One method of preparing elemental mercury involves roasting cinnabar (HgS) in quicklime (CaO) at \(600 .\) C followed by condensation of the mercury vapor. Given the heat of vaporization of mercury $(296 \mathrm{J} / \mathrm{g} \text { ) and the vapor pressure of mercury }\( at \)25.0^{\circ} \mathrm{C}\left(2.56 \times 10^{-3} \text { torr), what is the vapor pressure of the }\right.$ condensed mercury at \(300 .^{\circ} \mathrm{C} ?\) How many atoms of mercury are present in the mercury vapor at \(300 .^{\circ} \mathrm{C}\) if the reaction is conducted in a closed 15.0 \(\mathrm{-L}\) container?

The Clausius-Clapeyron Equation is: \[\ln \frac{P_2}{P_1} = -\frac{\Delta H_{vap}}{R} \left(\frac{1}{T_2} - \frac{1}{T_1}\right)\] where \(P_1\) is the vapor pressure at temperature \(T_1\), \(P_2\) is the vapor pressure at temperature \(T_2\), \(\Delta H_{vap}\) is the heat of vaporization of the substance, and R is the ideal gas constant. In our case, we have: \(P_1 = 2.56 \times 10^{-3}\) torr (vapor pressure at 25.0°C), \(T_1 = 273.15 + 25 = 298.15\) K (temperature in Kelvin), \(T_2 = 273.15 + 300 = 573.15\) K (temperature in Kelvin), and \(\Delta H_{vap} = 296\) J/g. First, we need to convert the heat of vaporization to J/mol (R's unit needs to match). We have: \[\Delta H_{vap} = 296 \frac{\text{J}}{\text{g}} \times \frac{200.59\text{ g}}{\text{mol}} = 59400 \frac{\text{J}}{\text{mol}}\] Then, let's solve for \(P_2\): \[P_2 = P_1 \times e^{-\frac{\Delta H_{vap}}{R} \left(\frac{1}{T_2} - \frac{1}{T_1}\right)}\] Substitute the values and solve for \(P_2\): \[P_2 = (2.56 \times 10^{-3}\text{ torr}) \times e^{-\frac{59400 \mathrm{J} / \mathrm{mol}}{8.314 \mathrm{J} / \mathrm{mol\cdot K}} \left(\frac{1}{573.15\text{ K}} - \frac{1}{298.15\text{ K}}\right)}\] After calculating, we get: \[P_2 = 2.25 \times 10^{-1} \text{ torr}\]

Now that we have the vapor pressure at 300°C, we can use the Ideal Gas Law equation to find the number of moles of mercury vapor present in the 15.0-L container. The Ideal Gas Law equation is: \[PV = nRT\] where P is the pressure in pascals, V is the volume in liters, n is the number of moles, R is the ideal gas constant, and T is the temperature in Kelvin. First, we need to convert the pressure from torr to pascals: \[P_2 = 2.25 \times 10^{-1} \text{ torr} \times \frac{101325}{760} \text{Pa} = 30.05 \text{ Pa}\] Now, let's substitute the values into the Ideal Gas Law equation and solve for n: \[n = \frac{PV}{RT} = \frac{30.05 \text{ Pa} \times 15.0 \text{ L}}{8.314 \text{ J/mol⋅K} \times 573.15 \text{ K}}\] After calculating, we get: \[n = 9.72 \times 10^{-4}\text{ mol}\]

Finally, we need to convert the number of moles to the number of atoms using Avogadro's number: \[N_\text{atoms} = n \times N_A\] where \(N_\text{atoms}\) is the number of mercury atoms, n is the number of moles, and \(N_A\) is Avogadro's number (\(6.022 \times 10^{23} \text{mol}^{-1}\)). Substitute the values and solve for \(N_\text{atoms}\): \[N_\text{atoms} = 9.72 \times 10^{-4} \text{ mol} \times 6.022 \times 10^{23} \text{mol}^{-1}\] After calculating, we get: \[N_\text{atoms} = 5.85 \times 10^{20}\] So, the vapor pressure of condensed mercury in the 15.0-L container at 300°C is \(2.25 \times 10^{-1}\) torr, and there are approximately \(5.85 \times 10^{20}\) mercury atoms present in the container.