Question

Benzene, a known carcinogen, was once widely used as a solvent. A sample of benzene vapor in a flask of constant volume exerts a pressure of \(325 \mathrm{~mm} \mathrm{Hg}\) at \(80^{\circ} \mathrm{C}\). The flask is slowly cooled. (a) Assuming no condensation, use the ideal gas law to calculate the pressure of the vapor at \(50^{\circ} \mathrm{C} ;\) at \(60^{\circ} \mathrm{C}\). (b) Compare your answers in (a) to the equilibrium vapor pressures of benzene: \(269 \mathrm{~mm} \mathrm{Hg}\) at \(50^{\circ} \mathrm{C}\), \(389 \mathrm{~mm} \mathrm{Hg}\) at \(60^{\circ} \mathrm{C}\) (c) On the basis of your answers to (a) and (b), predict the pressure exerted by the benzene at \(50^{\circ} \mathrm{C}\); at \(60^{\circ} \mathrm{C}\).

Short answer

Based on the ideal gas law calculations, the pressure exerted by benzene at \(50^{\circ} \mathrm{C}\) would be close to the equilibrium vapor pressure at this temperature, which is \(269 \mathrm{~mm} \mathrm{Hg}\), due to some condensation occurring. At \(60^{\circ} \mathrm{C}\), the pressure exerted by benzene would be the calculated pressure, \(306.02 \mathrm{~mm} \mathrm{Hg}\), as there would be no condensation occurring and the ideal gas law still applies.

Step-by-step solution

(a) Calculate the pressure of benzene vapor at \(50^{\circ} \mathrm{C}\) and \(60^{\circ} \mathrm{C}\) using the ideal gas law

We can use the ideal gas law given by: \(PV = nRT\) where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature. Since the volume is constant and we are comparing ratios of pressure and temperature, we can write the equation as: \(P_1/T_1 = P_2/T_2\) Given, \(P_1 = 325 \mathrm{~mm} \mathrm{Hg}\) and \(T_1 = 80^{\circ} \mathrm{C}\). We need to find \(P_2\) for \(T_2 = 50^{\circ} \mathrm{C}\) and \(T_2 = 60^{\circ} \mathrm{C}\). First, we need to convert the temperatures to Kelvin by adding 273.15: \(T_1 = 80 + 273.15 = 353.15 \mathrm{K}\) \(T_{2_1} = 50 + 273.15 = 323.15 \mathrm{K}\) \(T_{2_2} = 60 + 273.15 = 333.15 \mathrm{K}\) Now, we can find \(P_2\) for each temperature using the equation: For \(T_{2_1} = 323.15 \mathrm{K}\): \(P_2 = P_1 \cdot \frac{T_{2_1}}{T_1} = 325 \cdot \frac{323.15}{353.15} = 297.54 \mathrm{~mm} \mathrm{Hg}\) For \(T_{2_2} = 333.15 \mathrm{K}\): \(P_2 = P_1 \cdot \frac{T_{2_2}}{T_1} = 325 \cdot \frac{333.15}{353.15} = 306.02 \mathrm{~mm} \mathrm{Hg}\) So, the pressure at \(50^{\circ} \mathrm{C}\) is approximately \(297.54 \mathrm{~mm} \mathrm{Hg}\) and at \(60^{\circ} \mathrm{C}\) is approximately \(306.02 \mathrm{~mm} \mathrm{Hg}\).

(b) Compare the calculated pressures to the given equilibrium vapor pressures

The calculated pressure at \(50^{\circ} \mathrm{C}\) is \(297.54 \mathrm{~mm} \mathrm{Hg}\), and the given equilibrium vapor pressure is \(269 \mathrm{~mm} \mathrm{Hg}\). The calculated pressure is higher than the equilibrium vapor pressure. The calculated pressure at \(60^{\circ} \mathrm{C}\) is \(306.02 \mathrm{~mm} \mathrm{Hg}\), and the given equilibrium vapor pressure is \(389 \mathrm{~mm} \mathrm{Hg}\). The calculated pressure is lower than the equilibrium vapor pressure.

(c) Predict the pressure exerted by benzene at \(50^{\circ} \mathrm{C}\) and \(60^{\circ} \mathrm{C}\)

At \(50^{\circ} \mathrm{C}\), since the calculated pressure is higher than the equilibrium vapor pressure, we can expect some condensation to occur. Therefore, the pressure exerted by benzene at this temperature would be close to the equilibrium vapor pressure, \(269 \mathrm{~mm} \mathrm{Hg}\). At \(60^{\circ} \mathrm{C}\), since the calculated pressure is lower than the equilibrium vapor pressure, there would be no condensation occurring, and the ideal gas law still applies. Therefore, the pressure exerted by benzene at this temperature would be the calculated pressure, \(306.02 \mathrm{~mm} \mathrm{Hg}\).