Physical Chemistry
Physical Chemistry
2nd Edition
ISBN: 9781133958437
Author: Ball, David W. (david Warren), BAER, Tomas
Publisher: Wadsworth Cengage Learning,
bartleby

Concept explainers

bartleby

Videos

Textbook Question
Book Icon
Chapter 17, Problem 17.56E

Use the Sackur-Tetrode equation to derive the relationship Δ S = R ln ( V 2 / V 1 ) for an isothermal change and Δ S = C v ln ( T 2 / T 1 ) for an isochoric change.

Expert Solution & Answer
Check Mark
Interpretation Introduction

Interpretation:

The relationship ΔS=Rln(V2/V1) for an isothermal change and ΔS=Cvln(T2/T1) for an isochoric change is to be derived using Sackur-Tetrode equation.

Concept introduction:

The Sackur-Tetrode equation is given below.

S=Nk{ln[(2πmkTh2)3/2kTp]+52}

Where,

T is the temperature.

m is the mass.

N is the number of particles in a system.

This is the best way to calculate the absolute entropies using the statistical thermodynamics applied to gaseous systems.

Answer to Problem 17.56E

The relationship ΔS=Rln(V2/V1) for an isothermal change and ΔS=Cvln(T2/T1) for an isochoric change has been derived using Sackur-Tetrode equation.

Explanation of Solution

The Sackur-Tetrode equation is given below.

S=Nk{ln[(2πmkTh2)3/2kTp]+52}

Where,

T is the temperature.

m is the mass.

N is the number of particles in a system.

For an isothermal change ΔT=0. The change in entropy can be represented as,

ΔS=S2S1=Nk{ln[(2πmkTh2)3/2kTp2]+52}Nk{ln[(2πmkTh2)3/2kTp1]+52}=Nkln[(2πmkTh2)3/2kTp2]Nkln[(2πmkTh2)3/2kTp1]

The pressure can be substituted in terms of P=NkTV using ideal gas equation as shown below.

ΔS=Nkln[(2πmkTh2)3/2kTV2NkT]Nkln[(2πmkTh2)3/2kTV1NkT]=Nkln[(2πmkTh2)3/2V2N]Nkln[(2πmkTh2)3/2V1N]

The given equation is rearranged as shown below.

ΔS=Nk[ln(2πmkTh2)3/21N+lnV2]Nk[ln(2πmkTh2)3/21N+lnV1]

The common constant terms are cancelled as shown below.

ΔS=Nkln(2πmkTh2)3/21N+NklnV2Nkln(2πmkTh2)3/21NNklnV1ΔS=NklnV2NklnV1

The term NkR, that is, the product of number of particles and Boltzmann constant is equivalent to gas constant. The identity of logarithm (lnAB=lnAlnB) is used to modify the given equation. The equation becomes,

ΔS=RlnV2RlnV1ΔS=Rln(V2V1)

Thus, the given equation represents the relationship between entropy and volume at isothermal conditions.

The Sackur-Tetrode equation is given below.

S=Nk{ln[(2πmkTh2)3/2kTp]+52}

For an isochoric change ΔV=0. The pressure can be substituted in terms of P=NkTV using ideal gas equation as shown below.

The change in entropy can be represented as,

ΔS=S2S1ΔS=Nkln[(2πmkT2h2)3/2kT2VNkT2]Nkln[(2πmkT1h2)3/2kT1VNkT1]ΔS=Nkln[(2πmkT2h2)3/2VN]Nkln[(2πmkT1h2)3/2VN]

The given equation is rearranged as shown below.

ΔS=Nk[ln[{(2πmkh2)3/2VN}]+ln(T2)3/2]Nk[ln{(2πmkh2)3/2VN}+ln(T1)3/2]

The common constant terms are cancelled as shown below.

ΔS=Nkln[(2πmkh2)3/2VN]+Nkln(T2)3/2Nkln[(2πmkh2)3/2VN]Nkln(T1)3/2ΔS=Nkln(T2)3/2Nkln(T1)3/2

The term NkR, that is, the product of number of particles and Boltzmann constant is equivalent to gas constant. The identity of logarithm (lnAB=lnAlnB) is used to modify the given equation. The equation becomes,

ΔS=Rln(T2)3/2Rln(T1)3/2ΔS=Rln(T2T1)3/2ΔS=32Rln(T2T1)

For a monoatomic gas Cv=32R, is substituted in the given equation.

ΔS=Cvln(T2T1)

Thus, given equation represents the relationship between entropy and temperature at isochoric conditions.

Conclusion

The relationship ΔS=Rln(V2/V1) for an isothermal change and ΔS=Cvln(T2/T1) for an isochoric change has been derived using Sackur-Tetrode equation.

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
We know that trivalent cations (Cr3+, Mn3+, V3+) with a large difference between octahedral and tetrahedral EECC, form exclusively normal spinels. Bivalent cations (Ni2+ and Cu2+) with high EECC, form inverse spinels. Is this statement correct?
(b) Draw the product A that would be formed through the indicated sequence of steps from the given starting material. MeO (1) Br₂, hv (2) NaOEt, EtOH, A (3) BH3:THF (4) H₂O2, HO- B H₂C. CH₂ OH Edit
Small changes in secondary; tertiary primary; secondary primary; tertiary tertiary; secondary protein structure may lead to big changes in protein structures.

Chapter 17 Solutions

Physical Chemistry

Knowledge Booster
Background pattern image
Chemistry
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, chemistry and related others by exploring similar questions and additional content below.
Recommended textbooks for you
Text book image
Chemistry: Principles and Practice
Chemistry
ISBN:9780534420123
Author:Daniel L. Reger, Scott R. Goode, David W. Ball, Edward Mercer
Publisher:Cengage Learning
Text book image
Chemistry: The Molecular Science
Chemistry
ISBN:9781285199047
Author:John W. Moore, Conrad L. Stanitski
Publisher:Cengage Learning
Text book image
Chemistry & Chemical Reactivity
Chemistry
ISBN:9781133949640
Author:John C. Kotz, Paul M. Treichel, John Townsend, David Treichel
Publisher:Cengage Learning
Text book image
Chemistry for Engineering Students
Chemistry
ISBN:9781337398909
Author:Lawrence S. Brown, Tom Holme
Publisher:Cengage Learning
Text book image
Principles of Modern Chemistry
Chemistry
ISBN:9781305079113
Author:David W. Oxtoby, H. Pat Gillis, Laurie J. Butler
Publisher:Cengage Learning
Text book image
Chemistry & Chemical Reactivity
Chemistry
ISBN:9781337399074
Author:John C. Kotz, Paul M. Treichel, John Townsend, David Treichel
Publisher:Cengage Learning
Solutions: Crash Course Chemistry #27; Author: Crash Course;https://www.youtube.com/watch?v=9h2f1Bjr0p4;License: Standard YouTube License, CC-BY