Chemistry: Atoms First Julia Burdge & Jason Overby Chapter 11 Gases Kent L. McCorkle Cosumnes River College Sacramento, CA Copyright (c) The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 11.1 Properties of Gases Gases differ from solids and liquids in the following ways: 1) A sample of gas assumes both the shape and volume of the container. 2) Gases are compressible. 3) The densities of gases are much smaller than those of liquids and solids and are highly variable depending on temperature and pressure. 4) Gases form homogeneous mixtures (solutions) with one another in any proportion. Gas Pressure 11.4 The Gas Laws (a) Boyle’s law states that the pressure of a fixed amount of gas at a constant temperature is inversely proportional to the volume of the gas. 1 V P P1V1 = P2V2 at constant temperature (b) (c) P (mmHg) 760 1520 2280 V (mL) 100 50 33 The Gas Laws Calculate the volume of a sample of gas at 5.75 atm if it occupies 5.14 L at 2.49 atm. (Assume constant temperature.) Solution: Step 1: Use the relationship below to solve for V2: P1V1 = P2V2 P1 V1 2.49 atm 5.14 L V2 2.26 L P2 5.75 atm Worked Example 11.3 If a skin diver takes a breath at the surface, filling his lungs with 5.82 L of air, what volume will the air in his lungs occupy when he dives to a depth where the pressure of 1.92 atm? (Assume constant temperature and that the pressure at the surface is exactly 1 atm.) Strategy Use P1V1 = P2V2 to solve for V2. Solution P1 = 1.00 atm, V1 = 5.82 L, and P2 = 1.92 atm. V2 = P1 × V1 1.00 atm × 5.82 L = = 3.03 L P2 1.92 atm Think About It At higher pressure, the volume should be smaller. Therefore, the answer makes sense. The Gas Laws Charles’s and Gay-Lussac’s law, (or simply Charles’s Law) states that the volume of a gas maintained at constant pressure is directly proportional to the absolute temperature of the gas. Higher temperature Lower temperature The Gas Laws Charles’s and Gay-Lussac’s law, (or simply Charles’s Law) states that the volume of a gas maintained at constant pressure is directly proportional to the absolute temperature of the gas. V T V1 V2 T1 T2 at constant pressure Worked Example 11.4 A sample of argon gas that originally occupied 14.6 L at 25°C was heated to 50.0°C at constant pressure. What is its new volume? Strategy Use V1/T1 = V2/T2 to solve for V2. Remember that temperatures must be expressed in kelvin. Solution T1 = 298.15 K, V1 = 14.6 L, and T2 = 323.15 K. V2 = V1 × T2 14.6 L × 323.15 K = = 15.8 L T1 298.15 K Think About It When temperature increases at constant pressure, the volume of a gas sample increases. The Gas Laws Avogadro’s law states that the volume of a sample of gas is directly proportional to the number of moles in the sample at constant temperature and pressure. V n V1 V2 n1 n2 at constant temperature and pressure Worked Example 11.5 If we combine 3.0 L of NO and 1.5 L of O2, and they react according to the balanced equation 2NO(g) + O2(g) → 2NO2(g), what volume of NO2 will be produced? (Assume that the reactants and products are all at the same temperature and pressure.) Strategy Apply Avogadro’s law to determine the volume of a gaseous product. Solution Because volume is proportional to the number of moles, the balanced equation determines in what volume ratio the reactants combine and the ratio of product volume to reactant volume. The amounts of reactants given are stoichiometric amounts. According to the balanced equation, the volume of NO2 formed will be equal to the volume of NO that reacts. Therefore, 3.0 L of NO2 will form. Think About It When temperature increases at constant pressure, the volume of a gas sample increases. The Gas Laws A sample of gas originally occupies 29.1 L at 0.0°C. What is its new volume when it is heated to 15.0°C? (Assume constant pressure.) Solution: Step 1:Use the relationship below to solve for V2: (Remember that temperatures must be expressed in kelvin. V1 V2 T1 T2 V1 T2 29.1 L 288.15 K V2 30.7 L T1 273.15 K The Gas Laws What volume in liters of water vapor will be produced when 34 L of H2 and 17 L of O2 react according to the equation below: 2H2(g) + O2(g) → 2H2O(g) Assume constant pressure and temperature. Solution: Step 1:Because volume is proportional to the number of moles, the balanced equation determines in what volume ratio the reactants combine and the ratio of product volume to reactant volume. The amounts of reactants given are stoichiometeric amounts. 34 L of H2O will form. The Gas Laws Cooling at constant volume: pressure decreases Heating at constant volume: pressure increases The Gas Laws Cooling at constant pressure: volume decreases Heating at constant pressure: volume increases The Gas Laws The presence of additional molecules causes an increase in pressure. The Gas Laws The combined gas law can be used to solve problems where any or all of the variables changes. P1V1 P2V2 n1T1 n2T2 The Gas Laws The volume of a bubble starting at the bottom of a lake at 4.55°C increases by a factor of 10 as it rises to the surface where the temperature is 18.45°C and the air pressure is 0.965 atm. Assume the density of the lake water is 1.00 g/mL. Determine the depth of the lake. Solution: Step 1:Use the combined gas law to find the pressure at the bottom of the lake; assume constant moles of gas. P1V1 P2V2 n1T1 n2T2 P2V2T1 0.965 atm10 V1 277.7 K P1 9.19 atm T2V1 291.6 K V1 Worked Example 11.6 If a child releases a 6.25-L helium balloon in the parking lot of an amusement park where the temperature is 28.50°C and the air pressure is 757.2 mmHg, what will the volume of the balloon be when it has risen to an altitude where the temperature is -34.35°C and the air pressure is 366.4 mmHg? Strategy In this case, because there is a fixed amount of gas, we use P1V1/T1 = P2V2/T2. The only value we don’t know is V2. Temperatures must be expressed in kelvins. We can use any units of pressure, as long as we are consistent. Solution T1 = 301.65 K, T2 = 238.80 K. V2 = P1T2V1 757.2 mmHg × 238.80 K × 6.25 L = = 10.2 L P2T1 366.4 mmHg × 301.65 K Think About It Note that the solution is essentially multiplying the original volume by the ratio of P1 and P2, and by the ratio of T2 to T1. The effect of decreasing external pressure is to increase the balloon volume. The effect of decreasing temperature is to decrease the volume. In this case, the effect of decreasing pressure predominates and the balloon volume increases significantly. 11.5 The Ideal Gas Equation The gas laws can be combined into a general equation that describes the physical behavior of all gases. 1 V P V T V n Boyle’s law Charles’s law Avogadro’s law nT V P nT V R P rearrangement PV = nRT R is the proportionality constant, called the gas constant. The Ideal Gas Equation The ideal gas equation (below) describes the relationship among the four variables P, V, n, and T. PV = nRT An ideal gas is a hypothetical sample of gas whose pressurevolume-temperature behavior is predicted accurately by the ideal gas equation. The Ideal Gas Equation The gas constant (R) is the proportionality constant and its value and units depend on the units in which P and V are expressed. PV = nRT The Ideal Gas Equation Standard Temperature and Pressure (STP) are a special set of conditions where: Pressure is 1 atm Temperature is 0°C (273.15 K) The volume occupied by one mole of an ideal gas is then 22.41 L: PV = nRT (1 mol)(0.08206 L atm/K mol)(273.15 K) V = 22.41 L 1 atm Worked Example 11.7 Calculate the volume of a mole of ideal gas at room temperature (25°C) and 1 atm. Strategy Convert the temperature in °C to kelvins, and use the ideal gas equation to solve for the unknown volume. Solution The data given are n = 1 mol, T = 298.15 K, and P = 1.00 atm. Because the pressure is expressed in atmospheres, we use R = 0.08206 L∙atm/K∙mol in order to solve for volume in liters. V= nRT (1 mol)(0.08206 L∙atm/K∙mol)(298.15 K) = = 24.5 L P 1 atm Think About It With the pressure held constant, we should expect the volume to increase with increased temperature. Room temperature is higher than the standard temperature for gases (0°C), so the molar volume at room temperature (25°C) should be higher than the molar volume at 0°C–and it is. 11.7 Gas Mixtures When two or more gases are placed in a container, each gas behaves as though it occupies the container alone. 1.00 mole of N2 in a 5.00 L container at 0°C exerts a pressure of 4.48 atm. PN2 (1mol)(0.08206 L atm/K mol)(273.15 K) 4.48 atm 5.00 L Addition of 1.00 mole of O2 in the same container exerts an additional 4.48 atm of pressure. PO2 (1mol)(0.08206 L atm/K mol)(273.15 K) 4.48 atm 5.00 L The total pressure of the mixture is the sum of the partial pressures (Pi): Ptotal = PN2 + PO2 = 4.48 atm + 4.48 atm = 8.96 atm Gas Mixtures Dalton’s law of partial pressure states that the total pressure exerted by a gas mixture is the sum of the partial pressures exerted by each component of the mixture: Ptotal = Pi Gas Mixtures Determine the partial pressures and the total pressure in a 2.50-L vessel containing the following mixture of gases at 15.8°C: 0.0194 mol He, 0.0411 mol H2, and 0.169 mol Ne. Solution: Step 1:Since each gas behaves independently, calculate the partial pressure of each using the ideal gas equation: PHe 0.0194 mol He 0.08206 Latm/mol K 288.95 K 0.184 atm 2.50 L PH2 PNe 0.0411 mol H2 0.08206 Latm/mol K 288.95 K 0.390 atm 2.50 L 0.169 mol Ne 0.08206 Latm/mol K 288.95 K 1.60 atm 2.50 L Gas Mixtures Determine the partial pressures and the total pressure in a 2.50-L vessel containing the following mixture of gases at 15.8°C: 0.0194 mol He, 0.0411 mol H2, and 0.169 mol Ne. Solution: Step 2:Use the equation below to calculate total pressure. Ptotal = Pi Ptotal = 0.184 atm + 0.390 atm + 1.60 atm = 2.17 atm Gas Mixtures Each component of a gas mixture exerts a pressure independent of the other components. The total pressure is the sum of the partial pressures. Worked Example 11.12 A 1.00-L vessel contains 0.215 mole of N2 gas and 0.0118 mole of H2 gas at 25.5°C. Determine the partial pressure of each component and the total pressure in the vessel. Strategy Use the ideal gas equation to find the partial pressure of each component ofAbout the mixture, and sum the twoinpartial pressures to find Think It The total pressure the vessel can also be the total pressure. determined by summing the number of moles of mixture components (0.215 + 0.0118 = 0.227 mol) and solving the ideal gas Solution T = 298.65 equation for PtotalK: (0.215 mol)(0.08206 L∙atm/K∙mol)(298.65 K) PPN2 = = (0.227 mol)(0.08206 L∙atm/K∙mol)(298.65 K)= 5.27 atm = 5.56 atm 1.00 L total 1.00 L (0.0118 mol)(0.08206 L∙atm/K∙mol)(298.65 K) PH2 = = 0.289 atm 1.00 L Ptotal = PN2 + PH2 = 5.27 atm + 0.289 atm = 5.56 atm Gas Mixtures The relative amounts of the components in a gas mixture can be specified using mole fractions. ni i = ntotal Χi is the mole fraction. ni is the moles of a certain component ntotal is the total number of moles. There are three things to remember about mole fractions: 1) The mole fraction of a mixture component is always less than 1. 2) The sum of mole fractions for all components of a mixture is always 1. 3) Mole fractions are dimensionless. Worked Example 11.13 In 1999, the FDA approved the use of nitric oxide (NO) to treat and prevent lung disease, which occurs commonly in premature infants. The nitric oxide used in this therapy is supplied to hospitals in the form of a N2/NO mixture. Calculate the mole fraction of NO in a 10.00-L gas cylinder at room temperature (25°C) that contains 6.022 mol N2 and in which the total pressure is 14.75 atm. Strategy UseAbout the ideal gascheck equation calculate the total of moles in the Think It To yourtowork, determine χNnumber by subtracting 2 cylinder. N2 mole from the total and to determine moles of NO. Divide χNOSubtract from 1.moles Usingof each fraction the total pressure, moles NO by total get moleoffraction. calculate themoles partialtopressure each component using χi = Pi/Ptotal and verify that they sum to the total pressure. Solution The temperature is 298.15 K. total moles = PV (14.75 atm)(10.00 L) = = 6.029 mol RT (0.08206 L∙atm/K∙mol)(298.15 K) mol NO = total moles – N2 = 6.029 – 6.022 = 0.007 mol NO χNO = nNO 0.007 mol NO = = 0.001 ntotal 6.029 mol 11.8 Reactions with Gaseous Reactants and Products What mass (in grams) of Na2O2 is necessary to consume 1.00 L of CO2 at STP? 2Na2O2(s) + 2CO2(g) → 2Na2CO3(s) + O2(g) Solution: Step 1:Convert 1.00 L of CO2 at STP to moles using the ideal gas equation. PV = nRT (1 atm)(1.00 L) = n(0.08206 Latm/mol K)(273.15 K) n = 0.04461 moles CO2 Reactions with Gaseous Reactants and Products What mass (in grams) of Na2O2 is necessary to consume 1.00 L of CO2 at STP? 2Na2O2(s) + 2CO2(g) → 2Na2CO3(s) + O2(g) Solution: Step 2:Determine the stoichiometric amount of Na2O2. 2 mol Na 2O2 77.98 g 0.04461 moles CO 2 × × = 3.48 g 2 mol CO2 mol Worked Example 11.14 Sodium peroxide (Na2O2) is used to remove carbon dioxide from (and add oxygen to) the air supply in spacecrafts. It works by reacting with CO2 in the air to produce sodium carbonate (Na2CO3) and O2. 2Na2O2(s) + 2CO2(g) → 2Na2CO3(s) + O2(g) What volume (in liters) of CO2 (at STP) will react with a kilogram of Na2O2? Strategy Convert the given mass of Na2O2 to moles, use the balanced equation to determine the stoichiometric amount of CO2, and then use the ideal gas equation to convert moles of CO2 to liters. Worked Example 11.14 (cont.) Solution The molar mass of Na2O2 is 77.98 g/mol (1 kg = 1000 g). (Treat the specified mass of NaO2 as an exact number.) 1000 g Na2O2 × 1 mol Na2O2 = 12.82 mol Na2O2 77.98 g Na2O2 12.82 mol Na2O2 × VCO2 = 2 mol CO2 = 12.82 mol Na2O2 2 mol Na2O2 (12.82 mol CO2)(0.08206 L∙atm/K∙mol)(273.15 K) = 287.4 L CO2 1 atm Think About It The answer seems like an enormous volume of CO2. If you check the cancellation of units carefully in ideal gas equation problems, however, with practice you will develop a sense of whether such a calculated volume is reasonable. Reactions with Gaseous Reactants and Products Although there are no empirical gas laws that focus on the relationship between n and P, it is possible to rearrange the ideal gas equation to find the relationship. PV = nRT rearrangement V nP RT The change in pressure in a reaction vessel can be used to determine how many moles of gaseous reactant are consumed: V n P RT Worked Example 11.15 Another air-purification method for enclosed spaces involves the use of “scrubbers” containing aqueous lithium hydroxide, which react with carbon dioxide to produce lithium carbonate and water: 2LiOH(aq) + CO2(g) → Li2CO3(s) + H2O(l) Consider the air supply in a submarine with a total volume of 2.5×105 L. The pressure of 0.9970 atm, and the temperature 25°C. If the pressure in the Think About It Careful units isdioxide essential. Note that submarine drops to 0.9891 atm ascancellation the result ofofcarbon being consumed by this amount of CO2 corresponds to 162 moles or 3.9 kg of LiOH. an aqueous lithium hydroxide scrubber, how many moles of CO2 are consumed. (It’s a good idea to verify this yourself. Strategy Use Δn = ΔP×(V/RT) to determine Δn, the number of moles CO2 consumed. Solution ΔP = 0.9970 atm – 0.9891 atm = 7.9×10-3 atm, V = 2.5×105 L, and T = 298.15 K. 2.5×105 L -3 ΔnCO2 = 7.9×10 atm × = 81 moles CO2 (0.08206 L∙atm/K∙mol) × (298.15 K) consumed Reactions with Gaseous Reactants and Products The volume of gas produced by a chemical reaction can be measured using an apparatus like the one shown below. When gas is collected over water in this manner, the total pressure is the sum of two partial pressures: Ptotal = Pcollected gas + PH2O Reactions with Gaseous Reactants and Products The vapor pressure of water is known at various temperatures. Gas Mixtures Calculate the mass of O2 produced by the decomposition of KClO3 when 821 mL of O2 is collected over water at 30.0°C and 1.015 atm. Solution: Step 1:Use Table 11.5 to determine the vapor pressure of water at 30.0°C. Reactions with Gaseous Reactants and Products Calculate the mass of O2 produced by the decomposition of KClO3 when 821 mL of O2 is collected over water at 30.0°C and 1.015 atm. Solution: Step 2:Convert the vapor pressure of water at 30.0°C to atm and then use Dalton’s law to calculate the partial pressure of O2. PH O @ 30.0 C = 31.8 torr × 2 1 atm = 0.041842 atm 760 torr Ptotal = PO2 + PH2O PO2 = Ptotal – PH2O PO2 = 1.015 atm – 0.041842 atm = 0.973158 atm Reactions with Gaseous Reactants and Products Calculate the mass of O2 produced by the decomposition of KClO3 when 821 mL of O2 is collected over water at 30.0°C and 1.015 atm. Solution: Step 3:Convert to moles of O2 using the ideal gas equation and then find mass. PV = nRT n 0.973158 atm0.821 L PV 0.032117 moles O2 RT 0.08206 Latm/mol K 303.15 K 32.00 g 0.032117 moles O2 1.03 g O2 mole Worked Example 11.16 Calcium metal reacts with water to produce hydrogen gas: Ca(s) + H2O(l) → Ca(OH)2(aq) + H2(g) Determine the mass of H2 produced at 25°Cand 0.967 atm when 525 mL of the gas is collected over water as shown in Figure 11.23. Strategy Use Dalton’s law of partial pressures to determine the partial pressure Think Check unit cancellation carefully, of H2, use the About ideal gasIt equation to determine moles of H2, and remember then use the molar densities gases are careful relatively low. The mass of mass ofthat H2 the to convert toof mass. (Pay attention to units. Atmospheric approximately half a liter ofwhereas hydrogen or near room of temperature pressure is given in atmospheres, theatvapor pressure water is tabulated in torr.)and 1 atm should be a very small number. Solution PH2 = Ptotal – PH2O = 0.967 atm – 0.0313 atm = 0.936 atm moles of H2 = (0.9357 atm)(0.525 L) = 2.01×10-2 mol (0.08206 L∙atm/K∙mol)(298.15 K) moles of H2 = (2.008×10-2 mol)(2.016 g/mol) = 0.0405 g H2 11.6 Real Gases The van der Waals equation is useful for gases that do not behave ideally. Experimentally measured pressure Container volume an2 P 2 V nb nRT V corrected corrected pressure term volume term