Question

How‌ ‌do‌ ‌you‌ ‌calculate‌ ‌microstate‌ ‌in‌ ‌chemistry?

Answer 1

‌The‌ ‌microstate‌ ‌is‌ ‌defined‌ ‌as‌ ‌the‌ ‌specific‌ ‌configuration‌ ‌of‌ ‌the‌ ‌thermodynamic‌ ‌system.‌ ‌This‌ ‌is‌ ‌a‌ ‌system‌ ‌which‌ ‌tend‌ ‌to‌ ‌occupy‌ ‌with‌ ‌certain‌ ‌probability‌ ‌in‌ ‌the‌ ‌course‌ ‌of‌ ‌its‌ ‌thermal‌ ‌fluctuation.‌ ‌It‌ ‌is‌ ‌the‌ ‌arrangement‌ ‌of‌ ‌the‌ ‌every‌ ‌molecule‌ ‌in‌ ‌the‌ ‌system‌ ‌at‌ ‌the‌ ‌single‌ ‌instant.‌ ‌

Complete answer:
So when it is asked to us calculate the number of the microstate we usually calculate it at the 298.15 temperature with the help of the tabulated standard molar entropies. The Boltzmann formulation of entropy states the formula which is given below:
$S = {k_B}\ln \Omega$
Here the $\Omega$tends to represent the number of the microsate and the ${k_B}$ is the Boltzmann constant which is equal to $1.38065 \times {10^{ - 23}}J/K$.
There is another formula also which is a bit complicated for $\Omega$ which works for the systems where the available number of states have been larger than the number of the occupied ones. The formula is the following:
$\Omega = \exp \\{ \sum\limits_{i = 1}^N {[{N_i}\ln (\dfrac{{{g_i}}}{{{N_i}}}) + {N_i}} ]\\}$
But for the simplicity we can calculate the value of the number of microstate indirectly by the following formula:
$\Omega = {e^{\dfrac{S}{{{k_B}}}}} = \exp (S/{k_B})$
So let us take the example of oxygen which has its ${S^o}$ equal to 205.15$Jmo{l^{ - 1}}$. The temperature is 298.15K and the pressure is 1 bar. So under such conditions the number of microstates will be. Let us substitute the values in the formula we get,
$\Omega = \exp [\dfrac{{205.15}}{{1.38065 \times {{10}^{ - 23}} \times 6.023 \times {{10}^{23}}}}]$
On solving this we get,
$\Omega = 5.197 \times {10^{10}}$
So the value of the microstate is $\Omega = 5.197 \times {10^{10}}$.

Note: The macrostate has been defined as the macroscopic properties of the system which are temperature, pressure, volume. The microstate is the definite arrangement of the energy in which the content of energy of the definite oscillators is explained. There are microstates and macrostates present in the system.