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Display information for equation id:math.2555.97 on revision:2555

* Page found: Das ideale Fermigas (eq math.2555.97)

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TeX (original user input): 

\begin{align}

& \bar{N}=V{{N}_{C}}\xi \left[ 1-{{2}^{-\frac{3}{2}}}\frac{{\bar{N}}}{V{{N}_{C}}} \right] \\

& \Rightarrow \xi ={{e}^{\frac{\mu }{kT}}}\approx \frac{{\bar{N}}}{V{{N}_{C}}}\left[ 1+{{2}^{-\frac{3}{2}}}\frac{{\bar{N}}}{V{{N}_{C}}} \right] \\

& \Rightarrow U\approx V{{N}_{C}}\frac{3}{2}kT{{e}^{\frac{\mu }{kT}}}\left[ 1-{{e}^{\frac{\mu }{kT}}}\frac{1}{{{2}^{\frac{5}{2}}}} \right]\approx \frac{3}{2}kT\bar{N}\left[ 1+{{2}^{-\frac{3}{2}}}\frac{{\bar{N}}}{V{{N}_{C}}} \right]\left[ 1-\frac{1}{{{2}^{\frac{5}{2}}}}\frac{{\bar{N}}}{V{{N}_{C}}} \right] \\

\end{align}

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N ¯ = V ⁢ N C ⁢ ξ ⁢ [ 1 - 2 - 3 2 ⁢ N ¯ V ⁢ N C ] ⇒ ξ = e μ k ⁢ T ≈ N ¯ V ⁢ N C ⁢ [ 1 + 2 - 3 2 ⁢ N ¯ V ⁢ N C ] ⇒ U ≈ V ⁢ N C ⁢ 3 2 ⁢ k ⁢ T ⁢ e μ k ⁢ T ⁢ [ 1 - e μ k ⁢ T ⁢ 1 2 5 2 ] ≈ 3 2 ⁢ k ⁢ T ⁢ N ¯ ⁢ [ 1 + 2 - 3 2 ⁢ N ¯ V ⁢ N C ] ⁢ [ 1 - 1 2 5 2 ⁢ N ¯ V ⁢ N C ] absent ¯ 𝑁 𝑉 subscript 𝑁 𝐶 𝜉 delimited-[] 1 superscript 2 3 2 ¯ 𝑁 𝑉 subscript 𝑁 𝐶 absent ⇒ absent 𝜉 superscript 𝑒 𝜇 𝑘 𝑇 ¯ 𝑁 𝑉 subscript 𝑁 𝐶 delimited-[] 1 superscript 2 3 2 ¯ 𝑁 𝑉 subscript 𝑁 𝐶 absent ⇒ absent 𝑈 𝑉 subscript 𝑁 𝐶 3 2 𝑘 𝑇 superscript 𝑒 𝜇 𝑘 𝑇 delimited-[] 1 superscript 𝑒 𝜇 𝑘 𝑇 1 superscript 2 5 2 3 2 𝑘 𝑇 ¯ 𝑁 delimited-[] 1 superscript 2 3 2 ¯ 𝑁 𝑉 subscript 𝑁 𝐶 delimited-[] 1 1 superscript 2 5 2 ¯ 𝑁 𝑉 subscript 𝑁 𝐶 {\displaystyle{\displaystyle\begin{aligned} \par&\displaystyle\bar{N}=V{{N}_{C% }}\xi\left[1-{{2}^{-\frac{3}{2}}}\frac{{\bar{N}}}{V{{N}_{C}}}\right]\\ \par&\displaystyle\Rightarrow\xi={{e}^{\frac{\mu}{kT}}}\approx\frac{{\bar{N}}}% {V{{N}_{C}}}\left[1+{{2}^{-\frac{3}{2}}}\frac{{\bar{N}}}{V{{N}_{C}}}\right]\\ \par&\displaystyle\Rightarrow U\approx V{{N}_{C}}\frac{3}{2}kT{{e}^{\frac{\mu}% {kT}}}\left[1-{{e}^{\frac{\mu}{kT}}}\frac{1}{{{2}^{\frac{5}{2}}}}\right]% \approx\frac{3}{2}kT\bar{N}\left[1+{{2}^{-\frac{3}{2}}}\frac{{\bar{N}}}{V{{N}_% {C}}}\right]\left[1-\frac{1}{{{2}^{\frac{5}{2}}}}\frac{{\bar{N}}}{V{{N}_{C}}}% \right]\\ \par\end{aligned}}}

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N¯=VNCξ[1−2−32N¯VNC]⇒ξ=eμkT≈N¯VNC[1+2−32N¯VNC]⇒U≈VNC32kTeμkT[1−eμkT1252]≈32kTN¯[1+2−32N¯VNC][1−1252N¯VNC]

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