/*! 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 22 What happens to the internal ene... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 9: Problem 22

What happens to the internal energy of a gas when it expands (with no heat flow)?

Short Answer

Expert verified
Answer: When a gas expands without any heat flow, its internal energy decreases. This is because the work done by the system during adiabatic expansion is at the expense of its internal energy.

Step by step solution

01

Recall the First Law of Thermodynamics

The first law of thermodynamics states that the change in internal energy (鈭哢) of a system is equal to the difference between the heat (Q) added to the system and the work (W) done by the system on its surroundings. Mathematically, we can represent it as: 鈭哢 = Q - W Since there is no heat flow in this expansion (Q = 0), the equation becomes: 鈭哢 = -W
02

Relate Work Done to the Adiabatic Expansion

For an ideal gas undergoing an adiabatic expansion, the work done by the gas is given by the formula: W = (P1V1 - P2V2) / (纬 - 1) where P1 and V1 are the initial pressure and volume, P2 and V2 are the final pressure and volume, and 纬 (gamma) is the adiabatic index or the ratio of specific heat capacities (Cp/Cv).
03

Determine the Internal Energy Change

Now that we have related work to the adiabatic expansion, we can substitute the expression for work into the equation for internal energy change: 鈭哢 = - (P1V1 - P2V2) / (纬 - 1)
04

Analyze the Internal Energy Change

When a gas expands adiabatically, its pressure decreases and its volume increases. Therefore, P2 < P1 and V2 > V1. From the equation in step 3, we can see that if the term (P1V1 - P2V2) is positive, the internal energy change (鈭哢) will be negative. Thus, during adiabatic expansion, the internal energy of the gas decreases and the work done by the system is at the expense of its internal energy. So, the internal energy of the gas decreases when it expands without any heat flow.

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!

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

Synthetic natural gas (SNG), sometimes called substitute natural gas, is a methane-containing mixture produced from the gasification of coal or oil shale directly at the site of the mine or oil field. One reaction for the production of SNG is: $$4 \mathrm{CO}(g)+8 \mathrm{H}_{2}(g) \rightarrow 3 \mathrm{CH}_{4}(g)+\mathrm{CO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(g)$$ Use the following thermochemical equations to determine \(\Delta H^{\circ}\) for the reaction as written. $$\begin{aligned}\mathrm{C}(\text { graphite })+2 \mathrm{H}_{2}(g) & \rightarrow \mathrm{CH}_{4}(g) & \Delta H^{\circ} &=-74.8 \mathrm{kJ} \\\\\mathrm{C}(\text { graphite })+\frac{1}{2} \mathrm{O}_{2}(g) & \rightarrow \mathrm{CO}(g) & \Delta H^{\circ} &=-110.5 \mathrm{kJ}\end{aligned}$$ $$\begin{array}{ll}\mathrm{CO}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \rightarrow \mathrm{CO}_{2}(g) & \Delta H^{\circ}=-283.0 \mathrm{kJ} \\\\\mathrm{H}_{2}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \rightarrow \mathrm{H}_{2} \mathrm{O}(g) & \Delta H^{\circ}=-285.8 \mathrm{kJ}\end{array}$$

An insulated container holds \(50.0 \mathrm{g}\) of water at \(25.0^{\circ} \mathrm{C} .\) A \(7.25 \mathrm{g}\) sample of copper that had been heated to \(100.1^{\circ} \mathrm{C}\) is dropped into the water. What is the final shared temperature of the copper and the water?

The heavier (more dense) hydrocarbons in camp stove fuel are hexanes \(\left(\mathrm{C}_{6} \mathrm{H}_{14}\right)\). a. Calculate the fuel value of \(\mathrm{C}_{6} \mathrm{H}_{14},\) given that \(\Delta H_{\text {comb }}^{\circ}=\) \(-4163 \mathrm{kJ} / \mathrm{mol}\). b. How much heat is released during the combustion of \(1.00 \mathrm{kg}\) of \(\mathrm{C}_{6} \mathrm{H}_{14} ?\) c. How many grams of \(\mathrm{C}_{6} \mathrm{H}_{14}\) are needed to heat \(1.00 \mathrm{kg}\) of water from \(25.0^{\circ} \mathrm{C}\) to \(85.0^{\circ} \mathrm{C} ?\) Assume that all of the heat released during combustion is used to heat the water. d. Assume white gas is \(25 \% \mathrm{C}_{5}\) hydrocarbons (see Problem 9.109) and \(75 \%\) C \(_{6}\) hydrocarbons; how many grams of white gas are needed to heat \(1.00 \mathrm{kg}\) of water from \(25.0^{\circ} \mathrm{C}\) to \(85.0^{\circ} \mathrm{C} ?\)

Lightweight camping stoves typically use white gas, a mixture of \(\mathrm{C}_{5}\) and \(\mathrm{C}_{6}\) hydrocarbons. a. Calculate the fuel value of \(\mathrm{C}_{5} \mathrm{H}_{12},\) given that \(\Delta H_{\text {comb }}^{\circ}=\) \(-3535 \mathrm{kJ} / \mathrm{mol}\) b. How much heat is released during the combustion of \(1.00 \mathrm{kg}\) of \(\mathrm{C}_{5} \mathrm{H}_{12} ?\) c. How many grams of \(\mathrm{C}_{5} \mathrm{H}_{12}\) must be burned to heat \(1.00 \mathrm{kg}\) of water from \(20.0^{\circ} \mathrm{C}\) to \(90.0^{\circ} \mathrm{C} ?\) Assume that all the heat released during combustion is used to heat the water.

During a strenuous workout, an athlete generates \(233 \mathrm{kJ}\) of thermal energy. What mass of water would have to evaporate from the athlete's skin to dissipate this energy?
See all solutions

Recommended explanations on Chemistry Textbooks

Chemical Analysis

Read Explanation

Ionic and Molecular Compounds

Read Explanation

Kinetics

Read Explanation

Organic Chemistry

Read Explanation

Chemical Reactions

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.