The plot of molar heat capacity of a collection of harmonic oscillators as a function of T / θ V is to be plotted. Also, the vibrational heat capacity of ethyne at 298 K is to be calculated. Concept introduction: According to the equipartition theorem, the contribution of each degree of freedom to the heat capacity is 1 2 R . The molar heat capacity is calculated by using the formula shown below. C V , m / R = f 2

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Answer 1

Question

Chapter 13, Problem 13E.4AE

(i)

Interpretation Introduction

Interpretation:

The plot of molar heat capacity of a collection of harmonic oscillators as a function of $T/θV$ is to be plotted. Also, the vibrational heat capacity of ethyne at $298 K$ is to be calculated.

Concept introduction:

According to the equipartition theorem, the contribution of each degree of freedom to the heat capacity is $12 R$ . The molar heat capacity is calculated by using the formula shown below.

$CV,m/R=f2$

(i)

Expert Solution

Answer to Problem 13E.4AE

The plot of molar heat capacity of a collection of harmonic oscillators as a function of $x=T/θV$ is shown below.

Atkins' Physical Chemistry, Chapter 13, Problem 13E.4AE , additional homework tip 1

The vibrational heat capacity of ethyne is $14.93 J/mol⋅K_$ .

Explanation of Solution

The molar heat capacity is calculated by using the formula shown below.

$CV,m/R=f2$ (1)

The value of $f$ is expressed by the formula shown below.

$f=(θVT)×(e−θV/2T1−e−θV/T)$ (2)

Therefore, the value of molar heat capacity is calculated by using the formula shown below.

$CV,m/R=((θVT)×(e−θV/2T1−e−θV/T))2$ (3)

Now, first the value of $θV$ is calculate by using the formula shown below.

$θV=hcν¯k$ (4)

Where,

$h$ is the Planck constant $(6.626×10−34 J⋅s)$ .
$c$ is the speed of light $(2.998×108 m/s=2.998×1010 cm/s)$ .
$ν¯$ is the wavenumber.
$k$ is the Boltzmann constant with value $1.38×10−23 J/K$ .

The values of $θV$ for different values of wavenumber for the given data are to be calculated by using equation (4) as shown below.

For $ν¯=612 cm−1$ .

$θV=6.626×10−34 J⋅s×2.998×1010cm/s×612 cm−11.38×10−23 J/K=1.215×10−201.38×10−23K=880.4 K$

Therefore, the value of $θV/T$ is calculated as shown below.

$θV/T=880.4 K298 K=2.96$

For $ν¯=729 cm−1$ .

$θV=6.626×10−34 J⋅s×2.998×1010cm/s×729 cm−11.38×10−23 J/K=1.448×10−201.38×10−23K=1049.275 K$

Therefore, the value of $θV/T$ is calculated as shown below.

$θV/T=1049.275 K298 K=3.52$

For $ν¯=1974 cm−1$ .

$θV=6.626×10−34 J⋅s×2.998×1010cm/s×1974 cm−11.38×10−23 J/K=3.92×10−201.38×10−23K=2840.57 K$

Therefore, the value of $θV/T$ is calculated as shown below.

$θV/T=2840.57 K298 K=9.53$

For $ν¯=3287 cm−1$ .

$θV=6.626×10−34 J⋅s×2.998×1010cm/s×3287 cm−11.38×10−23 J/K=6.530×10−201.38×10−23K=4731.88 K$

Therefore, the value of $θV/T$ is calculated as shown below.

$θV/T=4731.88 K298 K=15.88$

For $ν¯=3374 cm−1$ .

$θV=6.626×10−34 J⋅s×2.998×1010cm/s×3374 cm−11.38×10−23 J/K=6.702×10−201.38×10−23K=4856.52 K$

Therefore, the value of $θV/T$ is calculated as shown below.

$θV/T=4856.52 K298 K=16.30$

Now, substitute the value of $θV/T$ in equation (3) to calculate the molar heat capacity when $ν¯=612 cm−1$ .

$CV,m/R=((2.96)×(e−1.4771−e−2.96))2=((2.96)×(0.2281−0.051))2=(2.96×0.2280.949)2=0.505$

Similarly, substitute the value of $θV/T$ in equation (3) to calculate the molar heat capacity when $ν¯=729 cm−1$ .

$CV,m/R=((3.52)×(e−1.7601−e−3.52))2=((3.52)×(0.1721−0.03))2=(3.52×0.1720.97)2=0.389$

Similarly, substitute the value of $θV/T$ in equation (3) to calculate the molar heat capacity when $ν¯=1974 cm−1$ .

$CV,m/R=((9.53)×(e−4.7661−e−9.53))2=((9.53)×(8.514×10−31−(7.263×10−5)))2=(9.53×8.514×10−30.999)2=0.007$

Similarly, substitute the value of $θV/T$ in equation (3) to calculate the molar heat capacity when $ν¯=3287 cm−1$ .

$CV,m/R=((15.88)×(e−7.941−e−15.88))2=((15.88)×(3.562×10−41−(1.268×10−7)))2=(15.88×3.562×10−40.999)2=3.20×10−5$

Similarly, substitute the value of $θV/T$ in equation (3) to calculate the molar heat capacity when $ν¯=3374 cm−1$ .

$CV,m/R=((16.30)×(e−8.141−e−16.30))2=((16.30)×(3×10−41−(8.3368×10−8)))2=(16.30×3×10−40.999)2=2.3×10−5$

Therefore, the values of $ν¯(cm−1)$ , $θV/T$ and $CV,m/R$ at $298 K$ are shown below in the Table.

$ν¯(cm−1)$	$θV/T$	$CV,m/R$
$612$	$2.96$	$0.505$
$612$	$2.96$	$0.505$
$729$	$3.52$	$0.389$
$729$	$3.52$	$0.389$
$1974$	$9.53$	$0.007$
$3287$	$15.88$	$3.20×10−5$
$3374$	$16.30$	$2.3×10−5$

Now, the plot of molar heat capacity of a collection of harmonic oscillators as a function of $x=T/θV$ is shown below.

Atkins' Physical Chemistry, Chapter 13, Problem 13E.4AE , additional homework tip 2

Figure 1

The vibrational heat capacity of ethyne is calculated by the formula shown below.

$CV,m/R$ (5)

The value of gas constant is $8.314 J/mol⋅K$ .

The sum of all the values of $CV,m/R$ at different wavenumbers is calculated as shown below.

$CV,m/R=0.505+0.505+0.389+0.389+0.007+3.20×10−5+2.3×10−5CV,m/R=1.7960$

Therefore, the vibrational heat capacity of ethyne is calculated as shown below.

$CV,m=1.7960×R=1.7960×8.314 J/mol⋅K=14.93 J/mol⋅K_$

The vibrational heat capacity of ethyne is $14.93 J/mol⋅K_$ .

(ii)

Interpretation Introduction

Interpretation:

The plot of molar heat capacity of a collection of harmonic oscillators as a function of $T/θV$ is to be plotted. Also, the vibrational heat capacity of ethyne at $500 K$ is to be calculated.

Concept introduction:

According to the equipartition theorem, the contribution of each degree of freedom to the heat capacity is $12 R$ . The molar heat capacity is calculated by using the formula shown below.

$CV,m/R=f2$

(ii)

Expert Solution

Answer to Problem 13E.4AE

The plot of molar heat capacity of a collection of harmonic oscillators as a function of $x=T/θV$ is shown below.

Atkins' Physical Chemistry, Chapter 13, Problem 13E.4AE , additional homework tip 3

The vibrational heat capacity of ethyne is $25.465 J/mol⋅K_$ .

Explanation of Solution

The molar heat capacity is calculated by using the formula shown below.

$CV,m/R=f2$ (1)

The value of $f$ is expressed by the formula shown below.

$f=(θVT)×(e−θV/2T1−e−θV/T)$ (2)

Therefore, the value of molar heat capacity is calculated by using the formula shown below.

$CV,m/R=((θVT)×(e−θV/2T1−e−θV/T))2$ (3)

Now, first the value of $θV$ is calculate by using the formula shown below.

$θV=hcν¯k$ (4)

Where,

$h$ is the Planck constant $(6.626×10−34 J⋅s)$ .
$c$ is the speed of light $(2.998×108 m/s=2.998×1010 cm/s)$ .
$ν¯$ is the wavenumber.
$k$ is the Boltzmann constant with value $1.38×10−23 J/K$ .

The values of $θV$ for different values of wavenumber for the given data are to be calculated by using equation (4) as shown below.

For $ν¯=612 cm−1$ .

$θV=6.626×10−34 J⋅s×2.998×1010cm/s×612 cm−11.38×10−23 J/K=1.215×10−201.38×10−23K=880.4 K$

Therefore, the value of $θV/T$ is calculated as shown below.

$θV/T=880.4 K500 K=1.76$

For $ν¯=729 cm−1$ .

$θV=6.626×10−34 J⋅s×2.998×1010cm/s×729 cm−11.38×10−23 J/K=1.448×10−201.38×10−23K=1049.275 K$

Therefore, the value of $θV/T$ is calculated as shown below.

$θV/T=1049.275 K500 K=2.09$

For $ν¯=1974 cm−1$ .

$θV=6.626×10−34 J⋅s×2.998×1010cm/s×1974 cm−11.38×10−23 J/K=3.92×10−201.38×10−23K=2840.57 K$

Therefore, the value of $θV/T$ is calculated as shown below.

$θV/T=2840.57 K500 K=5.68$

For $ν¯=3287 cm−1$ .

$θV=6.626×10−34 J⋅s×2.998×1010cm/s×3287 cm−11.38×10−23 J/K=6.530×10−201.38×10−23K=4731.88 K$

Therefore, the value of $θV/T$ is calculated as shown below.

$θV/T=4731.88 K500 K=9.46$

For $ν¯=3374 cm−1$ .

$θV=6.626×10−34 J⋅s×2.998×1010cm/s×3374 cm−11.38×10−23 J/K=6.702×10−201.38×10−23K=4856.52 K$

Therefore, the value of $θV/T$ is calculated as shown below.

$θV/T=4856.52 K500 K=9.71$

Now, substitute the value of $θV/T$ in equation (3) to calculate the molar heat capacity when $ν¯=612 cm−1$ .

$CV,m/R=((1.76)×(e−0.88041−e−1.76))2=((1.76)×(0.4141−0.1720))2=(1.76×0.4140.828)2=0.77$

Similarly, substitute the value of $θV/T$ in equation (3) to calculate the molar heat capacity when $ν¯=729 cm−1$ .

$CV,m/R=((2.09)×(e−1.0491−e−2.09))2=((2.09)×(0.3501−0.123))2=(2.09×0.3500.877)2=0.70$

Similarly, substitute the value of $θV/T$ in equation (3) to calculate the molar heat capacity when $ν¯=1974 cm−1$ .

$CV,m/R=((5.68)×(e−2.8401−e−5.68))2=((5.68)×(0.0581−(3.413×10−3)))2=(5.68×0.0580.99)2=0.110$

Similarly, substitute the value of $θV/T$ in equation (3) to calculate the molar heat capacity when $ν¯=3287 cm−1$ .

$CV,m/R=((9.46)×(e−4.7311−e−9.46))2=((9.46)×(8.817×10−31−(7.790×10−5)))2=(9.46×8.817×10−30.999)2=0.007$

Similarly, substitute the value of $θV/T$ in equation (3) to calculate the molar heat capacity when $ν¯=3374 cm−1$ .

$CV,m/R=((9.71)×(e−4.8561−e−9.71))2=((9.71)×(7.7815×10−31−(6.067×10−5)))2=(9.71×7.7815×10−30.999)2=5.7×10−3$

$CV,m/R≈0.006$

Therefore, the values of $ν¯(cm−1)$ , $θV/T$ and $CV,m/R$ at $500 K$ are shown below in the Table.

$ν¯(cm−1)$	$θV/T$	$CV,m/R$
$612$	$1.76$	$0.77$
$612$	$1.76$	$0.77$
$729$	$2.09$	$0.70$
$729$	$2.09$	$0.70$
$1974$	$5.68$	$0.110$
$3287$	$9.46$	$0.007$
$3374$	$9.71$	$0.006$

Now, the plot of molar heat capacity of a collection of harmonic oscillators as a function of $x=T/θV$ is shown below.

Atkins' Physical Chemistry, Chapter 13, Problem 13E.4AE , additional homework tip 4

Figure 2

The vibrational heat capacity of ethyne is calculated by the formula shown below.

$CV,m/R$ (5)

The value of gas constant is $8.314 J/mol⋅K$ .

The sum of all the values of $CV,m/R$ at different wavenumbers is calculated as shown below.

$CV,m/R=0.77+0.77+0.70+0.70+0.110+0.007+0.006CV,m/R=3.063$

Therefore, the vibrational heat capacity of ethyne is calculated as shown below.

$CV,m=3.063×R=3.063×8.314 J/mol⋅K=25.465 J/mol⋅K_$

The vibrational heat capacity of ethyne is $25.465 J/mol⋅K_$ .

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Answer 2

Question

Chapter 13, Problem 13E.4AE

(i)

Interpretation Introduction

Interpretation:

The plot of molar heat capacity of a collection of harmonic oscillators as a function of $T/θV$ is to be plotted. Also, the vibrational heat capacity of ethyne at $298 K$ is to be calculated.

Concept introduction:

According to the equipartition theorem, the contribution of each degree of freedom to the heat capacity is $12 R$ . The molar heat capacity is calculated by using the formula shown below.

$CV,m/R=f2$

(i)

Expert Solution

Answer to Problem 13E.4AE

The plot of molar heat capacity of a collection of harmonic oscillators as a function of $x=T/θV$ is shown below.

Atkins' Physical Chemistry, Chapter 13, Problem 13E.4AE , additional homework tip 1

The vibrational heat capacity of ethyne is $14.93 J/mol⋅K_$ .

Explanation of Solution

The molar heat capacity is calculated by using the formula shown below.

$CV,m/R=f2$ (1)

The value of $f$ is expressed by the formula shown below.

$f=(θVT)×(e−θV/2T1−e−θV/T)$ (2)

Therefore, the value of molar heat capacity is calculated by using the formula shown below.

$CV,m/R=((θVT)×(e−θV/2T1−e−θV/T))2$ (3)

Now, first the value of $θV$ is calculate by using the formula shown below.

$θV=hcν¯k$ (4)

Where,

$h$ is the Planck constant $(6.626×10−34 J⋅s)$ .
$c$ is the speed of light $(2.998×108 m/s=2.998×1010 cm/s)$ .
$ν¯$ is the wavenumber.
$k$ is the Boltzmann constant with value $1.38×10−23 J/K$ .