Chapter 18, Problem 95P

(a)

To determine

The proof that for limit $T>>T_{\text{E}}$ the expression of specific heat from Einstein model and Dulong petit law is same.

Answer to Problem 95P

It is proved that by both Einstein model and Dulong petit for the limit $T>>T_{\text{E}}$ specific heat is same that is $3R$ .

Explanation of Solution

Formula used:

The expression for specific heat from Einstein model is given by,

  $c_{\text{V}}{}^{\prime }=3R{\left(\frac{T_{\text{E}}}{T}\right)}^2\frac{e^{T_{\text{E}}/T}}{{\left(e^{T_{\text{E}}/T}-1\right)}^2}$

The expansion of exponential is given by,

  $e^x=1+x+\frac{x^2}{2!}+\dots$

Calculation:

The expression for specific heat from Einstein model is written as,

  $\begin{array}{c}{c}_{\text{V}}{}^{\prime }=3R{\left(\frac{T_{\text{E}}}{T}\right)}^2\frac{e^{T_{\text{E}}/T}}{{\left(e^{T_{\text{E}}/T}-1\right)}^2}\\ =3R{\left(\frac{T_{\text{E}}}{T}\right)}^2\frac{1}{\frac{e^{2T_{\text{E}}/T}-2e^{T_{\text{E}}/T}+1}{e^{T_{\text{E}}/T}}}\\ =3R{\left(\frac{T_{\text{E}}}{T}\right)}^2\frac{1}{e^{T_{\text{E}}/T}-2+e^{-T_{\text{E}}/T}}\end{array}$

The expansion of exponential terms is written as,

  $\begin{array}{c}{e}^{T_{\text{E}}/T}-2+e^{-T_{\text{E}}/T}=\left(1+\frac{T_{\text{E}}}{T}+\frac{1}{2}{\left(\frac{T_{\text{E}}}{T}\right)}^2+\dots \right)-2+\left(1-\frac{T_{\text{E}}}{T}+\frac{1}{2}{\left(\frac{T_{\text{E}}}{T}\right)}^2+\dots \right)\\ ={\left(\frac{T_{\text{E}}}{T}\right)}^2\left(\because T>T_{\text{E}}\right)\end{array}$

This implies,

  $\begin{array}{c}{c}_{\text{v}}{}^{\prime }=3R{\left(\frac{T_{\text{E}}}{T}\right)}^2\frac{1}{{\left(\frac{T_{\text{E}}}{T}\right)}^2}\\ =3R\end{array}$

Conclusion:

Therefore, it is proved that by both Einstein model and Dulong petit for the limit $T>>T_{\text{E}}$ specific heat is same that is $3R$ .

(b)

To determine

The increase in internal energy.

Answer to Problem 95P

The change in internal energy is $4.62\text{J}$ .

Explanation of Solution

Given:

For diamond $T_{\text{E}}$ is $1060\text{K}$ .

The initial temperature is $\text{300}\text{K}$ .

The final temperature is $\text{600}\text{K}$ .

Formula used:

The expression for change in internal energy is given by,

  $\Delta U=\frac{1}{2}n\left(c_{\text{v1}}+c_{\text{v2}}\right)\left(T_2-T_1\right)+\frac{1}{2}n\left(c_{\text{v2}}+c_{\text{v3}}\right)\left(T_3-T_2\right)+\frac{1}{2}n\left(c_{\text{v3}}+c_{\text{v4}}\right)\left(T_4-T_3\right)$

Calculation:

The change in internal energy is calculated as,

  $\begin{array}{c}\Delta U=\frac{1}{2}n\left(c_{\text{v1}}+c_{\text{v2}}\right)\left(T_2-T_1\right)+\frac{1}{2}n\left(c_{\text{v2}}+c_{\text{v3}}\right)\left(T_3-T_2\right)+\frac{1}{2}n\left(c_{\text{v3}}+c_{\text{v4}}\right)\left(T_4-T_3\right)\\ =\left(\begin{array}{l}\frac{1}{2}\left(1.00\text{mol}\right)\left(9.65\frac{\text{J}}{\text{mol}\cdot \text{K}}+14.33\frac{\text{J}}{\text{mol}\cdot \text{K}}\right)\left(400\text{K}-300\text{K}\right)\\ +\frac{1}{2}\left(1.00\text{mol}\right)\left(14.33\frac{\text{J}}{\text{mol}\cdot \text{K}}+17.38\frac{\text{J}}{\text{mol}\cdot \text{K}}\right)\left(500\text{K}-400\text{K}\right)+\\ \frac{1}{2}\left(1.00\text{mol}\right)\left(17.38\frac{\text{J}}{\text{mol}\cdot \text{K}}+19.35\frac{\text{J}}{\text{mol}\cdot \text{K}}\right)\left(600\text{K}-500\text{K}\right)\end{array}\right)\\ =\left(\left(4.62\times {10}^3\text{kJ}\right)\left(\frac{{10}^{-3}\text{J}}{1\text{kJ}}\right)\right)\\ =4.62\text{J}\end{array}$

Conclusion:

Therefore, the change in internal energy is $4.62\text{J}$ .