Evaluate the integrand in equation $7.112$as a power series in x, keeping terms through $x^4$• Then carry out the integral to find a more accurate expression for the energy in the high-temperature limit. Differentiate this expression to obtain the heat capacity, and use the result to estimate the percent deviation of $C_v$from$3NkatT=T_{D}andT=2T_{D.}$

We need to find calculate the integrand in the equation$7.112$as a power series in$x$keeping terms through $x^4$.

Equation is given as:

$U=\frac{9Nk{T}^4}{T_D^3}{\int }_0^{x_{\max }}\frac{x^3}{e^x-1}dx$

where,

$x=\frac{h{c}_{s}n}{2LkT}{x}_{\max }=\frac{T_D}{T}$

at low-temperature limit $T⩽T_D$, the value of is very small so we can expand the exponential in the denominator using:

$e^x\approx 1+x+x^2/2+x^3/6$

therefore:

$$\begin{gathered} U=\frac{9Nk{T}^4}{T_D^3}{\int }_0^{x_{\max }}\frac{x^3}{1+x+\frac{1}{2}{x}^2+\frac{1}{6}{x}^3-1} \\ U=\frac{9Nk{T}^4}{T_D^3}{\int }_0^{x_{\max }}\frac{x^3}{x+\frac{1}{2}{x}^2+\frac{1}{6}{x}^3} \\ U=\frac{9Nk{T}^4}{T_D^3}{\int }_0^{x_{\max }}{x}^2{\left[1+\frac{1}{2}x+\frac{1}{6}{x}^2\right]}^{-1} \end{gathered}$$

now we can use ${(1+y)}^{-1}\approx 1+y+y^2$, where $y=-\frac{1}{2}x-\frac{1}{6}{x}^2$we get :

$$\begin{gathered} U=\frac{9Nk{T}^4}{T_D^3}{\int }_0^{x_{\max }}{x}^2\left[1-\frac{1}{2}x-\frac{1}{6}{x}^2+{\left(-\frac{1}{2}x-\frac{1}{6}{x}^2\right)}^2\right] \\ U=\frac{9Nk{T}^4}{T_D^3}{\int }_0^{x_{\max }}{x}^2\left[1-\frac{1}{2}x-\frac{1}{6}{x}^2+\frac{1}{4}{x}^2+\frac{1}{36}{x}^4+\frac{1}{6}{x}^3\right] \end{gathered}$$

neglect the $x$with power of $3and4$inside the bracket because $x$ is very small ,so:

$U=\frac{9Nk{T}^4}{T_D^3}{\int }_0^{x_{\max }}\left[x^2-\frac{1}{2}{x}^3+\frac{1}{12}{x}^4\right]$

now the function inside the integration is easy to be integrated, so integrate from $0$to $x_{max}$to get:

$$\begin{gathered} U=\frac{9Nk{T}^4}{T_D^3}{\left[\frac{1}{3}{x}^3-\frac{1}{8}{x}^4+\frac{1}{60}{x}^5\right]}_0^{x_{\max }} \\ U=\frac{9Nk{T}^4}{T_D^3}\left[\frac{1}{3}{x}_{\max }^3-\frac{1}{8}{x}_{\max }^4+\frac{1}{60}{x}_{\max }^5\right] \end{gathered}$$

but $x_{max}=\raisebox{1ex}{$T$}\!\left/ \!\raisebox{-1ex}{$T_D$}\right.$, so,

$$\begin{gathered} U=\frac{9Nk{T}^4}{T_D^3}\left[\frac{1}{3}{\left(\frac{T_D}{T}\right)}^3-\frac{1}{8}{\left(\frac{T_D}{T}\right)}^4+\frac{1}{60}{\left(\frac{T_D}{T}\right)}^5\right] \\ U=9Nk{T}_D\left[\frac{1}{3}\left(\frac{T}{T_D}\right)-\frac{1}{8}+\frac{1}{60}\left(\frac{T_D}{T}\right)\right] \end{gathered}$$

the heat capacity at constant volume is the just the partial derivative of the energy with respect to the temperature, that is:

$C_v=\frac{\partial U}{\partial T}$

thus,

localid="1650887002266" $$\begin{gathered} C_v=9Nk{T}_D\frac{\partial }{\partial T}\left[\frac{1}{3}\left(\frac{T}{T_D}\right)-\frac{1}{8}+\frac{1}{60}\left(\frac{T_D}{T}\right)\right] \\ {C}_v=9Nk{T}_D\left[\frac{1}{3}\left(\frac{1}{T_D}\right)-\frac{1}{60}\left(\frac{T_D}{T^2}\right)\right] \\ {C}_v=3Nk\left[1-\frac{1}{20}{\left(\frac{T_D}{T}\right)}^2\right] \end{gathered}$$

Now calculating the deviation from the asymptotic value $3NkT,atT=T_D$,

substitute into the above equation we get:

$\begin{array}{c}{\left.C_V\right|}_{T=T_D}=3Nk\left[1-\frac{1}{20}{\left(\frac{T_D}{T_D}\right)}^2\right]\\ {\left.C_V\right|}_{T=T_D}=\left(0.95\right)3Nk\end{array}$

so it deviates by $5\%AtT=2T_D$.

$\begin{array}{c}{\left.C_V\right|}_{T=2T_D}=3Nk\left[1-\frac{1}{20}{\left(\frac{T_D}{2T_D}\right)}^2\right]\\ {\left.C_V\right|}_{T=2T_D}=\left(0.9875\right)3Nk\end{array}$

so it deviates by$1.25\%$