We will approach this problem step by step. Here’s the breakdown of the tasks:

Given formula:

$\ln\left(\frac{p}{\text{torr}}\right) = 23.7 - \frac{7769}{T} - \ln T$ Where:

• $p$ is the vapor pressure in torr,
• $T$ is the temperature in Kelvin (K).

We'll solve each part:

Part a: Calculate the vapor pressure at 25 °C

1. Convert 25°C to Kelvin: $T = 25 + 273.15 = 298.15\,\text{K}$
2. Substitute into the given equation: $\ln\left(\frac{p}{\text{torr}}\right) = 23.7 - \frac{7769}{298.15} - \ln(298.15)$
3. Solve for $p$:
   • First, calculate the right-hand side.
   • Then, exponentiate both sides to solve for $p$.

Part b: Calculate the boiling temperature at $p = 1$ bar

To solve this, we'll use the following approach:

1. Convert 1 bar to torr: $1\,\text{bar} = 750.062\,\text{torr}$
2. Substitute $p = 750.062 \, \text{torr}$ into the equation: $\ln\left(\frac{750.062}{\text{torr}}\right) = 23.7 - \frac{7769}{T} - \ln T$
3. Solve for $T$:
   • Rearrange and solve this equation numerically for $T$.

Part c: Calculate the enthalpy of vaporization at $T = 473 \, \text{K}$ and $673 \, \text{K}$

The enthalpy of vaporization, $\Delta H_{\text{vap}}$, can be derived using the Clausius-Clapeyron equation: $\Delta H_{\text{vap}} = -R \left( \frac{d\ln p}{d(1/T)} \right)$ Where:

• $R = 8.314 \, \text{J/mol·K}$ is the gas constant,
• $\frac{d\ln p}{d(1/T)}$ is the slope of the line obtained by plotting $\ln p$ vs. $1/T$.

To calculate $\Delta H_{\text{vap}}$ at $T = 473 \, \text{K}$ and $673 \, \text{K}$:

1. Differentiate the vapor pressure equation: $\frac{d}{dT}\left( 23.7 - \frac{7769}{T} - \ln T \right)$ Simplifying this gives: $\frac{d\ln p}{dT} = \frac{7769}{T^2} - \frac{1}{T}$
2. Use this derivative to calculate the enthalpy at the given temperatures: $\Delta H_{\text{vap}} = R \left( \frac{7769}{T} - 1 \right)$ Plug in $T = 473 \, \text{K}$ and $T = 673 \, \text{K}$.

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I will now calculate each part.Here are the results for each part:

Part a: Vapor pressure at 25°C

The vapor pressure of mercury at $25^\circ \text{C}$ (298.15 K) is approximately: $p \approx 0.00032 \, \text{torr}$

Part b: Boiling temperature at $p = 1$ bar

The boiling temperature of mercury at $p = 1 \, \text{bar}$ (750.062 torr) is approximately: $T \approx 742 \, \text{K}$

Part c: Enthalpy of vaporization

The enthalpy of vaporization at:

• $T = 473 \, \text{K}$ is approximately $0.128 \, \text{kJ/mol}$,
• $T = 673 \, \text{K}$ is approximately $0.088 \, \text{kJ/mol}$.

If you would like more details or have any questions, feel free to ask!

Here are five related questions to expand your understanding:

1. How does the vapor pressure change with increasing temperature?
2. What is the significance of the Clausius-Clapeyron equation in this context?
3. How does enthalpy of vaporization vary with temperature?
4. Why is mercury’s boiling temperature at 1 bar significantly higher than room temperature?
5. What are some practical applications of calculating vapor pressure and enthalpy of vaporization?

Tip: Enthalpy of vaporization generally decreases as temperature increases because less energy is required to overcome intermolecular forces at higher temperatures.