Question

Use Lee Kesler Correlation and Abbott Correlation to calculate Z and molar volume v for SF6 at 75oC and 30 bar. Compare the results from the two different correlations and comments

A spherical tank of diameter of 10 meters contains liquid toluene at 1 atmosphere at 25 oC. An initially filled tank is drained from the bottom such that the liquid level drops to half. As we drain the tank, toluene vaporizes. The enthalpy of vaporization of toluene at its normal boiling point, Tn =383.8 K, is 33.74 kJ/mol. Calculate the number of moles toluene evaporated from liquid to vapor phase. If the draining is isothermal, calculate the amount of heat provided by the atmosphere to the toluene in the tank.

Answer 1

P = 30 bar , T= 75 C=348 K

For SF6, Pc=36.079 bar, Tc= 318.6 K

Thus, Pr= 30/36.079 = 0.831, Tr=348/318.6=1.092

From Lee Kesler correlation, we have

Z=0.75

molar volume v = ZRT/P = 0.75 x 8.314 x 348/(30,00 kPa) = 0.7233 L

* Using ideal gas correlation:

Z=1

Molar volume v= RT/P = 8.314 x 348/(30,00 kPa) = 0.9644 L

*Using Van der waal correlation

Constants: a= 7.857 bar L²/mol² and b=0.08786 L/mol

(P + a/v²)(v-b)=RT

It becomes:

Pv³ - (Pb+RT)v² + av - ab=0

Substituting values and solving, we get

v=0.7407 L

z=Pv/RT = 0.768

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Initial volume = 4/3 πr³ = 523.6 m³

Volume lost = 0.5 x 523.6 m³ = 261.8 m³

Since this is liquid, we obtain the density of tolune from literature as 867 kg/m³

Thus, mass of toluene vaporized = 261.8 m³ x 867 kg/m³ = 226980 kg

Moles vaporized = mass / molar wt = 226980 kg/92.14 g/mol = 2463.426 kmol

Enthalpy and activation energy values are generally independent of temperature. Thus, we assume it to be same.

Heat given to toluene = enthalpy of vaporization x moles vaporized = 33.74 kJ/mol x 2463.426 x 10³ mol

= 8.3116 x 10⁷ kJ