Student Solutions Manual for Ball's Physical Chemistry, 2nd
Student Solutions Manual for Ball's Physical Chemistry, 2nd
2nd Edition
ISBN: 9798214169019
Author: David W. Ball
Publisher: Cengage Learning US
Question
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Chapter 18, Problem 18.59E
Interpretation Introduction

Interpretation:

The set of equations (or a small program) to evaluate the constant-volume heat capacity for a moleculeis to be stated. The graph of the result is to be plotted. The trend for the same is to be stated. The heat capacity versus temperature (say from 298K to 1000K) for H2O and CH4 is to be determined using these algorithm.

Concept introduction:

The heat capacity at constant volume for nonlinear polyatomic molecule is given by the formula,

CV=Nk[32+32+j=13N*6(θjT)2eθj/T(1eθj/T)2]

Where,

N is the Avogadro number.

T is the temperature.

k is the Boltzmann constant.

θj is the rotational temperature.

Expert Solution & Answer
Check Mark

Answer to Problem 18.59E

The set of equations (or a small program) to evaluate the constant-volume heat capacity for a molecule are,

Nk

(θjT)2

eθj/T

1eθj/T

(1eθj/T)2

eθj/T(1eθj/T)2

(θjT)2eθj/T(1eθj/T)2

The plot between CV and T for CH4 is,

Student Solutions Manual for Ball's Physical Chemistry, 2nd, Chapter 18, Problem 18.59E , additional homework tip  1

The plot between CV and T for H2O is,

Student Solutions Manual for Ball's Physical Chemistry, 2nd, Chapter 18, Problem 18.59E , additional homework tip  2

Explanation of Solution

The heat capacity at constant volume for nonlinear polyatomic molecule is given by the formula,

CV=Nk[32+32+j=13N*6(θjT)2eθj/T(1eθj/T)2]…(1)

The set of equations(or a small program) to evaluate the constant-volume heat capacity for a molecule are shown below.

Nk

(θjT)2

eθj/T

1eθj/T

(1eθj/T)2

eθj/T(1eθj/T)2

(θjT)2eθj/T(1eθj/T)2

The vibrational temperatures for CH4 are 7.54K.

Substitute the value of vibrational temperatures for CH4 and 298K in equation (1).

CV=(6.02×1023mol1)(1.381×1023JK1){32+32+(7.54K298K)2exp(7.54K298K)[1exp(7.54K298K)]2}

The value of CV at different temperature is shown below.

T(θjT)2eθj/T(1eθj/T)2CV

29888804243994320275925348121104453765637707923398158404776330964513102448200704124631171.04E+08

498248004190297021.58E+08548300304279020972.32E+08598357604395657453.29E+08648419904545524984.53E+08

698487204734406156.1E+08748559504968547698.05E+087986368041.25E+081.04E+098487191041.6E+081.33E+09

8988064042.01E+081.67E+099488987042.5E+082.08E+099989960043.07E+082.55E+09

The plot between CV and T is shown below.

Student Solutions Manual for Ball's Physical Chemistry, 2nd, Chapter 18, Problem 18.59E , additional homework tip  3

Figure 1

The three vibrational temperatures for H2O are 13.4K, 20.9K and 40.1K.

Substitute the value of vibrational temperatures for H2O and 298K in equation (1).

CV=(6.02×1023mol1)(1.381×1023JK1){32+32+(13.4K298K)2exp(13.4K298K)[1exp(13.4K298K)]2+(20.9K298K)2exp(20.9K298K)[1exp(20.9K298K)]2+(40.1K298K)2exp(40.1K298K)[1exp(40.1K298K)]2}

The value of CV at different temperature is shown below.

T(θjT)2eθj/T(1eθj/T)2CV

298    2.997914     24.912674482.99907724.922335982.99948224.925697482.99966924.92725

898      2.99977       24.92809

The plot between CV and T for H2O is shown below.

Student Solutions Manual for Ball's Physical Chemistry, 2nd, Chapter 18, Problem 18.59E , additional homework tip  4

Figure 2

Conclusion

The set of equations (or a small program) to evaluate the constant-volume heat capacity for a molecule are,

Nk

(θjT)2

eθj/T

1eθj/T

(1eθj/T)2

eθj/T(1eθj/T)2

(θjT)2eθj/T(1eθj/T)2

The plot between CV and T for CH4 and H2O is shown in Figure 1 and Figure 2.

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Chapter 18 Solutions

Student Solutions Manual for Ball's Physical Chemistry, 2nd

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