Problem 4.2 s of 5% n-butane (C4H10), 90% iso-octane (C8H18) and 5% n-hexane (C
ID: 717569 • Letter: P
Question
Problem 4.2 s of 5% n-butane (C4H10), 90% iso-octane (C8H18) and 5% n-hexane (C Ha) mixture consist data for the three compounds are found: following und Mol. wt. Boiling pt. Vapor pressure at 37.g°C Compoun -butane 58.12 n-bexane 86.17 68.7°C io-octane 114.22 -0.5°C 355.7 kPa 34.2 kPa 11.8 kPa 99.2°C The mixture compositi on is given in weight-%. Atrnospheric pressure is 1.01325 bar. 1. Calculate the mixture vapor pressure at 20°C 2. Calculate the mixture stoichiometric air/fuel-ratio Pe- miz 24.1 kPa bar; AF 15.11)Explanation / Answer
Given,
5% n-butane, 90% iso- octane and 5% n-hexane.
weight is in weight% i.e weight of solute / weight of solution× 100.
So, 5% means 5g n-butane in the 100 g solution
Similarly, 90g iso-octane in the 100g solution
Also, 5g n-hexane in the 100g solution.
Given, molar mass of n-butane= 58.12
Molar mass of n-hexane = 86.17
Molar mass of iso-octane = 114.22
No. Of moles of n-butane= mass of butane/ molar mass of butane.
No. Of moles= 5/58.12= 0.0860
No.of moles of n-hexane= 5/86.17= 0.0580
No. Of moles of iso-octane= 90/114.22= 0.7879
Now mole fraction= no. Of moles of one component/ no. Of moles of all components.
Therefore, mole fraction,
X(n-butane)= 0.0860/0.9319= 0.0922
X(n-hexane)= 0.0580/0.9319= 0.0622
X(iso-octane)= 0.7879/0.9319= 0.8454
For Calculating P2,
The formula looks like this: ln(P1/P2) = (Hvap/R)((1/T2) - (1/T1)). In this formula, the variables refer to:
Hvap: The enthalpy of vaporization of the liquid.
R: The real gas constant, or 8.314 J/(K × Mol).
Calculating, P2 for n-butane, n-hexane and iso-octane from above putting, P1 = 355.7 kPa, 34.2 kPa and 11.8 kPa respectively.
Also, del H vaporization of n-butane, n-hexane and iso-octane are 22.44, 28.85 and 35.1 KJ/mol.
T1: The temperature at which the vapor pressure is known (or the starting temperature.)
T2: The temperature at which the vapor pressure is to be found (or the final temperature.)
P1 and P2: The vapor pressures at the temperatures T1 and T2, respectively.
For n-butane,
Del H of vaporization= 22.44 KJ/mol
T1= 37.5+273 K= 310.5 K
P1= 355.7 kPa.
T2= 20+273 K= 293 K
R= 8.314 J/ K mol
Do, the calculations yourself.