Question

One mole of n-butane and one mole of n-pentane are charged into a container. The container is heatedto 180 ^oF where the pressure reads 100 psia. Determine the quantities and composition of the phases in the container.

Answer 1

given

n butane 1 mole & n pentane 1 mole

T=180^oF=82.22^oC & P = 100 psia =5171.49mmHg

to determine quantities & composition of phases in the container

by using antoine equation

log₁₀P^sat [mmHg]= A-B/(T[^oC]+C)

comp.	A	B	C
n butane	6.80776	935.77	238.789
n pentane	6.85296	1064.84	232.012

log₁₀P^sat [mmHg]= 6.80776-935.77/(82.22+238.789)                                        n butane

log₁₀P^sat [mmHg]=3.89267039

P_butane^sat=7810.34809mmHg

log₁₀P^sat [mmHg]= 6.85296-1064.84/(82.22+232.012)                                     n pentane

log₁₀P^sat [mmHg]=3.36425357

P_pentane^sat= 2912.41708mmHg

now by applying Raoult's law for butane

y_iP=x_iP_i^sat

y/x=P_butane^sat/P         

y/x= 1.51027037        &

x+y =1

solving above two equations we get

1=2.51027037x

x_butane=0.39836346 & y_butane=0.60163654

similarly solving for Pentane

y/x=P_pentane^sat/P

y/x=0.56316788

as x+y=1

(1-x)/x=0.56316788

1=1.56316788x

x_pentane=0.63972655

so y_pentane =0.36027345

as pentane & butane was in quantity 1 mole each their mole fraction will be same as their moles

so total quantity of gas phase =y_pentane+y_butane =0.36027345+0.60163654=0.96190999moles

total quantity in liquid phase =x_pentane+x_butane=0.63972655+0.39836346=1.03809001 moles

now we are asked composition of each phase

gas phase =0.96190999moles

moles fraction of butane in gas phase =moles of butane / total moles of gas phase =0.60163654/0.96190999 =0.62546033

so gas phase contain 62.54% butane & remaining 37.45% pentane

similarly

total moles in liquid phase are =1.03809001 moles

mole fraction of butane in liquid phase =0.39836346/1.03809001=0.38371958

so liquid phase contain 38.37% butane & remaining 61.62 %   pentane