Methyl alcohol, \(\mathrm{CH}_{3} \mathrm{OH}\), has a normal boiling point of \(64.7^{\circ} \mathrm{C}\) and has a vapor pressure of \(203 \mathrm{~mm} \mathrm{Hg}\) at \(35^{\circ} \mathrm{C}\). Estimate (a) its heat of vaporization \(\left(\Delta H_{\text {vap }}\right)\). (b) its vapor pressure at \(40.0^{\circ} \mathrm{C}\).

The Clausius-Clapeyron equation is given by the following formula: $$ \ln \left(\frac{P_{2}}{P_{1}}\right)=-\frac{\Delta H_{\text {vap }}}{R}\left(\frac{1}{T_{2}}-\frac{1}{T_{1}}\right) $$ Where \(P_1\) and \(P_2\) are the vapor pressures at temperatures \(T_1\) and \(T_2\), respectively, \(\Delta H_{\text{vap}}\) is the heat of vaporization, and R is the gas constant (\(8.314 \mathrm{~J~ mol^{-1} K^{-1}}\)). We are given the normal boiling point \((64.7^{\circ} \mathrm{C})\), the vapor pressure at \((35^{\circ} \mathrm{C})\), and are asked to find the vapor pressure at \((40^{\circ} \mathrm{C})\).

To use the Clausius-Clapeyron equation, we need to convert the temperature values from degrees Celsius to Kelvin. $$ T_{1} = 35 + 273.15 = 308.15\,\mathrm{K} $$ $$ T_{2} = 40 + 273.15 = 313.15\,\mathrm{K} $$ $$ T_{\text{normal}} = 64.7 + 273.15 = 337.85\,\mathrm{K} $$

We will use the boiling point and its corresponding vapor pressure \((P_\text{normal} = 760\, \mathrm{mmHg})\) to compute the heat of vaporization. Rearrange the Clausius-Clapeyron equation for \(\Delta H_{\text{vap}}\) and quickly convert pressure to the same units as the gas constant: $$ \Delta H_{\text {vap }}=-\frac{R\;\ln \left(\frac{P_{2}}{P_{1}}\right)}{\frac{1}{T_{2}}-\frac{1}{T_{1}}} = -\frac{8.314 \mathrm{~J~ mol^{-1} K^{-1}}\;\ln \left(\frac{760}{203}\right)}{\frac{1}{337.85\,\mathrm{K}}-\frac{1}{308.15\,\mathrm{K}}} $$ Calculate the heat of vaporization: $$ \Delta H_{\text {vap }} \approx 37.52 \times 10^{3}\,\mathrm{J ~mol^{-1}} $$

Now that we have the heat of vaporization, we can use the Clausius-Clapeyron equation to find the vapor pressure at \(40^{\circ} \mathrm{C}\) \((313.15\,\mathrm{K})\). Rearrange the equation for \(P_2\): $$ P_{2} = P_{1} \cdot e^{\left(-\frac{\Delta H_{\text {vap }}}{R}\left(\frac{1}{T_{2}}-\frac{1}{T_{1}}\right)\right)} $$ Plug in the known values and compute the vapor pressure: $$ P_{2} = 203 \mathrm{~mm~ Hg} \cdot e^{\left(-\frac{37.52 \times 10^{3}\,\mathrm{J ~mol^{-1}}}{8.314 \mathrm{~J~ mol^{-1} K^{-1}}}\left(\frac{1}{313.15\,\mathrm{K}}-\frac{1}{308.15\,\mathrm{K}}\right)\right)} $$ $$ P_{2} \approx 239.57\, \mathrm{mmHg} $$ In conclusion, the heat of vaporization of methyl alcohol is approximately \(37.52 \times 10^{3}\,\mathrm{J ~mol^{-1}}\), and its vapor pressure at \(40^{\circ} \mathrm{C}\) is approximately \(239.57\, \mathrm{mmHg}\).