91影视

The multiplicity of an Einstein solid is given as:

$惟(N,q)鈮�{\left(\frac{q+N}{q}\right)}^q{\left(\frac{q+N}{N}\right)}^N$

The entropy is given as:

$S=k\mathrm{濒苍惟}$

By substituting the given value in the above equation, we get,

$$\begin{gathered} S=k\ln \left\{{\left(\frac{q+N}{q}\right)}^q{\left(\frac{q+N}{N}\right)}^N\right\} \\ S=k[q\ln \left(\frac{q+N}{q}\right)+N\ln \left(\frac{q+N}{N}\right)] \\ S=k\left[q\ln \right(q+N)鈭�q\ln q+N\ln (q+N)鈭�N\ln N] \\ S=k\left[\right(q+N\left)\ln \right(q+N)鈭�q\ln q鈭�N\ln N] \end{gathered}$$

The multiplicity for Einstein solid can be given as:

$$\begin{gathered} \mathrm{惟}=\frac{鈱�q+N鈭�1}{鈱�q鈱�N鈭�1} \\ \mathrm{惟}=\frac{鈱�q+N鈭�1}{鈱�q鈱�N鈭�1}脳\frac{N(q+N)}{N(q+N)} \\ \mathrm{惟}=\frac{鈱�q+N}{鈱�q鈱�N}脳\frac{N}{(q+N)} \end{gathered}$$

But,

Stirling approximation of large n is given as:

localid="1647403887405" $鈱�n=\sqrt{2蟺n}{n}^n{e}^{-n}$

Hence,

$$\begin{gathered} \mathrm{惟}=\frac{N}{(q+N)}脳\frac{\sqrt{2蟺(q+N)}(q+N{)}^{q+N}{e}^{鈭�(q+N)}}{\sqrt{2蟺\left(q\right)}\left(q{)}^q{e}^{鈭�\left(q\right)}\sqrt{2蟺\left(N\right)}\right(N{)}^N{e}^{鈭�\left(N\right)}} \\ \mathrm{惟}=\frac{N}{(q+N)}脳\sqrt{\frac{q+N}{qN}}脳\frac{(q+N{)}^{q+N}{e}^{鈭�(q+N)}}{q^q{N}^N{e}^{鈭�(q+N)}} \\ \mathrm{惟}=\sqrt{\frac{N}{2蟺q(q+N)}{\left(\frac{q+N}{q}\right)}^q{\left(\frac{q+N}{N}\right)}^N} \end{gathered}$$

For large $q$and $N$, when compared to the power terms, the square root terms are small. As they have a negligible effect, they can be neglected. Hence, the approximate value can be written as:

$$\begin{gathered} \mathrm{惟}=\sqrt{\frac{N}{2蟺q(q+N)}}{\left(\frac{q+N}{q}\right)}^q{\left(\frac{q+N}{N}\right)}^N \\ \mathrm{惟}鈮�{\left(\frac{q+N}{q}\right)}^q{\left(\frac{q+N}{N}\right)}^N \end{gathered}$$

Hence, the entropy of an Einstein solid containing $N$oscillators and $q$energy units is given as:

$S=k\left[\right(q+N\left)\ln \right(q+N)鈭�q\ln q鈭�N\ln N]$

For large values of $q$and $N$, the square root terms are small when compared to the power terms. Hence they are neglected as they have a very negligible effect on the entropy. Therefore the approximate value of multiplicity is given as:

$\mathrm{惟}鈮�{\left(\frac{q+N}{q}\right)}^q{\left(\frac{q+N}{N}\right)}^N$

The total energy of the system is given as:

$U=q蔚$

The entropy of the system is given as:

$S=k\left[\right(q+N\left)\ln \right(q+N)-q\ln q-N\ln N]$

The temperature can be given as:

$$\begin{gathered} \frac{1}{T}=\frac{鈭�S}{鈭�U} \\ \frac{{\displaystyle 1}}{{\displaystyle T}}=\frac{鈭�\left[k\right[(q+N)\ln (q+N)-q\ln q-N\ln N\left]\right]}{蔚鈭�q} \\ \frac{{\displaystyle 1}}{{\displaystyle T}}=\frac{k}{系}\left[\ln \right(q+N)+1-\ln q-1-0] \\ \frac{{\displaystyle 1}}{{\displaystyle T}}=\frac{k}{系}\ln \left[\frac{q+N}{q}脳\frac{系}{系}\right] \\ \frac{系}{kT}=\ln \left(\frac{U+N蔚}{U}\right) \end{gathered}$$

The temperature as a function of its energy can be given as:

$\frac{系}{kT}=\ln \left(\frac{U+N蔚}{U}\right)$

The temperature as a function of energy is given as:

$\frac{系}{kT}=\ln \left(\frac{U+N蔚}{U}\right)$

As given:

$\begin{array}{r}\frac{系}{kT}=\ln 鈦�\left(\frac{U+N蔚}{U}\right)\end{array}$

On simplifying, we get,

$$\begin{gathered} e^{\frac{系}{kT}}=\frac{U+N系}{U} \\ U{e}^{\frac{系}{kT}}鈭�U=N蔚 \\ U=\frac{N系}{\left(e^{\frac{系}{kT}}鈭�1\right)} \end{gathered}$$

Now, differentiating the energy with respect to temperature to get the heat capacity as:

$$\begin{gathered} C=\frac{鈭�U}{鈭�T} \\ C=\frac{鈭�}{鈭�T}\left[\frac{N蔚}{\left(e^{\frac{系}{kT}}-1\right)}\right] \\ C=N蔚\left(-{\left(e^{\frac{系}{kT}}-1\right)}^{-2}\right)e^{\frac{系}{kT}}\left(\frac{系}{k{T}^2}\right) \\ C=\frac{{系}^{2}N{e}^{\frac{系}{kT}}}{k{T}^2{\left(e^{\frac{系}{kT}}-1\right)}^2} \end{gathered}$$

The energy as a function of temperature can be given as:

$U=\frac{N系}{\left(e^{\frac{系}{kT}}鈭�1\right)}$

Expression of Heat capacity is:

$C=\frac{{系}^{2}N{e}^{\frac{系}{kT}}}{k{T}^2{\left(e^{\frac{系}{kT}}-1\right)}^2}$

Heat capacity in terms of temperature is given as:

$C=\frac{{系}^{2}N{e}^{\frac{系}{kT}}}{k{T}^2{\left(e^{\frac{系}{kT}}-1\right)}^2}$

For very small values of$x,e^x鈮�1+x$.

By expanding the exponential for higher temperatures, we get

$e^{\frac{系}{kT}}鈮�1+\frac{蔚}{kT}$

The heat capacity for very large values of $\mathrm{T}$can be given as:

$$\begin{gathered} C=\frac{{系}^{2}N{e}^{\frac{系}{kT}}}{k{T}^2{(e^{\frac{系}{kT}}鈭�1)}^2} \\ C鈮�\frac{{系}^{2}N(1+\frac{系}{kT})}{k{T}^2{\left(\frac{系}{kT}\right)}^2} \\ C鈮�Nk(1+\frac{系}{kT}) \\ C鈮�Nk \end{gathered}$$

The derived expression is analogous to the expression of heat capacity at high temperatures.

So, $C鈮�Nk$is approximately the expected result.

The heat capacity is given as:

$C=\frac{{系}^{2}N{e}^{\frac{系}{kT}}}{k{T}^2{\left(e^{\frac{系}{kT}}-1\right)}^2}$

The heat capacity for low temperature can be given as:

$$\begin{gathered} C=\frac{{系}^{2}N{e}^{\frac{系}{kT}}}{k{T}^2{\left(e^{\frac{蔚}{kT}}-1\right)}^2} \\ C=\frac{{系}^{2}kN{e}^{\frac{蔚}{kT}}}{k^2{T}^2{\left(e^{\frac{蔚}{kT}}-1\right)}^2} \\ \frac{C}{Nk}=\frac{{\left(\frac{蔚}{kT}\right)}^2{e}^{\frac{蔚}{kT}}}{{\left(e^{\frac{蔚}{kT}}-1\right)}^2} \end{gathered}$$

Based on the derived equation, the graph of $\frac{C}{Nk}$versus $t=\frac{kT}{系}$can be made as follows:

For lead, heat capacity is maximum at$T=80\mathrm{K}$

$$\begin{gathered} {蔚}_{\text{lead}}=kT \\ {蔚}_{\text{lead}}=\left(1.38脳{10}^{-23}\right)脳80 \\ {蔚}_{\text{lead}}=1.1脳{10}^{-21}\mathrm{J} \\ {蔚}_{\text{lead}}=6.875脳{10}^{-3}\mathrm{eV} \end{gathered}$$

For aluminum, heat capacity is maximum at$T=300\mathrm{K}$

$$\begin{gathered} {蔚}_{\text{aluminium}}=kT \\ {蔚}_{\text{aluminium}}=\left(1.38脳{10}^{-23}\right)脳300 \\ {蔚}_{\text{aluminium}}=4.14脳{10}^{-21}\mathrm{J} \\ {蔚}_{\text{aluminium}}=2.587脳{10}^{-2}\mathrm{eV} \end{gathered}$$

For diamond, heat capacity is maximum at$T=2230\mathrm{K}$

$$\begin{gathered} {蔚}_{\text{diamond}}=kT \\ {蔚}_{\text{diamond}}=\left(1.38脳{10}^{-23}\right)脳2230 \\ {蔚}_{\text{diamond}}=3.0774脳{10}^{-20}\mathrm{J} \\ {蔚}_{\text{diamond}}=0.19233\mathrm{eV} \end{gathered}$$

The heat capacity for low temperatures can be given as:

$\frac{C}{Nk}=\frac{{\left(\frac{蔚}{kT}\right)}^2{e}^{\frac{蔚}{kT}}}{{\left(e^{\frac{蔚}{kT}}-1\right)}^2}$

The graph can be made as:

The values of $蔚$are as follows:

$\begin{array}{l}{蔚}_{\text{lead}}=6.875脳{10}^{鈭�3}\mathrm{eV}\\ {蔚}_{\text{aluminum}}=2.587脳{10}^{鈭�2}\mathrm{eV}\\ {蔚}_{\text{diamond}}=0.19233\mathrm{eV}\end{array}$

The graph of heat capacity versus temperature for lead, aluminum, and diamond is approximately anomalous as in figure 1.14.

The heat capacity is given as:

$C=\frac{{系}^{2}N{e}^{\frac{系}{kT}}}{k{T}^2{\left(e^{\frac{系}{kT}}-1\right)}^2}$

Let,

$x=\frac{蔚}{kT}$

The heat capacity can be modified as:

$\begin{array}{l}C=\frac{{蔚}^{2}N{e}^{\frac{蔚}{kT}}}{k{T}^2{(e^{\frac{系}{kT}}鈭�1)}^2}\\ \frac{C}{Nk}=\frac{{\left(\frac{系}{kT}\right)}^2{e}^{\frac{系}{kT}}}{{(e^{\frac{系}{kT}}鈭�1)}^2}\\ \frac{C}{Nk}=\frac{(x{)}^2{e}^x}{{(e^x鈭�1)}^2}\end{array}$

Now,

The Taylor expansion is given as:

$e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+....$

Hence, the above equation becomes:

$\begin{array}{l}\frac{C}{Nk}=\frac{(x{)}^2(1+x+\frac{x^2}{2!}+\frac{x^3}{3!})}{{(x+\frac{x^2}{2!}+\frac{x^3}{3!})}^2}\\ \frac{C}{Nk}=\frac{(1+x+\frac{x^2}{2!}+\frac{x^3}{3!})}{{(1+\frac{x}{2!}+\frac{x^2}{3!})}^2}\\ \frac{C}{Nk}=(1+x+\frac{x^2}{2!}+\frac{x^3}{3!}){(1+\frac{x}{2!}+\frac{x^2}{3!})}^{鈭�2}\end{array}$

Using the Taylor series, the above equation becomes:

$\frac{C}{Nk}=\left(1+x+\frac{x^2}{2}+\frac{x^3}{6}\right)\left(\frac{5x^2}{12}-x+1\right)$

Solving further and neglecting the higher powers than 2, we get,

$C鈮�Nk\left(1-\frac{{\left(\frac{系}{kT}\right)}^2}{12}\right)$

Hence, the required expression is:

$C鈮�Nk\left(1-\frac{{\left(\frac{系}{kT}\right)}^2}{12}\right)$