Question

Nitrogen is bubbled through a liquid mixture that contains initially equimolar amounts of benzene (B) and toluene (T). The system pressure is 3 atm and the temperature is 80C. The nitrogen flow rate is 10.0 standard liters per second. The gas leaving the bubbler is saturated with B and T vapors.

a) Estimate the initial rates (in mol/min) at which B and T leave the bubbler.

b) How will the mole fractions of B and T in the liquid change with time (increase or decrease or remain the same)? Explain your answer.

c) How will the mole fractions of B and T in the exiting gas change with time (increase or decrease or remain the same)? Explain your answer.

Answer 1

Nitrogen gas is bubbled through the system in which a liquid mixture of equilimolar amount of benzene and toluene is mixed at 80 and 3 bar pressure -

P = 3 BAR and T = 80 +273 = 353K

nitrogen volumetric flow rate = 10 L /min

By using ideal gas equation PV = n.RT

N= n = PV/RT =

initially the moles of benzene = 1 ( assumed ) and the moles of toluene = 1 ( assumed)

Now if suppose the vapour pressure of benzene = 760 mm Hg at 80

vapour pressure of toluene = 295 mm Hg at 80

Then partial pressure of benzene = Hg partial pressure of toluene

Mole fraction is X

We know that p=X.P where P is the total pressure

Now using this concept for toluene pT

295=X.2250 mmhg

X=0.13

Now using this concept for toluene pB

760=X.2250 mmhg

X=0.33

As time is past , in liquid the mole fraction will remain same.

But when exiting gas is there that means the molecules will get the kinetic energy so presure will be incresed and so total pressure will be increaded.

mole fraction will be decreased.