Chapter 12, Problem 12.11P

(a)

Interpretation Introduction

Interpretation:

The standard molar entropy values of $NeandXe$ has to be compared.

Concept Introduction:

The entropy of distinguishable particle in terms of partition function can be calculated using following formula.

$S=\frac{U-U\left(0\right)}{T}+Nk\ln q$

The entropy of indistinguishable particle in terms of partition function can be calculated using following formula.

$S=\frac{U-U\left(0\right)}{T}+Nk\ln q-Nk\left(\ln N-1\right)$

Where,

    $\begin{array}{l}S=entropy\\ U-U\left(0\right)=changein\mathrm{int}ernalenergy\\ k=plank\text{'}scons\tan t\\ q=partitionfunction\end{array}$

So from the above equation it is clear that the entropy depends on partition functions.

The expression for the translational partition function is given below.

$q^T=\frac{{\left(2\pi mkT\right)}^{\frac{3}{2}}V}{h^3}$

Where,

    $\begin{array}{c}{q}^T=translationalpartitionfunction\\ m=molecularmass\\ T=temperature\\ h=plank\text{'}scons\tan t\\ V=volumeats\tan dardconditions\\ k=boltzmanncons\tan t\end{array}$

The expression for rotational partition function is given below.

$q^R=\frac{kT}{\sigma hB}$

Where,

    $\begin{array}{c}{q}^R=rotationalpartitionfunction\\ k=boltzmanncons\tan t\\ T=temperature\\ h=plank\text{'}scons\tan t\\ \sigma =symmetrynumber\\ B=rotationalcons\tan t\end{array}$

The translational partition function increases with increase in molecular mass and in the case of rotational partition function which depends on moment of inertia, also increases with increase in molecular mass.

(b)

Interpretation Introduction

Interpretation:

The standard molar entropy values of  $H_{2}Oand{D}_{2}O$ has to be compared.

Concept Introduction:

The entropy of distinguishable particle in terms of partition function can be calculated using following formula.

$S=\frac{U-U\left(0\right)}{T}+Nk\ln q$

The entropy of indistinguishable particle in terms of partition function can be calculated using following formula.

$S=\frac{U-U\left(0\right)}{T}+Nk\ln q-Nk\left(\ln N-1\right)$

Where,

    $\begin{array}{l}S=entropy\\ U-U\left(0\right)=changein\mathrm{int}ernalenergy\\ k=plank\text{'}scons\tan t\\ q=partitionfunction\end{array}$

So from the above equation it is clear that the entropy depends on partition functions.

The expression for the translational partition function is given below.

$q^T=\frac{{\left(2\pi mkT\right)}^{\frac{3}{2}}V}{h^3}$

Where,

    $\begin{array}{c}{q}^T=translationalpartitionfunction\\ m=molecularmass\\ T=temperature\\ h=plank\text{'}scons\tan t\\ V=volumeats\tan dardconditions\\ k=boltzmanncons\tan t\end{array}$

The expression for rotational partition function is given below.

$q^R=\frac{kT}{\sigma hB}$

Where,

    $\begin{array}{c}{q}^R=rotationalpartitionfunction\\ k=boltzmanncons\tan t\\ T=temperature\\ h=plank\text{'}scons\tan t\\ \sigma =symmetrynumber\\ B=rotationalcons\tan t\end{array}$

The translational partition function increases with increase in molecular mass and in the case of rotational partition function which depends on moment of inertia, also increases with increase in molecular mass.

(c)

Interpretation Introduction

Interpretation:

The standard molar entropy values of $C\left(diamond\right)$ and $C\left(graphite\right)$ has to be compared.

Concept Introduction:

The entropies of solids influenced greatly by the arrangement of atoms in it.  Entropy will be less for those solids with more ordered arrangement.