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(1/3)"
E)"$\Delta$S = -3 ln 3"

Question

Quizplus · Accepted Answer

The entropy change, ΔS, when the multiplicity of a system changes can be calculated using the formula ΔS = k ln(Ω_final/Ω_initial). Here, if the multiplicity is reduced to one-third, Ω_final/Ω_initial = 1/3, leading to ΔS = k ln(1/3).

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

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