Download presentation

Published byNoe Brockus Modified over 3 years ago

1
**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.

2
Properties of Gases 11.1 Gases differ from solids and liquids in the following ways: A sample of gas assumes both the shape and volume of the container. Gases are compressible. The densities of gases are much smaller than those of liquids and solids and are highly variable depending on temperature and pressure. Gases form homogeneous mixtures (solutions) with one another in any proportion.

3
Gas Pressure

4
**at constant temperature**

The Gas Laws 11.4 (a) (b) (c) P (mmHg) 760 1520 2280 V (mL) 100 50 33 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. P1V1 = P2V2 at constant temperature

5
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

6
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 = = = 3.03 L P1 × V1 P2 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.

7
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

8
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

9
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 = K, V1 = 14.6 L, and T2 = K. V2 = = = 15.8 L V1 × T2 T1 14.6 L × K K Think About It When temperature increases at constant pressure, the volume of a gas sample increases.

10
**at constant temperature and pressure**

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

11
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.

12
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.

13
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.

14
**The Gas Laws Cooling at constant volume: pressure decreases**

Heating at constant volume: pressure increases

15
**The Gas Laws Cooling at constant pressure: volume decreases**

Heating at constant pressure: volume increases

16
The Gas Laws The presence of additional molecules causes an increase in pressure.

17
The Gas Laws The combined gas law can be used to solve problems where any or all of the variables changes.

18
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 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.

19
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 mmHg, what will the volume of the balloon be when it has risen to an altitude where the temperature is °C and the air pressure is 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 = K, T2 = K. V2 = = = 10.2 L P1T2V1 P2T1 757.2 mmHg × K × 6.25 L 366.4 mmHg × 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.

20
The Ideal Gas Equation 11.5 The gas laws can be combined into a general equation that describes the physical behavior of all gases. Boyle’s law Charles’s law Avogadro’s law PV = nRT rearrangement R is the proportionality constant, called the gas constant.

21
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.

22
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

23
The Ideal Gas Equation Standard Temperature and Pressure (STP) are a special set of conditions where: Pressure is 1 atm Temperature is 0°C ( K) The volume occupied by one mole of an ideal gas is then L: PV = nRT

24
**(1 mol)(0.08206 L∙atm/K∙mol)(298.15 K)**

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 = K, and P = 1.00 atm. Because the pressure is expressed in atmospheres, we use R = L∙atm/K∙mol in order to solve for volume in liters. V = = = 24.5 L nRT P (1 mol)( L∙atm/K∙mol)( K) 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.

25
**Ptotal = PN2 + PO2 = 4.48 atm + 4.48 atm = 8.96 atm**

Gas Mixtures 11.7 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. Addition of 1.00 mole of O2 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 (Pi): Ptotal = PN2 + PO2 = 4.48 atm atm = 8.96 atm

26
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:

27
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: mol He, mol H2, and mol Ne. Solution: Step 1: Since each gas behaves independently, calculate the partial pressure of each using the ideal gas equation:

28
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: mol He, mol H2, and mol Ne. Solution: Step 2: Use the equation below to calculate total pressure. Ptotal = atm atm atm = 2.17 atm

29
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.

30
Worked Example 11.12 A 1.00-L vessel contains mole of N2 gas and mole of H2 gas at 25.5°C. Determine the partial pressure of each component and the total pressure in the vessel. Think About It The total pressure in the vessel can also be determined by summing the number of moles of mixture components ( = mol) and solving the ideal gas equation for Ptotal: (0.227 mol)( L∙atm/K∙mol)( K) 1.00 L = 5.56 atm Ptotal = 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. Solution T = K Ptotal = PN2 + PH2 = 5.27 atm atm = 5.56 atm (0.215 mol)( L∙atm/K∙mol)( K) 1.00 L PN2 = = 5.27 atm ( mol)( L∙atm/K∙mol)( K) 1.00 L PH2 = = atm

31
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: The mole fraction of a mixture component is always less than 1. The sum of mole fractions for all components of a mixture is always 1. Mole fractions are dimensionless. Χi is the mole fraction. ni is the moles of a certain component ntotal is the total number of moles.

32
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 L gas cylinder at room temperature (25°C) that contains mol N2 and in which the total pressure is atm. Think About It To check your work, determine χN2 by subtracting χNO from 1. Using each mole fraction and the total pressure, calculate the partial pressure of each component using χi = Pi/Ptotal and verify that they sum to the total pressure. Strategy Use the ideal gas equation to calculate the total number of moles in the cylinder. Subtract moles of N2 from the total to determine moles of NO. Divide moles NO by total moles to get mole fraction. Solution The temperature is K. total moles = = mol NO = total moles – N2 = – = mol NO χNO = = = 0.001 PV RT (14.75 atm)(10.00 L) ( L∙atm/K∙mol)( K) = mol nNO ntotal 0.007 mol NO 6.029 mol

33
**Reactions with Gaseous Reactants and Products**

11.8 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( Latm/mol K)( K) n = moles CO2

34
**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.

35
**2Na2O2(s) + 2CO2(g) → 2Na2CO3(s) + O2(g)**

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.

36
**(12.82 mol CO2)(0.08206 L∙atm/K∙mol)(273.15 K)**

Worked Example (cont.) Solution The molar mass of Na2O2 is g/mol (1 kg = 1000 g). (Treat the specified mass of NaO2 as an exact number.) 1000 g Na2O2 × 12.82 mol Na2O2 × VCO2 = 1 mol Na2O2 77.98 g Na2O2 = mol Na2O2 2 mol CO2 2 mol Na2O2 = mol Na2O2 (12.82 mol CO2)( L∙atm/K∙mol)( K) 1 atm = L CO2 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.

37
**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 The change in pressure in a reaction vessel can be used to determine how many moles of gaseous reactant are consumed:

38
**2LiOH(aq) + CO2(g) → Li2CO3(s) + H2O(l)**

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 atm, and the temperature 25°C. If the pressure in the submarine drops to atm as the result of carbon dioxide being consumed by an aqueous lithium hydroxide scrubber, how many moles of CO2 are consumed. Think About It Careful cancellation of units is essential. Note that this amount of CO2 corresponds to 162 moles or 3.9 kg of LiOH. (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 = atm – atm = 7.9×10-3 atm, V = 2.5×105 L, and T = K. ΔnCO2 = 7.9×10-3 atm × 2.5×105 L ( L∙atm/K∙mol) × ( K) = 81 moles CO2 consumed

39
**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

40
**Reactions with Gaseous Reactants and Products**

The vapor pressure of water is known at various temperatures.

41
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 atm. Solution: Step 1: Use Table 11.5 to determine the vapor pressure of water at 30.0°C.

42
**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 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. Ptotal = PO2 + PH2O PO2 = Ptotal – PH2O PO2 = atm – atm = atm

43
**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 atm. Solution: Step 3: Convert to moles of O2 using the ideal gas equation and then find mass. PV = nRT

44
**Ca(s) + H2O(l) → Ca(OH)2(aq) + H2(g)**

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 atm when 525 mL of the gas is collected over water as shown in Figure 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. Strategy Use Dalton’s law of partial pressures to determine the partial pressure of H2, use the ideal gas equation to determine moles of H2, and then use the molar mass of H2 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.) Solution PH2 = Ptotal – PH2O = atm – atm = atm moles of H2 = moles of H2 = (2.008×10-2 mol)(2.016 g/mol) = g H2 ( atm)(0.525 L) ( L∙atm/K∙mol)( K) = 2.01×10-2 mol

45
**Experimentally measured pressure**

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

Similar presentations

OK

Mass and gaseous volume relationships in chemical reactions

Mass and gaseous volume relationships in chemical reactions

© 2018 SlidePlayer.com Inc.

All rights reserved.

To make this website work, we log user data and share it with processors. To use this website, you must agree to our Privacy Policy, including cookie policy.

Ads by Google