Chapter 11: Problem 37
The vapor pressure of liquid \(\mathrm{X}\) is lower than that of liquid \(Y\) at \(20^{\circ} \mathrm{C}\), but higher at \(60^{\circ} \mathrm{C}\). What can you deduce about the relative magnitude of the molar heats of vaporization of \(\mathrm{X}\) and \(\mathrm{Y}\) ?
Short Answer
Step by step solution
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Molar Heat of Vaporization
It plays a vital role in understanding how substances behave under heat and pressure changes. Generally speaking, a higher \( \Delta H_{vap} \) suggests that a substance requires more energy to escape into the vapor phase. Therefore, this means it may be less volatile at lower temperatures.
This concept ties directly into how substances interact with temperature increases, such as the shift observed between the 20°C and 60°C conditions for liquids X and Y in the problem.
Clausius-Clapeyron Equation
The variables in the equation include:
- \( \Delta H_{vap} \) - Molar heat of vaporization
- \( R \) - Universal gas constant
- \( T_1 \) and \( T_2 \) - Temperatures in Kelvin
- \( P_1 \) and \( P_2 \) - Vapor pressures at \( T_1 \) and \( T_2 \) respectively
Thermodynamic Equilibrium
This concept is crucial because vapor pressure is defined under these conditions, allowing us to study the system without interference from external changes.
In the problem's context, understanding thermodynamic equilibrium helps explain why changes in vapor pressures signify differences in other properties, like the molar heat of vaporization of the substances in question.
Volatility
Factors affecting volatility include:
- Intermolecular forces: Weaker forces increase volatility as less energy is needed to break these bonds.
- Molar heat of vaporization: Lower \( \Delta H_{vap} \) often means higher volatility since less energy is required for vaporization.