ISITEKS: UES

DARI:
untuk zat padat pada suhu mendekati 0 kelvin kapasitas pada volum tetap
$\large {\color{DarkOrange} c_{V}}$ adalah fungsi T saja menurut Debye

$\large {\color{DarkOrange} d\bar{S}=diferensial\ eksak}$

JADIKAN:

zat padat ini mengikuti UES

JAWAB:

$\large {\color{DarkOrange} d\bar{S}=\frac{d\bar{E}+pd\bar{V}}{T}}$

        $\LARGE {\color{DarkOrange} =\frac{\left ( \frac{\partial \bar{E}}{\partial T} \right )_{\bar{V}}dT+\left ( \frac{\partial \bar{E}}{\partial \bar{V}} \right )_{T}d\bar{V}+pd\bar {V}}{T}}$

        $\large {\color{DarkOrange} =\frac{1}{T}\left ( \frac{\partial \bar{E}}{\partial T} \right )_{\bar{V}}dT+\frac{\left [ p+\left ( \frac{\partial \bar{E}}{\partial \bar{V}}_{T} \right ) \right ]}{T}d\bar{V}}$

       $\large {\color{DarkOrange} =\frac{c_{V}}{T}dT+\frac{\left [ p+\left ( \frac{\partial \bar{E}}{\partial \bar{V}} \right )_{T} \right ]}{T}d\bar{V}}$

      $\large {\color{DarkOrange} d\bar{S}=dif\ eksak}$

       $\large {\color{DarkOrange} \frac{\partial \frac{c_{V}}{T}}{\partial \bar{V}}=\frac{\partial \left ( p+\left ( \frac{\partial \bar{E}}{\partial \bar{V}} \right )_{T} \right )/T}{\partial T}}$

       $\large {\color{DarkOrange} \frac{\partial \frac{fungsi\ T}{T}}{\partial \bar{V}}=\frac{\partial \left ( p+\left ( \frac{\partial \bar{E}}{\partial \bar{V}} \right )_{T} \right )/T}{\partial T}}$

     $\large {\color{DarkOrange} 0=\frac{\partial \left ( p+\left ( \frac{\partial \bar{E}}{\partial \bar{V}} \right )_{T} \right )/T}{\partial T}}$

$\large {\color{DarkOrange} \frac{p+\frac{\partial \bar{E}}{\partial \bar{V}}}{T}=f\left ( \bar{V} \right )}$


   $\large {\color{DarkOrange} \left ( p+\frac{\partial \bar{E}}{\partial \bar{V}} \right )\frac{1}{f\left ( \bar{V} \right )}=T}$

   $\large {\color{DarkOrange} \frac{1}{f\left ( \bar{V} \right )}\ dinamakan\ g\left ( \bar{V} \right )}$

$\large {\color{DarkOrange} \left ( p+\frac{\partial \bar{E}}{\partial \bar{V}} \right )\cdot g\left ( \bar{V} \right )=T}$

$\large {\color{DarkOrange} diambil\ g\left ( \bar{V} \right )=a+b\bar{V}+c\bar{V}^{2}+...}$

yang harus memenuhi syarat batas pada    $\large {\color{DarkOrange} \bar{V}=\infty }$       $\large {\color{DarkOrange} p\bar{V}=RT}$

Maka

$\large {\color{DarkOrange} p\cdot \left ( a+b\bar{V}+c\bar{V}^{2}+... \right )=T}$

$\large {\color{DarkOrange} p\cdot \left ( a+b\bar{V}+c\bar{V}^{2}+... \right )=\frac{p\bar{V}}{R}}$

$\large {\color{DarkOrange} \frac{p}{\bar{V}}\left ( a+b\bar{V}+c\bar{V}^{2}+... \right )=\frac{p}{R}}$

$\large {\color{DarkOrange} p\left ( \frac{a}{\bar{V}}+b+c\bar{V}^{2}+... \right )=\frac{p}{R}}$

$\large {\color{DarkOrange} karena\ \bar{V}=\infty \rightarrow \frac{a}{\bar{V}}=0\rightarrow p\left ( b+c\bar{V}^{2}+... \right )}$

$\large {\color{DarkOrange} p\left ( b+c\bar{V}^{2}+... \right )=\frac{p}{R}}$

$\large {\color{DarkOrange} maka\ b=\frac{1}{R}\ \ ,c=0\ \ ,d=0\ \ ,...}$

$\large {\color{DarkOrange} \left ( p+\frac{\partial \bar{E}}{\partial \bar{V}} \right )\left ( a+\frac{\bar{V}}{R} \right )=T}$

$\large {\color{DarkOrange} syarat\ batas\ yang\ lain\ adalah\ pada\ T=0 ,\bar{V}=\bar{V}_{0}\ ;maka}$

     $\large {\color{DarkOrange} \left ( p+\frac{\partial \bar{E}}{\partial \bar{V}} \right )\left ( a+\frac{\bar{V}_{0}}{R} \right )=0}$ , $\large {\color{DarkOrange} karena\ untuk\ \frac{\partial \bar{E}}{\partial \bar{V}}=0, }$

$\large {\color{DarkOrange} \frac{\partial \bar{E}}{\partial \bar{V}}=0\ ,p+\frac{\partial \bar{E}}{\partial \bar{V}}=p\ bisa\ tidak\ nol,}$

$\large {\color{DarkOrange} a+\frac{\bar{V}_{0}}{R}=0}$                                                          $\large {\color{DarkOrange} a=-\frac{\bar{V}_{0}}{R}}$

$\large {\color{DarkOrange} \left ( p+\frac{\partial \bar{E}}{\partial \bar{V}} \right )\left ( \frac{\bar{V}}{R}-\frac{\bar{V_{0}}}{R} \right )=T}$

$\large {\color{DarkOrange} \left ( p+\frac{\partial \bar{E}}{\partial \bar{V}} \right )\left ( \bar{V}-\bar{V}_{0} \right )=RT}$

ISITEKS

Rabu, 25 September 2013

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