Textbook Question
The change of state from liquid H2O to gaseous H2O has ∆H = +9.72 kcal/mol(+40.7 kJ/mol) and ∆S = -26.1 cal/(mol • K) [-109 J/(mol •K)].
a. Is the change from liquid to gaseous H2O favored or unfavored by ∆H? By ∆S?
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Ch.8 Gases, Liquids and Solids
McMurry8th EditionFundamentals of GOBISBN: 9780134015187
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Table of contents
- Ch.1 Matter and Measurements27
- Ch.2 Atoms and the Periodic Table20
- Ch.3 Ionic Compounds8
- Ch.4 Molecular Compounds23
- Ch.5 Classification & Balancing of Chemical Reactions12
- Ch.6 Chemical Reactions: Mole and Mass Relationships28
- Ch.7 Chemical Reactions: Energy, Rate and Equilibrium64
- Ch.8 Gases, Liquids and Solids24
- Ch.9 Solutions52
- Ch.10 Acids and Bases25
- Ch.11 Nuclear Chemistry22
- Ch.12 Introduction to Organic Chemistry: Alkanes57
- Ch.13 Alkenes, Alkynes, and Aromatic Compounds47
- Ch.14 Some Compounds with Oxygen, Sulfur, or a Halogen50
- Ch.15 Aldehydes and Ketones45
- Ch.16 Amines42
- Ch.17 Carboxylic Acids and Their Derivatives57
- Ch.18 Amino Acids and Proteins82
- Ch.19 Enzymes and Vitamins63
- Ch.20 Carbohydrates44
- Ch.21 The Generation of Biochemical Energy65
- Ch.22 Carbohydrate Metabolism45
- Ch.23 Lipids42
- Ch.24 Lipid Metabolism33
- Ch.25 Protein and Amino Acid Metabolism24
- Ch.26 Nucleic Acids and Protein Synthesis45
Chapter 8, Problem 7
A local weather station reports the barometric pressure as 29.5 inHg (inches of Hg). Convert this pressure to torr and to atm.
Verified step by step guidance1
Step 1: Understand the problem. The goal is to convert the given pressure of 29.5 inHg (inches of mercury) into two different units: torr and atm (atmospheres). To do this, we need the appropriate conversion factors.
Step 2: Recall the conversion factors. The relationships between the units are as follows: 1 inHg = 25.4 mmHg (millimeters of mercury), and 1 mmHg = 1 torr. Additionally, 1 atm = 760 torr.
Step 3: Convert inches of mercury (inHg) to torr. First, use the conversion factor 1 inHg = 25.4 mmHg to convert 29.5 inHg to mmHg. Then, since 1 mmHg = 1 torr, the value in mmHg will be numerically equal to the value in torr. Use the formula: \( \text{Pressure in torr} = \text{Pressure in inHg} \times 25.4 \).
Step 4: Convert torr to atm. Use the conversion factor 1 atm = 760 torr. Divide the pressure in torr by 760 to find the pressure in atm. Use the formula: \( \text{Pressure in atm} = \frac{\text{Pressure in torr}}{760} \).
Step 5: Summarize the process. You have now converted the pressure from inHg to torr and then from torr to atm using the appropriate conversion factors. Ensure all units are properly labeled in your final answers.
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Barometric Pressure
Barometric pressure is the pressure exerted by the weight of the atmosphere at a given point. It is commonly measured in units such as inches of mercury (inHg), torr, and atmospheres (atm). Understanding barometric pressure is essential for meteorology and various scientific applications, as it influences weather patterns and altitude measurements.
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01:15
Pressure Units Concept 1
Unit Conversion
Unit conversion is the process of converting a quantity expressed in one set of units to another. In this context, converting barometric pressure from inches of mercury to torr and atmospheres requires knowledge of the relationships between these units. For example, 1 inHg is equivalent to 25.4 torr and approximately 0.968 atm, which are crucial for accurate scientific communication.
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01:56Conversion Factors (Simplified) Concept 1
Torr and Atmosphere
Torr and atmosphere (atm) are both units of pressure. The torr is defined as 1/760 of an atmosphere, making it a convenient unit for measuring low pressures, such as those found in weather reports. Understanding these units is important for interpreting pressure readings and performing calculations in various scientific fields, including physics and engineering.
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02:09Dalton's Law: Partial Pressure (Simplified) Example 1
Related Practice
Textbook Question
The change of state from liquid H2O to gaseous H2O has ∆H = +9.72 kcal/mol(+40.7 kJ/mol) and ∆S = -26.1 cal/(mol • K) [-109 J/(mol •K)].
b. What are the values of ∆H and ∆S (in kcal/mol and kJ/mol) for the change from gaseous to liquid H2O?
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Textbook Question
Would you expect the boiling points to increase or decrease in the following series? Explain.
a. Kr, Ar, Ne
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Textbook Question
Determine the percent composition of air in the lungs from the following composition in partial pressures: PN2= 573 mmHg, PO2 = 100 mmHg, PCO2 = 40 mmHg, and PH2O = 47 mmHg; all at 37 °C and 1 atm pressure.
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Textbook Question
Compare the ∆Hvap values for water, isopropyl alcohol, ether, and ammonia, and order them from lowest to highest. Explain the rank order based on intermolecular attractive forces.
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Textbook Question
Assume that you have a sample of gas at 350 K in a sealed container, as represented in part (a). Which of the drawings (b)–(d) represents the gas after the temperature is lowered from 350 K to 150 K and if the gas has a boiling point of 200 K? Which drawing represents the gas at 150 K if the gas has a boiling point of 100 K?
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