The fundamental equation of system A is S = (R^2/v_0 theta)^1/3 (NVU)^1/3 and si

Question

The fundamental equation of system A is S = (R^2/v_0 theta)^1/3 (NVU)^1/3 and similarly ofr system B. The two systems are separated by a rigid, impermeable, adiabatic wall. System A has a volume of 9 times 10^-6 m^3 and a mole number of 3 moles. System B has a volume of 4 times 10^-6 m^3 and a mole number of 2 moles. The total energy of the composite system is 80 J. Plot the entropy as a function of U_A/(U_A + U_B). If the internal wall is now made dia thermal 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 v_0 theta, and R are positive constants.)

Explanation / Answer

The total entropy of the systems is the sum of A and B
S = SA + SB
= (R²/Vo)^(1/3) · (NA·VA·UA)^(1/3) + (R²/Vo)^(1/3) · (NB·VB·UB)^(1/3)
= (R²/Vo)^(1/3) · [ (NA·VA·UA)^(1/3) + (NB·VB·UB)^(1/3) ]

The total energy is the sum of the energies of A and B. According to the first law of thermodynamics the total energy of the system is constant:
U = UA + UB = constant
=>
UB = U - UA

Insert this into the entropy equation an rearrange to
S = (R²·U/Vo)^(1/3) · [ (NA·VA·(UA/U) )^1/3 + (NB·VB·(1 - (UA/U)))^(1/3) ]
which solely a function of UA/ U = UA/(UA + UB)
To plot the equation set
X = UA/ U and
K= (R²·U/Vo·U)^(1/3)
and insert the values of NA,VA,NB and VB:
S = K · [ 0.03·X^(1/3) + 0.02·(1-X)^(1/3)]

At equilibrium the total entropy of the system reaches a maximum. Therefore
dS/dUA = 0
<=>
dS/dX = 0
<=>
(K/3) · [ 0.03·X^(-2/3) - 0.02·(1-X)^(-2/3)] = 0
<=>
[(1-X)/X]^(-2/3) = 1.5
<=>
X = 0.6475

UA = U · X = 80J · 0.6475 = 51.8J
UB = U · (1-X) = 80J · 0.3525 = 28.2J

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