# What is the molar entropy of a system with 150000 microstates at a certain temperature?

Mar 28, 2017

Use the formula

S = k_blogΩ

Where ${k}_{b}$ is the Boltzmann constant which is equal to (1.38065×10^(−23) J)/K.

Ω is the number of microstates that is 150,000

Plug in the variables

S = (1.38065×10^(−23) J)/K log 150000

$S = 7.14637 \times {10}^{-} 23 J$

May 25, 2018

Boltzmann's formulation of entropy is given by:

$S = {k}_{B} \ln \Omega$

where ${k}_{B} = 1.38065 \times {10}^{- 23} \text{J/molecule"cdot"K}$ is the Boltzmann constant.

If the system has $150000$ microstates, we say that $\Omega = 150000$. Therefore:

S = 1.38065 xx 10^(-23) "J"/("molecule" cdot "K") ln(150000)

$= 1.646 \times {10}^{- 22} \text{J/molecule"cdot"K}$

Using units that we're more familiar with...

$\textcolor{b l u e}{S} = 1.646 \times {10}^{- 22} \text{J"/(cancel"molecule" cdot "K") xx (6.0221413 xx 10^23 cancel"molecules")/"1 mol}$

$= \textcolor{b l u e}{\text{99 J/mol"cdot"K}}$

We don't really know what temperature this is at, but it is on the right order of magnitude. If this was at $\text{300 K}$, it would be about 62% of the molar entropy of $\text{O} \left(g\right)$.