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Display information for equation id:math.2556.18 on revision:2556

* Page found: Das ideale Bosegas (eq math.2556.18)

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Hash: 66e6c349020cf8df221c1713c5bcf8a3

TeX (original user input): 

\begin{align}
  & \bar{N}=\left( 2s+1 \right)\frac{V}{{{\lambda }^{3}}}{{e}^{\frac{\mu }{kT}}}\left[ 1+\frac{1}{{{2}^{\frac{3}{2}}}}\frac{\bar{N}{{\lambda }^{3}}}{V\left( 2s+1 \right)} \right] \\ 
 & \Rightarrow {{e}^{\frac{\mu }{kT}}}\approx \frac{\bar{N}{{\lambda }^{3}}}{V\left( 2s+1 \right)}\left[ 1-\frac{1}{{{2}^{\frac{3}{2}}}}\frac{\bar{N}{{\lambda }^{3}}}{V\left( 2s+1 \right)} \right] \\ 
\end{align}

TeX (checked): 

{\begin{aligned}&{\bar {N}}=\left(2s+1\right){\frac {V}{{\lambda }^{3}}}{{e}^{\frac {\mu }{kT}}}\left[1+{\frac {1}{{2}^{\frac {3}{2}}}}{\frac {{\bar {N}}{{\lambda }^{3}}}{V\left(2s+1\right)}}\right]\\&\Rightarrow {{e}^{\frac {\mu }{kT}}}\approx {\frac {{\bar {N}}{{\lambda }^{3}}}{V\left(2s+1\right)}}\left[1-{\frac {1}{{2}^{\frac {3}{2}}}}{\frac {{\bar {N}}{{\lambda }^{3}}}{V\left(2s+1\right)}}\right]\\\end{aligned}}

LaTeXML (experimentell; verwendet MathML) rendering

MathML (34.185 KB / 3.811 KB) :

N ¯ = ( 2 ⁢ s + 1 ) ⁢ V λ 3 ⁢ e μ k ⁢ T ⁢ [ 1 + 1 2 3 2 ⁢ N ¯ ⁢ λ 3 V ⁢ ( 2 ⁢ s + 1 ) ] ⇒ e μ k ⁢ T ≈ N ¯ ⁢ λ 3 V ⁢ ( 2 ⁢ s + 1 ) ⁢ [ 1 - 1 2 3 2 ⁢ N ¯ ⁢ λ 3 V ⁢ ( 2 ⁢ s + 1 ) ] missing-subexpression ¯ 𝑁 2 𝑠 1 𝑉 superscript 𝜆 3 superscript 𝑒 𝜇 𝑘 𝑇 delimited-[] 1 1 superscript 2 3 2 ¯ 𝑁 superscript 𝜆 3 𝑉 2 𝑠 1 missing-subexpression ⇒ absent superscript 𝑒 𝜇 𝑘 𝑇 ¯ 𝑁 superscript 𝜆 3 𝑉 2 𝑠 1 delimited-[] 1 1 superscript 2 3 2 ¯ 𝑁 superscript 𝜆 3 𝑉 2 𝑠 1 {\displaystyle{\displaystyle\begin{aligned} &\displaystyle\bar{N}=\left(2s+1% \right)\frac{V}{{{\lambda}^{3}}}{{e}^{\frac{\mu}{kT}}}\left[1+\frac{1}{{{2}^{% \frac{3}{2}}}}\frac{\bar{N}{{\lambda}^{3}}}{V\left(2s+1\right)}\right]\\ &\displaystyle\Rightarrow{{e}^{\frac{\mu}{kT}}}\approx\frac{\bar{N}{{\lambda}^% {3}}}{V\left(2s+1\right)}\left[1-\frac{1}{{{2}^{\frac{3}{2}}}}\frac{\bar{N}{{% \lambda}^{3}}}{V\left(2s+1\right)}\right]\\ \end{aligned}}}

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MathML (4.022 KB / 496 B) :

N¯=(2s+1)Vλ3eμkT[1+1232N¯λ3V(2s+1)]⇒eμkT≈N¯λ3V(2s+1)[1−1232N¯λ3V(2s+1)]

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