/*! 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 9 Suppose 1.00 mol of oxygen is he... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 15: Problem 9

Suppose 1.00 mol of oxygen is heated at constant pressure of 1.00 atm from \(10.0^{\circ} \mathrm{C}\) to \(25.0^{\circ} \mathrm{C} .\) (a) How much heat is absorbed by the gas? (b) Using the ideal gas law, calculate the change of volume of the gas in this process. (c) What is the work done by the gas during this expansion? (d) From the first law, calculate the change of internal energy of the gas in this process.

Short Answer

Expert verified
Question: A 1.00 mol oxygen gas (O2) is heated from 10.0掳C to 25.0掳C under constant pressure of 1.00 atm. The heat capacity of O2 under constant pressure (Cp) is 29.4 J/mol.K. Calculate (a) the heat absorbed by the gas, (b) the change in volume, (c) the work done during expansion, and (d) the change in internal energy. Answer: (a) The heat absorbed by the gas is 441 J. (b) The change in volume is 0.00206 m^3. (c) The work done during expansion is -208.73 J. (d) The change in internal energy is 232.27 J.

Step by step solution

01

Part (a): Calculate the heat absorbed by the gas

To find the heat absorbed (q) by the gas, use the formula: q = n * Cp * 螖T where n is the number of moles, Cp is the heat capacity at constant pressure, and 螖T is the change in temperature (T2 - T1). 螖T = T2 - T1 = 25.0 - 10.0 = 15.0 K q = (1.00 mol) * (29.4 J/mol.K) * (15.0 K) q = 441 J So, the gas absorbs 441 J of heat.
02

Part (b): Calculate the change in volume

We can use the ideal gas law to find the change in volume (螖V): PV = nRT First, convert pressure from atm to Pa by multiplying it with the conversion factor (1 atm = 101325 Pa). Then, calculate the initial and final volumes using the ideal gas law: P = 1.00 atm * 101325 Pa/atm = 101325 Pa R = 8.314 J/mol.K Initial volume (V1): V1 = nRT1 / P V1 = (1.00 mol)(8.314 J/mol.K)(283.15 K) / 101325 Pa V1 = 0.02241 m^3 Final volume (V2): V2 = nRT2 / P V2 = (1.00 mol)(8.314 J/mol.K)(298.15 K) / 101325 Pa V2 = 0.02447 m^3 Change in volume (螖V): 螖V = V2 - V1 螖V = 0.02447 - 0.02241 螖V = 0.00206 m^3 The change in volume is 0.00206 m^3.
03

Part (c): Calculate the work done

To find the work done (w) by the gas during expansion, we use the formula: w = -P(螖V) w = -101325 Pa * 0.00206 m^3 w = -208.73 J The work done by the gas during expansion is -208.73 J.
04

Part (d): Calculate the change in internal energy

To find the change in internal energy (螖U), we use the first law of thermodynamics: 螖U = q + w 螖U = 441 J + (-208.73 J) 螖U = 232.27 J The change in internal energy of the gas is 232.27 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!

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

A balloon contains \(160 \mathrm{L}\) of nitrogen gas at \(25^{\circ} \mathrm{C}\) and 1.0 atm. How much energy must be added to raise the temperature of the nitrogen to \(45^{\circ} \mathrm{C}\) while allowing the balloon to expand at atmospheric pressure?
A heat pump delivers heat at a rate of \(7.81 \mathrm{kW}\) for $10.0 \mathrm{h} .\( If its coefficient of performance is \)6.85,$ how much heat is taken from the cold reservoir during that time?
An ideal gas is in contact with a heat reservoir so that it remains at a constant temperature of \(300.0 \mathrm{K}\). The gas is compressed from a volume of \(24.0 \mathrm{L}\) to a volume of 14.0 L. During the process, the mechanical device pushing the piston to compress the gas is found to expend \(5.00 \mathrm{kJ}\) of energy. How much heat flows between the heat reservoir and the gas and in what direction does the heat flow occur?

Show that the coefficient of performance for a reversible refrigerator is \(1 /\left[\left(T_{\mathrm{H}} / T_{\mathrm{C}}\right)-1\right]\).
A large block of copper initially at \(20.0^{\circ} \mathrm{C}\) is placed in a vat of hot water \(\left(80.0^{\circ} \mathrm{C}\right) .\) For the first $1.0 \mathrm{J}$ of heat that flows from the water into the block, find (a) the entropy change of the block, (b) the entropy change of the water, and (c) the entropy change of the universe. Note that the temperatures of the block and water are essentially unchanged by the flow of only \(1.0 \mathrm{J}\) of heat.
See all solutions

Recommended explanations on Physics Textbooks

Physics of Motion

Read Explanation

Force

Read Explanation

Waves Physics

Read Explanation

Atoms and Radioactivity

Read Explanation

Fluids

Read Explanation

Modern 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.