91影视

If $T<T_c$,

$C_V=5.0306{\left(\frac{2蟺m}{h^2}\right)}^{\frac{3}{2}}V{\left(K_{B}T\right)}^{\frac{3}{2}}{K}_B$

$U=\frac{2}{\sqrt{蟺}}\left(\frac{2蟺m}{h^2}\right)V路{\left(K_{B}T\right)}^{\frac{5}{2}}(1.7833)$

$=2.0122{\left(\frac{2蟺m}{h^2}\right)}^{\frac{3}{2}}V路{\left(K_{B}T\right)}^{\frac{5}{2}}$

Here,

$h$= Planck's constant,

$K_B$= Boltzmann's constant,

$V$= volume of the box,

$T$= temperature,

$m$= mass of the particle,

We have,

$S={鈭�}_0^{T}5.0306{\left(\frac{2蟺m}{h^2}\right)}^{\frac{3}{2}}V路K_B^{\frac{5}{2}}\frac{T^{\text{'}\frac{3}{2}}}{T^{\text{'}}}d{T}^{\text{'}}$

$=5.0306{\left(\frac{2蟺m}{h^2}\right)}^{\frac{3}{2}}V路K_B^{\frac{5}{2}}{鈭�}_0^T{T}^{\text{'}\frac{1}{2}}d{T}^{\text{'}}$

$={\left.5.0306{\left(\frac{2蟺m}{h^2}\right)}^{\frac{3}{2}}V路K_B^{\frac{5}{2}}\frac{T^{\frac{3}{2}}}{\frac{3}{2}}\right|}_0^T$

$=5.0306路\frac{2}{3}路{\left(\frac{2蟺m}{h^2}\right)}^{\frac{3}{2}}V路{\left(K_{B}T\right)}^{\frac{3}{2}}{K}_B$

$=5.0306路\frac{2}{3}路{\left(\frac{2蟺m}{h^2}\right)}^{\frac{3}{2}}V路{\left(K_{B}T\right)}^{\frac{3}{2}}{K}_B$

$S=3.3537{\left(\frac{2蟺m}{h^2}\right)}^{\frac{3}{2}}V路{\left(K_{B}T\right)}^{\frac{3}{2}}{K}_B$

We have,

$F=U-TS$

$=2.0122{\left(\frac{2蟺m}{h^2}\right)}^{\frac{3}{2}}V路{\left(K_{B}T\right)}^{\frac{5}{2}}-T\left(5.03056脳\frac{2}{3}{\left(\frac{2蟺m}{h^2}\right)}^{\frac{3}{2}}V路{\left(K_{B}T\right)}^{\frac{3}{2}}{K}_B\right)$

$={\left(\frac{2蟺m}{h^2}\right)}^{\frac{3}{2}}V路{\left(K_{B}T\right)}^{\frac{5}{2}}[2.0122-3.3537]$

$=-1.3415{\left(\frac{2蟺m{K}_{B}T}{h^2}\right)}^{\frac{3}{2}}V路\left(K_{B}T\right)$

$鈭�F=-1.3415{\left(\frac{2蟺m}{h^2}\right)}^{\frac{3}{2}}V路{\left(K_{B}T\right)}^{\frac{5}{2}}$

We have,

$P=-{\left(\frac{鈭�F}{鈭�V}\right)}_{N,T}$

We get to know that pressure is independent of volume and a function of temperature '$T$' only as for condensing gas.

Further reduction in the volume would condense more particles in ground state in the limit$T<T_c$