Question

A pure species (decadane) has an equation of state: ??/?? = 1-?′? + ?′? 2 . Here, B’ and C’ are functions of temperature only and their values are: 1.2 x 10-7 Pa-1 and 3.2 x 10-14 Pa-2 respectively at 300 K. Calculate decadane’s fugacity and fugacity coefficient at 25 bar and 300 K. (20 points). What kind of intermolecular interactions exist at these conditions? Explain

Answer 1

I think decadane is a typing mistake for dodecane as the chemical with name 'decadane' doesn't exist

I'm hoping that its Dodecane as the compound given is not clear. Anyway, it doesn't make a difference for fugacity calculations. However, for intermolecular interactions, both decane and dodecane have same intermolecular forces (i.e Van der Waals forces)

I tried really my best to explain the "intermolecular interactions" part. Hope you'll understand it

Fugacity and Fugacity coefficient calculations are shown in the below images

Dodecane is an oily liquid alkane hydrocarbon. It is a straight chain paraffin with a boiling point of 212.2 °C. It is a non-polar compound

Both the melting points and boiling points of alkanes are characteristic of the intermolecular forces found between the molecules. The attractions between one molecule and its neighbors are London dispersion forces (part of Van der Waals forces). These forces will be very small for a molecule with low molecular weight but will increase as the size of the molecules increase. As the melting and boiling points of the alkanes increase with the molecular size, there will be an increase in the London dispersion forces. (i.e. the intermolecular forces are stronger in larger hydrocarbons).

As Dodecane (C12H26) has a relatively higher B.P, it will have strong London dispersion forces. Therefore, London dispersion forces / Van der Waals forces exist in Dodecane.

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