Consider a sealed flask with a movable piston that contains \(5.25 \mathrm{~L}\) of \(\mathrm{O}_{2}\) saturated with water vapor at \(25^{\circ} \mathrm{C}\). The piston is depressed at constant temperature so that the gas is compressed to a volume of \(2.00 \mathrm{~L}\). (Use the table in Appendix 1 for the vapor pressure of water at various temperatures.) (a) What is the vapor pressure of water in the compressed gas mixture? (b) How many grams of water condense when the gas mixture is compressed?

First, we will determine the initial pressure of the flask, using the vapor pressure of water at 25°C. The vapor pressure of water at 25°C from the table in Appendix 1 is 23.76 mmHg. Thus, we assume that the total pressure in the flask is the same as the vapor pressure of water. Now we can use the Ideal Gas Law to find the number of moles of oxygen gas in the flask. The Ideal Gas Law is given by: PV = nRT We can convert the pressure to atmospheres by dividing it by 760 mmHg/atm. P = 23.76 mmHg * (1 atm / 760 mmHg) = 0.0313 atm Let nO2 be the number of moles of \(\ce{O2}\). We can now solve for nO2: nO2 = PV / RT = (0.0313 atm * 5.25 L) / (0.0821 L atm/mol K * (273 + 25) K) = 0.00755 mol To find the mole fraction of oxygen gas and water vapor, we'll first find the number of moles of water vapor present which can be found using nH2O = (P_H2O * V) / RT Where P_H2O is the partial pressure of water vapor which is same as the vapor pressure of water at 25°C, V = 5.25 L, R is the gas constant, 0.0821 L atm/mol K and T = 25°C + 273 = 298 K. nH2O = (0.0313 atm * 5.25 L) / (0.0821 L atm/mol K * 298 K) = 0.00755 mol Now we can find the mole fractions of oxygen and water vapor: Mole fraction of O2 (xO2) = nO2 / (nO2 + nH2O) = 0.00755 / (0.00755 + 0.00755) = 0.5 Mole fraction of H2O (xH2O) = nH2O / (nO2 + nH2O) = 0.00755 / (0.00755 + 0.00755) = 0.5

Since the temperature remains constant during the compression, we can use Boyle's Law for each component, P1V1 = P2V2 Final Pressure of O2 (P'O2) = (Initial Pressure of O2 * Initial Volume) / Final Volume = 0.01565 atm * 5.25 L / 2.00 L = 0.0409 atm Final Pressure of H2O (P'H2O) = (Initial Pressure of H2O * Initial Volume) / Final Volume = 0.01565 atm * 5.25 L / 2.00 L = 0.0409 atm Final total pressure of gas mixture (Pfinal) = P'O2 + P'H2O = 0.0409 atm + 0.0409 atm = 0.0818 atm

We already know the final total pressure of the gas mixture. Now we can use the mole fraction of water vapor to find the partial pressure of water vapor in the compressed gas mixture. P'H2O = xH2O * Pfinal = 0.5 * 0.0818 atm = 0.0409 atm Thus, the vapor pressure of water in the compressed gas mixture is 0.0409 atm.

Now that we know the final partial pressure of water vapor in the compressed gas mixture, we can find the moles of water vapor present after compression. nH2O_final = (P'H2O * Final Volume) / RT = (0.0409 atm * 2 L) / (0.0821 L atm/mol K * 298 K) = 0.00335 mol The moles of water vapor that condense can be calculated by subtracting the moles of water vapor after compression from the initial moles of water vapor. n_condensed = nH2O - nH2O_final = 0.00755 mol - 0.00335 mol = 0.00420 mol Now, we can convert the moles of water vapor that condense to grams using the molecular weight of water, 18.015 g/mol. grams_condensed = n_condensed * molecular_weight = 0.00420 mol * 18.015 g/mol = 0.0757 g So, 0.0757 grams of water condense when the gas mixture is compressed.