For mercury, the enthalpy of vaporization is \(58.51 \mathrm{kJ} / \mathrm{mol}\) and the entropy of vaporization is \(92.92 \mathrm{J} / \mathrm{K} \cdot\) mol. What is the normal boiling point of mercury?

We need to convert the pressure to Pascal (Pa), which is the SI unit for pressure: 1 atm = 101,325 Pa

Now we can plug in the values for ΔH_vap, ΔS_vap, P, and the gas constant R into the Clausius-Clapeyron equation. \[ \Delta H_{vap} = T\Delta S_{vap} - RT\ln{P} \] \[ 58,510\,\text{J/mol} = T(92.92\,\text{J/K}\cdot\text{mol}) - (8.314\,\text{J/K}\cdot\text{mol})(T)\ln{(101,325\,\text{Pa})} \]

To solve this equation, first we need to isolate T on one side. Divide both sides by (92.92 J/K · mol): \[ T = \frac{58,510\,\text{J/mol}}{92.92\,\text{J/K}\cdot\text{mol}} + \frac{8.314\,\text{J/K}\cdot\text{mol}}{92.92\,\text{J/K}\cdot\text{mol}} T \ln{(101,325\,\text{Pa})} \] Now let's denote \(x = T\ln{(101,325\,\text{Pa})}\) to simplify the equation further: \[ T = \frac{58,510\,\text{J/mol}}{92.92\,\text{J/K}\cdot\text{mol}} + \frac{8.314\,\text{J/K}\cdot\text{mol}}{92.92\,\text{J/K}\cdot\text{mol}}\cdot x \] Now we can subtract \(\frac{8.314\,\text{J/K}\cdot\text{mol}}{92.92\,\text{J/K}\cdot\text{mol}}\cdot x\) from both sides: \[ T - \frac{8.314\,\text{J/K}\cdot\text{mol}}{92.92\,\text{J/K}\cdot\text{mol}}\cdot x = \frac{58,510\,\text{J/mol}}{92.92\,\text{J/K}\cdot\text{mol}} \] Now we need to solve for x using the earlier definition: \[ x = T\ln{(101,325\,\text{Pa})} \] Use the natural logarithm to isolate T: \[ T = \frac{x}{\ln{(101,325\,\text{Pa})}} \] Now substitute this back into our simplified equation: \[ \frac{x}{\ln{(101,325\,\text{Pa})}} - \frac{8.314\,\text{J/K}\cdot\text{mol}}{92.92\,\text{J/K}\cdot\text{mol}}\cdot x = \frac{58,510\,\text{J/mol}}{92.92\,\text{J/K}\cdot\text{mol}} \] Solve for x: \[ x = 674.63\,\text{K} \] Now plug x back into the equation for T: \[ T = \frac{674.63\,\text{K}}{\ln{(101,325\,\text{Pa})}} = 356.94\,\text{K} \]

The normal boiling point of mercury is approximately 356.94 K.