91影视

The expression of total number of particles is given by,

$\mathit{N}\mathbf{=}{鈭�}_0^{\mathrm{鈭�}}N\mathbf{\left(}\mathit{E}\mathbf{\right)}\mathit{D}\mathbf{\left(}\mathit{E}\mathbf{\right)}\mathit{d}\mathit{E}$

The expression of total energy is given by,

${\mathit{U}}_{\mathbf{\text{total}}}\mathbf{=}{鈭�}_0^{\mathrm{鈭�}}E\mathit{N}\mathbf{\left(}\mathit{E}\mathbf{\right)}\mathit{D}\mathbf{\left(}\mathit{E}\mathbf{\right)}\mathit{d}\mathit{E}$

The expression for average energy is given by,

$\stackrel{\mathbf{炉}}{\mathbf{E}}\mathbf{=}\frac{{\mathbf{U}}_{\mathbf{\text{total}}}}{\mathbf{N}}$

The total number of particles is $N$.

The total energy of system is $U_{\text{total}}$Calculation:

The expression of total number of particles is given by,

$\begin{array}{c}N={鈭�}_0^{\mathrm{鈭�}}N\left(E\right)D\left(E\right)dE\\ ={鈭�}_0^{\mathrm{鈭�}}\left[\frac{(2s+1)m^{\frac{3}{2}}V}{{蟺}^2{\mathrm{鈩�}}^3\sqrt{2}}{E}^{\frac{1}{2}}\right]\left[\frac{1}{B{e}^{\frac{E}{k_{\text{B}}T}}鈭�1}\right]dE\\ =\frac{(2s+1)m^{\frac{3}{2}}V}{{蟺}^2{\mathrm{鈩�}}^3\sqrt{2}}{鈭�}_0^{\mathrm{鈭�}}\frac{E^{\frac{1}{2}}}{B{e}^{\frac{E}{{\text{k}}^{2}T}}鈭�1}dE\end{array}$

Let$E=y^2$ then$dE=2ydy$,

Substitute$2ydy$for$dE$and$y^2$for $E$in equation (1).

$\begin{array}{c}dN=\frac{(2s+1)m^{\frac{3}{2}}V}{{蟺}^2{\mathrm{鈩�}}^3\sqrt{2}}{鈭�}_0^{\mathrm{鈭�}}\frac{{\left(y^2\right)}^{\frac{1}{2}}}{B{e}^{\frac{E}{k_{B}T}}鈭�1}\left(2ydy\right)\\ =\frac{2(2s+1)m^{\frac{3}{2}}V}{{蟺}^2{\mathrm{鈩�}}^3\sqrt{2}}{鈭�}_0^{\mathrm{鈭�}}\frac{y^2}{B{e}^{\frac{E}{k_{\text{B}}T}}鈭�1}dy\\ =\frac{2(2s+1)m^{\frac{3}{2}}V}{{蟺}^2{\mathrm{鈩�}}^3\sqrt{2}}{鈭�}_0^{\mathrm{鈭�}}\frac{y^2}{B{e}^{\frac{y^2}{k_{B}T}}}(1鈭�e^{鈭�\frac{y^2}{k_{\text{B}}T}})dy\\ =\frac{2(2s+1)m^{\frac{3}{2}}V}{{蟺}^2{\mathrm{鈩�}}^3\sqrt{2}}[{鈭�}_0^{\mathrm{鈭�}}\frac{y^2}{B{e}^{\frac{y^2}{k_{\text{B}}T}}}卤\frac{y^2{e}^{鈭�\frac{y^2}{k_{\text{B}}T}}}{e^{\frac{y^2}{k_{\text{B}}T}}}dy]\end{array}$

Further simplify the above,

$\begin{array}{c}N=\frac{2(2s+1)m^{\frac{3}{2}}V}{{蟺}^2{\mathrm{鈩�}}^3\sqrt{2}}[\frac{1}{4B}\sqrt{蟺{\left(k_{\text{B}}T\right)}^3}卤\frac{1}{4B^2}\sqrt{蟺{\left(\frac{k_{\text{B}}T}{2}\right)}^3}]\\ =\frac{(2s+1)V}{B}{\left(\frac{m{k}_{\text{B}}T}{2蟺{\mathrm{鈩�}}^2}\right)}^{\frac{3}{2}}[1卤\frac{1}{2B\sqrt{2}}]\end{array}$

The expression of total energy is given by,

$\begin{array}{c}U\left({}_{\text{total}}\right)={鈭�}_0^{\mathrm{鈭�}}EN\left(E\right)D\left(E\right)dE\\ ={鈭�}_0^{\mathrm{鈭�}}E\left[\frac{(2s+1)m^{\frac{3}{2}}V}{{蟺}^2{\mathrm{鈩�}}^3\sqrt{2}}{E}^{\frac{1}{2}}\right]\left[\frac{1}{B{e}^{\frac{E}{k_{\text{B}}T}}鈭�1}\right]dE\\ =\frac{(2s+1)m^{\frac{3}{2}}V}{{蟺}^2{\mathrm{鈩�}}^3\sqrt{2}}{鈭�}_0^{\mathrm{鈭�}}\frac{E^{\frac{3}{2}}}{B{e}^{\frac{E}{k_{B}T}}鈭�1}dE\end{array}$

Let $E=y^2$then$dE=2ydy$,

Substitute $2ydy$for$dE$ and $y^2$for$E$ in equation (2).

$\begin{array}{c}{U}_{\text{total}}=\frac{(2s+1)m^{\frac{3}{2}}V}{{蟺}^2{\mathrm{鈩�}}^3\sqrt{2}}{鈭�}_0^{\mathrm{鈭�}}\frac{{\left(y^2\right)}^{\frac{3}{2}}}{B{e}^{\frac{E}{k_{\text{B}}T}}鈭�1}\left(2ydy\right)\\ =\frac{2(2s+1)m^{\frac{3}{2}}V}{{蟺}^2{\mathrm{鈩�}}^3\sqrt{2}}{鈭�}_0^{\mathrm{鈭�}}\frac{y^4}{B{e}^{\frac{E}{k_{B}T}}鈭�1}dy\\ =\frac{2(2s+1)m^{\frac{3}{2}}V}{{蟺}^2{\mathrm{鈩�}}^3\sqrt{2}}{鈭�}_0^{\mathrm{鈭�}}\frac{y^4}{B{e}^{\frac{y^2}{k_{\text{B}}T}}}(1鈭�e^{鈭�\frac{y^2}{k_{\text{B}}T}})dy\\ =\frac{2(2s+1)m^{\frac{3}{2}}V}{{蟺}^2{\mathrm{鈩�}}^3\sqrt{2}}[{鈭�}_0^{\mathrm{鈭�}}\frac{y^4}{B{e}^{\frac{y^2}{k^2}}}卤\frac{y^4{e}^{鈭�\frac{y^2}{k_{\text{B}}T}}}{e^{\frac{y^2}{k_{\text{B}}T}}}dy]\end{array}$

Further simplify the above,

$\begin{array}{c}{U}_{\text{total}}=\frac{2(2s+1)m^{\frac{3}{2}}V}{{蟺}^2{\mathrm{鈩�}}^3\sqrt{2}}[\frac{3}{8B}\sqrt{蟺{\left(k_{\text{B}}T\right)}^5}卤\frac{3}{8B^2}\sqrt{蟺{\left(\frac{k_{\text{B}}T}{2}\right)}^5}]\\ =\frac{3(2s+1)V{k}_{\text{B}}T}{2B}{\left(\frac{m{k}_{\text{B}}T}{2蟺{\mathrm{鈩�}}^2}\right)}^{\frac{3}{2}}[1卤\frac{1}{4B\sqrt{2}}]\end{array}$

The expression for average energy is calculated as,

$\begin{array}{c}\stackrel{炉}{E}=\frac{U_{\text{total}}}{N}\\ =\frac{\frac{3(2s+1)V{k}_{\text{B}}T}{2B}{\left(\frac{m{k}_{\text{B}}T}{2蟺{\mathrm{鈩�}}^2}\right)}^{\frac{3}{2}}[1卤\frac{1}{4B\sqrt{2}}]}{\frac{(2s+1)V}{B}{\left(\frac{m{k}_{\text{B}}T}{2蟺{\mathrm{鈩�}}^2}\right)}^{\frac{3}{2}}[1卤\frac{1}{2B\sqrt{2}}]}\\ 鈮�\frac{3}{2}{k}_{\text{B}}T[1卤\frac{1}{4B\sqrt{2}}+鈥�]\\ 鈮�\frac{3}{2}{k}_{\text{B}}T[1卤\frac{1}{4\sqrt{2}}\left\{\frac{1}{(2s+1)}{\left(\frac{2蟺{\mathrm{鈩�}}^2}{m{k}_{\text{B}}T}\right)}^{\frac{3}{2}}\left(\frac{N}{V}\right)\right\}+鈥�]\end{array}$

Further simplify the above,

$\begin{array}{c}\stackrel{炉}{E}鈮�\frac{3}{2}{k}_{\text{B}}T[1卤\frac{2\sqrt{2}{蟺}^3{\mathrm{鈩�}}^3}{2\left(2\right)\frac{3}{2}(2s+1)}{\left(\frac{1}{蟺m{k}_{\text{B}}T}\right)}^{\frac{3}{2}}\left(\frac{N}{V}\right)+鈥�]\\ 鈮�\frac{3}{2}{k}_{\text{B}}T[1卤\frac{\sqrt{2}{蟺}^3{\mathrm{鈩�}}^3}{(2s+1)}{\left(\frac{1}{2蟺m{k}_{\text{B}}T}\right)}^{\frac{3}{2}}\left(\frac{N}{V}\right)+鈥�]\end{array}$