Let k be the Boltzmann constant. If the configuration of the molecules in a gas changes so that the multiplicity is reduced to one-third its previous value, the entropy of the gas changes by:
A)$\Delta$ S = 0
B) $\Delta$ S = 3k ln 2
C) $\Delta$ S = -3k ln 2
D) $\Delta$ S = k ln 3
E) $\Delta$ S = -3 ln 3

Question

Quizplus · Accepted Answer

The change in entropy is given by ΔS = k ln(W_final/W_initial). If the multiplicity is reduced to one-third, W_final = W_initial/3, so ΔS = k ln(1/3) = -k ln 3.

Let K Be the Boltzmann Constant Δ\DeltaΔ S = 0 B) Δ\DeltaΔ

Let K Be the Boltzmann Constant $\Delta$ S = 0
B) $\Delta$