Question

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

Answer 1

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= 7.14637xx10^-23J`

Answer 2

Boltzmann's formulation of entropy is given by:

`S = k_BlnOmega`

where `k_B = 1.38065 xx 10^(-23) "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 xx 10^(-22) "J/molecule"cdot"K"`

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

`color(blue)(S) = 1.646 xx 10^(-22) "J"/(cancel"molecule" cdot "K") xx (6.0221413 xx 10^23 cancel"molecules")/"1 mol"`

`= color(blue)("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 `"300 K"`, it would be about `62%` of the molar entropy of `"O"(g)`.