By subtracting $\mu N$from localid="1648229964064" $U,H,F$,or$G$,one can obtain four new thermodynamic potentials. Of the four, the most useful is the grand free energy (or grand potential),

$\Phi \equiv U-TS-\mu N.$

(a) Derive the thermodynamic identity for $\Phi$, and the related formulas for the partial derivatives of$\Phi$with respect to$T,V$, and $\mu N$

(b) Prove that, for a system in thermal and diffusive equilibrium (with a reservoir that can supply both energy and particles), $\Phi$tends to decrease.

(c) Prove that$\varphi =-PV.$

(d) As a simple application, let the system be a single proton, which can be "occupied" either by a single electron (making a hydrogen atom, with energy $-13.6\mathrm{eV}$) or by none (with energy zero). Neglect the excited states of the atom and the two spin states of the electron, so that both the occupied and unoccupied states of the proton have zero entropy. Suppose that this proton is in the atmosphere of the sun, a reservoir with a temperature of $5800\mathrm{K}$and an electron concentration of about $2\times {10}^{19}$per cubic meter. Calculate $\Phi$for both the occupied and unoccupied states, to determine which is more stable under these conditions. To compute the chemical potential of the electrons, treat them as an ideal gas. At about what temperature would the occupied and unoccupied states be equally stable, for this value of the electron concentration? (As in Problem 5.20, the prediction for such a small system is only a probabilistic one.)

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

Given:

The expression for the grand potential.

$\Phi =U-TS-\mu N\dots ..\left(1\right)$

Here, $\Phi$is grand potential

$U$is the internal energy

$T$is the absolute temperature

$S$is the entropy

$\mu$is the chemical potential and

$N$is the number of molecules.

Formula:

Write the expression for the internal energy of the system.

$U=TS-PV+\mu N\dots ..\left(2\right)$

Calculation:

Write the expression for the infinitesimal change of$\Phi$

$d\Phi =dU-TdS-SdT-\mu dN-Nd\mu \dots ..\left(3\right)$

Write the expression for the infinitesimal change of$U$

$dU=TdS-PdV+\mu dN\dots ..\left(4\right)$

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

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

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

${\left(\frac{\partial \Phi }{\partial \mu }\right)}_{V,T}=-N$

Rearrange expression (5) for constant $\mu$and $T$for which $d\mu =dT=0$.

${\left(\frac{\partial \Phi }{\partial V}\right)}_{\mu ,T}=-P$

Rearrange expression (5) for constant $\mu$and $V$for which $d\mu =dV=0$.

${\left(\frac{\partial \Phi }{\partial T}\right)}_{\mu ,V}=-S$

Hence,the related derivatives for grand potential are

${\left(\frac{\partial \Phi }{\partial \mu }\right)}_{V,T}=-N,{\left(\frac{\partial \Phi }{\partial V}\right)}_{\mu ,T}=-P\text{and}{\left(\frac{\partial \Phi }{\partial T}\right)}_{\mu ,V}=-S$

The reason for decrease in $\varphi$for a system being at thermal and diffusive equilibrium.

Given:

The expression for the grand potential is.

$\Phi =U-TS-\mu N$

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

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

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 $dV=0$.

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

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

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

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

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

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

Substitute $d\varphi$for $(dU-TdS-\mu dN)$in above expression.

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

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

The relation of$\Phi =-PV$

Given:

Write the grand potential's expression.

$\Phi =U-TS-\mu N$

Formula:

The internal energy of the system expression is

$U=TS-PV+\mu N$

The Gibbs energy of the system expression is

$G=U-TS+PV$

Rearrange expression (1).

$\Phi =(U-TS+PV)-PV-\mu N$

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

$\Phi =G-PV-\mu N$

Substitute $\mu N$for G in the above expression.

$$\begin{gathered} \Phi =\mu N-PV-\mu N \\ =-PV \end{gathered}$$

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 $2\times {10}^{19}{\mathrm{m}}^{-3}$.

Formula:

The grand potential of the expression is

$\Phi =U-TS-\mu N$

And

The chemical potential expression is

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

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\times {10}^{-23}{\mathrm{JK}}^{-1}$

$T=5800k$

$m=9.11\times {10}^{-31}kg$

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

$v=1.064\times {10}^{27}$

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

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

For unoccupied state, substitute 0 for U, S and N in the expression $\Phi =U-TS-\mu N$

so, ${\Phi }_{\text{unoccupied}}=0$

For occupied state, substitute 0 for U, S and N in the expression $\Phi =U-TS-\mu N$

so,${\Phi }_{\text{occupied}}=U_{\mathrm{g}}-\mu$

Now ${\Phi }_{\text{occupied}}=U_{\mathrm{g}}-\mu$

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

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

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

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

${\Phi }_{\text{unoccupied}}={\Phi }_{\text{occupied}}$

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

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

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

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

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

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.