Question

If I roll two 6 sided dice, one yellow and one clear, what are the possible microstates, and what is the Gibbs entropy at 298.15 K? What is the entropy if the dice are indistinguishable? Calculate the entropy if one can only measure the sum of the dice instead of their individual values.

Answer 1

Consider a die represents a particle with 6 states and any one of those states comes up randomly. So one die has 6 equally probable outcomes. so for two die rolled randomly the total number of possibilities or microstates are, X = 6X6 = 36.

the :

(1,1)(1,2)(1,3)(1,4)(1,5)(1,6)(2,1)(2,2)(2,3)(2,4)(2,5)(2,6)(3,1)(3,2)(3,3)(3,4)(3,5)(3,6)(4,1)(4,2)(4,3)(4,4)(4,5)(4,6)(5,1)(5,2)(5,3)(5,4)(5,5)(5,6)(6,1)(6,2)(6,3)(6,4)(6,5)(6,6).

Probability cases are: one yellow on dice 1 and one clear on dice 2, or one yellow on dice 2 and one clear on dice 1

P(one yellow and one clear) = 2/36 = 1/18 = 0.055

Entropy , S = p(x) ×botzman constt. X ln(#microstates) = 0.055 Xbotzman constt. X ln (36) =0.055 X 1.38 × 10-23 X ln 36 J/K = 2.7 X 10^-24 J/K

B) what is the Gibbs entropy at 298.15 K?

G = 0, as the states dont change

C) the entropy if the dice are indistinguishable

p(x) = 1/36

Entropy , S = p(x) ×botzman constt. X ln(#microstates) = 1.37X10^-24 J/K

D)measure the sum of the dice

sum starts from 2 and max sum obtainable = 12

no of probabilities = 11

p(x) = 11/36

S = p(x) ×botzman constt. X ln(#microstates) = 1.511 X 10^-23 J/K