RVP values for kerosenes are much lower than for gasolines as would be expected from their higher boiling range. An equation of the form:

P = Aexp(-B/T)

where P is the vapour pressure, T the temperature and A and B are constants has sometimes been applied to the vapour pressure of kerosenes. Such an equation applies rigorously only to pure hydrocarbon compounds, being an integrated form of the Clausius-Clapeyron equation. In this case B is the heat of vaporisation divided by the gas constant. That complex mixtures of hydrocarbons when examined experimentally often conform to such a functional form is well known. In reference several kerosene jet fuels were examined for temperature dependence of vapour pressure and the results fitted to an equation of the form above. One of the fuels was found to fit the equation:

P/kPa = 6905exp(-2196.53/T)

Putting T = 311K which is the temperature of an RVP measurement: P= 5.9 kPa

A similar fuel was found to conform to: P/kPa = 2779.43exp(-1773.77/T) giving a RVP of 9.3 kPa. Work cited in reference gives, for another kerosene jet fuel:

P/kPa = 653436.08exp(-4243.3/T) giving an RVP of 0.78 kPa (6 mm Hg). That some kerosenes do have RVP values as low as this is well known.

When with results of the type outlined above ln(P/kPa) is plotted against 103K/T the result is a line of slope -B (units K). As we have seen, for a pure compound:

**B = ΔH _{vap}/R**

So a steep slope would signify a high heat of vaporisation. We expect intuitively that the same would apply to the kerosenes. In the example equations given above the constant in the exponential varies by a factor in excess of two. That the one with the highest value of B is the one with the lowest RVP is significant.

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