Question

Problem 2 2800 2600 OA 2400 11 2200 w Pressure (kPa) 2000 16001L 1400 1200+ 0 0.1 0.2 0.8 0.9 0.3 0.4 0.5 0.6 0.7 Mole fraction n-pentane Consider the n-pentane (1) + methanol (2) system at 422.6 K in the figure above. Numbered black dots are provided on the Pxy diagram, and they are all at 422.6 K. For each of the points, report the following information: (a) Equilibrium phases (liquid, vapor, dr both liquid and vapor) (b) Composition of the equilibrium phases (c) Amount of the equilibrium phases (if possible), assuming 1 mole of the system (overall)

Answer 1

Note: Since the graph is not precisely drawn for further decimal points, some values are approximated.

Here, we are seeing a P-x-y (Pressure-liquid composition- vapour composition) at constant temperature(422.6K), graph of a minimum boiling point azeotrope mixture of n-pentane(1) and methanol(2)

As one can understand,

the upper curve is the P-x curve (as the pressure increases mixture has more tendency to go into liquid state) and hence the area above the curve represents the mixture in the liquid state. On the p-x curve it is saturated liquid.

the lower curve is the P-y curve and hence the area below this curve represents the mixture in vapour state. On the p-y curve it is saturated vapor.

So, when the cordinate is in between the curve(that is, at a particular pressure and composition), the mixture seperates into corresponding saturated liquid and a saturated vapour, that is saturated liquid and saturated vapour coexist in equillibrium in the system.

2800 2600 Pressure (kPa) 1600 1400 1200- © 0.1 0.16.2 0.3 0.39. 4 0.5054 0.6 0.90.72 0981 0.89 0.94 Mole fraction n-pentane 0.94

Now I will create a table and asnwer all the questions in the table,

No. of Mixture	Equillibrium Phases	Composition of the equilibrium phases (mole fraction of n-peptane)	Amount of equillibrium phases (for overall 1 mole)(mol)
1	Both saturated lquid and saturated vapor	Saturated liquid-0.16 Saturated vapor-0.39	Saturated liquid-0.3914 Saturated vapor-0.6086
2	Both saturated lquid and saturated vapor	Saturated liquid-0.89 Saturated vapor-0.72	Saturated liquid-0.47 Saturated vapor-0.53
3	Superheated vapour	0.54	1
4	Azeotropic mixture (saturated liquid boiling or saturated vapour condensing)	0.54	1
5	Saturated vapour	0.3	1
6	Both saturated lquid and saturated vapor	Saturated liquid-0.94 Saturated vapor-0.81	Saturated liquid-0.692 Saturated vapor-0.308
7	Pure n-peptane saturated liquid boiling or saturated vapour condensing ( at boiling point)	1.0	1

To calculate the amount of the equillibrium phases, I will use inverse lever arm rule, I will show a sample calculation for the 1st mixture,

The 1st mixture separates into saturated liquid M and saturated vapour N.

,

M Molesof liquid N1 N" MolesofvaporM1

We also know, ,

M + N = 1

M N 0.39 -0.3 0,3-0,16 = 0.643or M = 0.643N

so,

0.643N + N=1

N - 1 N = = 0.6086mol

M = 0.3914mol

So The saturated liquid is 0.3914 mol and saturated vapour is 0.6086 mol