Question

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1 (Lecture LOI) 50 m3 of a 35-C gas mixture with composition 98 miole percent air(3), 2 mole percent toluene(1) is contacted with 400 kg of 35°C pure n-decane(2). The vapor-liquid system is allowed to come to equilibrium isothermally at 35 oC at a constant pressure of 1.0 atm. What are the final compositions of the liquid and vapor phases? You should neglect dissolution of air in the liquid, but you must account for evaporation of n-decane. Thus, the vapor phase contains all three components, but the liquid phase contains only toluene(1) and n-decane(2). Assume that the vapor phase behaves as an ideal gas, and that Raoult's law holds for both toluene and n-decane

Answer 1

Gas Pressure= 1.0 atm

Initial Gas Volume= 50 m³

Temperature= 273.15+ 35 = 308.15 K

Molecular weight of n-decane= 142

$\large PV=N_{total}RT$

RI totalPV

8.314 308.15 total 1.01325 105 50

N_total= 1977.5 moles

Initial moles of toluene in gas phase = 1977.5*.02 = 39.55 moles

Initial moles of air in gas phase = 1977.5*.98 = 1937.95 moles

Air always stays in gas phase

Therefore, moles of air after equilibrium = 1937.95 moles

Initial moles of n-decane in the liquid= 400*1000/142 = 2816.9 moles

At equilibrium, let n_t be the moles of toluene transferred from gas phase to liquid phase and n_d be the moles of n-decane transferred from liquid phase to gas phase

Therefore at equilibrium:

Gas Phase:

Toluene =39.55 - n_t

Air =1937.95

n-decane = n_d

Total Moles at equilibrium in gas phase = 39.55 - n_t + 1937.95 + n_d = 1977.5 + n_d -n_t

Liquid Phase:

Toluene = n_t

n-decane = 2816.9 - n_d

Total Moles at equilibrium in liquid phase = n_t + 2816.9 - n_d = 2816.9 +  n_t - n_d

Condition	Gas phase				Liquid Phase
	Air	Toluene	N-Decane	Total	Air	Toluene	N-Decane	Total
Initial Moles	1937.95	39.55	0	1977.5	0	0	2816.9	2816.9
Moles Transferred	0	- nt	+ nd	nd -nt	0	+ nt	- nd	nt - nd
Moles At Equilibrium	1937.95	39.55-nt	nd	1977.5 + nd -nt	0	nt	2816.9-nd	2816.9 +  nt - nd
Mole Fraction at equilibrium	1937.95 197 d nt	39.55- 197 d nt	1977.5+d nt	1	0	2816.9+nt-nd	2816.9-nd 2816.9+nt-nd	1

The Antoine constants for toluene and n-decane are given below:

log10P(bar) = A T(K) O

Compound	A	B	C	Vapor Pressure at 35oC (atm)
Toluene	4.08245	1346.382	-53.508	0.0616
N-Decane	0.21021	440.616	-156.896	0.0020

Rault's Law:

$Py_i=x_iP_i^{sat}$

$\therefore 1*y_i=x_iP_i^{sat}$

$\therefore y_i=x_iP_i^{sat}$

For toluene:

$y_{toluene}=x_{toluene}P_{toluene}^{sat}$

39.55 nt 7n 0.0616 1977.5 +nd-n 2816.9+nt-nd t n

For n-decane

$\therefore y_i=x_iP_i^{sat}$

$\therefore \frac{n_d}{1977.5+n_d-n_t}=\frac{2816.9-n_d}{2816.9+n_t-n_d}*0.0020$

Two equations and two unknowns.

We solve the above two equations simultaneously

$\therefore$n_t = 37.96 moles

$\therefore$n_d = 3.75 moles

Condition	Gas phase				Liquid Phase
	Air	Toluene	N-Decane	Total	Air	Toluene	N-Decane	Total
Initial Moles	1937.95	39.55	0	1977.50	0	0	2816.9	2816.90
Moles Transferred	0	-37.96	3.75	-34.21	0	37.96	-3.75	34.21
Moles At Equilibrium	1937.95	1.59	3.75	1943.29	0.00	37.96	2813.15	2851.106
Mole Fraction	0.9973	0.0008	0.0019	1.0000	0.0000	0.0133	0.9867	1.0000