19.35 A box divided into identical compartments A and Bcontains two distinguisha

Question

19.35

A box divided into identical compartments A and Bcontains two distinguishable particles in A and five distinguishable particles in B. There are seven energy units in the box, initially shared by the two particles in A. The system is closed, but particles can exchange energy through collisions with the partition.

Part A

Calculate the number of basic states initially.

Part B

Calculate the number of basic states once the system has reached equilibrium

19.35

A box divided into identical compartments A and Bcontains two distinguishable particles in A and five distinguishable particles in B. There are seven energy units in the box, initially shared by the two particles in A. The system is closed, but particles can exchange energy through collisions with the partition.

Part A

Calculate the number of basic states initially.

Part B

Calculate the number of basic states once the system has reached equilibrium

Explanation / Answer

Ans:- This case is related to the entropy of a closed system initially shared by the two particles in A. The system is closed, but particles can exchange energy through collisions with the partition. So the basic state of a entropy system in this state explanation in terms of the properties of A & B particles those  make up the whole system, so that the position and velocity of the relation between different A & B of a system randomly distribution of particle energy in the system, each particle in the system share equal equipartition of energy and the all particle A & B are uniformly distribute in the equipartition of space.

So that box divided into identical compartments A and B contains two distinguishable particles in A and five distinguishable particles in B. Since the no of basic state initially

W = gi ( gi + Ni - 1) / Ni

    = 7(7 + 2 -1)/ 2

   = 7 X 8 X 7 X 6 X 4 X 3 X 2 / 2

= 1753920

the total energy of the system in equilibrium. so that parical function is

Z = gi e-i /KT = e-1 /KT + e-2 /KT   = 2 e-KT    becouse = 1=2

so that the equilibrium so we assume A & B is 1,2

Ni /gi = N / z e-i /KT

N1 = (N / 2e-i /KT) .e-i /KT = N / 2 = N2

The particles are A & B uniformly distributed in the two cells.

U = NKT2 = d/dT Iog Z = NKT2 d/dT (log 2 -  / KT) = N KT2 *  /KT2 = N

let we assume in A & B in equalibrium state

Z = g1e-1 /KT + g2e-2 /KT = 1+ e-/KT  

N1 = N /z e-1 /KT    = N/ 1+ e-/KT   

N2 = N /e-1 /KT   =N/ 1+ e-/KT

U = NKT2 d/dT log [ 1 + e- /KT] = N / 1 + e-/KT

So that we are prove number of basic states of the system has reached equilibrium

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