We have learned that the enthalpy of vaporization of a liquid is generally a function of temperature. If we wish to take this temperature variation into account, we cannot use the Clausius–Clapeyron equation in the form given in the text (that is, equation 12.2). Instead, we must go back to the differential equation upon which the Clausius–Clapeyron equation is based and reintegrate it into a new expression. Our starting point is the following equation describing the rate of change of vapor pressure with temperature in terms of the enthalpy of vaporization, the difference in molar volumes of the vapor (V_g), and liquid (V_l), and the temperature.

Because in most cases the volume of one mole of vapor greatly exceeds the molar volume of liquid, we can treat the V_l term as if it were zero. Also, unless the vapor pressure is unusually high, we can treat the vapor as if it were an ideal gas; that is, for one mole of vapor, PV = RT Make appropriate substitutions into the above expression, and separate the P and dP terms from the T and dT terms. The appropriate substitution for Δ_vapH means expressing it as a function of temperature. Finally, integrate the two sides of the equation between the limits P₁ and P₂ on one side and T₁ and T₂ on the other.

(a) Derive an equation for the vapor pressure of C₂H₄(l) as a function of temperature, if Δ_vapH = 15,971 + 14.55 T - 0.160 T² (in J mol^-1).

(b) Use the equation derived in (a), together with the fact that the vapor pressure of C₂H₄(l) at 120 K is 10.16 Torr, to determine the normal boiling point of ethylene.

Eq. 12.2

Transcribed Image Text:

dP dT AvapH T(Vg - V₁)