Chapter 9: Problem 29
Use the following data to extimate the normal boiling point and the molur entalpy of vaporization of lead. \begin{tabular}{l|lllll} \(f / \mathrm{K}\) & 1500 & 1600 & 1700 & 1800 & 1900 \\ \hline\(H / \mathrm{kar}\) & \(19.72\) & \(48.4 \mathrm{~K}\) & \(107.2\) & \(217.7\) & \(408.2\) \end{tabular}
Short Answer
Step by step solution
Arrange the Given Data
Convert Temperatures to Inverse
Calculate the Natural Logarithm of Enthalpy
Perform Linear Regression
Calculate the Molar Enthalpy of Vaporization
Estimate the Normal Boiling Point
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Normal Boiling Point
The boiling point can change with pressure variations, which is why the term "normal" specifies the standard atmospheric condition of 1 atm. In practical terms:
- At the normal boiling point, the heat supplied is equivalent to the amount needed to break intermolecular forces without changing temperature.
- The normal boiling point represents a key phase change.
Linear Regression in Thermodynamics
- Data points are plotted, such as \(\ln H\) versus \(1/T\).
- A straight line best fitting these points is drawn, known as the regression line.
- The slope of this line \(m\) carries physical meaning, often representing negative enthalpy of vaporization divided by the gas constant \(R\).
- The intercept \(c\) provides additional reference, often linked to pressure relationships.
Phase Change Enthalpy Calculations
- Perform a linear regression analysis, plotting \(\ln H\) versus \(1/T\).
- Determine the slope \(m\), which is related to the enthalpy change by the equation \(\ H = -\frac{\Delta H_{vap}}{R} \times \frac{1}{T} + C\).
- Use the slope to calculate \(\Delta H_{vap}\) with the relation \(\Delta H_{vap} = -m \times R\).