Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 20: Q58P (page 607)

Repeat Problem 57, with the pressure now kept constant.

Short Answer

Expert verified

The change in the entropy of the gas at constant pressure is 5.98 J/K

Step by step solution

01

The given data

The number of moles is n = 1.00 mol .

The gas ismonatomic.

The initial temperature isTi=300K.

The final temperature isTf=400K.

The pressure is constant.

02

Understanding the concept of entropy change

By finding the equation for heat absorbed at constant pressure and using Equation 20-1, we can find the change in the entropy of the gas at constant pressure.

Formula:

The energy absorbed as heat by the gas at constant pressure, dQ=52nRdT (1)

The change in entropy of the gas, ∆S=∫dQT (2)

03

Calculation of the entropy change

Since the pressureis kept constant, substituting the value of equation (1) in equation (2), we can get the change in entropy as given:

∆S=∫TiTf5nR2dTT=52nRlnTfTi=52×1.00mol×8.31Jmol·K×400K300K=5.98J/K

Hence, the value of the entropy change is 5.98 J/K

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

Construct a table like Table 20 - 1for eight molecules

In four experiments, 2.5 molof hydrogen gas undergoes reversible isothermal expansions, starting from the same volume but at different temperatures. The corresponding P-Vplots are shown in Fig. 20-21. Rank the situations according to the change in the entropy of the gas, greatest first.

An inventor has built an engine X and claims that its efficiency X is greater than the efficiency of an ideal engine operating between the same two temperatures. Suppose you couple engine X to an ideal refrigerator (Fig. 20-34a) and adjust the cycle of engine X so that the work per cycle it provides equals the work per cycle required by the ideal refrigerator. Treat this combination as a single unit and show that if the inventor’s claim were true(ifεx>ε), the combined unit would act as a perfect refrigerator (Fig. 20-34b), transferring energy as heat from the low-temperature reservoir to the high-temperature reservoir without the need for work.

In an experiment, 200 gof aluminum (with a specific heat of 900J/kg.K) at 100°Cis mixed with 50.0gof water at 20.0°C, with the mixture thermally isolated.(a)What is the equilibrium temperature?(b)What is the entropy changes of the aluminum, (c)What is the entropy changes of the water, and (d)What is the entropy changes of the aluminum – water system?

A brass rod is in thermal contact with a constant-temperature reservoir at130°Cat one end and a constant-temperature reservoir at 24.0°C at the other end. (a) Compute the total change in entropy of the rod–reservoirs system when 5030 Jof energy is conducted through the rod, from one reservoir to the other. (b) Does the entropy of the rod change?
See all solutions

Recommended explanations on Physics Textbooks

Work Energy and Power

Read Explanation

Fluids

Read Explanation

Magnetism

Read Explanation

Particle Model of Matter

Read Explanation

Electrostatics

Read Explanation

Nuclear Physics

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