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Chemistry: Atoms First Julia Burdge & Jason Overby Copyright (c) The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Chapter 11 Gases Kent L. McCorkle Cosumnes River College Sacramento, CA

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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. 11.1

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Gas Pressure

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The Gas Laws 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. 11.4 (a)(b)(c) P (mmHg)76015202280 V (mL)1005033 P 1 V 1 = P 2 V 2 at constant temperature

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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 V 2 : P 1 V 1 = P 2 V 2

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Worked Example 11.3 Strategy Use P 1 V 1 = P 2 V 2 to solve for V 2. 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.) Solution P 1 = 1.00 atm, V 1 = 5.82 L, and P 2 = 1.92 atm. V 2 = = = 3.03 L P 1 × V 1 P 2 1.00 atm × 5.82 L 1.92 atm Think About It At higher pressure, the volume should be smaller. Therefore, the answer makes sense.

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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 temperatureLower temperature

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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. at constant pressure

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Worked Example 11.4 Strategy Use V 1 /T 1 = V 2 /T 2 to solve for V 2. Remember that temperatures must be expressed in kelvin. 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? Solution T 1 = 298.15 K, V 1 = 14.6 L, and T 2 = 323.15 K. V 2 = = = 15.8 L V 1 × T 2 T 1 14.6 L × 323.15 K 298.15 K Think About It When temperature increases at constant pressure, the volume of a gas sample increases.

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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. at constant temperature and pressure

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Worked Example 11.5 Strategy Apply Avogadro’s law to determine the volume of a gaseous product. If we combine 3.0 L of NO and 1.5 L of O 2, and they react according to the balanced equation 2NO(g) + O 2 (g) → 2NO 2 (g), what volume of NO 2 will be produced? (Assume that the reactants and products are all at the same temperature and pressure.) 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 NO 2 formed will be equal to the volume of NO that reacts. Therefore, 3.0 L of NO 2 will form. Think About It When temperature increases at constant pressure, the volume of a gas sample increases.

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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 V 2 : (Remember that temperatures must be expressed in kelvin.

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The Gas Laws What volume in liters of water vapor will be produced when 34 L of H 2 and 17 L of O 2 react according to the equation below: 2H 2 (g) + O 2 (g) → 2H 2 O(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 H 2 O will form.

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The Gas Laws Heating at constant volume: pressure increases Cooling at constant volume: pressure decreases

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The Gas Laws Heating at constant pressure: volume increases Cooling at constant pressure: volume decreases

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The Gas Laws The presence of additional molecules causes an increase in pressure.

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The Gas Laws The combined gas law can be used to solve problems where any or all of the variables changes.

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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.

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Worked Example 11.6 Strategy In this case, because there is a fixed amount of gas, we use P 1 V 1 /T 1 = P 2 V 2 /T 2. The only value we don’t know is V 2. Temperatures must be expressed in kelvins. We can use any units of pressure, as long as we are consistent. 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? Think About It Note that the solution is essentially multiplying the original volume by the ratio of P 1 and P 2, and by the ratio of T 2 to T 1. 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. Solution T 1 = 301.65 K, T 2 = 238.80 K. V 2 = = = 10.2 L P1T2V1P2T1P1T2V1P2T1 757.2 mmHg × 238.80 K × 6.25 L 366.4 mmHg × 301.65 K

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The Ideal Gas Equation The gas laws can be combined into a general equation that describes the physical behavior of all gases. 11.5 Boyle’s law Avogadro’s lawCharles’s law PV = nRT rearrangement R is the proportionality constant, called the gas constant.

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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 pressure- volume-temperature behavior is predicted accurately by the ideal gas equation.

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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

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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

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Worked Example 11.7 Strategy Convert the temperature in °C to kelvins, and use the ideal gas equation to solve for the unknown volume. Calculate the volume of a mole of ideal gas at room temperature (25°C) and 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. 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 = = = 24.5 L nRT P (1 mol)(0.08206 L∙atm/K∙mol)(298.15 K) 1 atm

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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 N 2 in a 5.00 L container at 0°C exerts a pressure of 4.48 atm. Addition of 1.00 mole of O 2 in the same container exerts an additional 4.48 atm of pressure. The total pressure of the mixture is the sum of the partial pressures (P i ): P total = P N 2 + P O 2 = 4.48 atm + 4.48 atm = 8.96 atm 11.7

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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:

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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 H 2, and 0.169 mol Ne. Solution: Step 1:Since each gas behaves independently, calculate the partial pressure of each using the ideal gas equation:

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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 H 2, and 0.169 mol Ne. Solution: Step 2:Use the equation below to calculate total pressure. P total = 0.184 atm + 0.390 atm + 1.60 atm = 2.17 atm

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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.

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Worked Example 11.12 Strategy Use the ideal gas equation to find the partial pressure of each component of the mixture, and sum the two partial pressures to find the total pressure. A 1.00-L vessel contains 0.215 mole of N 2 gas and 0.0118 mole of H 2 gas at 25.5°C. Determine the partial pressure of each component and the total pressure in the vessel. Solution T = 298.65 K P total = P N 2 + P H 2 = 5.27 atm + 0.289 atm = 5.56 atm (0.215 mol)(0.08206 L∙atm/K∙mol)(298.65 K) 1.00 L = 5.27 atm P N 2 = (0.0118 mol)(0.08206 L∙atm/K∙mol)(298.65 K) 1.00 L = 0.289 atm P H 2 = Think About It The total pressure in the vessel can also be determined by summing the number of moles of mixture components (0.215 + 0.0118 = 0.227 mol) and solving the ideal gas equation for P total : (0.227 mol)(0.08206 L∙atm/K∙mol)(298.65 K) 1.00 L = 5.56 atm P total =

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Gas Mixtures The relative amounts of the components in a gas mixture can be specified using mole fractions. 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. Χ i is the mole fraction. n i is the moles of a certain component n total is the total number of moles.

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Worked Example 11.13 Strategy Use the ideal gas equation to calculate the total number of moles in the cylinder. Subtract moles of N 2 from the total to determine moles of NO. Divide moles NO by total moles to get mole fraction. 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 N 2 /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 N 2 and in which the total pressure is 14.75 atm. Solution The temperature is 298.15 K. total moles = = mol NO = total moles – N 2 = 6.029 – 6.022 = 0.007 mol NO χ NO = = = 0.001 (14.75 atm)(10.00 L) (0.08206 L∙atm/K∙mol)(298.15 K) = 6.029 mol PV RT n NO n total 0.007 mol NO 6.029 mol Think About It To check your work, determine χ N 2 by subtracting χ NO from 1. Using each mole fraction and the total pressure, calculate the partial pressure of each component using χ i = P i /P total and verify that they sum to the total pressure.

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Reactions with Gaseous Reactants and Products What mass (in grams) of Na 2 O 2 is necessary to consume 1.00 L of CO 2 at STP? 2Na 2 O 2 (s) + 2CO 2 (g) → 2Na 2 CO 3 (s) + O 2 (g) Solution: Step 1:Convert 1.00 L of CO 2 at STP to moles using the ideal gas equation. 11.8 PV = nRT (1 atm)(1.00 L) = n(0.08206 Latm/mol K)(273.15 K) n = 0.04461 moles CO 2

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Reactions with Gaseous Reactants and Products What mass (in grams) of Na 2 O 2 is necessary to consume 1.00 L of CO 2 at STP? 2Na 2 O 2 (s) + 2CO 2 (g) → 2Na 2 CO 3 (s) + O 2 (g) Solution: Step 2:Determine the stoichiometric amount of Na 2 O 2.

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Worked Example 11.14 Strategy Convert the given mass of Na 2 O 2 to moles, use the balanced equation to determine the stoichiometric amount of CO 2, and then use the ideal gas equation to convert moles of CO 2 to liters. Sodium peroxide (Na 2 O 2 ) is used to remove carbon dioxide from (and add oxygen to) the air supply in spacecrafts. It works by reacting with CO 2 in the air to produce sodium carbonate (Na 2 CO 3 ) and O 2. 2Na 2 O 2 (s) + 2CO 2 (g) → 2Na 2 CO 3 (s) + O 2 (g) What volume (in liters) of CO 2 (at STP) will react with a kilogram of Na 2 O 2 ?

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Worked Example 11.14 (cont.) Solution The molar mass of Na 2 O 2 is 77.98 g/mol (1 kg = 1000 g). (Treat the specified mass of NaO 2 as an exact number.) 1000 g Na 2 O 2 × 12.82 mol Na 2 O 2 × V CO 2 = = 12.82 mol Na 2 O 2 1 mol Na 2 O 2 77.98 g Na 2 O 2 2 mol CO2 2 mol Na 2 O 2 = 12.82 mol Na 2 O 2 (12.82 mol CO 2 )(0.08206 L∙atm/K∙mol)(273.15 K) 1 atm = 287.4 L CO 2 Think About It The answer seems like an enormous volume of CO 2. 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.

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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 The change in pressure in a reaction vessel can be used to determine how many moles of gaseous reactant are consumed: rearrangement

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Worked Example 11.15 Strategy Use Δn = ΔP×(V/RT) to determine Δn, the number of moles CO 2 consumed. 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) + CO 2 (g) → Li 2 CO 3 (s) + H 2 O(l) Consider the air supply in a submarine with a total volume of 2.5×10 5 L. The pressure of 0.9970 atm, and the temperature 25°C. If the pressure in the submarine drops to 0.9891 atm as the result of carbon dioxide being consumed by an aqueous lithium hydroxide scrubber, how many moles of CO 2 are consumed. Solution ΔP = 0.9970 atm – 0.9891 atm = 7.9×10 -3 atm, V = 2.5×10 5 L, and T = 298.15 K. Δn CO 2 = 7.9×10 -3 atm × 2.5×10 5 L (0.08206 L∙atm/K∙mol) × (298.15 K) = 81 moles CO 2 consumed Think About It Careful cancellation of units is essential. Note that this amount of CO 2 corresponds to 162 moles or 3.9 kg of LiOH. (It’s a good idea to verify this yourself.

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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: P total = P collected gas + P H 2 O

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Reactions with Gaseous Reactants and Products The vapor pressure of water is known at various temperatures.

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Gas Mixtures Calculate the mass of O 2 produced by the decomposition of KClO 3 when 821 mL of O 2 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.

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Reactions with Gaseous Reactants and Products Calculate the mass of O 2 produced by the decomposition of KClO 3 when 821 mL of O 2 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 useDalton’s law to calculate the partial pressure of O 2. P total = P O 2 + P H 2 O P O 2 = P total – P H 2 O P O 2 = 1.015 atm – 0.041842 atm = 0.973158 atm

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Reactions with Gaseous Reactants and Products Calculate the mass of O 2 produced by the decomposition of KClO 3 when 821 mL of O 2 is collected over water at 30.0°C and 1.015 atm. Solution: Step 3:Convert to moles of O 2 using the ideal gas equation and then find mass. PV = nRT

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Worked Example 11.16 Strategy Use Dalton’s law of partial pressures to determine the partial pressure of H 2, use the ideal gas equation to determine moles of H 2, and then use the molar mass of H 2 to convert to mass. (Pay careful attention to units. Atmospheric pressure is given in atmospheres, whereas the vapor pressure of water is tabulated in torr.) Calcium metal reacts with water to produce hydrogen gas: Ca(s) + H 2 O(l) → Ca(OH) 2 (aq) + H 2 (g) Determine the mass of H 2 produced at 25°Cand 0.967 atm when 525 mL of the gas is collected over water as shown in Figure 11.23. Solution P H 2 = P total – P H 2 O = 0.967 atm – 0.0313 atm = 0.936 atm moles of H 2 = moles of H 2 = (2.008×10 -2 mol)(2.016 g/mol) = 0.0405 g H 2 (0.9357 atm)(0.525 L) (0.08206 L∙atm/K∙mol)(298.15 K) = 2.01×10 -2 mol Think About It Check unit cancellation carefully, and remember that the densities of gases are relatively low. The mass of approximately half a liter of hydrogen at or near room temperature and 1 atm should be a very small number.

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Real Gases The van der Waals equation is useful for gases that do not behave ideally. 11.6 Experimentally measured pressure corrected pressure term corrected volume term Container volume

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