## Kb and nR

$\Delta S = \frac{q_{rev}}{T}$

bonnie_schmitz_1F
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### Kb and nR

In class we said that ∆S = KB ln (v2/v1) and that ∆S = nR ln (v2/v1). In this case, Boltzmann's constant and R times the number of moles must be equal. However, the number of moles can change for different gases, so can someone explain why these equations are equal?

Matthew Tran 1H
Posts: 165
Joined: Fri Sep 28, 2018 12:16 am

### Re: Kb and nR

Keep in mind that in the Boltzmann equation, W (which is proportional to V) depends on the number of particles, which is in the exponent of the term inside of the ln. This term can then be moved down as a coefficient using log properties. When there is 1 mol of particles, the coefficient is $N_{A}$, or Avogadro's constant. $N_{A}$ x $k_{b}$ is equal to R. However, the units of R are $\frac{J}{K*mol}$, so you need n, the number of moles, as the coefficient, whereas for the Boltzmann equation the implied coefficient is the number of particles as I mentioned earlier. Hope this helps!