Xe=0:.1:1;
Xw=1-Xe;
for m=1:11
for T=273.15:.01:450;
Pe=exp(18.5242-3578.91/(T-50.5)); % Vapor pressure of Ethanol
Pw=exp(18.3036-3816.44/(T-46.13)); % Vapor pressure of Water
Ke=Pe/760; % Ethanol
Kw=Pw/760; % Water
sum=Ke*Xe(m)+Kw*Xw(m);
if sum>1
break
end
end
Temp(m)=T;
ye(m)=Ke*Xe(m);
end...