Question

Vapor pressure measurements at several different temperatures are shown for mercury. Determine graphically the molar heat of vaporization for mercury. $$ \begin{array}{lccccc} T\left({ }^{\circ} \mathrm{C}\right) & 200 & 250 & 300 & 320 & 340 \\ \hline P(\mathrm{mmHg}) & 17.3 & 74.4 & 246.8 & 376.3 & 557.9 \end{array} $$

Short answer

Graphically determine ln P vs. 1/T; calculate slope to find \(\Delta H_{vap}\) using \(R\).

Step-by-step solution

Convert Temperatures to Kelvin

First, convert the given temperatures in degrees Celsius to Kelvin by adding 273.15 to each temperature. This is because absolute temperature in Kelvin is required for the calculations.\[ T(K) = T(°C) + 273.15 \]For the temperatures given:- \( T(200°C) = 473.15 \, K \)- \( T(250°C) = 523.15 \, K \)- \( T(300°C) = 573.15 \, K \)- \( T(320°C) = 593.15 \, K \)- \( T(340°C) = 613.15 \, K \)

Calculate Natural Logarithm of Pressures

Convert the pressures from mmHg to natural logarithms, since the Clausius-Clapeyron equation involves the natural logarithm of pressure.\[ P(\mathrm{mmHg}) = \{17.3, 74.4, 246.8, 376.3, 557.9\} \]Calculating \( \ln P \):- \( \ln 17.3 \approx 2.85 \)- \( \ln 74.4 \approx 4.31 \)- \( \ln 246.8 \approx 5.51 \)- \( \ln 376.3 \approx 5.93 \)- \( \ln 557.9 \approx 6.32 \)

Plot ln(P) Against 1/T

Plot the natural logarithm of pressure (\(\ln P\)) on the y-axis against the inverse of temperature in Kelvin (1/T) on the x-axis. This plot should result in a straight line based on the Clausius-Clapeyron relation:\[\ln P = -\frac{\Delta H_{vap}}{R} \cdot \frac{1}{T} + C\]where \(\Delta H_{vap}\) is the molar heat of vaporization and \(R\) is the ideal gas constant (8.314 J/mol·K).

Determine Slope of the Line

From the plot of \(\ln P\) versus \(1/T\), determine the slope of the line, which should be negative based on the equation from the Clausius-Clapeyron relation. The slope \(m\) will be:\[m = -\frac{\Delta H_{vap}}{R}\]

Calculate the Molar Heat of Vaporization

Using the slope obtained from the plot, calculate the molar heat of vaporization (\(\Delta H_{vap}\)) of mercury by rearranging the formula:\[\Delta H_{vap} = -m \cdot R\]Substitute the calculated slope into this equation using \(R = 8.314 \, \text{J/mol} \, \text{K}\) to find \(\Delta H_{vap}\).

Key concepts

Clausius-Clapeyron Equation

The Clausius-Clapeyron equation is pivotal for understanding the relationship between vapor pressure and temperature. It helps calculate the molar heat of vaporization, which is the amount of energy needed to convert one mole of a liquid into vapor at constant temperature. This equation is an expression of the phase equilibrium condition of substances with pressure and temperature. In a typical format, the equation is given by: \[ \ln P = -\frac{\Delta H_{vap}}{R} \cdot \frac{1}{T} + C \]Here,

• \(P\) is the vapor pressure of the substance.
• \(\Delta H_{vap}\) represents the molar heat of vaporization.
• \(R\) is the ideal gas constant, valued at 8.314 J/mol·K.
• \(T\) is the absolute temperature in Kelvin.
• \(C\) is a constant that depends on the substance and its state.

The slope of the line from plotting \(\ln P\) versus \(1/T\) yields information about \(\Delta H_{vap}\). This graphical method allows students to visually understand how changes in temperature influence vapor pressure.

Natural Logarithm

The natural logarithm (often abbreviated as \(\ln\)) is a mathematical function used in the Clausius-Clapeyron equation to linearize data, making it easier to analyze graphically. In this context, \(\ln\) represents the logarithm to the base \(e\), where \(e\) is approximately 2.718. When you take the natural logarithm of vapor pressure, you transform exponential relationships into linear ones, allowing for simpler analysis through straight-line graphs.Here's a quick reminder about natural logarithms:

• It is the power to which \(e\) must be raised to obtain a given number.
• Converting pressure to its natural logarithm helps interpret non-linear relationships.
• Using \(\ln\) makes equations more manageable for calculations.

In the problem, converting given pressures into \(\ln P\) supports understanding pressure trends against temperature.

Kelvin Conversion

Understanding temperatures in the Kelvin scale is essential because it provides an absolute reference frame. In scientific calculations involving gas laws and thermodynamics, Kelvin is preferred over Celsius or Fahrenheit because it eliminates negative temperature values, which can complicate calculations. To convert Celsius to Kelvin, simply add 273.15 to the Celsius value. For instance:

• If the temperature is 200°C, converting to Kelvin would result in a temperature of 473.15 K.
• This conversion is critical, especially when applying the Clausius-Clapeyron equation, which requires absolute temperatures for accuracy.

Understanding Kelvin conversion ensures accurate determination of vapor pressures at different temperatures, providing a clear scientific foundation for further calculations.

Vapor Pressure

Vapor pressure is the pressure exerted by a vapor in equilibrium with its liquid or solid form at a given temperature. Understanding this concept is crucial for studying the phase changes of substances like mercury and their responses to temperature changes.When a liquid is in a closed container, molecules escape from the liquid into the vapor phase. When equilibrium is reached, the number of molecules escaping equals the number returning, and the pressure exerted by this vapor is known as vapor pressure.Key points to remember about vapor pressure:

• It increases with temperature because more molecules have enough energy to escape into the vapor phase.
• Vapor pressure is unique to each substance and varies depending on molecular interactions and temperature.
• Vapor pressure data at various temperatures support analyzing and understanding heat of vaporization calculations.

This concept is fundamental when utilizing the Clausius-Clapeyron equation, as calculating \(\ln P\) for plotting requires an understanding of vapor pressure behavior.

Graphical Analysis

Graphical analysis transforms mathematical data into a visual format, making complex relationships and trends more accessible. For the vapor pressure and temperature relationship, plotting \(\ln P\) against \(1/T\) creates a straight-line graph, simplifying the understanding of the Clausius-Clapeyron equation.Here's how graphical analysis aids learning:

• Visually represents the data and highlights the linear relationship between \(\ln P\) and \(1/T\).
• The slope of the line indicates the heat of vaporization, serving as a practical tool for calculations.
• It helps detect any anomalies or deviations in the underlying data, ensuring accuracy.

This visual approach helps students intuitively grasp how temperature shifts affect vapor pressure and assists in calculating the substance's molar heat of vaporization effectively.