/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 144 The molar enthalpy of vaporizati... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

The molar enthalpy of vaporization of water at \(373 \mathrm{~K}\) and \(1.00\) atm is \(40.7 \mathrm{~kJ} / \mathrm{mol}\). What fraction of this energy is used to change the internal energy of the water, and what fraction is used to do work against the atmosphere? (Hint: Assume that water vapor is an ideal gas.)

Short Answer

Expert verified
The fraction of the 40.7 kJ/mol energy used for changing the internal energy of water is approximately 1.076, and the fraction used for working against the atmosphere is approximately 0.076.

Step by step solution

01

Recall the definition of enthalpy change and the first law of thermodynamics

Enthalpy change, ΔH, is the sum of the change in internal energy, ΔU, and the work done against the atmosphere, @PΔV@. This can be written as: \( ΔH = ΔU + PΔV \) The first law of thermodynamics states that the change in internal energy of a system is equal to the heat absorbed, q, minus the work done by the system, w. This can be expressed as: \( ΔU = q - w \)
02

Determine the work done against the atmosphere

For an ideal gas, the work done against the atmosphere (w) can be calculated using the formula: \( w = -nRT \) Where: - n is the number of moles - R is the gas constant (\(8.314~\mathrm{J~mol^{-1}~K^{-1}}\)) - T is the temperature in Kelvin We know that we are working with 1 mole of water, so n = 1. The temperature is given as 373 K. Substituting these values, we get: \( w = -1 \cdot 8.314 \cdot 373 \) \( w = -3097.722 \mathrm{~J} \) \( w = -3.098 \mathrm{~kJ} \) The work done against the atmosphere is -3.098 kJ.
03

Calculate the change in internal energy

We can now use the enthalpy change equation to find the change in internal energy: \( ΔH = ΔU + PΔV \) \( 40.7 = ΔU - 3.098 \) Solve for ΔU: \( ΔU = 40.7 + 3.098 \) \( ΔU = 43.798 \mathrm{~kJ} \) The change in internal energy of the water is 43.798 kJ.
04

Calculate the fraction of energy for the change in internal energy and work against the atmosphere

Now, we can calculate the fraction of the total energy used to change the internal energy and do work against the atmosphere: Fraction of energy for change in internal energy: \( \frac{ΔU}{ΔH} = \frac{43.798}{40.7} = 1.076 \) Fraction of energy for work against the atmosphere: \( \frac{-w}{ΔH} = \frac{-(-3.098)}{40.7} = 0.076 \) The fraction of the 40.7 kJ/mol energy used for changing the internal energy of water is approximately 1.076, and the fraction used for working against the atmosphere is approximately 0.076.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

First Law of Thermodynamics
The first law of thermodynamics, often called the law of energy conservation, establishes a fundamental principle crucial to the science of physics and chemistry. It states that the energy of a closed system is constant; energy can neither be created nor destroyed, but only transformed from one form to another. In the exercise, this law is applied to the phase change of water from liquid to gas.

When water vaporizes, the energy used in the process goes into changing the internal energy of the water and doing work against external pressure, such as the atmosphere. This is a direct application of the first law, as the molar enthalpy of vaporization is the energy input (heat absorbed, q) required to change the state of one mole of water without changing its temperature. Understanding this concept is essential for explaining why two different fractions of energy are calculated in the solution—the internal energy change and the work done on the surroundings.
Enthalpy Change
Enthalpy change, denoted as ΔH, is a measure of the total heat content change in a system at constant pressure. It is a fundamental concept in thermodynamics and is particularly significant when analyzing phase changes, such as vaporization. In the given exercise, the molar enthalpy of vaporization refers to the amount of heat, ΔH, needed to turn one mole of liquid water into water vapor at a constant temperature and pressure.

The exercise improvement advice here is to clarify that enthalpy change encompasses both the internal energy change (\(ΔU\)) and the work done by expanding the gas against constant atmospheric pressure (\(PΔV\)). The enthalpy change during the vaporization process provides us with the total amount of energy needed to overcome intermolecular forces within the liquid (related to ΔU) and to push back the atmosphere to make room for the expanding gas (related to work done, \(PΔV\)).
Internal Energy of Water
Internal energy of a system, denoted as U, is the sum of the kinetic and potential energies of all particles within the system. It's a microscopic energy on the molecular level. For water, this includes the energy associated with the motion of molecules (kinetic) as well as the energies involved in interactions between molecules (potential).

During vaporization, when water changes from liquid to gas, the internal energy increases due to the greater freedom of movement and distance between water molecules in the gaseous state compared to the liquid state. In the exercise, this increment is represented as the change in internal energy, ΔU. Understanding internal energy is key to grasping why vaporization requires energy: it is needed to break the strong hydrogen bonds between water molecules in the liquid state. The solution shows us how to calculate this internal energy change, highlighting its significant role in the process of vaporization.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

You are asked to help set up a historical display in the park by stacking some cannonballs next to a Revolutionary War cannon. You are told to stack them by starting with a triangle in which each side is composed of four touching cannonballs. You are to continue stacking them until you have a single ball on the top centered over the middle of the triangular base. a. How many cannonballs do you need? b. What type of closest packing is displayed by the cannonballs? c. The four corners of the pyramid of cannonballs form the corners of what type of regular geometric solid?

The critical point of \(\mathrm{NH}_{3}\) is \(132^{\circ} \mathrm{C}\) and \(111 \mathrm{~atm}\), and the critical point of \(\mathrm{N}_{2}\) is \(-147^{\circ} \mathrm{C}\) and \(34 \mathrm{~atm}\). Which of these substances cannot be liquefied at room temperature no matter how much pressure is applied? Explain.

When a person has a severe fever, one therapy used to reduce the fever is an "alcohol rub." Explain how the evaporation of alcohol from a person's skin removes heat energy from the body.

What type of solid will each of the following substances form? a. \(\mathrm{CO}_{2}\) g. \(\mathrm{KBr}\) b. \(\mathrm{SiO}_{2}\) h. \(\mathrm{H}_{2} \mathrm{O}\) c. \(\mathrm{Si}\) i. \(\mathrm{NaOH}\) d. \(\mathrm{CH}_{4}\) j. \(\underline{U}\) e. Ru k. \(\mathrm{CaCO}_{3}\) f. \(\mathrm{I}_{2}\) 1\. \(\mathrm{PH}_{3}\)

General Zod has sold Lex Luthor what Zod claims to be a new copper-colored form of kryptonite, the only substance that can harm Superman. Lex, not believing in honor among thieves, decided to carry out some tests on the supposed kryptonite. From previous tests, Lex knew that kryptonite is a metal having a specific heat capacity of \(0.082 \mathrm{~J} / \mathrm{g} \cdot{ }^{\circ} \mathrm{C}\) and a density of \(9.2 \mathrm{~g} / \mathrm{cm}^{3}\) Lex Luthor's first experiment was an attempt to find the specific heat capacity of kryptonite. He dropped a \(10 \mathrm{~g} \pm 3 \mathrm{~g}\) sample of the metal into a boiling water bath at a temperature of \(100.0^{\circ} \mathrm{C} \pm 0.2^{\circ} \mathrm{C}\). He waited until the metal had reached the bath temperature and then quickly transferred it to \(100 \mathrm{~g} \pm\) \(3 \mathrm{~g}\) of water that was contained in a calorimeter at an initial temperature of \(25.0^{\circ} \mathrm{C} \pm 0.2^{\circ} \mathrm{C}\). The final temperature of the metal and water was \(25.2^{\circ} \mathrm{C}\). Based on these results, is it possible to distinguish between copper and kryptonite? Explain. When Lex found that his results from the first experiment were inconclusive, he decided to determine the density of the sample. He managed to steal a better balance and determined the mass of another portion of the purported kryptonite to be \(4 \mathrm{~g} \pm 1 \mathrm{~g}\). He dropped this sample into water contained in a \(25-\mathrm{mL}\) graduated cylinder and found that it displaced a volume of \(0.42 \mathrm{~mL} \pm 0.02 \mathrm{~mL}\). Is the metal copper or kryptonite? Explain. Lex was finally forced to determine the crystal structure of the metal General Zod had given him. He found that the cubic unit cell contained four atoms and had an edge length of 600\. pm. Explain how this information enabled Lex to identify the metal as copper or kryptonite. Will Lex be going after Superman with the kryptonite or seeking revenge on General Zod? What improvements could he have made in his experimental techniques to avoid performing the crystal structure determination?

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.