Question

Fuel tanks for barbecues contain propane and n-butane. At 120oF, in an essentially full tank of liquid that contains liquid and vapor in equilibrium and exhibits a pressure of 100 psia, what is the overall mole fraction of butane?

Answer 1

Let,

mole fraction of butane in liquid phase = Xb

mole fraction of propane in liquid phase(Xp) = 1 - Xb

mole fraction of butane in vapour phase = Yb

mole fraction of propane in vapour phase(Yp) = 1 - Yb

By Raoult's law

P(total) = Pb * Xb + Pp * Xp

For vapour phase

Yb = Pb * Xb / P(total)       and Yp = Pp * Xp / P(total)

Now,

Vapour pressure of propane at 120 degreeF = 1.24 x 10^4 mmHg = 238.97 psia

Vapour pressure of butane at 120 degreeF = 3608.11 mmHg = 69.77 psia

So,

P(total) = Pb * Xb + Pp * Xp

P(total) = Pb * Xb + Pp * ( 1- Xb)

100 = 69.77 * Xb + 238.97( 1 - Xb)

100 = 69.77Xb + 238.97 - 238.97Xb

169.2Xb = 138.97

Xb = 0.8213

Xb = 0.8213 and Xp = 1 - 0.8213 = 0.1787

Now,

Yb = Pb * Xb / P(total) = (69.77 * 0.8213) / 100 = 0.5730

Yp = 1 - 0.5730 = 0.4270

So,

Overall mole fraction of butane = mole fraction of butane in liquid phase + mole fraction of butane in vapour phase

Overall mole fraction of butane = Xb + Yb = 0.8213 + 0.5730 = 1.3943