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clear;
clc;
// Stoichiometry
// Chapter 5
// Energy Balances
// Example 5.9
// Page 220
printf("Example 5.9, Page 220 \n \n");
// solution
// (a)
T = 305.15 //K
Pv1 = 10^(4.0026-(1171.530/(305.15-48.784))) // bar
// (b)
T = 395.15
Pv2 = 10^(3.559-(643.748/(395.15-198.043))) // bar
printf(" (a) \n \n V.P. of n-hexane at 305.15K = "+string(Pv1)+" bar. \n \n \n (b) \n \n V.P. of water at 395.15K = "+string(Pv2)+" bar.")
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