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// Theory and Problems of Thermodynamics
// Chapter 5
//Second Law of Thermodynamcis
// Example 16
clear ;clc;
//Given data
N = 1 //number of moles of air to be seperated
P = 0.1*1e6 //pressure of of air before seperation in Pa
T = 300 //temperature of air before seperation in K
R = 8.314 //gas constant
// Calculations for determining the minimum work required
// From first and second law of thermodynamics
// dU = T dS - P dV
// initial and final temperatures are identical => dU = 0
S_mix = -R*(0.21*log(0.21) + 0.79*log(0.79)) // entorphy for mixing
S_sep = - S_mix // entrophy change for seperation
W = T * S_sep // minimum work to be done
// Output Results
mprintf('Minimum work to be done for seperation = %6.1f kJ', -W)
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