Question

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?

Short answer

Answer: The vapor pressure of water in the compressed gas mixture is 62.5 mmHg. To find the amount of water that condenses during the process, calculate the difference in moles of water vapor in the initial and final states, and then multiply by the molar mass of water.

Step-by-step solution

Determine the initial vapor pressure of water

We are given the initial temperature of the flask: \(25^{\circ} \mathrm{C}\). We can now look up the vapor pressure of water at this temperature using the table provided in Appendix 1, which gives us the vapor pressure of water at \(25^{\circ} \mathrm{C}\) as \(23.76 \ mmHg\).

Calculate the final vapor pressure of water in the compressed gas mixture

The compression of the gas mixture causes the partial pressures of both oxygen and water vapor to increase. Since the temperature is held constant, we can use the ideal gas law, which states that \(pV=nRT\) where \(p\) is pressure, \(V\) is volume, \(n\) is the amount of substance, \(R\) is the gas constant, and \(T\) is temperature. For water vapor, we will only focus on the pressure and volume, so we can simplify the relationship as follows: \(p_{1}V_{1} = p_{2}V_{2}\) where indices 1 and 2 represent the initial and final states respectively. The initial conditions : \(V_1 = 5.25 \mathrm{~L}\) and \(p_1 = 23.76 \ mmHg\). The final volume is given: \(V_2 = 2.00 \mathrm{~L}\). We can now solve for the final vapor pressure, \(p_2\). \(p_{2}V_{2} = p_{1}V_{1}\) \(\Rightarrow p_{2} = \dfrac{p_{1}V_{1}}{V_{2}}\) \(p_{2} = \dfrac{23.76 \ mmHg \times 5.25 \mathrm{~L}}{2 \mathrm{~L}}\) \(p_{2} = 62.5 \ mmHg\) So, the vapor pressure of water in the compressed gas mixture is \(62.5 \ mmHg\).

Determine the amount of water that condenses

To find the amount of water that condenses, we first calculate the difference in moles of water vapor in the initial and final states. Since our temperature is constant, we can use the ideal gas law in its simplified form as shown in step 2. For the initial state, let's denote the moles of water vapor as \(n_1\). Thus, we have: \(n_1 = \dfrac{p_1 V_1}{RT} = \dfrac{23.76 \ mmHg \times 5.25 \mathrm{~L}}{R \times (25+273)K}\) For the final state, the moles of water vapor, \(n_2\), can be expressed as: \(n_2 = \dfrac{p_2 V_2}{RT} = \dfrac{62.5 \ mmHg \times 2.00 \mathrm{~L}}{R \times (25+273)K}\) Keep in mind that we need to convert \(mmHg\) to \(atm\) by dividing by \(760\) i.e., \(1 \ atm = 760 \ mmHg\). So, \(n_1 = \dfrac{23.76/760 \ atm \times 5.25 \mathrm{~L}}{R \times (25+273)K}\) and \(n_2 = \dfrac{62.5/760 \ atm \times 2.00 \mathrm{~L}}{R \times (25+273)K}\) Now calculate the difference in moles, which represents the moles of water that condense: \(\Delta n = n_1 - n_2\) To find the mass of water that condenses, we simply multiply by the molar mass of water, \(M_w = 18.015 \ g/mol\): \(m = \Delta n \times M_w\) Once we plug in our values and carry out the calculations, we can find the mass of water that condenses when the gas mixture is compressed.