Problem 5 - A Modern Course in Statistical Physics

Question

Consider the vaporization curve for liquid mercury. The latent heat of vaporization, \(L\) (in \(\mathrm{J} / \mathrm{mol}\) ) varies slowly with pressure, but has significant variation with temperature and can be written \(L=(7724-0.9768 T) R\), where \(R\) is the gas constant and \(T\) is measured in kelvin. It is known that at atmospheric pressure ( \(P_{\mathrm{am}}=1.013 \times 10^{5} \mathrm{~Pa}\) ) mercury begins to vaporize at temperature \(T=\) \(630 \mathrm{~K}\). (a) Plot the vaporization curve for mercury between temperatures \(T=500 \mathrm{~K}\) and \(T=650 \mathrm{~K}\). (b) At what pressure does mercury begin to vaporize when \(T=530 \mathrm{~K}\) ? (Hint: The volume of the liquid mercury can be neglected relative to that of the vapor and the vapor can be treated as an ideal gas.)

Step-by-Step Solution

Answer

Plot the vaporization curve from 500K to 650K, calculate pressure at T=530K using integrated Clausius-Clapeyron relation.

Step 1: Write the given latent heat of vaporization formula

The latent heat of vaporization, as given, is: \[ L = (7724 - 0.9768T)R \]

Step 2: Recall the Clausius-Clapeyron equation

The Clausius-Clapeyron equation relates the pressure and temperature during a phase change: \[ \frac{dP}{dT} = \frac{L}{T \Delta V} \] Since the volume of liquid mercury is negligible, assume \( \Delta V \) is approximately the volume of gas phase: \[ V = \frac{RT}{P} \]

Step 3: Substitute the expression for volume into Clausius-Clapeyron equation

Substitute \( V = \frac{RT}{P} \) into Clausius-Clapeyron equation:\[ \frac{dP}{dT} = \frac{L}{T \left( \frac{RT}{P} \right)} = \frac{LP}{R T^2 }\]

Step 4: Separate variables and integrate

Separate the variables for integration: \[ \frac{dP}{P} = \frac{L}{R} \cdot \frac{dT}{T^2} \]Integrate from \( P_{am} \) to \( P \) and \( T_{am} = 630 \) K to \( T \): \[ \int_{P_{am}}^{P} \frac{dP}{P} = \int_{T_{am}}^{T} \frac{L}{R} \cdot \frac{dT}{T^2} \]

Step 5: Evaluate the integrals

Evaluate both sides: \[ \bigg[ \ln P \bigg]_{P_{am}}^{P} = \int_{630}^{T} \frac{7724 - 0.9768T}{RT^2} dT \] For the left side result: \[ \bigg[ \ln P \bigg]_{P_{am}}^{P} = \ln \frac{P}{P_{am}} \] For the right side, split the integral: \[ \int_{630}^{T} \frac{7724}{RT^2} dT - \int_{630}^{T} \frac{0.9768}{RT} dT = \frac{7724}{R} \left[ -\frac{1}{T} \right]_{630}^{T} - \frac{0.9768}{R} \left[ \ln T \right]_{630}^{T} \]

Step 6: Solve and simplify

Solve the right side: \[ \frac{7724}{R} \left( -\frac{1}{T} + \frac{1}{630} \right) - \frac{0.9768}{R} (\ln T - \ln 630) = -\frac{7724}{R} \left( \frac{T - 630}{T \cdot 630} \right) - \frac{0.9768}{R} \ln \frac{T}{630} \] Simplify the expression: \[ \ln \frac{P}{P_{am}} = -\frac{7724(T - 630)}{R \cdot T \cdot 630} - \frac{0.9768}{R} \ln \frac{T}{630} \] Use boundary values for specifying equation and follow hints assumptions for plotting and P calculation.

Step 7: Plot the vaporization curve for T=[500,650]K

These equations can be coded or use software like MATLAB or Excel for drawing the vaporization curve for T=[500,650]K.

Step 8: Calculate vaporization pressure at T=530K

Substitute \(T = 530 \)K into the simplified equation and solve for \( P \) relative to \( P_{am} \).

Key Concepts

Clausius-Clapeyron equation

The Clausius-Clapeyron equation is a vital tool for understanding the relationship between pressure and temperature during a phase change, particularly between liquid and vapor phases. The equation is given by:
\[ \frac{dP}{dT} = \frac{L}{T \triangle V} \]
Here,

• P is the pressure,
• T is the temperature,
• L is the latent heat of vaporization,
• \(\triangle V\) is the change in volume during the phase transition.

For mercury, given the negligible volume of its liquid phase, we approximate \(\triangle V\) using the gas phase volume:
\[ V = \frac{RT}{P} \]
Substituting this into the Clausius-Clapeyron equation, we get:
\[ \frac{dP}{dT} = \frac{L P}{R T^2} \]
This differential form helps in analyzing how slight changes in temperature affect the vapor pressure. By integrating this expression, we can determine the relationship between pressure and temperature more explicitly, which is essential for solving problems involving phase changes, like those of mercury vaporization.

Ideal gas law

The ideal gas law provides a good approximation for the behavior of gases under many conditions. It is expressed as:
\[ PV = nRT \]
Where,

• P is the pressure,
• V is the volume,
• n is the number of moles,
• R is the gas constant,
• T is the temperature in Kelvin.

In scenarios like the vaporization of mercury, where the gaseous phase dominates, we use the ideal gas law to approximate the volume of vapor. This substitution simplifies the Clausius-Clapeyron equation and enables us to integrate and analyze the phase transition between liquid and gas precisely. By considering the behavior of mercury vapor as an ideal gas, we facilitate the calculation of pressures and temperatures relevant to the vaporization process.

Phase change

Phase changes, such as from liquid to vapor, involve significant energy changes characterized by the latent heat of vaporization. When a substance like mercury vaporizes, it absorbs heat ( L ), which allows the molecules to break free from the liquid phase. The latent heat equation for mercury is given by:
\[ L = (7724 - 0.9768T)R \]
Here, the latent heat decreases slightly with increasing temperature. Understanding phase changes involves examining how temperature and pressure interplay to allow a substance to move between phases.
These transitions are best understood by the Clausius-Clapeyron equation and the ideal gas law, which together provide a comprehensive framework for calculating the conditions under which vaporization or condensation occurs.

• The energy required for phase change (latent heat)
• The influence of ambient pressure and temperature
• Behavior differences between the liquid and gas phases

By integrating these concepts, you can understand how substances like mercury behave under different conditions and predict where phase changes will occur. This understanding is crucial for plotting vaporization curves and solving related thermodynamic problems.