Question

A gas mixture containing 60% N2 and the balance n-hexane flows through a pipe at a rate of 200m3/hr. The pressure is 2atm absolute and the temperature is 100 C. a) What is the molar flow rate of the gas in kmol/hr? b) Is the gas saturated? Show calculations (hint: use vapor pressure of hexane at 100 C) c) If 80% of n-hexane is condensed (and leaves as a liquid), how much of vapor is leaving and what is its composition.

Answer 1

For calcularting the molar flow rate, we need to calculate the denisty of mixture at 100 deg.c in kmol/m3

Mixture contains 60% N2 and 40 n-hexane

Basis ; 1 mole of mixture this contains 0.6mole N2 and 0.4moles hexane

Density of mixture at STP =1/22.4 moles/m3 ( since 1 mole of any gas contains 22.4 m3 at STP)=0.045 moles/m3

STP refer to 1 atm and 273.15K

at 2 atm and 100 deg.c =100+273.15 kg

Density of the mixture= 0.045*2/1*(273.15/373.15) (increase in pressure reduces the volume and hence increases the density and increase in temperature increases the volume and decreases the density)=0.066 moles/m3

Molar flow rate= volumetric flow rate* density ( moles/m3)= 200*0.066=13.2 moles/hr

Moles of n- hexane= 13.2*0.4= 5.28 moles/hr, moles of nitrogen= 13.2-5.28 =7.92 moles/hr

partial pressure of n-hexane= Total pressure* Moles of n-hexane/ total moles= 2*5.28/13.2=0.8 atm

Vapor pressure of n-hexane at 100 deg.c= 1852 mm Hg

Since the partial pressure of n-hexane is 0.8*760 mm =608 mmHg < vapor pressure of n-hexane at 100 deg.c, the gas is not saturated.

when 80% of n- hexane is condensed, moles of n-hexane remaining= 5.28*0.2=1.056 moles/hr, moles of nitrogen remain the same = 7.92moles/hr

total moles after condensation= 1.056 moles/hr (n-hexane)+ 7.92 moles/ hr (moles of nitrogen )=8.976 moles/hr

Mole % of vapor leaving : Mole fractions   n-hexane= 1.056/8.976=0.118 Nitrogen= 1-0.118=0.882