101.265 kN/m2 is essentially 101.3 kN/m2 and hence, at this pressure, the boiling or the bubble point of the equimolar mixture is 365.1 K which lies between the boiling points of pure benzene,

353.3 K, and pure toluene, 383.8 K. Example 11.5

What is the dew point of a equimolar mixture of benzene and toluene at 101.3 kN/m2?

Solution

From Raoult's Law, equations 11.1 and 11.2:

Since the total pressure is 101.3 kN/m2, Pb = Pt = 50.65 kN/m2 and hence:

It now remains to estimate the saturation vapour pressures as a function of temperature, using the data of Example 11.3, and then determine, by a process of trial and error, when (xB + xT) = 1.0. The data, with pressures in kN/m2 are:

T(K) |
P ° |
Xb |
P 0 |
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