The vapor pressure of a substance is measured over a range of temperatures. a plot of the natural log of the vapor pressure versus the inverse of the temperature (in kelvin) produces a straight line with a slope of −3.58×103k. what is the enthalpy of vaporization of the substance? 29.8 kj/mol2.32×1023kj/mol0.431 kj/mol294 kj/mol

ΔHvap = 431 J/mol

Explanation:

The relation between the vapor pressure and the absolute temperature is given by the Clausius-Clapeyron equation.

where,

p: vapor pressure

ΔHvap: enthalpy of vaporization

T: absolute temperature

C: constant

We can see that this corresponds to a linear equation with slope -ΔHvap/R and intercept C. If the slope is -3.58 × 10³ K,

-3.58 × 10³ K = -ΔHvap/R

3.58 × 10³ K = ΔHvap/(8.314 J/K.mol)

ΔHvap = 431 J/mol

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