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

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

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Answer 1 · 2017-03-28T15:21:36

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 · 2018-05-25T20:35:43

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 $"O"(g)$ .

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

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