Energi Gas Monoatomik ~ Max Planck

Assalamu'alaikum Wr. Wb.
Sahabat Fisika yang berbahagia, kalian pasti mengenal energi gas monoatomik dibawah kan..??? hehehe

$E=\frac{3}{2}NKT$

Saya yakin, kita hanya mengenal tanpa pernah membuktikan. Baik untuk kali ini saya akan memposting bagaimana cara kita membuktikan rumus di atas.

Kita sudah mengenal tingkat energi gas monoatomik pada kotak yaitu:

$\large E_{r}=\frac{n^{2}\pi ^{2}\pounds ^{2}}{2mL^{2}}$

Jadi kalau dalam tiga dimensi menjadi,

$\large E_{r}=\frac{\pi ^{2}\pounds ^{2}}{2m}(\frac{n_{x}^{2}}{L_{x}^{2}}+\frac{n_{y}^{2}}{L_{y}^{2}}+\frac{n_{z}^{2}}{L_{z}^{2}})$

Keterangan:

$\large E_{r}=$ Tingkat Energi
$\large \pounds=$ Ketetapan Planck
$\large n=$ Bilangan bulat (1,2,3,.....)
$\large L=$ Panjang sisi kotak
$\large m=$ Masa gas monoatomik

Untuk sampai ke energi gas monoatomik, kita membutuhkan fungsi partisi (Z) yang dituliskan,

$\large Z=\sum_{r}\mathrm{e^{\mathit{-\beta E_{r}}}}$

Atau dalam tiga dimensi,

$\large Z=\sum_{n_{x}n_{y}n_{z}}\mathrm{e^{\mathit{-\beta E_{n_{x}n_{y}n_{z}}}}}$

$\large =\sum_{n_{x}}\sum_{n_{y}}\sum_{n_{z}}\exp \left [ \frac{-\beta \pi ^{2}\pounds ^{2}}{2m}\left ( \frac{n_{x}^{2}}{L_{x}^{2}}+\frac{n_{y}^{2}}{L_{y}^{2}}+\frac{n_{z}^{2}}{L_{z}^{2}} \right ) \right ]=Z_{x}Z_{y}Z_{z}$

Kita ambil fungsi partisi pada sumbu x,

$\large Z_{x}=\sum_{n_{x}}\exp \left [ \frac{-\beta \pi ^{2}\pounds ^{2}}{2m}\left ( \frac{n_{x}^{2}}{L_{x}^{2}}\right ) \right ]$

Kemudian kita integralkan dengan batas ( $\large 0-\infty$ ), sehingga

$\large Z_{x}=\int_{0}^{\infty }\exp \left [ -\beta \frac{\pi^{2}\pounds^{2}}{2m}\left ( \frac{n_{x}^{2}}{L_{x}^{2}} \right ) \right ]dn_{x}$

$\large =\int_{0}^{\infty }\left ( \frac{1}{\beta \frac{\pi^{2}\pounds^{2}}{2m}\left ( \frac{n_{x}^{2}}{L_{x}^{2}} \right )} \right )^{\frac{1}{2}}\left ( \beta \frac{\pi^{2}\pounds^{2}}{2m}\left ( \frac{n_{x}^{2}}{L_{x}^{2}} \right ) \right )^{\frac{1}{2}}\exp \left [ -\beta \frac{\pi^{2}\pounds^{2}}{2m}\left ( \frac{n_{x}^{2}}{L_{x}^{2}} \right ) \right ]dn_{x}$

Kita ambil keadaan pada tingkat energi $n_{x}=1$ , maka:

$\large Z_{x}=\int_{0}^{\infty }\left ( \frac{1}{\beta \frac{\pi^{2}\pounds^{2}}{2m}\left ( \frac{1^{2}}{L_{x}^{2}} \right )} \right )^{\frac{1}{2}}\left ( \beta \frac{\pi^{2}\pounds^{2}}{2m}\left ( \frac{1^{2}}{L_{x}^{2}} \right ) \right )^{\frac{1}{2}}\exp \left [ -\beta \frac{\pi^{2}\pounds^{2}}{2m}\left ( \frac{1^{2}}{L_{x}^{2}} \right ) \right ]dn_{x}$

$\large =\int_{0}^{\infty }\left ( \frac{1}{\beta \frac{\pi^{2}\pounds^{2}}{2mL_{x}^{2}}\left \right } \right )^{\frac{1}{2}}\left ( \beta \frac{\pi^{2}\pounds^{2}}{2mL_{x}^{2}}\left \right \right )^{\frac{1}{2}}\exp \left [ -\beta \frac{\pi^{2}\pounds^{2}}{2mL_{x}^{2}}\left \right \right ]dn_{x}$

kita misalkan $\beta \frac{\pi^{2}\pounds^{2}}{2mL_{x}^{2}}=x^{2}$ , maka

$\large Z_{x}=\left ( \frac{2mL_{x}^{2}}{\beta \pi^{2}\pounds^{2}\left \right } \right )^{\frac{1}{2}} \right )\int_{0}^{\infty }\left \left ( x^{2}\right )^{\frac{1}{2}}\mathbf{e}^{-x}dn_{x}$

$\large =\left ( \frac{2mL_{x}^{2}}{\beta \pi^{2}\pounds^{2}\left \right } \right )^{\frac{1}{2}} \right )\int_{0}^{\infty }x\mathbf{e}^{-x}dn_{x}$

Karena $\int_{0}^{\infty }x\mathbf{e}^{-x}dn_{x}$ adalah fungsi gamma yang nilainya ( $\frac{1}{2}\sqrt{\pi}$ ), maka,
$\large Z_{x}=\left ( \frac{2mL_{x}^{2}}{\beta \pi^{2}\pounds^{2}\left \right } \right )^{\frac{1}{2}} \right )\frac{1}{2}\sqrt{\pi}$
$\large =\left ( \frac{2mL_{x}^{2}\pi}{4\beta \pi^{2}\pounds^{2}\left \right } \right )^{\frac{1}{2}} \right )=\left ( \frac{m}{2\beta \pi\pounds^{2}\left \right } \right )^{\frac{1}{2}} \right )L_{x}$
jadi,

$\large Z_{x}=\left ( \frac{m}{2\beta \pi\pounds^{2}\left \right } \right )^{\frac{1}{2}} \right )L_{x}$ ,

$\large Z_{y}=\left ( \frac{m}{2\beta \pi\pounds^{2}\left \right } \right )^{\frac{1}{2}} \right )L_{y}$

$\large Z_{z}=\left ( \frac{m}{2\beta \pi\pounds^{2}\left \right } \right )^{\frac{1}{2}} \right )L_{z}$

Selanjutnya kita cari nilai Z dengan cara:

$\large Z=Z_{x}Z_{y}Z_{z}$

$\large =\left ( \frac{m}{2\beta \pi\pounds^{2}\left \right } \right )^{\frac{1}{2}} \right )L_{x}\left ( \frac{m}{2\beta \pi\pounds^{2}\left \right } \right )^{\frac{1}{2}} \right )L_{y}\left ( \frac{m}{2\beta \pi\pounds^{2}\left \right } \right )^{\frac{1}{2}} \right )L_{z}$

$\large =\left ( \frac{m}{2\beta \pi\pounds^{2}\left \right } \right )^{\frac{1}{2}+\frac{1}{2}+\frac{1}{2}} \right )L_{x}L_{y}L_{z}$

Karena $\large L_{x}L_{y}L_{z}$ =Volume, maka

$\large Z=\left ( \frac{m}{2 \pi\pounds^{2}\left \right } \right )^{\frac{3}{2}} \right )\frac{V}{\beta^{\frac{3}{2}}}$

$\large \ln Z=\frac{3}{2}\ln \left ( \frac{m}{2 \pi\pounds^{2}} \right )+\ln V-\frac{3}{2}\ln\beta$

Dalam fisika statistik, energi rata-rata gas secara umum dinyatakan dengan persamaan,
$\large \overline{E}=-\frac{\partial \ln Z}{\partial \beta }$
Jadi dengan memasukkan nilai (ln Z ), kita dapatkan,

$\large \overline{E}=-\frac{\partial}{\partial \beta }\left [ \frac{3}{2}\ln\left ( \frac{m}{2 \pi\pounds^{2}} \right )+\ln V-\frac{3}{2}\ln \beta \right ]$
$\large =0+0+\frac{3}{2}\frac{\partial\ln \beta}{\partial \beta }=\frac{3}{2}\frac{1}{\beta }$

Karena $\large \beta =\frac{1}{KT}$ , maka

$\large E=\frac{3}{2}KT$

Bila dalam satu kotak terdapat N Gas monoatomik, maka

$\large E=\frac{3}{2}NKT$

Terbukti kan..???? hehehe

Itulah sedikit uraian tentang pembuktian energi gas monoatomik, semoga bermanfaat...!!!!

Max Planck

Belajar Fisika lewat Internet

Kamis, 21 Mei 2015

Energi Gas Monoatomik

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