Yes, you can find a relation between $E_F$ and other functions. However remember that this value is a parameter, not a function.

The average number of particles in state $k$ in Fermi-Dirac statistic is

$$<n_k> = \frac{1}{e^{\beta(E-\mu)}+1}$$

With $\beta = 1/kT$. At a very low temperatures, you find that $<n_k>$ can only have two values: 0, if $E_F > \mu$ or 1, if $E_F < \mu$, that is a state can be empty or occupied, so we can write $<n_k>=\theta(E_F-E)$. In this expression $\theta(x-a)$ is the Heaviside step function.

You can use this on the integrals to calculate density and pressure. If you remember, these integrals are:

$$\rho(\mu,T) = 2\pi g { \left ( \frac{2m}{h^2} \right ) }^{3/2} \int _0 ^{+\infty} E^{1/2}<n_k> dE$$ $$P(\mu,T) = \frac{2}{3}\pi g { \left ( \frac{2m}{h^2} \right ) }^{3/2} \int _0 ^{+\infty}E^{3/2}<n_k> dE$$

Where $g$ is the spin degeneration. This integrals can be solved easily with $<n_k>=\theta(E_F-E)$. With that, you will find expressions for a gas strongly degenerated, at $T\simeq 0$. Using that you find that you can write $E_F$ in terms of the density:

$$E_F = {\left ( \frac{6\pi}{g} \right ) }^{2/3} \frac{h^2}{2m}\rho^{2/3} = \mu(T=0)$$

Here you can see that $E_F$ is a constant value, and it equals the value of $\mu$ at $T\simeq0$. You can also write, from the second integral, $P(T=0)=\frac{2}{3}\rho E_F$, and using the expression $P = \frac{2U}{3V}$, valid for an ideal, non-relativistic quantum gas, you can write:

$$U(T=0) = \frac{3}{5}\rho E_F$$

So now you have the Fermi energy related with $U$. Remember, this is for a quantum gas where the temperature is really low. Note that all the equations are constants, calculated for $T\rightarrow 0$. If you want another aproximation, for higher $T$, the integrals on density and pressure can be approximated as follows:

$$I = \int _0 ^{+\infty} \frac{f(E)}{<n_k>} dE \simeq F(\mu) + \frac{\pi^2}{6}(kT)^2{\left ( \frac{df(E)}{dE} \right ) }_{E=\mu}$$

with

$$F(E) = \int _0 ^E f(x)dx$$

I'm not going to demonstrate this expression here because it is very long, however, I think you can find it in some classic books like McQuarrie or Pathria. Using this, you can calculate another expression for pressure and density, also for low $T$, but not for $T\simeq0$. To properly work with that you'll have to use lot of approximations during the calculus. You can find now $\mu(T)$

$$\mu(T) \simeq E_F \left [ 1- \frac{pi^2}{12} {\left ( \frac{kT}{E_F} \right ) }^2 \right ]$$

And use this find the dependence of some thermodynamics potentials with $T$:

$$U(T) = \frac{3}{5} N {E_F}\left [ 1+ \frac{5\pi^2}{12} {\left ( \frac{kT}{E_F} \right ) }^2 \right ]$$

$$S(T) = \frac{\pi^2Nk^2T}{2E_F}$$

I think this are some of the expressions you wanted. The first approximation works for $T\simeq 0$ and the second one for low $T$.

Yes, you can find a relation between $E_F$ and other functions. However remember that this value is a parameter, not a function.

The average number of particles in state $k$ in Fermi-Dirac statistic is

$$<n_k> = \frac{1}{e^{\beta(E-\mu)}+1}$$

With $\beta = 1/kT$. At a very low temperatures, you find that $<n_k>$ can only have two values: 0, if $E_F > \mu$ or 1, if $E_F < \mu$, that is a state can be empty or occupied, so we can write $<n_k>=\theta(E_F-E)$. In this expression $\theta(x-a)$ is the Heaviside step function.

You can use this on the integrals to calculate density and pressure. If you remember, these integrals are:

$$\rho(\mu,T) = 2\pi g { \left ( \frac{2m}{h^2} \right ) }^{3/2} \int _0 ^{+\infty} E^{1/2}<n_k> dE$$ $$P(\mu,T) = \frac{2}{3}\pi g { \left ( \frac{2m}{h^2} \right ) }^{3/2} \int _0 ^{+\infty}E^{3/2}<n_k> dE$$

Where $g$ is the spin degeneration. This integrals can be solved easily with $<n_k>=\theta(E_F-E)$. With that, you will find expressions for a gas strongly degenerated, at $T\simeq 0$. Using that you find that you can write $E_F$ in terms of the density:

$$E_F = {\left ( \frac{6\pi}{g} \right ) }^{2/3} \frac{h^2}{2m}\rho^{2/3} = \mu(T=0)$$

Here you can see that $E_F$ is a constant value, and it equals the value of $\mu$ at $T\simeq0$. You can also write, from the second integral, $P(T=0)=\frac{2}{3}\rho E_F$, and using the expression $P = \frac{2U}{3V}$, valid for an ideal, non-relativistic quantum gas, you can write:

$$U(T=0) = \frac{3}{5}\rho E_F$$

So now you have the Fermi energy related with $U$. Remember, this is for a quantum gas where the temperature is really low. Note that all the equations are constants, calculated for $T\rightarrow 0$. If you want another aproximation, for higher $T$, the integrals on density and pressure can be approximated as follows:

$$I = \int _0 ^{+\infty} \frac{f(E)}{<n_k>} dE \simeq F(\mu) + \frac{\pi^2}{6}(kT)^2{\left ( \frac{df(E)}{dE} \right ) }_{E=\mu}$$

with

$$F(E) = \int _0 ^E f(x)dx$$

I'm not going to demonstrate this expression here because it is very long, however, I think you can find it in some classic books like McQuarrie or Pathria. Using this, you can calculate another expression for pressure and density, also for low $T$, but not for $T\simeq0$. To properly work with that you'll have to use lot of approximations during the calculus. You can find now $\mu(T)$

$$\mu(T) \simeq E_F \left [ 1- \frac{\pi^2}{12} {\left ( \frac{kT}{E_F} \right ) }^2 \right ]$$

And use this find the dependence of some thermodynamics potentials with $T$:

$$U(T) = \frac{3}{5} N {E_F}\left [ 1+ \frac{5\pi^2}{12} {\left ( \frac{kT}{E_F} \right ) }^2 \right ]$$

$$S(T) = \frac{\pi^2Nk^2T}{2E_F}$$

I think this are some of the expressions you wanted. The first approximation works for $T\simeq 0$ and the second one for low $T$.

Yes, you can find a relation between $E_F$ and other functions. However remember that this value is a parameter, not a function.

The average number of particles in state $k$ in Fermi-Dirac statistic is

$$<n_k> = \frac{1}{e^{\beta(E-\mu)}+1}$$

With $\beta = 1/kT$. At a very low temperatures, you find that $<n_k>$ can only have two values: 0, if $E_F > \mu$ or 1, if $E_F < \mu$, that is a state can be empty or occupied, so we can write $<n_k>=\theta(E_F-E)$. In this expression $\theta(x-a)$ is the Heaviside step function.

You can use this on the integrals to calculate density and pressure. If you remember, these integrals are:

$$\rho(\mu,T) = 2\pi g { \left ( \frac{2m}{h^2} \right ) }^{3/2} \int _0 ^{+\infty} \frac{E^{1/2}}{<n_k>} dE$$ $$P(\mu,T) = \frac{2}{3}\pi g { \left ( \frac{2m}{h^2} \right ) }^{3/2} \int _0 ^{+\infty}\frac{E^{3/2}}{<n_k>} dE$$

Where $g$ is the spin degeneration. This integrals can be solved easily with $<n_k>=\theta(E_F-E)$. With that, you will find expressions for a gas strongly degenerated, at $T\simeq 0$. Using that you find that you can write $E_F$ in terms of the density:

$$E_F = {\left ( \frac{6\pi}{g} \right ) }^{2/3} \frac{h^2}{2m}\rho^{2/3} = \mu(T=0)$$

Here you can see that $E_F$ is a constant value, and it equals the value of $\mu$ at $T\simeq0$. You can also write, from the second integral, $P(T=0)=\frac{2}{3}\rho E_F$, and using the expression $P = \frac{2U}{3V}$, valid for an ideal, non-relativistic quantum gas, you can write:

$$U(T=0) = \frac{3}{5}\rho E_F$$

So now you have the Fermi energy related with $U$. Remember, this is for a quantum gas where the temperature is really low. Note that all the equations are constants, calculated for $T\rightarrow 0$. If you want another aproximation, for higher $T$, the integrals on density and pressure can be approximated as follows:

$$I = \int _0 ^{+\infty} \frac{f(E)}{<n_k>} dE \simeq F(\mu) + \frac{\pi^2}{6}(kT)^2{\left ( \frac{df(E)}{dE} \right ) }_{E=\mu}$$

with

$$F(E) = \int _0 ^E f(x)dx$$

I'm not going to demonstrate this expression here because it is very long, however, I think you can find it in some classic books like McQuarrie or Pathria. Using this, you can calculate another expression for pressure and density, also for low $T$, but not for $T\simeq0$. To properly work with that you'll have to use lot of approximations during the calculus. You can find now $\mu(T)$

$$\mu(T) \simeq E_F \left [ 1- \frac{pi^2}{12} {\left ( \frac{kT}{E_F} \right ) }^2 \right ]$$

And use this find the dependence of some thermodynamics potentials with $T$:

$$U(T) = \frac{3}{5} N {E_F}\left [ 1+ \frac{5\pi^2}{12} {\left ( \frac{kT}{E_F} \right ) }^2 \right ]$$

$$S(T) = \frac{\pi^2Nk^2T}{2E_F}$$

I think this are some of the expressions you wanted. The first approximation works for $T\simeq 0$ and the second one for low $T$.

Yes, you can find a relation between $E_F$ and other functions. However remember that this value is a parameter, not a function.

The average number of particles in state $k$ in Fermi-Dirac statistic is

$$<n_k> = \frac{1}{e^{\beta(E-\mu)}+1}$$

With $\beta = 1/kT$. At a very low temperatures, you find that $<n_k>$ can only have two values: 0, if $E_F > \mu$ or 1, if $E_F < \mu$, that is a state can be empty or occupied, so we can write $<n_k>=\theta(E_F-E)$. In this expression $\theta(x-a)$ is the Heaviside step function.

You can use this on the integrals to calculate density and pressure. If you remember, these integrals are:

$$\rho(\mu,T) = 2\pi g { \left ( \frac{2m}{h^2} \right ) }^{3/2} \int _0 ^{+\infty} E^{1/2}<n_k> dE$$ $$P(\mu,T) = \frac{2}{3}\pi g { \left ( \frac{2m}{h^2} \right ) }^{3/2} \int _0 ^{+\infty}E^{3/2}<n_k> dE$$

Where $g$ is the spin degeneration. This integrals can be solved easily with $<n_k>=\theta(E_F-E)$. With that, you will find expressions for a gas strongly degenerated, at $T\simeq 0$. Using that you find that you can write $E_F$ in terms of the density:

$$E_F = {\left ( \frac{6\pi}{g} \right ) }^{2/3} \frac{h^2}{2m}\rho^{2/3} = \mu(T=0)$$

Here you can see that $E_F$ is a constant value, and it equals the value of $\mu$ at $T\simeq0$. You can also write, from the second integral, $P(T=0)=\frac{2}{3}\rho E_F$, and using the expression $P = \frac{2U}{3V}$, valid for an ideal, non-relativistic quantum gas, you can write:

$$U(T=0) = \frac{3}{5}\rho E_F$$

So now you have the Fermi energy related with $U$. Remember, this is for a quantum gas where the temperature is really low. Note that all the equations are constants, calculated for $T\rightarrow 0$. If you want another aproximation, for higher $T$, the integrals on density and pressure can be approximated as follows:

$$I = \int _0 ^{+\infty} \frac{f(E)}{<n_k>} dE \simeq F(\mu) + \frac{\pi^2}{6}(kT)^2{\left ( \frac{df(E)}{dE} \right ) }_{E=\mu}$$

with

$$F(E) = \int _0 ^E f(x)dx$$

I'm not going to demonstrate this expression here because it is very long, however, I think you can find it in some classic books like McQuarrie or Pathria. Using this, you can calculate another expression for pressure and density, also for low $T$, but not for $T\simeq0$. To properly work with that you'll have to use lot of approximations during the calculus. You can find now $\mu(T)$

$$\mu(T) \simeq E_F \left [ 1- \frac{pi^2}{12} {\left ( \frac{kT}{E_F} \right ) }^2 \right ]$$

And use this find the dependence of some thermodynamics potentials with $T$:

$$U(T) = \frac{3}{5} N {E_F}\left [ 1+ \frac{5\pi^2}{12} {\left ( \frac{kT}{E_F} \right ) }^2 \right ]$$

$$S(T) = \frac{\pi^2Nk^2T}{2E_F}$$

I think this are some of the expressions you wanted. The first approximation works for $T\simeq 0$ and the second one for low $T$.