In conducting a separation of n-pentane (1) / n-heptane (2), we want to make the

Question

In conducting a separation of n-pentane (1) / n-heptane (2), we want to make the separation easy by having the partition coefficient K = y1 / x1 > 2, and p = 1.0 atm. It turns out the K depends upon the temperature, as given by Raoult’s law (with for instance the Antoine equation for vapor pressure). Typically, higher T makes for harder separation. It is also important that when we have K > 2, that the temperature is high enough that the operation does not proceed at a snail’s pace! The vapor pressures are given by Antoine’s equation (pv in kPa and T in C):

2451.88 T +232.014 2910.26 T +216.432 In P-13.7667- In p: = 13.8622 13.8622-29026 TSN. What is the maximum temperature we can use to have K > 2, when Xi = 0.47? What is the maximum temperature we can use to have K> 2, when xi 0.47?

Explanation / Answer

From Raoults law y1P= x1P1sat , y1,x1 are mole fraction of n-pentane. P1sat, P2sat are saturation pressures of n-pentane and n-heptane.

And hence y1/x1= P1sat/P     P11sat/P >2 (1)

P= x1p1sat +x2P2sat (2)

Given x1=0.47 and x2= 1-0.47= 0.53

From Eq.1 and Eq.2

P1sat>2*(x1P1sat + x2P2sat) > 0.94P1sat+1.06 P2sat

P1sat*(1-0.94)= 1.06 P2sat,   0.06P1sat> 1.06P2sat

P1sat> 1.06/0.06 P2sat

P1sat>17.7P2sat (3)

The maximum temperature is one which give P1sat/ P2sat > 17.7

The vapor pressure of n-pentane is greater than the vapor pressure of n-heptane at all temperatures. So the maximum temperature that can be allowed is the boiling point of n-heptane which is 98.42 deg.c

So the procedure involve assume some temperature, calculate P1sat and P2sat from the Antoine equations given and see that Equation 3 is satisfied or not. The boundaries for temperature are boiling point of n-pentane and n-heptane.

Boiling point of n-pentane= 36.1 deg.c and that of n-heptane =98.42 deg.c

Choose the temperature as 98.42 deg.c

lnP1sat =13.7667 – 2451.88/ (98.42+ 232.014), P1sat= 570 Kpa

lnP2sat= 13.8622- 2910.26/ (98.42+216.432) =101. 4 Kpa

P1sat/ P2sat= 570/101.4 = 5.62 This is not the required ratio or separation.

The temperature is reduced and the trials are continued for different temperatures using solver and the highest temperature that is obtained is -5.3 deg.C

Spread sheet Calculatios are also attached

T -5.3 deg.c 2.951841 0.078122 Psat 19.14116 1.081254 P1sat/P2sat 17.70273

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