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// Theory and Problems of Thermodynamics
// Chapter 5
//Second Law of Thermodynamcis
// Example 15
clear ;clc;
//Given data
N1 = 2 // number of moles of helium
P1 = 1.0*1e6 // pressure of helium compartment in Pa
T1 = 600 // temperature of helium compartment in K
N2 = 5 // number of moles of air
P2 = 2.0*1e6 // pressure of air compartment in Pa
T2 = 500 // temperature of air compartment in K
R = 8.314 // universal gas constant J/mol/K
C1 = 1.5 * R // specific heat(ischoric) of helium in kJ/kmol/K
C2 = 2.5 * R // specific heat(ischoric) of air in kJ/kmol/K
// Calculations for determining the entorphy change of given process
V1 = N1*T1*R/P1 // volume of helium before mixing
V2 = N2*T2*R/P2 // volume of air before mixing
// Consider the entire container as the system, then Q = 0; W = 0;
// Hence, del U = 0;
deff('y=Tf(T)', 'y = N1*C1*(T-T1) + N2*C2*(T-T2)')
T = fsolve(0,Tf) // Final temperature after mixing
V = V1 + V2 // final volume after mixing
N = N1 + N2 // total moles after mixing
P = N*T*R/V // Final pressure after mixing in Pa
// entropy change for helium from initial state to final state
S1 = N1*{2.5*R*log(T/T1) - R*log(P/P1)}
// entropy change for air from initial state to final state
S2 = N2*{3.5*R*log(T/T2) - R*log(P/P2)}
// entropy change for final state of individual to mixture
S3 = -7*R*{(N1/N)*log(N1/N) + (N2/N)*log(N2/N)}
// The Total change in enthalphy for entire process
S = S1 + S2 + S3
// Output Results
mprintf('The entrophy change for associated with the process = %6.4f kJ/K' ,S)
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