Question

1\. 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 \(45^{\circ} \mathrm{C} ?\)

Short answer

Question: Calculate the pressure exerted by the methyl alcohol vapor at 35°C and 45°C. Compare these calculated pressures to the given equilibrium vapor pressures and predict the physical state of methyl alcohol at these temperatures. Answer: The pressure exerted by the methyl alcohol vapor at 35°C and 45°C are approximately 237 mmHg and 244 mmHg, respectively. Since the calculated pressures are lower than the equilibrium vapor pressures at these temperatures, the vapor will begin to condense into a liquid at both 35°C and 45°C.

Step-by-step solution

Understand the values provided

We are given the initial pressure \(P_1 = 254 mmHg\), and initial temperature \(T_1 = 57^{\circ}C\). We need to find the pressure exerted by the methyl alcohol vapor at \(T_2 = 35^{\circ}C\), and \(T_3 = 45^{\circ}C\). To use the ideal gas law, we must first convert the temperatures to Kelvin: - \(T_1 = 57^{\circ}C + 273.15 = 330.15 K\) - \(T_2 = 35^{\circ}C + 273.15 = 308.15 K\) - \(T_3 = 45^{\circ}C + 273.15 = 318.15 K\)

Apply the ideal gas law

By assuming no condensation, we can assume that the amount of gas and the volume of the flask are constants. Therefore, we can rewrite the ideal gas law as: $$\frac{P_1}{T_1} = \frac{P_2}{T_2} = \frac{P_3}{T_3}$$ (a) To find the pressures \(P_2\) and \(P_3\), we can use the ideal gas law: - \(P_2 = P_1 \frac{T_2}{T_1} = 254 \frac{308.15}{330.15} \approx 237 \, mmHg\) - \(P_3 = P_1 \frac{T_3}{T_1} = 254 \frac{318.15}{330.15} \approx 244 \, mmHg\) So, the pressure of the vapor at \(35^{\circ}C\) is approximately \(237 \, mmHg\), and at \(45^{\circ}C\), it is approximately \(244 \, mmHg\). (b) Given equilibrium vapor pressures: \(P_{eq}(35^{\circ}C) = 203 \, mmHg\), and \(P_{eq}(45^{\circ}C) = 325 \, mmHg\) Comparing with our calculated pressures: \(P_2 < P_{eq}(35^{\circ}C)\), and \(P_3 < P_{eq}(45^{\circ}C)\) (c) Since the calculated pressures are lower than the equilibrium vapor pressures, we can conclude that the pressure exerted by the methyl alcohol in the flask is effectively equal to our calculated pressures: \(P(35^{\circ}C) \approx 237 \, mmHg\) and \(P(45^{\circ}C) \approx 244 \, mmHg\) (d) By comparing the computed pressures to the given equilibrium vapor pressures, we can deduce the physical states of methyl alcohol at various temperatures: - At \(35^{\circ}C\), \(P_2 < P_{eq}(35^{\circ}C)\): Vapor will begin to condense into a liquid. - At \(45^{\circ}C\), \(P_3 < P_{eq}(45^{\circ}C)\): Vapor will begin to condense into a liquid.