Elements Of Physical Chemistry
Elements Of Physical Chemistry
7th Edition
ISBN: 9780198796701
Author: ATKINS, P. W. (peter William), De Paula, Julio
Publisher: Oxford University Press
Question
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Chapter 2, Problem 2.6PR

(a)

Interpretation Introduction

Interpretation:

For chlorine the temperature at which the equipartition theory becomes valid has to be calculated.

Concept introduction:

Equipartition theorem:

According to the equipartition theorem, the total energy of a molecule is divided equally amongst the various degrees of freedom of the molecules.

(a)

Expert Solution
Check Mark

Answer to Problem 2.6PR

The temperature is 805.87K.

Explanation of Solution

According to the equipartition theorem,

Energy contribution by each of the translational degrees of freedom is 12kBT per molecule.

Energy contribution by each of the rotational degrees of freedom is 12kBT per molecule.

For vibrational degrees of freedom, due to collisions between the molecules each vibrational degrees of freedom possess both kinetic energy and potential energy i.e. each vibration involves two degrees of freedom and hence the energy contribution by each of the vibrational degrees of freedom is given by,

  Ev=P.E+K.E.Ev=12kBT+12kBTEv=kBT

Hence each vibrational degrees of freedom contributes kBT energy per molecule.

Now, the equipartition theorem is valid only if kBT is very much greater than the separation between the energy levels of the mode of the motion Δε.  Translational and rotational energy levels are sufficiently close together for most of the molecules at room temperature.

But the separation between the vibrational energy levels is much greater than kBT and as a result for vibrational degrees of freedom the contribution in total energy has to be calculated at higher temperature only.

It is given in the question that the energy from vibration is given by Δε1-e-ΔεkT.

According to the question for chlorine the separation between the vibrational energy levels is,

  Δε=6.70kJmol-1

Now the temperature at which equipartition theorem for chlorine is valid that has to be calculated.

Hence,

  Δε1-e-ΔεkT=kTΔεkT=1-e-ΔεkT

Chlorine is linear diatomic molecule and hence it will have degrees of freedom (3×2-5)=1

Now at higher temperature only vibrational degrees of freedom of chlorine can show equipartition

Now by expansion, e-ΔεkT=1ΔεkT+(ΔεkT)2(ΔεkT)3+....

But higher temperature all the higher terms can be neglected.

Hence the temperature is, T=Evk

Vibrational energy per mole, Ev=RT

Hence temperature is,

  Ev=RTT=6.70×103Jmol-18.314Jmol-1K-1T=805.87K

Hence the temperature is 805.87K.

(b)

Interpretation Introduction

Interpretation:

For a given exponential ex=1+x+x22!+x33!+.... it has to be shown that the exact expression reduced to the result obtained from the equipartition theorem.

Concept introduction:

Equipartition theorem:

According to the equipartition theorem, the total energy of a molecule is divided equally amongst the various degrees of freedom of the molecules.

(b)

Expert Solution
Check Mark

Explanation of Solution

According to the equipartition theorem,

Energy contribution by each of the translational degrees of freedom is 12kBT per molecule.

Energy contribution by each of the rotational degrees of freedom is 12kBT per molecule.

For vibrational degrees of freedom, due to collisions between the molecules each vibrational degrees of freedom possess both kinetic energy and potential energy i.e. each vibration involves two degrees of freedom and hence the energy contribution by each of the vibrational degrees of freedom is given by,

  Ev=P.E+K.E.Ev=12kBT+12kBTEv=kBT

Hence each vibrational degrees of freedom contributes kBT energy per molecule.

Now, the equipartition theorem is valid only if kBT is very much greater than the separation between the energy levels of the mode of the motion Δε.  Translational and rotational energy levels are sufficiently close together for most of the molecules at room temperature.

But the separation between the vibrational energy levels is much greater than kBT and as a result for vibrational degrees of freedom the contribution in total energy has to be calculated at higher temperature only.

It is given in the question that the energy from vibration is given by Δε1-e-ΔεkT.

Now the exponential form given is,

  ex=1+x+x22!+x33!+....

Hence expanding e-ΔεkT like the given exponential form,

  e-ΔεkT=1ΔεkT+(ΔεkT)2(ΔεkT)3+....

Now using this expansion

  Ev=Δε1-e-ΔεkTEv=Δε1-(1ΔεkT+(ΔεkT)2(ΔεkT)3+....)

Now only at higher temperature vibrational degrees of freedom gives full contribution to the total energy.

Now at higher temperature all the higher terms of the expansion will be very much lesser than one and become negligible

Hence the equation becomes,

  Ev=Δε1-(1ΔεkT)Ev=Δε1-1+ΔεkTEv=ΔεΔεkTEv=kT

Thus at higher temperature the exact expression for vibrational degrees of freedom reduces to the result obtained from equipartition theorem.

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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
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