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Chapter 11: Problem 44

Methanol, \(\mathrm{CH}_{3} \mathrm{OH}\), a colorless, volatile liquid, was formerly known as wood alcohol. It boils at \(65.0^{\circ} \mathrm{C}\) and has a heat of vaporization of \(37.4 \mathrm{~kJ} / \mathrm{mol}\). What is its vapor pressure at \(22.0^{\circ} \mathrm{C}\) ?

Short Answer

Expert verified
The vapor pressure of methanol at 22.0掳C is approximately 13.636 kPa.

Step by step solution

01

Identify the Formula for Vapor Pressure

To find the vapor pressure of methanol at a given temperature, we can use the Clausius-Clapeyron equation, given by:\[\ln\left(\frac{P_2}{P_1}\right) = \frac{\Delta H_{\text{vap}}}{R}\left(\frac{1}{T_1} - \frac{1}{T_2}\right)\]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 universal gas constant \((8.314 \: \text{J/mol} \cdot \text{K})\).
02

Convert Temperatures to Kelvin

Convert the given and known boiling point temperatures from Celsius to Kelvin:\[T_1 = 65.0^{\circ}C + 273.15 = 338.15\ K\]\[T_2 = 22.0^{\circ}C + 273.15 = 295.15\ K\]
03

Set Known Conditions

We know that the vapor pressure at the boiling point \(T_1\) is equal to 1 atmosphere (or 101.325 kPa since we often use SI units in physics), therefore \(P_1 = 101.325\ kPa\). The heat of vaporization \(\Delta H_{\text{vap}} = 37.4\ kJ/mol = 37400\ J/mol\).
04

Apply the Clausius-Clapeyron Equation

Use the Clausius-Clapeyron equation:\[\ln\left(\frac{P_2}{101.325}\right) = \frac{37400}{8.314}\left(\frac{1}{338.15} - \frac{1}{295.15}\right)\]
05

Solve for \(\ln(P_2)\)

Calculate the right-hand side of the equation:\[\frac{37400}{8.314} \times \left(\frac{1}{338.15} - \frac{1}{295.15}\right) \approx -2.034 \]This gives:\[\ln\left(\frac{P_2}{101.325}\right) = -2.034\]
06

Calculate \(P_2\)

Solve for \(P_2\) by exponentiating both sides:\[\frac{P_2}{101.325} = e^{-2.034}\]\[\Rightarrow P_2 = 101.325 \times e^{-2.034}\]\[\Rightarrow P_2 \approx 13.636\ kPa\]
07

Confirm the Answer

Check the calculations to ensure they are correct, ensuring unit consistency and re-performing any arithmetic if necessary.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Vapor Pressure
Vapor pressure is the pressure exerted by a vapor in equilibrium with its liquid form. It depends on temperature; as the temperature rises, so does the vapor pressure. This is because higher temperatures provide more energy for molecules to escape the liquid phase and enter the vapor phase. In the given exercise, we calculated the vapor pressure of methanol at a non-boiling temperature, demonstrating how we can determine it using known data such as boiling point vapor pressure and the Clausius-Clapeyron equation. It's important to note that understanding vapor pressure plays a crucial role in fields such as meteorology and food science.
Heat of Vaporization
The heat of vaporization is the amount of energy required to convert a liquid into vapor without changing its temperature. It is expressed in Joules per mole (J/mol) or kilojoules per mole (kJ/mol). Methanol, for example, has a heat of vaporization of 37.4 kJ/mol, meaning each mole of liquid methanol requires 37.4 kJ to become vapor. This value is crucial when using the Clausius-Clapeyron equation to find vapor pressures at different temperatures. It indicates how much energy is needed for molecules to overcome intermolecular forces and transition into the vapor phase.
Universal Gas Constant
The universal gas constant, denoted as \( R \), is a key component in the Clausius-Clapeyron equation. It represents the proportionality constant in the ideal gas law and is approximately \( 8.314 \, \text{J/mol} \cdot \text{K} \). This constant aids in relating temperature changes to vapor pressure changes, providing a bridge between macroscopic observations and molecular level interactions. By including \( R \) in the equation, we can apply standardized units ensuring our calculations are correct and comparable across different scenarios. This constant is fundamental in any thermodynamic equation involving gases.
Temperature Conversion
Temperature conversion is often necessary in chemistry problems, especially when dealing with equations like the Clausius-Clapeyron that require temperature in Kelvin for consistency and accuracy. Celsius temperatures can be converted to Kelvin by adding 273.15. For instance, the boiling point of methanol at \( 65.0^{\circ}C \) becomes \( 338.15 \, K \), and \( 22.0^{\circ}C \) converts to \( 295.15 \, K \). Converting to Kelvin ensures that all thermodynamic calculations are conducted on an absolute scale, in line with physical principles. This step is critical for accurate results when applying mathematical models to real-world phenomena.

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Most popular questions from this chapter

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