Answer:
Explanation:
given data :
xi=0.5 (i.e. mole fraction 50%)
two liquid mixtures are given i.e. propane and n-butane
generally raoults law is applicable to ideal mixtures
the basic equation of raoults law
ki=pi saturated (T)/p...............1
first we take propane liquid
the basic forula to calculate psat is
%log psat=A-B/T(K)+C
for this we need to calculate tv=B-C/A-ln(P)
Substitute the values from antoine data
B=803.997,A=3.92828,C=26.11,P=0.6*10^-3
TV= 44.74K
NOW WE WILL take n-butane and calculate tv
for butane antoine data is as follows
B=2292,A=13.98,C=-27.86
Tv=134.96 k
now calculating psat
%log psat=A-B/T(K)+C
Psat for propane is -11.29 bar
psat for n-butane is -21.27 bar
pure component of vapour pressure is
1. p1=xi*psat=0.5*-11.29=-5.645 bar
2.p2=xi*psat=0.5*-21.27=-10.635 bar
2. now we are calculating the composition of phase
to calculate the composition
first we need to add up all the pure component pressure
p=p1+p2=-5.645+10.635=4.99 bar
i am assuming yi as vapour composition
yi=p1/p=-5.645/4.99=-1.131
y2=p2/p=-10.635/4.99=-2.131
so these are the minimum values of vapour pressure and pure component vapour pressure
