Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 11: Problem 11

As a metal such as lead melts, what happens to (a) the average kinetic energy of the atoms and (b) the average distance between the atoms?

Short Answer

Expert verified
As a metal like lead melts, (a) the average kinetic energy of the atoms increases due to the temperature rise and faster atomic movement, and (b) the average distance between the atoms increases as the material transitions from a closely packed solid-state lattice to a less ordered liquid state with increased repulsion forces.

Step by step solution

01

Understand the concept of kinetic energy

Kinetic energy is the energy of motion. In the context of atoms, it directly relates to the speed at which the atoms are moving within the material.
02

Consider the effect of temperature

When a metal melts, the temperature increases, meaning the atoms have more energy. As the atoms gain more energy, they move faster. This results in an increase in the average kinetic energy of the atoms.
03

Final answer for (a)

As a metal such as lead melts, the average kinetic energy of the atoms increases. #b) The average distance between the atoms#
04

Understand the concept of atomic spacing

The average distance between the atoms refers to the space between individual atoms within the material.
05

Consider the transition from solid to liquid state

As a metal melts, it transitions from a solid state to a liquid state. In the solid state, the atoms are arranged in a fixed, closely packed lattice. In the liquid state, the atoms lose their fixed positions and flow freely, which means the arrangement becomes less ordered.
06

Account for the increase in energy

In the liquid state, the atoms have more energy and move more rapidly, causing the atoms to collide and repel each other. This leads to an increase in the average distance between the atoms due to the increased repulsion forces.
07

Final answer for (b)

As a metal such as lead melts, the average distance between the atoms increases.

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.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

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

For many years drinking water has been cooled in hot climates by evaporating it from the surfaces of canvas bags or porous clay pots. How many grams of water can be cooled from 35 to \(20^{\circ} \mathrm{C}\) by the evaporation of \(60 \mathrm{~g}\) of water? (The heat of vaporization of water in this temperature range is \(2.4 \mathrm{~kJ} / \mathrm{g} .\) The specific heat of water is \(4.18 \mathrm{~J} / \mathrm{g}-\mathrm{K} .)\)

Suppose the vapor pressure of a substance is measured at two different temperatures. (a) By using the Clausius-Clapeyron equation (Equation 11.1) derive the following relationship between the vapor pressures, \(P_{1}\) and \(P_{2}\), and the absolute temperatures at which they were measured, \(T_{1}\) and \(T_{2}\) : $$ \ln \frac{P_{1}}{P_{2}}=-\frac{\Delta H_{\mathrm{vap}}}{R}\left(\frac{1}{T_{1}}-\frac{1}{T_{2}}\right) $$ (b) Gasoline is a mixture of hydrocarbons, a component of which is octane $\left(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{3}\right)$. Octane has a vapor pressure of \(1.86 \mathrm{kPa}\) at \(25^{\circ} \mathrm{C}\) and a vapor pressure of \(19.3 \mathrm{kPa}\) at \(75^{\circ} \mathrm{C}\). Use these data and the equation in part (a) to calculate the heat of vaporization of octane. \((\mathbf{c})\) By using the equation in part (a) and the data given in part (b), calculate the normal boiling point of octane. Compare your answer to the one you obtained from Exercise 11.81 . (d) Calculate the vapor pressure of octane at \(-30^{\circ} \mathrm{C}\).

(a) How does the average kinetic energy of molecules compare with the average energy of attraction between molecules in solids, liquids, and gases? (b) Why does increasing the temperature cause a solid substance to change in succession from a solid to a liquid to a gas? (c) What happens to a gas if you put it under extremely high pressure?

The vapor pressure of a volatile liquid can be determined by slowly bubbling a known volume of gas through it at a known temperature and pressure. In an experiment, \(8.00 \mathrm{~L}\) of argon gas is passed through $11.7872 \mathrm{~g}\( of liquid hexane \)\mathrm{C}_{6} \mathrm{H}_{14}$ at \(30.0^{\circ} \mathrm{C}\). The mass of the remaining liquid after the experiment is \(4.875 \mathrm{~g}\). Assuming that the gas becomes saturated with hexane vapor and that the total gas volume and temperature remain constant, what is the vapor pressure of hexane in atm?

Which type of intermolecular force accounts for each of these differences? (a) Acetone, \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CO},\) boils at $56^{\circ} \mathrm{C} ;\( dimethyl sulfoxide or DMSO, (CH \)\left._{3}\right)_{2}$ SO, boils at \(189^{\circ}\) C. (b) \(\mathrm{CCl}_{4}\) is a liquid at atmospheric pressure and room temperature, whereas \(\mathrm{CH}_{4}\) is a gas under the same conditions. \((\mathbf{c})\) \(\mathrm{H}_{2} \mathrm{O}\) boils at \(100^{\circ} \mathrm{C}\) but \(\mathrm{H}_{2} \mathrm{~S}\) boils at \(-60^{\circ} \mathrm{C}\). (d) 1 -propanol boils at \(97^{\circ} \mathrm{C}\), whereas 2 -propanol boils at \(82.6^{\circ} \mathrm{C}\).
See all solutions

Recommended explanations on Chemistry Textbooks

Chemical Analysis

Read Explanation

Chemistry Branches

Read Explanation

Chemical Reactions

Read Explanation

Nuclear Chemistry

Read Explanation

Inorganic Chemistry

Read Explanation

Ionic and Molecular Compounds

Read Explanation
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.

Sign-up for free