \\cancel{g Ne}} \\times \\dfrac{1\\text{ mol Ne}}{20.0\\text{ \\cancel{g Ne}}}=0.25\\text{ mol Ne}\\)<\/p>\r\nThe total number of moles is the sum of the two mole amounts:\r\n

    total moles = 1.00 mol +\u00a00.025 mol = 1.25 mol<\/p>\r\nThe mole fractions are simply the ratio of each mole amount to the total number of moles, 1.25 mol:\r\n\r\n\\[\\begin{array}{rrcll}\r\n\\chi_{\\text{He}}&=&\\dfrac{1.00\\text{ \\cancel{mol}}}{1.25\\text{ \\cancel{mol}}}&=&0.800 \\\\ \\\\\r\n\\chi_{\\text{Ne}}&=&\\dfrac{0.25\\text{ \\cancel{mol}}}{1.25\\text{ \\cancel{mol}}}&=&0.200\r\n\\end{array}\\]\r\n\r\nThe sum of the mole fractions equals exactly 1.\r\n\r\nFor gases, there is another way to determine the mole fraction. When gases have the same volume and temperature (as they would in a mixture of gases), the number of moles is proportional to partial pressure, so the mole fractions for a gas mixture can be determined by taking the ratio of partial pressure to total pressure:\r\n\r\n\\[\\chi_{\\text{i}}=\\dfrac{P_{\\text{i}}}{P_{\\text{tot}}}\\]\r\n\r\nThis expression allows us to determine mole fractions without calculating the moles of each component directly.\r\n

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    Example 6.18<\/p>\r\n\r\n<\/header>\r\n

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    Problem<\/h1>\r\nA container has a mixture of He at 0.80 atm and Ne at 0.60 atm. What is the mole fraction of each component?\r\n

    Solution<\/h2>\r\nAccording to Dalton\u2019s law, the total pressure is the sum of the partial pressures:\r\n

    P<\/i>tot<\/sub> = 0.80 atm + 0.60 atm = 1.40 atm<\/p>\r\nThe mole fractions are the ratios of the partial pressure of each component to the total pressure:\r\n\r\n\\[\\begin{array}{rrrrl}\r\n\\chi_{\\text{He}}&=&\\dfrac{0.80\\text{ atm}}{1.40\\text{ atm}}&=&0.57 \\\\ \\\\\r\n\\chi_{\\text{Ne}}&=&\\dfrac{0.60\\text{ atm}}{1.40\\text{ atm}}&=&0.43\r\n\\end{array}\\]\r\n\r\nAgain, the sum of the mole fractions is exactly 1.\r\n

    Test Yourself<\/h1>\r\nWhat are the mole fractions when 0.65 atm of O2<\/sub> and 1.30 atm of N2<\/sub> are mixed in a container?\r\n

    Answer<\/h2>\r\n\\(\\chi_{\\ce{O2}}=0.33; \\chi_{\\ce{N2}}=0.67\\)\r\n\r\n<\/div>\r\n<\/div>\r\n
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    Food and Drink App: Carbonated Beverages<\/h1>\r\nCarbonated beverages \u2014 sodas, beer, sparkling wines \u2014 have one thing in common: they have CO2<\/sub> gas dissolved in them in such sufficient quantities that it affects the drinking experience. Most people find the drinking experience pleasant \u2014 indeed, in the United States alone, over 1.5 \u00d7 109<\/sup> gal of soda are consumed each year, which is almost 50 gal per person! This figure does not include other types of carbonated beverages, so the total consumption is probably significantly higher.\r\n\r\nAll carbonated beverages are made in one of two ways. First, the flat beverage is subjected to a high pressure of CO2<\/sub> gas, which forces the gas into solution. The carbonated beverage is then packaged in a tightly sealed package (usually a bottle or a can) and sold. When the container is opened, the CO2<\/sub> pressure is released, resulting in the well-known hiss<\/i>, and CO2<\/sub> bubbles come out of solution (Figure 6.5 \"Opening a Carbonated Beverage\"). This must be done with care: if the CO2<\/sub> comes out too violently, a mess can occur!\r\n\r\n[caption id=\"attachment_293\" align=\"aligncenter\" width=\"398\"]\"A<\/a> Figure 6.5 \"Opening a Carbonated Beverage.\" If you are not careful opening a container of a carbonated beverage, you can make a mess as the CO2 comes out of solution suddenly.[\/caption]\r\n\r\nThe second way a beverage can become carbonated is by the ingestion of sugar by yeast, which then generates CO2<\/sub> as a digestion product. This process is called fermentation<\/i>. The overall reaction is:\r\n

    C6<\/sub>H12<\/sub>O6<\/sub>(aq) \u2192\u00a02C2<\/sub>H5<\/sub>OH(aq) +\u00a02CO2<\/sub>(aq)<\/p>\r\nWhen this process occurs in a closed container, the CO2<\/sub> produced dissolves in the liquid, only to be released from solution when the container is opened. Most fine sparkling wines and champagnes are turned into carbonated beverages this way. Less-expensive sparkling wines are made like sodas and beer, with exposure to high pressures of CO2<\/sub> gas.\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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