# Some explicit computations for the monoatomic ideal gas

Now that we have fixed everything, in order to understand if what we have done until now makes sense let us use all this machinery to see what happens when we apply it to the monoatomic ideal gas (which is the simplest system we can think of). From what we have just seen above the phase space volume of the system is:

${\displaystyle \Omega (E,V,N)={\frac {V^{N}}{N!h^{3N}}}{\frac {(2\pi mE)^{3N/2}}{\Gamma \left({\frac {3}{2}}N+1\right)}}}$
and the entropy can be rewritten more shortly as:
${\displaystyle S(E,V,N)=Nk_{B}\left[{\frac {5}{2}}-\ln(\rho \lambda ^{3})\right]}$
where:
${\displaystyle \rho ={\frac {N}{V}}\quad \qquad \lambda ={\sqrt {{\frac {3h^{2}}{4\pi m}}\cdot {\frac {N}{E}}}}}$
Therefore, we can now compute:
${\displaystyle {\frac {1}{T}}={\frac {\partial S}{\partial E}}_{|V,N}={\frac {3}{2}}N{\frac {k_{B}}{E}}}$
and thus:
${\displaystyle E={\frac {3}{2}}Nk_{B}T}$
We have really obtained the expression of the energy for an ideal gas!

Now, going further:

${\displaystyle {\frac {P}{T}}={\frac {\partial S}{\partial V}}_{|E,N}={\frac {Nk_{B}}{V}}}$
from which we get the ideal gas law:
${\displaystyle PV=Nk_{B}T}$
The efforts we have done are therefore not useless, since we are really obtaining all the thermodynamics of the system from the fundamental postulate of statistical mechanics! This further confirms that this postulate is reasonable, and again justifies its introduction.

To conclude we can also compute the chemical potential for an ideal gas, which turns out to be:

${\displaystyle \mu =-T{\frac {\partial S}{\partial N}}_{|E,V}=k_{B}T\ln \left[{\frac {V}{N}}\left({\frac {4\pi mE}{3Nh^{2}}}\right)^{3/2}\right]=-k_{B}T\ln(\rho \lambda ^{3})}$

## The Maxwell-Boltzmann distribution

We could now ask: can we determine the probability distribution of the momentum, or equivalently of the velocity, of a single particle? Since we are only interested in the probability density of the momentum of a single particle, say the ${\displaystyle j}$-th one, we take the probability distribution of the whole system in phase space and then integrate over all the coordinates we do not care about, i.e. ${\displaystyle {\vec {q}}_{i}\quad \forall i}$ and ${\displaystyle {\vec {p}}_{i}\quad \forall i\neq j}$, with the constraint that ${\displaystyle {\vec {p}}_{j}}$ has some given value ${\displaystyle {\vec {p}}}$; this way we are left with the probability distribution of a single particle. We therefore have:

${\displaystyle \rho ({\vec {p}})=\left\langle \delta ({\vec {p}}_{j}-{\vec {p}})\right\rangle ={\frac {1}{\Omega (E,V,N)}}\int d\Gamma _{N}\delta \left(\sum _{i=1}^{N}{\frac {{\vec {p}}_{i}{}^{2}}{2m}}-E\right)\delta ({\vec {p}}_{j}-{\vec {p}})}$
where we have written ${\displaystyle d\Gamma _{N}}$ to show explicitly that it is relative to all the ${\displaystyle N}$ particles; of course ${\displaystyle d\Gamma _{N}=d\Gamma _{N-1}d{\vec {q}}_{j}d{\vec {p}}_{j}}$ and so integrating over ${\displaystyle {\vec {q}}_{j}}$ and ${\displaystyle {\vec {p}}_{j}}$ we get:
{\displaystyle {\begin{aligned}\rho ({\vec {p}})={\frac {1}{\Omega (E,V,N)}}V\int d\Gamma _{N-1}\delta \left[\sum _{i=1}^{N-1}{\frac {{\vec {p}}_{i}{}^{2}}{2m}}-\left(E-{\frac {{\vec {p}}_{\text{ }}{}^{2}}{2m}}\right)\right]=\\={\frac {V}{\Omega (E,V,N)}}\Omega \left(E-{\frac {{\vec {p}}_{\text{ }}{}^{2}}{2m}},V,N-1\right)\end{aligned}}}
where for simplicity we have rearranged the indices ${\displaystyle i}$ that label the particles so that the ${\displaystyle j}$-th one is the last (otherwise, the sum over ${\displaystyle i}$ should have been written as ${\textstyle \sum _{i=1,i\neq j}^{N}}$). Substituting the expressions of ${\displaystyle \Omega }$ for the ideal gas:
{\displaystyle {\begin{aligned}\rho ({\vec {p}})=V{\frac {N!h^{3N}\left({\frac {3}{2}}N\right)!}{V^{N}(2\pi mE)^{3N/2}}}\cdot {\frac {V^{N-1}\left[2\pi m\left(E-{\frac {{\vec {p}}_{\text{ }}{}^{2}}{2m}}\right)\right]^{3(N-1)/2}}{(N-1)!h^{3(N-1)}\left[{\frac {3}{2}}(N-1)\right]!}}=\\=h^{3}N{\frac {\left({\frac {3}{2}}N\right)!}{\left[{\frac {3}{2}}(N-1)\right]!}}\left[2\pi m\left(E-{\frac {{\vec {p}}_{\text{ }}{}^{2}}{2m}}\right)\right]^{-3/2}\left(1-{\frac {{\vec {p}}_{\text{ }}{}^{2}}{2mE}}\right)^{3N/2}\end{aligned}}}
Now, since ${\displaystyle N}$ is large we can use Stirling's approximation to see how:
${\displaystyle A:={\frac {\left({\frac {3}{2}}N\right)!}{\left[{\frac {3}{2}}(N-1)\right]!}}}$
behaves for large ${\displaystyle N}$. We have:
${\displaystyle \ln A\sim {\frac {3}{2}}N\ln \left({\frac {3}{2}}N\right)-{\frac {3}{2}}N-{\frac {3}{2}}(N-1)\ln \left[{\frac {3}{2}}(N-1)\right]+{\frac {3}{2}}(N-1)}$
and with some reshuffling we get to:
${\displaystyle \ln A\sim {\frac {3}{2}}N\ln \left({\frac {N}{N-1}}\right)+{\frac {3}{2}}\ln {\frac {3}{2}}+{\frac {3}{2}}\ln(N-1)}$
If ${\displaystyle N\to \infty }$, the first term vanishes because the argument of the logarithm tends to 1; in the end, for large ${\displaystyle N}$ we have:
${\displaystyle \ln A\sim {\frac {3}{2}}\ln {\frac {3}{2}}+{\frac {3}{2}}\ln(N-1)\sim {\frac {3}{2}}\ln {\frac {3}{2}}+{\frac {3}{2}}\ln N}$
and exponentiating:
${\displaystyle A\sim \left({\frac {3}{2}}\right)^{3/2}N^{3/2}}$
Therefore if we also consider that ${\displaystyle {\vec {p}}_{\text{ }}{}^{2}/2m\ll E}$, then:
${\displaystyle \rho ({\vec {p}})=h^{3}N^{5/2}\left({\frac {2}{3}}\cdot 2\pi mE\right)^{-3/2}\left(1-{\frac {{\vec {p}}_{\text{ }}{}^{2}}{2mE}}\right)^{3N/2}}$
Now, the last term becomes an exponential in the thermodynamic limit; in fact, defining ${\displaystyle \varepsilon =E/N}$ we have:
${\displaystyle \left(1-{\frac {{\vec {p}}_{\text{ }}{}^{2}}{2mE}}\right)^{3N/2}=\left(1-{\frac {{\vec {p}}_{\text{ }}{}^{2}}{2m\varepsilon N}}\right)^{3N/2}{\stackrel {M:=3N/2}{=}}\left(1-{\frac {3{\vec {p}}_{\text{ }}{}^{2}}{4m\varepsilon }}\cdot {\frac {1}{M}}\right)^{M}{\stackrel {M\to \infty }{\longrightarrow }}e^{-{\frac {3{\vec {p}}_{\text{ }}{}^{2}}{4m\varepsilon }}}}$
Therefore:
${\displaystyle \rho ({\vec {p}})=h^{3}N^{5/2}\left({\frac {2}{3}}\cdot 2\pi mE\right)^{-3/2}e^{-{\frac {3N}{4mE}}{\vec {p}}_{\text{ }}{}^{2}}}$
Since for the ideal gas ${\displaystyle E=3Nk_{B}T/2}$, we have:
${\displaystyle \rho ({\vec {p}})=h^{3}N(2\pi mk_{B}T)^{-3/2}e^{-{\frac {3N}{4mE}}{\vec {p}}_{\text{ }}{}^{2}}=(2\pi mk_{B}T)^{-3/2}e^{-{\frac {3N}{4mE}}{\vec {p}}_{\text{ }}{}^{2}+\ln N+3\ln h}}$
Now, ${\displaystyle \ln N\ll N}$ for large ${\displaystyle N}$ and ${\displaystyle h}$ is a constant so we can neglect these last two contributions in the exponential; therefore:
${\displaystyle \rho ({\vec {p}})=(2\pi mk_{B}T)^{-3/2}e^{-{\frac {{\vec {p}}_{\text{ }}{}^{2}}{2mk_{B}T}}}}$
This is the Mawxell-Boltzmann distribution for the momentum. If we want to express it in terms of the velocity ${\displaystyle {\vec {v}}}$, then from the fact that[1]${\displaystyle \rho ({\vec {v}})d{\vec {v}}=\rho ({\vec {p}})d{\vec {p}}=m^{3}\rho ({\vec {p}})d{\vec {v}}}$ we have:
${\displaystyle \rho ({\vec {v}})=\left({\frac {m}{2\pi k_{B}T}}\right)^{3/2}e^{-{\frac {m{\vec {v}}{}^{2}}{2k_{B}T}}}}$

1. In fact, what must remain invariant is the probability to find a particle with a momentum or velocity in a given interval. In other words, if ${\displaystyle \rho ({\vec {p}})dp}$ is the probability of finding a particle with momentum in ${\displaystyle [{\vec {p}},{\vec {p}}+d{\vec {p}}]}$, or equivalently in ${\displaystyle [m{\vec {v}},m{\vec {v}}+md{\vec {v}}]}$, we see that this is equal to the probability of finding a particle with velocity in ${\displaystyle [{\vec {v}},{\vec {v}}+d{\vec {v}}]}$, which is exactly ${\displaystyle \rho ({\vec {v}})d{\vec {v}}}$.