Sackur-Tetrode equation
We know entropy is a state function. Let’s calculate it for an ideal gas in terms of macroscopic quantities S ( U , V , N ) S(U,V,N) S ( U , V , N ) L L L
Assume the state of each of the N N N n ⃗ = { n i } \vec n = \{ n_i \} n = { n i } i ∈ [ 1 , 3 N ] i \in [1,3N] i ∈ [ 1 , 3 N ]
U = π 2 ℏ 2 2 m L 2 ( ∣ n ⃗ 1 ∣ 2 + ∣ n ⃗ 2 ∣ 2 + ⋯ ) .
U = \frac{\pi^2 \hbar^2}{2mL^2}\left( |\vec n_1|^2 + |\vec n_2|^2 + \cdots \right).
U = 2 m L 2 π 2 ℏ 2 ( ∣ n 1 ∣ 2 + ∣ n 2 ∣ 2 + ⋯ ) . It follows that the number of states with energy at most E E E
N ( E ) = ∑ n 1 ∑ n 2 ⋯ Θ ( E − π 2 ℏ 2 2 m L 2 ∑ j ∣ n ⃗ j ∣ 2 ) ≈ ∫ ∏ i N d 3 n i Θ ( E − π 2 ℏ 2 2 m L 2 ∑ j N ∣ n ⃗ j ∣ 2 ) .
\begin{align*}
N(E) &= \sum_{n_1} \sum_{n_2} \cdots \Theta \left(E - \frac{\pi^2 \hbar^2}{2mL^2} \sum_j |\vec n_j|^2 \right) \\
&\approx \int \prod_i^N \d^3 n_i\, \Theta\left(E - \frac{\pi^2 \hbar^2}{2mL^2} \sum_j^N |\vec n_j|^2 \right) \tag{for large \textit{E}}.
\end{align*}
N ( E ) = n 1 ∑ n 2 ∑ ⋯ Θ ( E − 2 m L 2 π 2 ℏ 2 j ∑ ∣ n j ∣ 2 ) ≈ ∫ i ∏ N d 3 n i Θ ( E − 2 m L 2 π 2 ℏ 2 j ∑ N ∣ n j ∣ 2 ) . ( for large E ) Above we assume E E E
Note
Why don’t I have particle in a box notes from QM?
Make a change of variables to write N ( E ) N(E) N ( E ) 3 N 3N 3 N
R = 2 E m L 2 π 2 ℏ 2 , n i = R z i N ( E ) = R 3 N ∫ 0 ∞ ∏ i 3 N d z i Θ ( 1 − ∑ j 3 N z j 2 ) = ( 1 2 ) 3 N R 3 N π 3 N / 2 ( 3 N / 2 ) ! .
\begin{align*}
R &= \sqrt\frac{2EmL^2}{\pi^2\hbar^2}, \\
n_i &= R z_i \\
N(E) &= R^{3N} \int_0^\infty \prod_i^{3N} \d z_i\, \Theta\left(1-\sum_{j}^{3N} z_j^2\right) \\
&= \left(\frac 12\right)^{3N} \frac{R^{3N} \pi^{3N/2}}{(3N/2)!}.
\end{align*}
R n i N ( E ) = π 2 ℏ 2 2 E m L 2 , = R z i = R 3 N ∫ 0 ∞ i ∏ 3 N d z i Θ ( 1 − j ∑ 3 N z j 2 ) = ( 2 1 ) 3 N ( 3 N /2 )! R 3 N π 3 N /2 . Then compute multiplicity Γ \Gamma Γ 1 / N ! 1/N! 1/ N !
Γ = 1 N ! d N d E δ E = 1 N ! ( π 4 ) 3 N / 2 3 N / 2 ( 3 N / 2 ) ! ( 2 m L 2 E π 2 ℏ 2 ) 3 N / 2 δ E E .
\begin{align*}
\Gamma = \frac{1}{N!} \frac{\d N}{\d E} \delta E = \frac{1}{N!} \left(\frac{\pi}{4}\right)^{3N/2} \frac{3N/2}{(3N/2)!} \left(\frac{2mL^2 E}{\pi^2 \hbar^2}\right)^{3N/2} \frac{\delta E}{E}.
\end{align*}
Γ = N ! 1 d E d N δ E = N ! 1 ( 4 π ) 3 N /2 ( 3 N /2 )! 3 N /2 ( π 2 ℏ 2 2 m L 2 E ) 3 N /2 E δ E . And apply Stirling’s approximation to find entropy S S S ln N \ln N ln N
S ( U , V , N ) = k ln [ 1 N ! ( π 4 ) 3 N / 2 3 N / 2 ( 3 N / 2 ) ! ( 2 m L 2 U π 2 ℏ 2 ) 3 N / 2 δ U U ] = k N ln ( 2 m L 2 U π 2 ℏ 2 ) 3 N / 2 + k ln 3 N 2 + k ln [ − 3 N 2 ln 3 N 2 + 3 N 2 ] − k N ln N = k N [ ln V N + ln ( U m 3 N π ℏ 2 ) 3 / 2 + 5 2 ] = k N [ ln V N + 3 2 ln U N + 3 2 ln m 3 π ℏ 2 + 5 2 ]
\begin{align*}
S(U,V,N) &= k \ln \left[ \frac{1}{N!} \left(\frac{\pi}{4}\right)^{3N/2} \frac{3N/2}{(3N/2)!} \left(\frac{2mL^2 U}{\pi^2 \hbar^2}\right)^{3N/2} \frac{\delta U}{U}\right] \\
&= k N \ln \left(\frac{2mL^2 U}{\pi^2 \hbar^2}\right)^{3N/2} + k \ln \frac{3N}{2} + k \ln \left[
-\frac{3N}{2} \ln \frac{3N}{2} + \frac{3N}{2}
\right] - k N \ln N \\
&= k N \left[ \ln \frac VN + \ln \left(\frac{Um}{3N\pi \hbar^2}\right)^{3/2} + \frac52 \right] \\
&= kN \left[ \ln \frac VN + \frac 32 \ln \frac UN + \frac 32 \ln \frac{m}{3\pi\hbar^2} + \frac 52 \right] \tag{!!}
\end{align*}
S ( U , V , N ) = k ln [ N ! 1 ( 4 π ) 3 N /2 ( 3 N /2 )! 3 N /2 ( π 2 ℏ 2 2 m L 2 U ) 3 N /2 U δ U ] = k N ln ( π 2 ℏ 2 2 m L 2 U ) 3 N /2 + k ln 2 3 N + k ln [ − 2 3 N ln 2 3 N + 2 3 N ] − k N ln N = k N [ ln N V + ln ( 3 N π ℏ 2 U m ) 3/2 + 2 5 ] = k N [ ln N V + 2 3 ln N U + 2 3 ln 3 π ℏ 2 m + 2 5 ] ( !! ) The final line (!!) is the Sackur-Tetrode equation. It can be shown that S S S