Problems 33 1103 The fundamental equation of system A is 2 1

Problems 33 1.10-3. The fundamental equation of system A is 2 1/3 1/3 (NVU) and similarly for system B. The two systems are separated by a rigid, imperme- able, adiabatic wall. System A has a volume of 9 x 10-6 m3 and a mole number of 3 moles. System B has a volume of 4 x 10 6 ms and a mole number of 2 moles. The total energy of the composite system is 80 J. Plot the entropy as a function of UA/(UA UB). If the internal wall is now made diathermal and the system is allowed to come to equilibrium, what are the internal energies of each of the individual systems? (As in Problem 1.10-1, the quantities vo, b. and R are positive constants.)

Solution

The total entropy of the systems is written as follows:
         S = S_A + S_B
            = (R²/Vo^)^(1/3) · (N_A·V_A·U_A)^(1/3) + (R²/Vo^)^(1/3) · (N_B·V_B·U_B)^(1/3)
            = (R²/Vo^)^(1/3) · [ (N_A·V_A·U_A)^1/3 + (N_B·V_B·U_B)^(1/3) ]

The total energy is

           U = U_A + U_B = constant
then,
           U_B = U - U_A

so, the entropy is,
         S = (R²·U/Vo^)^(1/3) · [ (N_A·V_A·(U_A/U) )^1/3 + (N_B·V_B·(1 - (U_A/U)))^(1/3) ]

this is a function of U_A/ U = U_A/(U_A + U_B), to plot the equation set,
               x = U_A/ U and K= (R²·U/Vo^·U)^(1/3)

thus, the entropy is,
S = K · [ 0.03·x^(1/3) + 0.02·(1-x)^(1/3)]

At equilibrium, we get
          dS/dU_A = 0
             dS/dx = 0
then,
       (K/3) · [ 0.03·x^(-2/3) - 0.02·(1-x)^(2/3)] = 0
                                                [(1-x)/x]^(2/3) = 1.5
after solving, we get
                            x = 1/(1+ 1.5^1.5) = 0.3525

therefore, the energies are,
U_A = U · x = 80 J * 0.3525 = 28.2 J
U_B = U · (1-x) = 80 J*0.6475 = 51.8 J