91Ó°ÊÓ

Derive the related formulas from the thermodynamic identity for grand potential.

Given:

The expression for the grand potential.

$\mathrm{Φ}=U-TS-\mathrm{μ}N…..\left(1\right)$

Here, $\mathrm{Φ}$is grand potential

$U$is the internal energy

$T$is the absolute temperature

$S$is the entropy

$\mathrm{μ}$is the chemical potential and

$N$is the number of molecules.

Formula:

Write the expression for the internal energy of the system.

$U=TS-PV+\mathrm{μ}N…..\left(2\right)$

Calculation:

Write the expression for the infinitesimal change of$\mathrm{Φ}$

$d\mathrm{Φ}=dU-TdS-SdT-\mathrm{μ}dN-Nd\mathrm{μ}…..\left(3\right)$

Write the expression for the infinitesimal change of$U$

$dU=TdS-PdV+\mathrm{μ}dN…..\left(4\right)$

Substitutelocalid="1648229955784" $(TdS-PdV+\mathrm{μ}dN)$forlocalid="1648229313031" $dU$in expression (3).

localid="1648229656341" $$\begin{gathered} d\mathrm{Φ}=TdS-PdV+\mathrm{μ}dN-TdS-SdT-\mathrm{μ}dN-Nd\mathrm{μ}.......\left(5\right) \\ =-PdV-SdT-Nd\mathrm{μ} \end{gathered}$$

Rearrange expression (5) for constant$V$and$T$for which $dV=dT=0.$

${\left(\frac{∂\mathrm{Φ}}{∂\mathrm{μ}}\right)}_{V,T}=-N$

Rearrange expression (5) for constant $\mathrm{μ}$and $T$for which $d\mathrm{μ}=dT=0$.

${\left(\frac{∂\mathrm{Φ}}{∂V}\right)}_{\mathrm{μ},T}=-P$

Rearrange expression (5) for constant $\mathrm{μ}$and $V$for which $d\mathrm{μ}=dV=0$.

${\left(\frac{∂\mathrm{Φ}}{∂T}\right)}_{\mathrm{μ},V}=-S$

Hence,the related derivatives for grand potential are

${\left(\frac{∂\mathrm{Φ}}{∂\mathrm{μ}}\right)}_{V,T}=-N,{\left(\frac{∂\mathrm{Φ}}{∂V}\right)}_{\mathrm{μ},T}=-P\text{and}{\left(\frac{∂\mathrm{Φ}}{∂T}\right)}_{\mathrm{μ},V}=-S$

The reason for decrease in $\mathrm{Ï•}$for a system being at thermal and diffusive equilibrium.

Given:

The expression for the grand potential is.

$\mathrm{Φ}=U-TS-\mathrm{μ}N$

Write an expression for a system's total entropy change when it comes into contact with a reservoir.

role="math" localid="1648230417137" $d{S}_{\mathrm{tot}}=dS+d{S}_{\mathrm{R}}…..\left(6\right)$

Here, $d{S}_{\text{tot}}$is the total change in entropy,

$dS$is the change of entropy of the system and

$d{S}_{\mathrm{R}}$is the change of entropy of the reservoir.

Calculation:

Rearrange expression (4) for constant volume for which $dV=0$.

$d{S}_{\mathrm{R}}=\frac{1}{T}d{U}_{\mathrm{R}}-\frac{1}{T}\mathrm{μ}d{N}_{\mathrm{R}}…..\left(7\right)$

Rearrange expression ( 7 ) Considering that the reservoir's change in entropy is equal to the system's negative change in entropy.

$d{S}_{\mathrm{R}}=-\frac{1}{T}dU+\frac{1}{T}\mathrm{μ}dN$

Where, $dU$is the change of internal energy and $dN$is the change in number of molecules of the system.

Substitute $\left(-\frac{1}{T}dU+\frac{1}{T}\mathrm{μ}dN\right)$for$d{S}_R$in expression (6).

role="math" localid="1648230741206" $$\begin{gathered} d{S}_{\mathrm{tot}}=dS-\frac{1}{T}dU+\frac{1}{T}\mathrm{μ}dN \\ =-\frac{1}{T}(dU-TdS-\mathrm{μ}dN) \end{gathered}$$

Substitute $d\mathrm{ϕ}$for $(dU-TdS-\mathrm{μ}dN)$in above expression.

$d{S}_{\mathrm{tot}}=-\frac{1}{T}d\mathrm{Φ}$

Hence $\mathrm{Ï•}$for a system being at thermal and diffusive equilibrium tends to decrease with the increase of $d{S}_{\mathrm{tot}}$.

The relation of$\mathrm{Φ}=-PV$

Given:

Write the grand potential's expression.

$\mathrm{Φ}=U-TS-\mathrm{μ}N$

Formula:

The internal energy of the system expression is

$U=TS-PV+\mathrm{μ}N$

The Gibbs energy of the system expression is

$G=U-TS+PV$

Rearrange expression (1).

$\mathrm{Φ}=(U-TS+PV)-PV-\mathrm{μ}N$

Substitute $G$for $(U-TS+PV)$in the above expression.

$\mathrm{Φ}=G-PV-\mathrm{μ}N$

Substitute $\mathrm{μ}N$for G in the above expression.

$$\begin{gathered} \mathrm{Φ}=\mathrm{μ}N-PV-\mathrm{μ}N \\ =-PV \end{gathered}$$

Hence its proved.

The temperature at which both occupied and unoccupied states are equally stable, as well as the value of grand potential for both occupied and unoccupied states.

Given:

The temperature of the system is $5800\mathrm{K}$and the electron concentration is $2×{10}^{19}{\mathrm{m}}^{-3}$.

Formula:

The grand potential of the expression is

$\mathrm{Φ}=U-TS-\mathrm{μ}N$

And

The chemical potential expression is

$\mathrm{μ}=-k_{B}T\ln \left[\frac{V}{N}{\left(\frac{2\mathrm{π}m{k}_{B}T}{h^2}\right)}^{3/2}\right]……\left(8\right)$

Where,$k_B$is Boltzmann Constant,

localid="1648234771752" $m$is the mass of the particle,

$h$is Planck's Constant.

Substitute $K_B=$$1.38×{10}^{-23}{\mathrm{JK}}^{-1}$

$T=5800k$

$m=9.11×{10}^{-31}kg$

$h=6.626×{10}^{-34}\mathrm{J}·\mathrm{s}$

$v=1.064×{10}^{27}$

And $N=2.0×{10}^{19}{\mathrm{m}}^{-3}$in expression (8)

localid="1648233393234" $$\begin{gathered} \mathrm{μ}=-\left(1.38×{10}^{-23}{\mathrm{JK}}^{-1}\right)(5800\mathrm{K})\ln \left[\frac{1.064×{10}^{27}}{2.0×{10}^{19}{\mathrm{m}}^{-3}}{\left(\frac{2\mathrm{π}\left(9.11×{10}^{-31}\mathrm{Kg}\right)\left(1.38×{10}^{-23}{\mathrm{JK}}^{-1}\right)(5800\mathrm{K})}{{\left(6.626×{10}^{-34}\mathrm{J}-\mathrm{s}\right)}^2}\right)}^{3/2}\right] \\ =-1.424×{10}^{-18}\mathrm{J} \\ =-1.424×{10}^{-18}\mathrm{J}\left(\frac{{10}^{19}\mathrm{eV}}{1.6\mathrm{J}}\right) \\ =-8.9\mathrm{eV} \end{gathered}$$

For unoccupied state, substitute 0 for U, S and N in the expression $\mathrm{Φ}=U-TS-\mathrm{μ}N$

so, ${\mathrm{Φ}}_{\text{unoccupied}}=0$

For occupied state, substitute 0 for U, S and N in the expression $\mathrm{Φ}=U-TS-\mathrm{μ}N$

so,${\mathrm{Φ}}_{\text{occupied}}=U_{\mathrm{g}}-\mathrm{μ}$

Now ${\mathrm{Φ}}_{\text{occupied}}=U_{\mathrm{g}}-\mathrm{μ}$

Where $U_g$is the internal energy of the ground state

Substitute $$\begin{gathered} u_g=-13.6\mathrm{eV} \\ \mathrm{μ}=-8.9eV \end{gathered}$$in above expression

$$\begin{gathered} {\mathrm{Φ}}_{\text{occupied}}=-13.6\mathrm{eV}+8.9\mathrm{eV} \\ =-4.7eV \end{gathered}$$

Write the condition for stability of both occupied and unoccupied states.

${\mathrm{Φ}}_{\text{unoccupied}}={\mathrm{Φ}}_{\text{occupied}}$

The expression for $U_{\mathrm{g}}=\mathrm{μ}N$

Now substitute $\mathrm{μ}N=-17.8\left(k_{B}T\right)$in above expression

role="math" localid="1648234514735" $T=-\frac{U_g}{\left(17.8k_B\right)}$

Now the value of $$\begin{gathered} u_g=-13.6\mathrm{eV} \\ {k}_B=8.6×{10}^{-5}{\mathrm{eVK}}^{-1} \end{gathered}$$in above expression

$$\begin{gathered} T=-\frac{(-13.6\mathrm{eV})}{(17.8)\left(8.6×{10}^{-5}{\mathrm{eVK}}^{-1}\right)} \\ =8866k \end{gathered}$$

Hence As a result, the grand potential for an unoccupied state is 0 and$-4.7eV$for an occupied state.$8866k$ is the required temperature.