Question

Mercury (Hg) vapor is toxic and readily absorbed from the lungs. At \(20 .^{\circ} \mathrm{C},\) mercury \(\left(\Delta H_{\mathrm{vap}}=59.1 \mathrm{~kJ} / \mathrm{mol}\right)\) has a vapor pressure of \(1.20 \times 10^{-3}\) torr, which is high enough to be hazardous. To reduce the danger to workers in processing plants, \(\mathrm{Hg}\) is cooled to lower its vapor pressure. At what temperature would the vapor pressure of \(\mathrm{Hg}\) be at the safer level of \(5.0 \times 10^{-5}\) torr?

Short answer

The temperature at which the vapor pressure of Hg is 5.0×10^{-5} torr is approximately -51.34°C.

Step-by-step solution

Understand the Problem

The goal is to determine the temperature at which the vapor pressure of mercury (Hg) is reduced to a safer level of 5.0 × 10^{-5} torr, given the vapor pressure at 20°C, which is 1.20 × 10^{-3} torr, and the enthalpy of vaporization ΔH_{vap} of 59.1 kJ/mol.

Use the Clausius-Clapeyron Equation

The Clausius-Clapeyron equation relates the vapor pressures and temperatures for a substance, given by: \[\frac{\text{d} \text{ln}(P)}{\text{d}T} = \frac{\triangle H_{vap}}{R T^2}\] which can be integrated to: \[ \text{ln} \frac{P_2}{P_1} = \frac{\triangle H_{vap}}{R} \bigg( \frac{1}{T_1} - \frac{1}{T_2} \bigg)\].

Identify Known Variables

Given: \[ P_1 = 1.20 \times 10^{-3} \text{ torr}, \ P_2 = 5.0 \times 10^{-5} \text{ torr}, \ T_1 = 20^\text{o}C = 293.15 \text{ K}, \ \triangle H_{vap} = 59.1 \text{ kJ/mol} = 59,100 \text{ J/mol}, \ R = 8.314 \text{ J/mol·K}\]. We need to find: \[T_2\].

Rearrange and Solve for T_2

Rewriting the Clausius-Clapeyron equation for \[T_2\]: \[ \frac{1}{T_2} = \frac{1}{T_1} - \frac{R}{\triangle H_{vap}} \text{ln} \frac{P_2}{P_1}\]. Substitute the known variables:\[ \frac{1}{T_2} = \frac{1}{293.15} - \frac{8.314}{59100} \text{ln} \bigg(\frac{5.0 \times 10^{-5}}{1.20 \times 10^{-3}}\bigg)\] \[ \frac{1}{T_2} = \frac{1}{293.15} - \frac{8.314}{59100} \text{ln} \bigg(\frac{5.0}{120.0}\bigg)\] \[ \frac{1}{T_2} = \frac{1}{293.15} - \frac{8.314}{59100} \text{ln} (4.17 \times 10^{-4})\].

Calculate ln (4.17 × 10^{-4})

Using a calculator, \[ \text{ln} (4.17 \times 10^{-4}) \thickapprox -7.786\].

Substitute Back to Find 1/T_2

Plugging back into the equation:\[ \frac{1}{T_2} = \frac{1}{293.15} - \frac{8.314 \times (-7.786)}{59100}\] \[ \frac{1}{T_2} = \frac{1}{293.15} + \frac{64.73}{59100} \thickapprox \frac{1}{293.15} + 0.001095\].

Calculate T_2

Evaluate the expression:\[ \frac{1}{T_2} \thickapprox 0.003412 + 0.001095 = 0.004507\] Thus, \[ T_2 = \frac{1}{0.004507} \thickapprox 221.81 \text{ K}\]. Converting to Celsius: \[ T_2 \thickapprox 221.81 - 273.15 = -51.34^\text{o}C\].

Key concepts

Understanding Vapor Pressure

Vapor pressure is an important concept to grasp when studying the behavior of liquids and gases at various temperatures. It refers to the pressure exerted by a vapor in equilibrium with its liquid phase at a particular temperature. Simply put, it's how eager the molecules of a liquid are to escape into the air.
For example, water at 25°C has a certain vapor pressure, whereas at boiling, the pressure is much higher due to increased molecular movement.
A key fact to remember is that vapor pressure increases with temperature because heating a liquid boosts the kinetic energy of its molecules, making them more likely to evaporate.

• High vapor pressure means the substance evaporates easily.
• Low vapor pressure means it evaporates less readily.

In the context of mercury (Hg), high vapor pressure at room temperature means it can be dangerous, thus understanding how to reduce it via cooling is crucial.

Delving into Enthalpy of Vaporization

The enthalpy of vaporization, denoted \( \Delta H_{\text{vap}} \, \), is the heat required to convert a liquid to a gas without a temperature change. It's measured in kJ/mol and reflects the energy needed to break intermolecular forces.
Mercury's enthalpy of vaporization is 59.1 kJ/mol. This high value indicates significant energy is required to transform mercury from a liquid to a gaseous state.
When using the Clausius-Clapeyron equation, this value helps determine how temperature influences vapor pressure. It essentially informs us how the energy dynamics of mercury change with temperature. Understanding this aids in evaluating and manipulating vapor pressures for safety purposes in industrial settings.

• Higher \ \Delta H_{\text{vap}} \ means stronger molecular forces.
• Lower \ \Delta H_{\text{vap}} \ indicates easier phase transition.

The Importance of Temperature Conversion

Temperature conversion is vital when applying formulas like the Clausius-Clapeyron equation. Temperature scales used often are Celsius (°C) and Kelvin (K).
Knowing how to convert between them ensures accurate calculation of the involved thermodynamic processes.
Celsius to Kelvin conversion is straightforward: simply add 273.15 to the Celsius value.

• Example: 20°C = 293.15 K
• Another Example: -51.34°C = 221.81 K

Using Kelvin is critical because thermodynamic equations, including the Clausius-Clapeyron, require absolute temperature values. Celsius doesn't start from absolute zero, while Kelvin does.
Proper temperature conversion eliminates potential errors in scientific calculations, ensuring precise results and safety in applications like controlling mercury vapor pressure in industrial environments.