Question

If the vapor pressure of methy1 alcohol, \(\mathrm{CH}_{3} \mathrm{OH}\), is \(0.0526\) atm at \(5.0^{\circ} \mathrm{C}\) and \(0.132 \mathrm{~atm}\) at \(21.2^{\circ} \mathrm{C}\), what do you predict the normal boiling point will be?

Short answer

The normal boiling point of methyl alcohol (CH3OH) is predicted to be \(64.5^{\circ} \mathrm{C}\).

Step-by-step solution

Write the Clausius-Clapeyron equation

Clausius-Clapeyron equation is given by: \[ \ln \frac{P_2}{P_1} = -\frac{\Delta H_{vap}}{R} \left( \frac{1}{T_2} - \frac{1}{T_1} \right) \] where: \(P_1\) and \(P_2\) are vapor pressures at temperatures \(T_1\) and \(T_2\), \(\Delta H_{vap}\) is the enthalpy of vaporization, \(R\) is the ideal gas constant (8.314 J/mol·K).

Convert the given temperatures to Kelvin

To convert the given temperature in degree Celsius to Kelvin, add 273.15 to each temperature: \(T_1 = 5.0^{\circ} \mathrm{C} + 273.15 = 278.15 \mathrm{K}\) \(T_2 = 21.2^{\circ} \mathrm{C} + 273.15 = 294.35 \mathrm{K}\)

Calculate the enthalpy of vaporization

The given vapor pressure values are: \(P_1 = 0.0526 \, \mathrm{atm}\) \(P_2 = 0.132 \, \mathrm{atm}\) First, rearrange the Clausius-Clapeyron equation to isolate \(\Delta H_{vap}\) on one side: \(\Delta H_{vap} = -R \left( \frac{1}{T_2} - \frac{1}{T_1} \right)^{-1} \ln \frac{P_2}{P_1}\) Now, substitute the values: \(\Delta H_{vap} = -8.314 \left( \frac{1}{294.35} - \frac{1}{278.15} \right)^{-1} \ln \frac{0.132}{0.0526}\) Calculate \(\Delta H_{vap}\): \(\Delta H_{vap} = 3.88 \times 10^4 \, \mathrm{J/mol}\)

Calculate the normal boiling point temperature

The normal boiling point is the temperature at which the vapor pressure of a substance is equal to 1 atm. So, we can rewrite the Clausius-Clapeyron equation using the normal boiling point temperature (\(T_{nb}\)) and vapor pressure equal to 1 atm (\(P_{nb} = 1 \, \mathrm{atm}\)): \[ \ln \frac{P_{nb}}{P_1} = -\frac{\Delta H_{vap}}{R} \left( \frac{1}{T_{nb}} - \frac{1}{T_1} \right) \] Rearrange the equation to isolate \(T_{nb}\) on one side: \(T_{nb} = \left( \frac{1}{T_1} + \frac{R}{\Delta H_{vap}} \ln \frac{P_{nb}}{P_1} \right)^{-1}\) Now, substitute the values: \(T_{nb} = \left( \frac{1}{278.15} + \frac{8.314}{3.88 \times 10^4} \ln \frac{1}{0.0526} \right)^{-1}\) Calculate \(T_{nb}\): \(T_{nb} = 337.7 \, \mathrm{K}\)

Convert the normal boiling point temperature to Celsius

Finally, convert the calculated normal boiling point temperature in Kelvin to degree Celsius by subtracting 273.15: \(T_{nb} = 337.7 \, \mathrm{K} - 273.15 = 64.5^{\circ} \mathrm{C}\) The normal boiling point of methyl alcohol (CH3OH) is predicted to be \(64.5^{\circ} \mathrm{C}\).

Key concepts

Clausius-Clapeyron Equation

The Clausius-Clapeyron equation is a fundamental principle in thermodynamics which describes how the pressure of a gas in equilibrium with its liquid (or solid) phase changes in response to temperature. In simpler terms, it relates the temperature and pressure at which a substance undergoes phase transitions, such as boiling.

This equation takes the form of: \[\begin{equation}\ln \frac{P_2}{P_1} = -\frac{\Delta H_{vap}}{R} \left( \frac{1}{T_2} - \frac{1}{T_1} \right)\end{equation}\]
Where:\

• \(P_1\) and \(P_2\) are vapor pressures at temperatures \(T_1\) and \(T_2\) respectively.
• \(\Delta H_{vap}\) is the enthalpy of vaporization, which is the energy required to transition from the liquid phase to the gaseous phase at constant temperature and pressure.
• \(R\) is the ideal gas constant, valued at 8.314 J/mol·K.

By using the Clausius-Clapeyron equation, students can calculate changes in vapor pressure with temperature or predict the boiling point of a substance at a given pressure, which is typically set to atmospheric pressure for 'normal' boiling point calculations.

Enthalpy of Vaporization

The enthalpy of vaporization, \(\Delta H_{vap}\), is an essential concept when discussing phase changes, particularly during the transition from liquid to gas. It represents the amount of heat energy required to vaporize one mole of a liquid at constant pressure. Physically, this energy is needed to overcome the intermolecular forces that hold the liquid together.

In the Clausius-Clapeyron equation, the enthalpy of vaporization plays a central role, allowing one to link how a substance's vapor pressure changes with temperature. For many substances, the enthalpy of vaporization can be looked up in chemical reference materials or calculated from experimental data using the mentioned equation.

To give a better understanding, the calculation method is often included in the step-by-step solutions to help students grasp the practical application of this thermodynamic value. Often, they might find absolute values in tables, but understanding the calculation process is crucial for mastering the concept.

Converting Celsius to Kelvin

Temperature conversions are frequently encountered in scientific problems, especially when working with equations like the Clausius-Clapeyron where absolute temperatures are needed. The Kelvin scale is an absolute temperature scale used by scientists to ensure accuracy in calculations where temperature changes are crucial.

The conversion between Celsius and Kelvin is straightforward: the Kelvin temperature is simply the Celsius temperature plus 273.15. Therefore, to convert:\

• \(T_{Kelvin} = T_{Celsius} + 273.15\)

Remember, unlike Celsius, where temperatures can be negative, Kelvin temperatures are always positive, reflecting the absolute nature of this scale. When students perform these conversions, it's an opportunity to reinforce understanding of different temperature scales and the importance of using Kelvin for certain scientific calculations.

Methyl Alcohol Properties

Methyl alcohol, also known as methanol or wood alcohol, has the chemical formula \(CH_3OH\). Methanol is a colorless, volatile, and flammable liquid with a distinct odor, and it is used in various industrial processes including as a solvent, antifreeze, and fuel. Understanding its physical properties, like boiling and melting points, is paramount in safely handling and using this chemical.

Its boiling point, which has been a point of discussion in the original problem, is a critical property that distinguishes methanol from other alcohols, like ethanol. Knowing that normal boiling point is the temperature at which a liquid's vapor pressure equals atmospheric pressure, and by applying the concepts of the Clausius-Clapeyron equation, enthalpy of vaporization, and temperature conversions, students can predict the normal boiling point of methanol under standard conditions.

Incorporating the properties of methanol into calculations helps provide context and enhances the practical relevance of the thermodynamic principles being applied.