Elements Of Physical Chemistry
Elements Of Physical Chemistry
7th Edition
ISBN: 9780198727873
Author: ATKINS, P. W. (peter William), De Paula, Julio
Publisher: Oxford University Press
Question
Book Icon
Chapter 2, Problem 2.5P

(a)

Interpretation Introduction

Interpretation:

The total contribution to the molar internal energy of gaseous CO2 molecules has to be estimated.

Concept Introduction:

According to equipartition theorem, each translational and rotational degree of freedom contributes 1/2RT to the molar internal energy and each active vibrational degree of freedom contributes RT to the molar internal energy.

(a)

Expert Solution
Check Mark

Explanation of Solution

For all atoms and molecules, there are three translational degrees of freedom. The average molar translational energy is therefore Etrans=3/2RT.

For linear molecules rotate about two axes, perpendicular to the internuclear axis. The average molar rotational energy is therefore Erot=2×1/2RT=RT.

For linear molecules, there are (3N5) vibrational modes.

Consider the molecule CO2.

CO2 with N=3 atoms, the total molar energy is therefore

  E(CO2)=Etrans+Erot+EvibE(CO2)=3RT2+RT+2×(3×35)RTE(CO2)=3RT2+RT+8RTE(CO2)=(212)×8.314JK1mol1×2000KE(CO2)=175kJK1mol1.

(b)

Interpretation Introduction

Interpretation:

The total contribution to the molar internal energy of gaseous C2Br2 molecules has to be estimated.

Concept Introduction:

According to equipartition theorem, each translational and rotational degree of freedom contributes 1/2RT to the molar internal energy and each active vibrational degree of freedom contributes RT to the molar internal energy.

(b)

Expert Solution
Check Mark

Explanation of Solution

For all atoms and molecules, there are three translational degrees of freedom. The average molar translational energy is therefore Etrans=3/2RT.

For linear molecules rotate about two axes, perpendicular to the internuclear axis. The average molar rotational energy is therefore Erot=2×1/2RT=RT.

For linear molecules, there are (3N5) vibrational modes.

Consider the molecule C2Br2.

C2Br2 with N=4 atoms, the total molar energy is therefore

  E(C2Br2)=Etrans+Erot+EvibE(C2Br2)=3RT2+RT+2×(3×45)RTE(C2Br2)=3RT2+RT+14RTE(C2Br2)=(332)×8.314JK1mol1×2000KE(C2Br2)=274kJK1mol1.

(c)

Interpretation Introduction

Interpretation:

The total contribution to the molar internal energy of gaseous NO2 molecules has to be estimated.

Concept Introduction:

According to equipartition theorem, each translational and rotational degree of freedom contributes 1/2RT to the molar internal energy and each active vibrational degree of freedom contributes RT to the molar internal energy.

(c)

Expert Solution
Check Mark

Explanation of Solution

For all atoms and molecules, there are three translational degrees of freedom. The average molar translational energy is therefore Etrans=3/2RT.

For nonlinear molecules rotate about three axes. The average molar rotational energy is therefore Erot=3×1/2RT=3/2RT.

For nonlinear molecules, there are (3N6) vibrational modes.

Consider the molecule NO2.

NO2 with N=3 atoms, the total molar energy is therefore

  E(NO2)=Etrans+Erot+EvibE(NO2)=3RT2+32RT+2×(3×36)RTE(NO2)=3RT2+32RT+6RTE(NO2)=9×8.314JK1mol1×2000KE(NO2)=150kJK1mol1.

(d)

Interpretation Introduction

Interpretation:

The total contribution to the molar internal energy of gaseous C2Br4 molecules has to be estimated.

Concept Introduction:

According to equipartition theorem, each translational and rotational degree of freedom contributes 1/2RT to the molar internal energy and each active vibrational degree of freedom contributes RT to the molar internal energy.

(d)

Expert Solution
Check Mark

Explanation of Solution

For all atoms and molecules, there are three translational degrees of freedom. The average molar translational energy is therefore Etrans=3/2RT.

For nonlinear molecules rotate about three axes. The average molar rotational energy is therefore Erot=3×1/2RT=3/2RT.

For nonlinear molecules, there are (3N6) vibrational modes.

Consider the molecule C2Br4.

C2Br4 with N=6 atoms, the total molar energy is therefore

  E(C2Br4)=Etrans+Erot+EvibE(C2Br4)=3RT2+32RT+2×(3×66)RTE(C2Br4)=3RT2+32RT+24RTE(C2Br4)=27×8.314JK1mol1×2000KE(C2Br4)=449kJK1mol1.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Chapter 2 Solutions

Elements Of Physical Chemistry

Ch. 2 - Prob. 2D.2STCh. 2 - Prob. 2E.1STCh. 2 - Prob. 2E.2STCh. 2 - Prob. 2E.3STCh. 2 - Prob. 2F.1STCh. 2 - Prob. 2F.2STCh. 2 - Prob. 2F.3STCh. 2 - Prob. 2F.4STCh. 2 - Prob. 2F.5STCh. 2 - Prob. 2F.6STCh. 2 - Prob. 2A.2ECh. 2 - Prob. 2A.3ECh. 2 - Prob. 2A.4ECh. 2 - Prob. 2A.5ECh. 2 - Prob. 2A.6ECh. 2 - Prob. 2A.7ECh. 2 - Prob. 2A.8ECh. 2 - Prob. 2B.1ECh. 2 - Prob. 2B.2ECh. 2 - Prob. 2B.3ECh. 2 - Prob. 2B.4ECh. 2 - Prob. 2B.5ECh. 2 - Prob. 2C.1ECh. 2 - Prob. 2C.2ECh. 2 - Prob. 2D.1ECh. 2 - Prob. 2D.2ECh. 2 - Prob. 2D.3ECh. 2 - Prob. 2D.4ECh. 2 - Prob. 2D.5ECh. 2 - Prob. 2D.6ECh. 2 - Prob. 2E.1ECh. 2 - Prob. 2E.2ECh. 2 - Prob. 2E.3ECh. 2 - Prob. 2E.4ECh. 2 - Prob. 2E.5ECh. 2 - Prob. 2E.6ECh. 2 - Prob. 2E.7ECh. 2 - Prob. 2E.8ECh. 2 - Prob. 2E.9ECh. 2 - Prob. 2F.1ECh. 2 - Prob. 2F.2ECh. 2 - Prob. 2F.3ECh. 2 - Prob. 2F.4ECh. 2 - Prob. 2F.5ECh. 2 - Prob. 2F.6ECh. 2 - Prob. 2F.7ECh. 2 - Prob. 2F.8ECh. 2 - Prob. 2F.9ECh. 2 - Prob. 2F.10ECh. 2 - Prob. 2.1DQCh. 2 - Prob. 2.2DQCh. 2 - Prob. 2.3DQCh. 2 - Prob. 2.4DQCh. 2 - Prob. 2.5DQCh. 2 - Prob. 2.6DQCh. 2 - Prob. 2.7DQCh. 2 - Prob. 2.8DQCh. 2 - Prob. 2.9DQCh. 2 - Prob. 2.10DQCh. 2 - Prob. 2.11DQCh. 2 - Prob. 2.12DQCh. 2 - Prob. 2.13DQCh. 2 - Prob. 2.14DQCh. 2 - Prob. 2.15DQCh. 2 - Prob. 2.16DQCh. 2 - Prob. 2.1PCh. 2 - Prob. 2.2PCh. 2 - Prob. 2.3PCh. 2 - Prob. 2.4PCh. 2 - Prob. 2.5PCh. 2 - Prob. 2.6PCh. 2 - Prob. 2.7PCh. 2 - Prob. 2.8PCh. 2 - Prob. 2.9PCh. 2 - Prob. 2.10PCh. 2 - Prob. 2.12PCh. 2 - Prob. 2.13PCh. 2 - Prob. 2.14PCh. 2 - Prob. 2.15PCh. 2 - Prob. 2.16PCh. 2 - Prob. 2.17PCh. 2 - Prob. 2.18PCh. 2 - Prob. 2.19PCh. 2 - Prob. 2.20PCh. 2 - Prob. 2.21PCh. 2 - Prob. 2.22PCh. 2 - Prob. 2.23PCh. 2 - Prob. 2.25PCh. 2 - Prob. 2.1PRCh. 2 - Prob. 2.2PRCh. 2 - Prob. 2.3PRCh. 2 - Prob. 2.4PRCh. 2 - Prob. 2.5PRCh. 2 - Prob. 2.6PRCh. 2 - Prob. 2.8PRCh. 2 - Prob. 2.9PRCh. 2 - Prob. 2.10PR
Knowledge Booster
Background pattern image
Recommended textbooks for you
Text book image
Chemistry
Chemistry
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Cengage Learning
Text book image
Chemistry
Chemistry
ISBN:9781259911156
Author:Raymond Chang Dr., Jason Overby Professor
Publisher:McGraw-Hill Education
Text book image
Principles of Instrumental Analysis
Chemistry
ISBN:9781305577213
Author:Douglas A. Skoog, F. James Holler, Stanley R. Crouch
Publisher:Cengage Learning
Text book image
Organic Chemistry
Chemistry
ISBN:9780078021558
Author:Janice Gorzynski Smith Dr.
Publisher:McGraw-Hill Education
Text book image
Chemistry: Principles and Reactions
Chemistry
ISBN:9781305079373
Author:William L. Masterton, Cecile N. Hurley
Publisher:Cengage Learning
Text book image
Elementary Principles of Chemical Processes, Bind...
Chemistry
ISBN:9781118431221
Author:Richard M. Felder, Ronald W. Rousseau, Lisa G. Bullard
Publisher:WILEY