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 \(\left(V_{g}\right),\) and liquid \(\left(V_{1}\right),\) and the temperature. $$\frac{d P}{d T}=\frac{\Delta H_{\mathrm{vap}}}{T\left(V_{\mathrm{g}}-V_{1}\right)}$$ Because in most cases the volume of one mole of vapor greatly exceeds the molar volume of liquid, we can treat the \(V_{1}\) 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, \(P V=R T\). Make appropriate substitutions into the above expression, and separate the \(P\) and \(d P\) terms from the \(T\) and \(d T\) terms. The appropriate substitution for \(\Delta H_{\text {vap }}\) means expressing it as a function of temperature. Finally, integrate the two sides of the equation between the limits \(P_{1}\) and \(P_{2}\) on one side and \(T_{1}\) and \(T_{2}\) on the other. (a) Derive an equation for the vapor pressure of \(\mathrm{C}_{2} \mathrm{H}_{4}(\mathrm{l})\) as a function of temperature, if \(\Delta H_{\mathrm{vap}}=\) \(15,971+14.55 T-0.160 T^{2}\left(\text { in } J m o l^{-1}\right)\) (b) Use the equation derived in (a), together with the fact that the vapor pressure of \(\mathrm{C}_{2} \mathrm{H}_{4}(1)\) at \(120 \mathrm{K}\) is 10.16 Torr, to determine the normal boiling point of ethylene.

Step by step solution

01

Start with the given equation

Start with the given equation \( \frac{d P}{d T}=\frac{\Delta H_{\mathrm{vap}}}{T(V_{\mathrm{g}}-V_{1})} \). Assuming \(V_1\) is approximately zero and using the ideal gas law \(P V = RT\), we have \(V_g = \frac{RT}{P}\). Substitute these into the equation, which simplifies to \( \frac{d P}{d T}=\frac{P \Delta H_{vap}}{RT^2} \)

02

Separate variables

Now we separate the \(P\) and \(d P\) terms from the \(T\) and \(dT\) terms to create an expression suitable for integrating. This gives \( \frac{d P}{P}=\frac{\Delta H_{vap}}{RT^2} dT. \)

03

Substitute for ΔHvap and integrate

Substitute \(\Delta H_{vap}\) as a function of temperature, so the expression becomes \( \frac{d P}{P}=\frac{15971 + 14.55T - 0.16T^2}{RT^2} dT \). Integrate both sides between \(P_1\) and \(P_2\) and between \(T_1\) and \(T_2\) respectively.

04

Solve for P2 (the boiling point pressure)

After integrating and simplifying, you should have an equation in the form of \( ln(P_2/P_1) = \int_{T_1}^{T_2} \frac{15971 + 14.55t - 0.16t^2}{Rt^2} dt \). Given that the vapor pressure of ethylene at 120K is 10.16 Torr (or about 1.3417 kPa), we can set \(P_1\) to this value and \(T_1\) to 120. Solving for \(P_2\) when \(T_2\) is x will give the vapor pressure at any temperature x.

05

Finding the normal boiling point

The normal boiling point of a substance is the temperature at which its vapor pressure is equal to 1 atmosphere. So this can be calculated by solving \( ln(P_2/P_1) = \int_{120}^{T_2} \frac{15971 + 14.55t - 0.16t^2}{Rt^2} dt \) where \(P_2 = 1\) atm, for \(T_2\).

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