Methyl alcohol can be used as a fuel instead of, or combined with, gasoline. A sample of methyl alcohol, \(\mathrm{CH}_{3} \mathrm{OH}\), in a flask of constant volume exerts a pressure of \(254 \mathrm{~mm} \mathrm{Hg}\) at \(57^{\circ} \mathrm{C}\). The flask is slowly cooled. (a) Assuming no condensation, use the ideal gas law to calculate the pressure of the vapor at \(35^{\circ} \mathrm{C}\); at \(45^{\circ} \mathrm{C}\). (b) Compare your answers in (a) with the equilibrium vapor pressures of methyl alcohol: \(203 \mathrm{~mm} \mathrm{Hg}\) at \(35^{\circ} \mathrm{C} ; 325 \mathrm{~mm} \mathrm{Hg}\) at \(45^{\circ} \mathrm{C}\). (c) On the basis of your answers to (a) and (b), predict the pressure exerted by the methyl alcohol in the flask at \(35^{\circ} \mathrm{C} ;\) at \(45^{\circ} \mathrm{C}\). (d) What physical states of methyl alcohol are present in the flask at \(35^{\circ} \mathrm{C} ?\) At \(45^{\circ} \mathrm{C} ?\)

#tag_title# Step 3: Calculate the Pressure at the Given Temperatures Using the Ideal Gas Law #tag_content# Since the volume is constant, rearrange the Ideal Gas Law equation to express the pressure ratios: $$\frac{P_1}{T_1} = \frac{P_2}{T_2} = \frac{P_3}{T_3}$$ Given the initial pressure \(P_1 = 0.959\,\mathrm{atm}\), we can calculate the pressures at the given temperatures: $$P_2 = \frac{T_2}{T_1}P_1 = \frac{308.15\,\mathrm{K}}{330.15\,\mathrm{K}} (0.959\,\mathrm{atm}) = 0.846\,\mathrm{atm}$$ $$P_3 = \frac{T_3}{T_1}P_1 = \frac{318.15\,\mathrm{K}}{330.15\,\mathrm{K}} (0.959\,\mathrm{atm}) = 0.922\,\mathrm{atm}$$ #tag_title# Step 4: Compare Calculated Pressures with Equilibrium Vapor Pressures #tag_content# Compare the calculated pressures with the given equilibrium vapor pressures at the respective temperatures: At \(35^{\circ}\mathrm{C}\), calculated pressure: \(0.846\,\mathrm{atm}\), given equilibrium vapor pressure: \(0.339\,\mathrm{atm}\) At \(45^{\circ}\mathrm{C}\), calculated pressure: \(0.922\,\mathrm{atm}\), given equilibrium vapor pressure: \(0.697\,\mathrm{atm}\) #tag_title# Step 5: Predict the Pressure Exerted by the Methyl Alcohol Vapor #tag_content# Since the calculated pressures are higher than the given equilibrium vapor pressures at both temperatures, the pressure of the methyl alcohol vapor will be equal to the equilibrium vapor pressures: At \(35^{\circ}\mathrm{C}\), pressure of methyl alcohol vapor: \(0.339\,\mathrm{atm}\) At \(45^{\circ}\mathrm{C}\), pressure of methyl alcohol vapor: \(0.697\,\mathrm{atm}\) #tag_title# Step 6: Determine the Physical States of Methyl Alcohol #tag_content# At both temperatures, the calculated pressures are higher than the given equilibrium vapor pressures, which means that methyl alcohol exists as a mixture of both gas and liquid states inside the closed flask. #Summary (Short Answer)# At \(35^{\circ}\mathrm{C}\), the pressure exerted by the methyl alcohol vapor is \(0.339\,\mathrm{atm}\), and at \(45^{\circ}\mathrm{C}\), it is \(0.697\,\mathrm{atm}\). Since the calculated pressures are higher than the given equilibrium vapor pressures at both temperatures, methyl alcohol exists in a mixture of both gas and liquid states inside the closed flask.