/*! 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 36 If your air conditioner is more ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 13: Problem 36

If your air conditioner is more than several years old, it may use the chlorofluorocarbon \(\mathrm{CCl}_{2} \mathrm{F}_{2}\) as the heat transfer fluid. The normal boiling point of \(\mathrm{CCl}_{2} \mathrm{F}_{2}\) is \(-29.8^{\circ} \mathrm{C},\) and the enthalpy of vaporization is \(20.11 \mathrm{kJ} / \mathrm{mol} .\) The gas and the liquid have molar heat capacities of \(117.2 \mathrm{J} / \mathrm{mol} \cdot \mathrm{K}\) and \(72.3 \mathrm{J} / \mathrm{mol} \cdot \mathrm{K},\) respec- tively. How much energy is evolved as heat when \(20.0 \mathrm{g}\) of \(\mathrm{CCl}_{2} \mathrm{F}_{2}\) is cooled from \(+40^{\circ} \mathrm{C}\) to \(-40^{\circ} \mathrm{C} ?\)

Short Answer

Expert verified
The total energy evolved is \(-4787.4\,\text{J}.\)

Step by step solution

01

Calculate the moles of CCl2F2

First, we determine the number of moles of \( \mathrm{CCl}_{2} \mathrm{F}_{2} \) using its molar mass. The molar mass of \( \mathrm{CCl}_{2} \mathrm{F}_{2} \) is calculated as follows: \[ \text{Molar mass} = (12.01 + 2 \times 35.45 + 2 \times 18.998) \ \text{g/mol} = 120.91 \ \text{g/mol}. \] Now, calculate the moles for the given mass (20.0 g): \[ \text{Moles} = \frac{20.0 \ \text{g}}{120.91 \ \text{g/mol}} = 0.165 \ \text{mol}. \]
02

Calculate cooling in the gas phase

Cool the gas from \(40^{\circ} \mathrm{C} \) to the boiling point \(-29.8^{\circ} \mathrm{C}\). Use the formula for heat transfer in the gas phase: \[ q = nC_{p,\text{gas}} \Delta T. \] Here, \( C_{p,\text{gas}} = 117.2 \ \mathrm{J/mol \cdot K} \), \( n = 0.165 \ \text{mol} \), and \( \Delta T = -29.8 - 40 = -69.8^{\circ} \mathrm{C} = -69.8 \ \text{K} \). \[ q = 0.165 \times 117.2 \times (-69.8) = -1344.9 \ \text{J}. \]
03

Phase Change (Condensation)

Now convert the gas to liquid at the boiling point. Use the enthalpy of vaporization: \( \Delta H_{\text{vap}} = 20.11 \ \mathrm{kJ/mol} \). The energy change for this phase transition is: \[ q = -n \Delta H_{\text{vap}} = -0.165 \times 20.11 \times 1000 = -3320.8 \ \text{J}. \]
04

Cooling in the liquid phase

Cool the liquid from the boiling point \(-29.8^{\circ} \mathrm{C}\) to \(-40^{\circ} \mathrm{C}\). Use the formula for heat transfer in the liquid phase: \[ q = nC_{p,\text{liquid}} \Delta T. \] Here, \( C_{p,\text{liquid}} = 72.3 \ \mathrm{J/mol \cdot K} \), \( n = 0.165 \ \text{mol} \), and \( \Delta T = -40 - (-29.8) = -10.2 \ \text{K} \). \[ q = 0.165 \times 72.3 \times (-10.2) = -121.7 \ \text{J}. \]
05

Sum all energy changes

Sum all calculated energy changes from each step to find the total energy evolved: \[ q_{\text{total}} = q_{\text{gas}} + q_{\text{phase}} + q_{\text{liquid}} = -1344.9 - 3320.8 - 121.7 = -4787.4 \ \text{J}. \]

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 91影视!

Key Concepts

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

Phase Change
Phase change refers to the transition of a substance from one state of matter to another, such as solid to liquid, liquid to gas, or vice versa. In this problem, we specifically deal with the transition from gas to liquid, known as condensation.
A phase change is significant because it involves a large amount of energy absorption or release. During condensation, energy is released as the molecules decrease their kinetic energy to form intermolecular bonds.
This is why we use enthalpy of vaporization to calculate the energy needed or released during these transitions. Understanding phase changes is crucial in thermodynamics, as they are involved in processes like refrigeration and heating.
Enthalpy of Vaporization
Enthalpy of vaporization (\( \Delta H_{\text{vap}} \)) is the amount of heat required to convert one mole of a liquid into a gas at constant temperature and pressure. For the reverse process鈥攃ondensation鈥攊t is the heat released when a gas turns back into a liquid.
In the exercise, \( \Delta H_{\text{vap}} \) for \(\text{CCl}_{2}\text{F}_{2}\) is given as 20.11 kJ/mol. This indicates that when the gas condenses, \(20.11 kJ/mol\) is released.
This concept is essential when trying to understand how much energy is absorbed or released during a phase transition, impacting systems like air conditioners, which rely on substantial energy changes to operate efficiently.
Heat Capacity
Heat capacity (\(C_p\)) is the amount of heat needed to raise the temperature of a given quantity of a substance by 1 Kelvin. It is specific to the state of matter of the substance, as seen with \(\text{CCl}_{2}\text{F}_{2}\) in this problem, which has different heat capacities for its liquid and gas states.
  • \(C_{p,\text{gas}} = 117.2 \text{ J/mol} \cdot \text{K}\) while \(C_{p,\text{liquid}} = 72.3 \text{ J/mol} \cdot \text{K}\).
These values influence the amount of energy released or absorbed during temperature changes within the gaseous and liquid states of the material.
Understanding heat capacity helps predict how much a substance's temperature will change when it absorbs or releases a given amount of energy, crucial for designing and operating thermodynamic systems effectively.

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

Which will show the highest conductivity at \(298 \mathrm{K}\) silicon or germanium?

Two identical swimming pools are filled with uniform spheres of ice packed as closely as possible. The spheres in the first pool are the size of grains of sand; those in the second pool are the size of oranges. The ice in both pools melts. In which pool, if either, will the water level be higher? (Ignore any differences in filling space at the planes next to the walls and bottom.)

Considering only the molecular orbitals formed by combinations of the \(2 s\) atomic orbitals, how many molecular orbitals can be formed by 1000 Li atoms? In the lowest energy state, how many of these orbitals will be populated by pairs of electrons and how many will be empty?

Most metals are shiny, that is, they reflect light. How does the bonding theory for metals explain this characteristic?

Boron phosphide, BP, is a semiconductor and a hard, abrasion-resistant material. It is made by reacting boron tribromide and phosphorus tribromide in a hydrogen atmosphere at high temperature \(\left(>750^{\circ} \mathrm{C}\right)\) (a) Write a balanced chemical equation for the synthesis of BP. (Hint: Hydrogen is a reducing agent.) (b) Boron phosphide crystallizes in a zinc-blend structure, formed from boron atoms in a face-centered cubic lattice and phosphorus atoms in tetrahedral holes. How many tetrahedral holes are filled with P atoms in each unit cell? (c) The length of a unit cell of BP is 478 pm. What is the density of the solid in \(\mathrm{g} / \mathrm{cm}^{3} ?^{-}\) (d) Calculate the closest distance between a \(\mathrm{B}\) and a P atom in the unit cell. (Assume the B atoms do not touch along the cell edge. The \(\mathrm{B}\) atoms in the faces touch the \(\mathrm{B}\) atoms at the corners of the unit cell.)
See all solutions

Recommended explanations on Chemistry Textbooks

The Earths Atmosphere

Read Explanation

Ionic and Molecular Compounds

Read Explanation

Kinetics

Read Explanation

Nuclear Chemistry

Read Explanation

Chemical Analysis

Read Explanation

Physical Chemistry

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.