91影视

The chemical energy per mole of a substance is its "chemical potential." Gibbs free energy is defined here as chemical energy, and the substance can either be a single, pure substance or a system of several substances.

(a)

From problem 4.38:

$$\begin{gathered} E_n=\left(n+\frac{3}{2}\right)\sout{h}蝇,n=0,1,2,...;d_n \\ =\frac{1}{2}\left(n+1\right)\left(n+2\right) \end{gathered}$$

From Eq. 5.103

$$\begin{gathered} n\left(鈭�\right)=e^{-\left(e-渭\right)/k_{B}T} \\ {N}_n=\underset{n=0}{\overset{鈭�}{鈭�}}{N}_n=\frac{1}{2}{e}^{\left(渭-\frac{3}{2}\sout{h}蝇\right)/kT}\underset{n=0}{\overset{鈭�}{鈭�}}\left(n+1\right)\left(n+2\right)x^n,\mathrm{Where},x\mathit{=}{e}^{\mathit{-}\sout{\mathit{h}}\mathit{蝇}\mathit{/}{\mathit{k}}_{\mathit{B}}\mathit{T}} \\ \mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\frac{\mathit{1}}{\mathit{1}\mathit{-}\mathit{x}}\mathit{=}\underset{\mathit{n}\mathit{=}\mathit{0}}{\overset{\mathit{鈭�}}{\mathit{鈭�}}}{x}^{\mathit{n}} \\ \mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{鈬�}\mathit{}\frac{\mathit{x}}{\mathit{1}\mathit{-}\mathit{x}}\mathit{=}\underset{\mathit{n}\mathit{=}\mathit{0}}{\overset{\mathit{鈭�}}{\mathit{鈭�}}}{x}^{\mathit{n}\mathit{+}\mathit{1}} \end{gathered}$$

Now,

$$\begin{gathered} 鈬�\frac{d}{dx}\left(\frac{x}{1-x}\right)=\underset{n=0}{\overset{鈭�}{鈭�}}\left(n+1\right)x^n \\ 鈬�\frac{x}{{\left(1-x\right)}^2}=\underset{n=0}{\overset{鈭�}{鈭�}}\left(n+1\right)x^n \\ \underset{n=0}{\overset{鈭�}{鈭�}}\left(n+1\right)\left(n+2\right)x^n=\frac{2}{{\left(1-x\right)}^3}.N=e^{渭/k_{B}T}\frac{1}{{\left(1-e^{-\sout{h}蝇/k_{B}T}\right)}^3} \\ 渭=k_{B}T\left[InN+3\mathrm{In}\left(1-{\mathrm{e}}^{-\sout{\mathrm{h}}\mathrm{蝇}/{\mathrm{k}}_{\mathrm{B}}\mathrm{T}}\right)+\frac{3}{2}h蝇\mathit{/}{k}_{\mathit{B}}T\right] \end{gathered}$$

In order to find energy, we have to calculate:

$E=\underset{n=0}{\overset{鈭�}{鈭�}}{N}_n{蔚}_n=\frac{\sout{h}蝇e^{\left(渭-3\sout{h}蝇/2\right)/k_{B}T}}{2}\underset{n=0}{\overset{鈭�}{鈭�}}\left(n+\frac{3}{2}\right)\left(n+1\right)\left(n+2\right)x^n,x=e^{\sout{h}蝇/2/k_{B}T}$

We use the same trick as before:

$$\begin{gathered} \underset{n=0}{\overset{鈭�}{鈭�}}\left(n+1\right)\left(n+2\right)x^n=\frac{2}{{\left(1-x\right)}^3}/.x^{3/2}\underset{n=0}{\overset{鈭�}{鈭�}}\left(n+1\right)\left(n+2\right)x^{n+3/2} \\ =\frac{2x^{3/2}}{\left(1-x\right)}/\frac{d}{dx}\underset{n=0}{\overset{鈭�}{鈭�}}\left(n+\frac{3}{2}\right)\left(n+1\right)\left(n+2\right)x^{n+1/2} \\ =\frac{3x^{1/2}\left(1+x\right)}{{\left(1-x\right)}^4}鈬�\underset{n=0}{\overset{鈭�}{鈭�}}\left(n+\frac{3}{2}\right)\left(n+1\right)\left(n+2\right)x^n \\ =\frac{3\left(1+x\right)}{{\left(1-x\right)}^4} \end{gathered}$$

So, energy is:

$$\begin{gathered} E=\frac{\sout{h}蝇e^{\left(渭-3\sout{h}蝇/2\right)k_{B}T}}{2}\frac{3\left(1+e^{\sout{h}蝇/2k_{B}T}\right)}{\left(1-e^{\sout{h}蝇/2k_{B}T}\right)},e^{\left(渭-3\sout{h}蝇/2\right)k_{B}T}=N{\left(1-e^{\sout{h}蝇/2k_{B}T}\right)}^3 \\ 鈬�E=\frac{3}{2}N\sout{h}蝇\frac{1+e^{\sout{h}蝇/2k_{B}T}}{1-e^{\sout{h}蝇/2k_{B}T}} \end{gathered}$$

(b)

For$k_{\mathrm{B}}T鈮�\sout{\mathrm{h}}$we have$e^{\sout{-h}蝇/2k_{B}T}鈫�0$

So energy is $E=\frac{3}{2}\sout{h}蝇N$. . All particles are in the ground state.

(c)

For$k_{\mathrm{B}}T鈮�\sout{\mathrm{h}}$we have$e^{\sout{h}蝇/2k_{B}T}鈫�0鈮�1-\frac{\sout{h}蝇}{k_{B}T}$. So energy is$E=3NkBT$

Equipartition theorem says that energy of a system in thermal equilibrium is equal to:

$E=n_f\frac{N{k}_{B}T}{2}$

Where$n_f$ is number of degrees of freedom. In our case, particle has 6 degrees of freedom (3 kinetic and 3 potential).