Question

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 \(40.0^{\circ} \mathrm{C}\).

Short answer

Question: Calculate the heat of vaporization and vapor pressure of methyl alcohol at 40°C using the given information: normal boiling point is 64.7°C and vapor pressure at 35°C is 203 mmHg. Answer: The heat of vaporization of methyl alcohol is approximately 37.52 x 10^3 J/mol, and its vapor pressure at 40°C is approximately 239.57 mmHg.

Step-by-step solution

Understand the Clausius-Clapeyron Equation

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 \((40^{\circ} \mathrm{C})\).

Convert given temperatures to Kelvin

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

Calculate the heat of vaporization (a)

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: $$ \Delta H_{\text {vap }} \approx 37.52 \times 10^{3}\,\mathrm{J ~mol^{-1}} $$

Calculate the vapor pressure at \(40^{\circ} \mathrm{C}\) (b)

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 \(239.57\, \mathrm{mmHg}\).

Key concepts

Vapor Pressure

Vapor pressure is a crucial concept in understanding how substances evaporate. It refers to the pressure exerted by the vapor of a liquid (or solid) in equilibrium with its non-vapor phases. In layman's terms, it's a measure of how readily molecules escape from a liquid into the gas phase.

Let's break it down further:

• When a liquid is placed in a closed container, some molecules gain enough energy to escape from the surface into the vapor state, forming gas molecules above the liquid.
• The vapor pressure is reached when the rate at which molecules enter the vapor equals the rate at which they return to the liquid.
• Factors influencing vapor pressure include the nature of the liquid (stronger intermolecular forces result in lower vapor pressures) and temperature (higher temperatures increase vapor pressures).

For instance, as demonstrated in this exercise, methyl alcohol exhibits a specific vapor pressure at a given temperature. This value is essential for calculations involving the Clausius-Clapeyron Equation to predict changes in vapor pressure with temperature variations.

Heat of Vaporization

The heat of vaporization (\(\Delta H_{\text{vap}}\)) is the amount of energy required to transform a given quantity of a substance from a liquid into a gas or vapor without a temperature change. In other words, it measures how much 'oomph' you need to make liquid molecules break free into the gas phase.

Here's what you need to know:

• This is a critical value for understanding how substances like alcohol vaporize. A high heat of vaporization indicates strong intermolecular forces holding the liquid molecules together.
• During vaporization, energy is used to overcome these forces rather than raising the temperature, which is why the phase change happens at a constant temperature.
• In our exercise, calculating the heat of vaporization relied on using known vapor pressures at different temperatures and applying the Clausius-Clapeyron Equation to derive the amount of energy required.

Knowing the heat of vaporization helps scientists and engineers understand and predict boiling points and the energy efficiency of heat exchange processes in the industry.

Temperature Conversion

Converting temperatures is a straightforward yet essential part of many scientific calculations, especially when using the Clausius-Clapeyron Equation. Temperatures are often measured in degrees Celsius but must be converted to Kelvin for scientific formulas.

Here's how to do it:

• The Kelvin scale is an absolute temperature scale used in scientific calculations, representing the molecular kinetic energy.
• To convert Celsius to Kelvin, simply add 273.15 to the Celsius temperature.
• For example, converting \(35^{\circ}C\) to Kelvin involves: \(35 + 273.15 = 308.15\,\mathrm{K}\).
• This conversion is necessary because the Kelvin scale ensures that temperature ratios are the same as energy ratios, which is crucial for equations involving gas laws!

By performing these conversions, we ensure that all units match and that the calculations are accurate, enabling precise predictions of vapor pressures and other thermodynamic properties.