Partial Pressure

9 Key excerpts on "Partial Pressure"

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  Chemistry 2e

  • Paul Flowers, Klaus Theopold, Richard Langley, William R. Robinson(Authors)
  • 2019(Publication Date)
  • Openstax

    (Publisher)

  Each individual gas in a mixture exerts the same pressure that it would exert if it were present alone in the container ( Figure 9.20). The pressure exerted by each individual gas in a mixture is called its Partial Pressure. This observation is summarized by Dalton’s law of Partial Pressures: The total pressure of a mixture of ideal gases is equal to the sum of the Partial Pressures of the component gases: In the equation P Total is the total pressure of a mixture of gases, P A is the Partial Pressure of gas A; P B is the Partial Pressure of gas B; P C is the Partial Pressure of gas C; and so on. 440 9 • Gases Access for free at openstax.org FIGURE 9.20 If equal-volume cylinders containing gasses at pressures of 300 kPa, 450 kPa, and 600 kPa are all combined in the same-size cylinder, the total pressure of the gas mixture is 1350 kPa. The Partial Pressure of gas A is related to the total pressure of the gas mixture via its mole fraction (X), a unit of concentration defined as the number of moles of a component of a solution divided by the total number of moles of all components: where P A , X A , and n A are the Partial Pressure, mole fraction, and number of moles of gas A, respectively, and n Total is the number of moles of all components in the mixture. EXAMPLE 9.14 The Pressure of a Mixture of Gases A 10.0-L vessel contains 2.50 10 −3 mol of H 2 , 1.00 10 −3 mol of He, and 3.00 10 −4 mol of Ne at 35 °C. (a) What are the Partial Pressures of each of the gases? (b) What is the total pressure in atmospheres? Solution The gases behave independently, so the Partial Pressure of each gas can be determined from the ideal gas equation, using : The total pressure is given by the sum of the Partial Pressures: 9.3 • Stoichiometry of Gaseous Substances, Mixtures, and Reactions 441 Check Your Learning A 5.73-L flask at 25 °C contains 0.0388 mol of N 2 , 0.147 mol of CO, and 0.0803 mol of H 2 .

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  Chemistry

  The Molecular Nature of Matter

  • James E. Brady, Neil D. Jespersen, Alison Hyslop(Authors)
  • 2014(Publication Date)
  • Wiley

    (Publisher)

  Furthermore, the temperature of each gaseous component is the same as the temperature of the entire mixture. Therefore, in a gas mixture, each of the components has the same volume and the same temperature. Partial Pressures In a mixture of nonreacting gases such as air, each gas contributes to the total pressure in proportion to the fraction (by moles) in which it is present (see Figure 10.10). This contri- bution to the total pressure is called the Partial Pressure of the gas. It is the pressure the gas would exert if it were the only gas in a container of the same size at the same temperature. The general symbol we will use for the Partial Pressure of a gas A is P A . For a particular gas, the formula of the gas may be put into the subscript, as in P O 2 . What John Dalton discovered about Partial Pressures is now called Dalton’s law of Partial Pressures: The total pressure of a mixture of gases is the sum of their individual Partial Pressures. In equa- tion form, the law is P total = P A + P B + P C + … (10.3) In dry CO 2 -free air at STP, for example, P O 2 is 159.12 torr, P N 2 is 593.44 torr, and P Ar is 7.10 torr. These Partial Pressures add up to 759.66 torr, just 0.34 torr less than 760 torr or 1.00 atm. The remaining 0.34 torr is contributed by several trace gases, including other noble gases. The explosive PETN, pentaerythritol tetranitrate, is one of the most powerful explosives known. It reacts with the following balanced equation C(CH 2 ONO 2 ) 4 (s) h 2CO( g) + 4H 2 O( g) + 3CO 2 ( g) + 2N 2 ( g) If 0.0250 mol of PETN reacts to fill a 30.0 L sphere at a temperature of 25 °C, what is the Partial Pressure of each gas and the total pressure in the flask? (Hint: Start by calculating the moles of each gas present.) Suppose a tank of oxygen-enriched air prepared for scuba diving has a volume of 17.00 L and a pressure of 237.0 atm at 25 °C.

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  General Chemistry: Atoms First

  • Young, William Vining, Roberta Day, Beatrice Botch(Authors)
  • 2017(Publication Date)
  • Cengage Learning EMEA

    (Publisher)

  This means that although individual Partial Pressures may differ, all gases in the mixture have the same volume (equal to the container volume) and temperature. Example Problem 11.4.1 Calculate pressure using Dalton’s law of Partial Pressures. A gas mixture is made up of O 2 (0.136 g) , CO 2 (0.230 g) , and Xe (1.35 g) . The mixture has a volume of 1.82 L at 22.0 °C . Calculate the Partial Pressure of each gas in the mixture and the total pressure of the gas mixture. Solution: You are asked to calculate the Partial Pressure of each gas in a mixture of gases and the total pressure of the gas mixture. You are given the identity and mass of each gas in the sample and the volume and tempera-ture of the gas mixture. The Partial Pressure of each gas is calculated from the ideal gas equation. P O 2 5 nRT V 5 a 0.136 g 3 1 mol O 2 32.00 g b (0.082057 L # atm/K # mol)(22.0 1 273.15 K) 1.82 L 5 0.0566 atm P CO 2 5 nRT V 5 a 0.230 g 3 1 mol CO 2 44.01 g b (0.082057 L # atm/K # mol)(22.0 1 273.15 K) 1.82 L 5 0.0695 atm c Copyright 2018 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. WCN 02-300 Unit 11 Gases 322 P Xe 5 nRT V 5 a 1.35 g 3 1 mol Xe 131.3 g b (0.082057 L # atm/K # mol)(22.0 1 273.15 K) 1.82 L 5 0.137 atm Notice that the three gases have the same volume and temperature but different pressures. The total pressure is the sum of the Partial Pressures for the gases in the mixture. P total 5 P O 2 1 P CO 2 1 P Xe 5 0.0566 atm 1 0.0695 atm 1 0.137 atm 5 0.263 atm Collecting a Gas by Water Displacement A common laboratory experiment involves collecting the gas generated during a chemi-cal reaction by water displacement (Interactive Figure 11.4.1). Because water can exist in gaseous form (as water vapor), the gas that is collected is a mixture of both the gas formed during the chemical reaction and water vapor.

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  Essential Equations for Anaesthesia

  Key Clinical Concepts for the FRCA and EDA

  • Edward T. Gilbert-Kawai, Marc D. Wittenberg(Authors)
  • 2014(Publication Date)
  • Cambridge University Press

    (Publisher)

  Using the universal gas equation as above, the remaining volume can be calculated. 10 Section 1: Physics & Part 1a: Gases Dalton’s law of Partial Pressures P TOTAL = P GAS A + P GAS B Definition of terms used P Total = total pressure P GAS A = Partial Pressure of gas A P GAS B = Partial Pressure of gas B Units Units of pressure. Explanation Dalton’s law (John Dalton, 1801) states that in a mixture of gases the total pressure is always equal to the sum of the individual Partial Pressures of the gases present. The pressure of each gas is determined by both the number of molecules present and the total volume occupied, and is independent of the presence of any other gases in a mixture. Clinical application/worked example 1. Calculate the alveolar Partial Pressure of oxygen (P A O 2 ) given the following conditions: F i O 2 = 21% Body temperature = 37  C Atmospheric pressure = 100 KPa P A CO 2 = 4 KPa The Partial Pressure of inspired oxygen (P i O 2 ) = F i O 2 × atmospheric pressure = 0.21 × 100 = 21 KPa. However, air in the lungs is saturated with water vapour and mixed with alveolar CO 2 . 11 At 37  C, in normal physiological circumstances, saturated vapour pressure (SVP) of water  6.3 KPa. So, using Dalton’s law: P A O 2 = P i O 2 - (P A CO 2 + P A H 2 O) = 21 - (4 + 6.3) = 10.7 KPa 2. What is the Partial Pressure of oxygen at the top of Everest? Atmospheric pressure (= P Total ) at sea level is approximately 101.3 KPa. Atmospheric pressure (=P Total ) at the top of Everest is approximately 33.7 KPa. The concentration of oxygen is 21%. Therefore, using Dalton’s law, and assuming all other gases being constant: At sea level, P TOTAL = P O2 + P other gases P O2 = P TOTAL – P other gases P O2 = 101.3 – 80.1 P O2 = 21.2 KPa On Everest, P TOTAL = P O2 + P other gases P O2 = P TOTAL – P other gases P O2 = 33.7 – 26.6 P O2 = 7.1 KPa 12 Section 1: Physics & Part 1a: Gases

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  Thermodynamics and Heat Power, Ninth Edition

  • Irving Granet, Jorge Alvarado, Maurice Bluestein(Authors)
  • 2020(Publication Date)
  • CRC Press

    (Publisher)

  Dalton’s law (or Dalton’s rule of additive pressures) is found to apply exactly if the ideal gas relation holds for the components and for the mixture. Dalton’s law can be stated as follows:

  FIGURE 7.1 Dalton’s law.

  1. 1. Each gas behaves as if it alone occupied the volume.
  2. 2. Each gas behaves as if it exists at the temperature of the mixture and filled the entire volume by itself.
  3. 3. The pressure of the gas mixture is the sum of the pressures of each of its components when each behaves as described in statements 1 and 2.

  Applying Dalton’s law to the situation shown in Figure 7.1 , where the mixture m and each of the components are all shown to occupy equal volumes, yields

  p m = p a + p b + p c

  (7.1)

  where p m is the total pressure of the mixture, and p a , p b , and p c represent the pressures that each of the constituent gases would exert if they alone occupied the total volume at the temperature of the mixture. These pressures are called the Partial Pressures of each gas. By assuming that the pressure–volume relationship for an ideal gas is applicable and denoting the molecular weight of the mixture as MWm , we have, in English units, per Section 6.2,

  p m V = m m R T = m m 1545 T MW m

  (7.2)

  where m m is the mass of the mixture equal to the sum of the masses of the individual gases. The Partial Pressure of each gas can be substituted for p m , the mass of each gas for m m , and the molecular weight of each gas for MWm in Equation 7.2. This yields a set of ideal gas equations for each component gas. Using each of these equations in Equation 7.1 yields

  m m MW m = m a MW a + m b MW b + m c MW c

  (7.2a)

  Before continuing, it will be convenient to recall certain terms that were discussed briefly in Chapters 1 and 6 . If the molecular weight of a substance is expressed in pounds, the resulting quantity is known as the pound molecular weight, a pound mole, or simply a mole. Thus, 32 lbm of oxygen constitutes 1 lb. mol of oxygen, and 28.02 lbm of nitrogen is 1 lb. mol of nitrogen. It has also been found that 1 mol of a substance, whether it be solid, liquid, or gas, contains the same number of molecules as 1 mol of any other substance. The number of molecules in 1 mol is known as the Avogadro number or Avogadro’s constant (6.02 × 1023 ). Thus, by application of the ideal gas equation of state, it is found that at a given pressure and temperature, 1 mol of gas will occupy a fixed volume, regardless of the gas. This volume is known as the molar volume, and at 14.7 psia and 32°F, it is 358 ft.3 . Table 7.1

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  Chemistry

  • Paul Flowers, Klaus Theopold, Richard Langley, William R. Robinson(Authors)
  • 2015(Publication Date)
  • Openstax

    (Publisher)

  Summary 9.1 Gas Pressure Gases exert pressure, which is force per unit area. The pressure of a gas may be expressed in the SI unit of pascal or kilopascal, as well as in many other units including torr, atmosphere, and bar. Atmospheric pressure is measured using a barometer; other gas pressures can be measured using one of several types of manometers. 9.2 Relating Pressure, Volume, Amount, and Temperature: The Ideal Gas Law The behavior of gases can be described by several laws based on experimental observations of their properties. The pressure of a given amount of gas is directly proportional to its absolute temperature, provided that the volume does not change (Amontons’s law). The volume of a given gas sample is directly proportional to its absolute temperature at constant pressure (Charles’s law). The volume of a given amount of gas is inversely proportional to its pressure when temperature is held constant (Boyle’s law). Under the same conditions of temperature and pressure, equal volumes of all gases contain the same number of molecules (Avogadro’s law). The equations describing these laws are special cases of the ideal gas law, PV = nRT, where P is the pressure of the gas, V is its volume, n is the number of moles of the gas, T is its kelvin temperature, and R is the ideal (universal) gas constant. 9.3 Stoichiometry of Gaseous Substances, Mixtures, and Reactions The ideal gas law can be used to derive a number of convenient equations relating directly measured quantities to properties of interest for gaseous substances and mixtures. Appropriate rearrangement of the ideal gas equation may be made to permit the calculation of gas densities and molar masses. Dalton’s law of Partial Pressures may be used to relate measured gas pressures for gaseous mixtures to their compositions. Avogadro’s law may be used in stoichiometric computations for chemical reactions involving gaseous reactants or products.

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  Chemical Property Estimation

  Theory and Application

  • Edward Baum(Author)
  • 2018(Publication Date)
  • CRC Press

    (Publisher)

  CHAPTER 6 Vapor Pressure 6.1 INTRODUCTION

  This chapter describes methods of estimating the vapor pressure of a pure chemical and of chemicals in mixtures. The (saturation) vapor pressure, PS (Pa), is the pressure of a pure chemical vapor that is in equilibrium with the pure liquid or solid. It is an important control of a chemical’s partitioning between air, water, and soil and of its volatilization rate. For vapor-solid equilibrium, PS is sometimes called the sublimation pressure. The vapor pressure of a chemical in a mixture of volatile chemicals is its partial vapor pressure.

  Many different units of pressure are widely used in addition to the pascal, the SI unit. They include the standard atmosphere (1 atm= 101.325 kPa), the bar (1 bar = 100 kPa), the torr (1 atm = 760 torr), and the millimeter of mercury (1 mmHg ≈ 1 torr). The mmHg and the torr are almost exactly equal and can be used interchangeably.

  The reported saturation vapor pressures of chemicals at ordinary temperatures range from 760 to below 1 × 10−9 torr. Many hazardous chemicals exhibit vapor pressures of less than 1 torr in the normal ambient temperature range of −40 to +40°C. Vapor pressures below 1 torr are difficult to measure, and reliable values are hard to find in the literature. Computational methods that offer reasonably accurate estimates of such low vapor pressures are particularly useful, therefore, and of particular interest to environmental specialists. The subject was reviewed by Burkhard et al. (1985) and Reid et al. (1987). The literature prior to 1981 was reviewed by Mackay et al. (1982) and by Grain (1990a).

  6.2 A VAPOR PRESSURE MODEL

  Saturation vapor pressure is a sensitive function of molecular structure and ambient temperature. Molecular structure determines the type and strength of the attractive intermolecular forces that a chemical exhibits. The attractive forces are, in order of increasing strength: the London dispersion forces exhibited by all molecules, polar interactions exhibited by all asymmetric molecules, and hydrogen bonding exhibited by molecules containing O–H, N–H, and F–H bonds. The energy required to escape the liquid phase and the vapor pressure depends on the sum of all forces exhibited by a chemical.

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  Student Solutions Manual for Chemistry, Third Canadian Edition

  • John A. Olmsted, Gregory M. Williams, Robert C. Burk(Authors)
  • 2016(Publication Date)
  • Wiley

    (Publisher)

  Chemistry, 3ce Student Solutions Manual Chapter 2 © 2016 John Wiley and Sons Canada, Ltd. 31 f f i i f f i i (5.00 bar)(20.0 L) , from which = 100. L 1.00 bar p V pV p V V p = = = 2.9 When some variables are held fixed, rearrange the ideal gas equation, pV = nRT, to collect fixed values on the right. (a) n, R, V are constant: i f i f = constant; = = p p p nR T V T T (b) = nRT V p (c) n, R, T are constant: pV = nRT = constant; p i V i = p f V f 2.11 When some conditions change but others remain fixed, rearrange the ideal gas equation so the constant terms are grouped on the right. In this problem, n and p are fixed: i f i f = constant, so = = V V V nR T p T T V i = 0.255 L Convert temperature to kelvins: i f i f f i = 25 + 273.15 = 298 K = 15 + 273.15 = 258 K (0.255 L)(258 K) = = = 0.221 L (298 K) T T VT V T − 2.13 The equation is valid only for a gas for which n and T are fixed. (a) n and T are fixed, so the equation is valid. (b) n can change, so the equation is not valid. (c) T changes, so the equation is not valid. (d) The equation is not valid for liquids. 2.15 According to Dalton's law of Partial Pressures, each gas exerts a pressure equal to the total pressure times its mole fraction. Mole fraction can be found from concentration in ppm: 2 6 7 2 NO 6 ppm = 10 0.78 molecules NO 1 atm = 758.4 Torr = 7.8 10 atm 760 Torr 10 molecules of air X p −           ×       2.17 According to Dalton’s law of Partial Pressures, each gas exerts a pressure equal to Chapter 2 Chemistry, 3ce Student Solutions Manual 32 © 2016 John Wiley & Sons Canada, Ltd. its mole fraction times the total pressure: p i = X i p total . Standard atmospheric conditions correspond to p total = 1 atm.

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  Chemistry

  The Molecular Nature of Matter

  • Neil D. Jespersen, Alison Hyslop(Authors)
  • 2021(Publication Date)
  • Wiley

    (Publisher)

  We learned about the SI unit of pressure, the pascal, in this chapter. The temperature unit is the kelvin. Calculate the value of the gas constant in the SI units, m 3 Pa mol −1 K −1 . 10.18 Develop an equation for density from the ideal gas law. Be sure your equation includes pressure, temperature, and molar mass. Dalton’s Law of Partial Pressures 10.19 What is Partial Pressure? 10.20 State Dalton’s law of Partial Pressures in the form of an equation. 10.21 Define mole fraction. How is the Partial Pressure of a gas related to its mole fraction and the total pressure? 10.22 Consider the diagrams below that illustrate three mixtures of gases A and B. If the total pressure of the mixture is 1.00 atm, which of the drawings corresponds to a mixture in which the Partial Pressure of A equals 0.600 atm? What are the Partial Pressures of A in the other mixtures? What are the Partial Pressures of B? Gas A Gas B 10.23 Explain how a gas is collected over water. Why does the tem- perature of the gas need to be known to determine the pressure of the gas collected over water? Why is the volume of the collection vessel not needed? 10.24 What is the difference between diffusion and effusion? State Graham’s law in the form of an equation. 10.25 Why does the rate of effusion relate to the square root of the density or the molecular mass of the gas? Kinetic Molecular Theory of Gases 10.26 Describe the model of a gas proposed by the kinetic theory of gases. 10.27 Use the kinetic molecular theory of gases to explain (a) Charles’ law and (b) Boyle’s law. 10.28 If the molecules of a gas at constant volume are somehow given a lower average kinetic energy, what two measurable properties of the gas will change and in what direction? 10.29 Explain how raising the temperature of a gas causes it to expand at constant pressure.

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