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

Question

Owen Ambrose · Accepted Answer

Apply 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 150,000 microstates.

Enter the variables.

#S = (1.38065×10^(−23) J)/K log 150000#

#S= 7.14637xx10^-23J#

Henry Antrim · Answer

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)#.

Mateo Aguilar · Answer

undefined The molar entropy of a system with 150,000 microstates at a certain temperature can be calculated using the formula:

$$ S = k \cdot \ln(W) $$

Where:
- $ S $ is the molar entropy.
- $ k $ is the Boltzmann constant ($1.380649 \times 10^{-23} \, \text{J/K}$).
- $ W $ is the number of microstates.

Given $ W = 150,000 $, the molar entropy can be calculated.