Question

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) S = 3kln2 B) S =

Answer 1

Answer:

ΔS =  - k ln (3)

Explanation:

Using the Boltzmann's expression of entropy, we have;

S = k ln Ω

Where;

S = Entropy

Ω = Multiplicity

From the question, the configuration of the molecules in a gas changes so that the multiplicity is reduced to one-third its previous value. This also causes a change in the entropy of the gas as follows;

ΔS = k ln (ΔΩ)

ΔS = kln(Ω₂)  - kln(Ω₁)

ΔS = kln(Ω₂ / Ω₁)              -------------(i)

Where;

Ω₂ = Final/Current value of the multiplicity

Ω₁ = Initial/Previous value of the multiplicity

Ω₂ = $\frac{1}{3}$ Ω₁         [since the multiplicity is reduced to one-third of the previous value]

Substitute these values into equation (i) as follows;

ΔS = k ln ($\frac{1}{3}$ Ω₁ / Ω₁)

ΔS = k ln ($\frac{1}{3}$)

ΔS = k ln (3⁻¹)

ΔS =  - k ln (3)

Therefore, the entropy changes by - k ln (3)