The total contribution to the molar internal energy of gaseous C O 2 molecules has to be estimated. Concept Introduction: According to equipartition theorem, each translational and rotational degree of freedom contributes 1 / 2 R T to the molar internal energy and each active vibrational degree of freedom contributes R T to the molar internal energy.

Question

Want to see more full solutions like this?

Answer 1

Question

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

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 (3N−5) 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×(3×3−5)RT E(CO2)= 3RT2+RT+8RT E(CO2)=(212)×8.314 JK−1mol−1×2000 KE(CO2)=175 kJK−1mol−1.

(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

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 (3N−5) 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×(3×4−5)RT E(C2Br2)= 3RT2+RT+14RT E(C2Br2)=(332)×8.314 JK−1mol−1×2000 KE(C2Br2)=274 kJK−1mol−1.

(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

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 (3N−6) 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×(3×3−6)RT E(NO2)= 3RT2+32RT+6RT E(NO2)=9×8.314 JK−1mol−1×2000 KE(NO2)=150 kJK−1mol−1.

(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

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 (3N−6) 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×(3×6−6)RT E(C2Br4)= 3RT2+32RT+24RT E(C2Br4)=27×8.314 JK−1mol−1×2000 KE(C2Br4)=449 kJK−1mol−1.

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!

Students have asked these similar questions

QUESTION: Fill in the answers in the empty green boxes 1. Step 2 2. Step 3 3. Step 4 (SUM) 4. Step 5 (df) (GIVEN) 5. Determine S y/x value *The data values have been provided in the worksheet attached in the first image*

If the symbol A is placed in a reaction, at what temperature does it take place?

By malonic or acetylacetic synthesis, synthesize 3-methyl-4-oxopentanoic acid (indicate the formulas of the compounds).

Answer 2

Question

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

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 (3N−5) 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×(3×3−5)RT E(CO2)= 3RT2+RT+8RT E(CO2)=(212)×8.314 JK−1mol−1×2000 KE(CO2)=175 kJK−1mol−1.

(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

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 (3N−5) 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×(3×4−5)RT E(C2Br2)= 3RT2+RT+14RT E(C2Br2)=(332)×8.314 JK−1mol−1×2000 KE(C2Br2)=274 kJK−1mol−1.

(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

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 (3N−6) 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×(3×3−6)RT E(NO2)= 3RT2+32RT+6RT E(NO2)=9×8.314 JK−1mol−1×2000 KE(NO2)=150 kJK−1mol−1.

(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

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 (3N−6) 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×(3×6−6)RT E(C2Br4)= 3RT2+32RT+24RT E(C2Br4)=27×8.314 JK−1mol−1×2000 KE(C2Br4)=449 kJK−1mol−1.

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!

Answer 3

Question

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

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 (3N−5) 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×(3×3−5)RT E(CO2)= 3RT2+RT+8RT E(CO2)=(212)×8.314 JK−1mol−1×2000 KE(CO2)=175 kJK−1mol−1.

(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

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 (3N−5) 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×(3×4−5)RT E(C2Br2)= 3RT2+RT+14RT E(C2Br2)=(332)×8.314 JK−1mol−1×2000 KE(C2Br2)=274 kJK−1mol−1.

(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

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 (3N−6) 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×(3×3−6)RT E(NO2)= 3RT2+32RT+6RT E(NO2)=9×8.314 JK−1mol−1×2000 KE(NO2)=150 kJK−1mol−1.

(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

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 (3N−6) 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×(3×6−6)RT E(C2Br4)= 3RT2+32RT+24RT E(C2Br4)=27×8.314 JK−1mol−1×2000 KE(C2Br4)=449 kJK−1mol−1.

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!

Answer 4

Question

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

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 (3N−5) 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×(3×3−5)RT E(CO2)= 3RT2+RT+8RT E(CO2)=(212)×8.314 JK−1mol−1×2000 KE(CO2)=175 kJK−1mol−1.

(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

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 (3N−5) 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×(3×4−5)RT E(C2Br2)= 3RT2+RT+14RT E(C2Br2)=(332)×8.314 JK−1mol−1×2000 KE(C2Br2)=274 kJK−1mol−1.

(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

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 (3N−6) 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×(3×3−6)RT E(NO2)= 3RT2+32RT+6RT E(NO2)=9×8.314 JK−1mol−1×2000 KE(NO2)=150 kJK−1mol−1.

(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

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 (3N−6) 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×(3×6−6)RT E(C2Br4)= 3RT2+32RT+24RT E(C2Br4)=27×8.314 JK−1mol−1×2000 KE(C2Br4)=449 kJK−1mol−1.

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!

Answer 5

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

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 (3N−5) 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×(3×3−5)RT E(CO2)= 3RT2+RT+8RT E(CO2)=(212)×8.314 JK−1mol−1×2000 KE(CO2)=175 kJK−1mol−1.

(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

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 (3N−5) 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×(3×4−5)RT E(C2Br2)= 3RT2+RT+14RT E(C2Br2)=(332)×8.314 JK−1mol−1×2000 KE(C2Br2)=274 kJK−1mol−1.

(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

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 (3N−6) 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×(3×3−6)RT E(NO2)= 3RT2+32RT+6RT E(NO2)=9×8.314 JK−1mol−1×2000 KE(NO2)=150 kJK−1mol−1.

(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

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 (3N−6) 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×(3×6−6)RT E(C2Br4)= 3RT2+32RT+24RT E(C2Br4)=27×8.314 JK−1mol−1×2000 KE(C2Br4)=449 kJK−1mol−1.

Answer 6

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

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 (3N−5) 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×(3×3−5)RT E(CO2)= 3RT2+RT+8RT E(CO2)=(212)×8.314 JK−1mol−1×2000 KE(CO2)=175 kJK−1mol−1.

Answer 7

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.

Answer 8

(a)

Expert Solution

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 (3N−5) 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×(3×3−5)RT E(CO2)= 3RT2+RT+8RT E(CO2)=(212)×8.314 JK−1mol−1×2000 KE(CO2)=175 kJK−1mol−1.

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

(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

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 (3N−5) 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×(3×4−5)RT E(C2Br2)= 3RT2+RT+14RT E(C2Br2)=(332)×8.314 JK−1mol−1×2000 KE(C2Br2)=274 kJK−1mol−1.

Answer 13

(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.

Answer 14

(b)

Expert Solution

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 (3N−5) 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×(3×4−5)RT E(C2Br2)= 3RT2+RT+14RT E(C2Br2)=(332)×8.314 JK−1mol−1×2000 KE(C2Br2)=274 kJK−1mol−1.

Answer 15

(b)

Expert Solution

Answer 16

(b)

Expert Solution

Answer 17

Expert Solution

Answer 18

(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

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 (3N−6) 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×(3×3−6)RT E(NO2)= 3RT2+32RT+6RT E(NO2)=9×8.314 JK−1mol−1×2000 KE(NO2)=150 kJK−1mol−1.

Answer 19

(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.

Answer 20

(c)

Expert Solution

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 (3N−6) 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×(3×3−6)RT E(NO2)= 3RT2+32RT+6RT E(NO2)=9×8.314 JK−1mol−1×2000 KE(NO2)=150 kJK−1mol−1.

Answer 21

(c)

Expert Solution

Answer 22

(c)

Expert Solution

Answer 23

Expert Solution

Answer 24

(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

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 (3N−6) 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×(3×6−6)RT E(C2Br4)= 3RT2+32RT+24RT E(C2Br4)=27×8.314 JK−1mol−1×2000 KE(C2Br4)=449 kJK−1mol−1.

Answer 25

(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.

Answer 26

(d)

Expert Solution

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 (3N−6) 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×(3×6−6)RT E(C2Br4)= 3RT2+32RT+24RT E(C2Br4)=27×8.314 JK−1mol−1×2000 KE(C2Br4)=449 kJK−1mol−1.

Answer 27

(d)

Expert Solution

Answer 28

(d)

Expert Solution

Answer 29

Expert Solution

Answer 30

Ch. 2 - Prob. 2A.1ST Ch. 2 - Prob. 2A.2ST Ch. 2 - Prob. 2B.1ST Ch. 2 - Prob. 2B.2ST Ch. 2 - Prob. 2B.3ST Ch. 2 - Prob. 2B.4ST Ch. 2 - Prob. 2B.5ST Ch. 2 - Prob. 2C.1ST Ch. 2 - Prob. 2C.2ST Ch. 2 - Prob. 2D.1ST

Explanation of Solution

Explanation of Solution

Explanation of Solution

Explanation of Solution

Want to see more full solutions like this?

Chapter 2 Solutions