The vapor pressure of mercury is given by ln(p/torr) = 23.7 − 7769/T − lnT. a) Calculate the vapor pressure at 25 °C; b) Calculate the boiling temperature at p = 1 bar; c) Calculate the enthalpy of vaporization at T = 473 K and 673 K.

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 \( T = 673 \, \text{K} \).

---

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.

This tutorial walks through solving for mercury's vapor pressure at 25°C, boiling temperature at 1 bar, and enthalpy of vaporization at T = 473 K and 673 K using thermodynamic principles like the Clausius-Clapeyron equation.

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