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Display information for equation id:math.2558.19 on revision:2558

* Page found: Das ideale Bosegas (eq math.2558.19)

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\begin{align}
  & U=\left( 2s+1 \right)\frac{4\pi V}{{{h}^{3}}}\int_{0}^{\infty }{{}}dp{{p}^{2}}\frac{\frac{{{p}^{2}}}{2m}}{\exp \left( \frac{\left( \frac{{{p}^{2}}}{2m}-\mu  \right)}{kT} \right)-1} \\ 
 & \frac{{{p}^{2}}}{2mkT}=y \\ 
 & \Rightarrow U=\frac{\left( 2s+1 \right)}{2}\frac{4\pi V}{{{h}^{3}}}{{\left( 2mkT \right)}^{\frac{3}{2}kT}}\int_{0}^{\infty }{{}}dy{{y}^{\frac{3}{2}}}\frac{\xi {{e}^{-y}}}{1-\xi {{e}^{-y}}} \\ 
 & \int_{0}^{\infty }{{}}dy{{y}^{\frac{3}{2}}}\frac{\xi {{e}^{-y}}}{1-\xi {{e}^{-y}}}\approx \xi \int_{0}^{\infty }{{}}dy{{y}^{\frac{3}{2}}}{{e}^{-y}}+{{\xi }^{2}}\int_{0}^{\infty }{{}}dy{{y}^{\frac{1}{2}}}{{e}^{-2y}}+.... \\ 
 & \int_{0}^{\infty }{{}}dy{{y}^{\frac{3}{2}}}{{e}^{-y}}=\frac{3}{4}\sqrt{\pi } \\ 
 & \int_{0}^{\infty }{{}}dy{{y}^{\frac{3}{2}}}{{e}^{-2y}}=\frac{1}{{{2}^{\frac{5}{2}}}}\frac{3}{4}\sqrt{\pi } \\ 
 & \Rightarrow U\approx \frac{3}{2}kTV\frac{\left( 2s+1 \right)}{{{h}^{3}}}{{\left( 2\pi mkT \right)}^{\frac{3}{2}}}\left[ \xi +\frac{1}{{{2}^{\frac{5}{2}}}}{{\xi }^{2}} \right] \\ 
 & \lambda :={{\left( \frac{{{h}^{2}}}{2\pi mkT} \right)}^{\frac{1}{2}}}={{\left( \frac{2s+1}{{{N}_{C}}} \right)}^{\frac{1}{3}}} \\ 
 & \Rightarrow U\approx \frac{3}{2}\left( 2s+1 \right)\frac{VkT}{{{\lambda }^{3}}}\xi \left[ 1+\frac{1}{{{2}^{\frac{5}{2}}}}\xi  \right]=\frac{3}{2}\left( 2s+1 \right)kT\frac{V}{{{\lambda }^{3}}}{{e}^{\frac{\mu }{kT}}}\left[ 1+\frac{1}{{{2}^{\frac{5}{2}}}}{{e}^{\frac{\mu }{kT}}} \right] \\ 
\end{align}

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U=(2s+1)4πVh30dpp2p22mexp((p22mμ)kT)1p22mkT=yU=(2s+1)24πVh3(2mkT)32kT0dyy32ξey1ξey0dyy32ξey1ξeyξ0dyy32ey+ξ20dyy12e2y+....0dyy32ey=34π0dyy32e2y=125234πU32kTV(2s+1)h3(2πmkT)32[ξ+1252ξ2]λ:=(h22πmkT)12=(2s+1NC)13U32(2s+1)VkTλ3ξ[1+1252ξ]=32(2s+1)kTVλ3eμkT[1+1252eμkT]
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