Question

Consider a 1 mole solid of substance a in contact with another 1 mole solid of substance b. Initially both solids are pure. After a full annealing, only a homogeneous a-b solid solution exists in the final state. Assume no reaction happens. Calculate the entropy difference between the two states, using the boltzmann relation. 1.

Answer 1

According to Boltzmann reaction entropy difference between two states is given by ΔS and is equal to 5.76 .

According to Boltzmann Reaction:

 S= kB lnW

 Where,

W is number of Arrangements possible

Now ,

 ΔS = S(final) - S(initial)

S(initial) = kBln1 (Since only one arrangement is possible)

             = 0

Again,

 S(final) = kBln2^n

Now,

  As A-B can exist as A-B or B-A .

Hence, two arrangements are possible and since 1 mole of species .

Hence, 2^n

S(final) = N kB ln2

           =R ln 2

           = 8.314 ×2.303× log 2

Hence,

S(final) = 5.76

ΔS = 5.76 -0

      =5.76

Thus, from the above conclusion we can say that, according to Boltzmann reaction entropy difference between two states is given by ΔS and is equal to 5.76 .

Learn more about Entropy here : brainly.com/question/17241209

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