A sample of gas collected over water at \(42^{\circ} \mathrm{C}\) occupies a volume of one liter. The wet gas has a pressure of \(0.986\) atm. The gas is dried, and the dry gas occupies \(1.04 \mathrm{~L}\) with a pressure of \(1.00 \mathrm{~atm}\) at \(90^{\circ} \mathrm{C}\). Using this information, calculate the vapor pressure of water at \(42^{\circ} \mathrm{C}\).

The Ideal Gas Law is given by: \(PV = nRT\) Where P is the pressure in atm, V is the volume in L, n is the number of moles of the gas, R is the ideal gas constant (0.0821 \(\frac{\mathrm{L}\cdot \mathrm{atm}}{\mathrm{mol}\cdot \mathrm{K}}\)), and T is the temperature in Kelvin.

First, convert the given temperatures from Celsius to Kelvin: \(T_{1} = 42 + 273.15 = 315.15\) K (for wet gas) \(T_{2} = 90 + 273.15 = 363.15\) K (for dry gas)

For the wet gas, the total pressure is the sum of the partial pressures of the dry gas and the water vapor: \(P_{wet} = P_{dry,1} + P_{H_2O,1}\) Using the Ideal Gas Law for the initial (wet) gas condition: \((P_{dry,1} + P_{H_2O,1})V_{1} = n_{dry}RT_{1}\) (1) And for the final (dry) gas condition: \(P_{dry,2}V_{2} = n_{dry}RT_{2}\) (2)

Rearrange equation (1) to express the moles of dry gas, \(n_{dry}\): \(n_{dry} = \frac{(P_{dry,1} + P_{H_2O,1})V_{1}}{RT_{1}}\) Do the same for the equation (2): \(n_{dry} = \frac{P_{dry,2}V_{2}}{RT_{2}}\) Now we can set these two equal since \(n_{dry}\) is the same for the initial and final conditions: \(\frac{(P_{dry,1} + P_{H_2O,1})V_{1}}{RT_{1}}=\frac{P_{dry,2}V_{2}}{RT_{2}}\)

Rearrange the equation to solve for \(P_{H_2O,1}\): \(P_{H_2O,1} = \frac{P_{dry,2}V_{2}RT_{1}}{(RT_{2})} - \frac{P_{dry,1}V_{1}}{(RT_{1})}\) We know that \(P_{wet} = 0.986 \mathrm{~atm}\) and \(P_{dry,2} = 1.00 \mathrm{~atm}\). We can find \(P_{dry,1}\) by multiplying the ratios of volumes: \(P_{dry,1} = P_{dry,2} \frac{V_{1}}{V_{2}} = 1.00 \mathrm{~atm} \frac{1.0 L}{1.04 L}\) Now, plug in all the given values and solve for \(P_{H_2O,1}\): \(P_{H_2O,1} = \frac{(1.00\mathrm{~atm})(1.04\mathrm{~L})(0.0821\frac{\mathrm{L}\cdot \mathrm{atm}}{\mathrm{mol}\cdot \mathrm{K}})(315.15\mathrm{~K})}{(0.0821\frac{\mathrm{L}\cdot \mathrm{atm}}{\mathrm{mol}\cdot \mathrm{K}})(363.15\mathrm{~K})} - \frac{(1.00 \mathrm{~atm})(\frac{1.0}{1.04})(1.0 \mathrm{~L})}{(0.0821\frac{\mathrm{L}\cdot \mathrm{atm}}{\mathrm{mol}\cdot \mathrm{K}})(315.15\mathrm{~K})}\) After calculating, we get the vapor pressure of water at \(42^{\circ} C\): \(P_{H_2O,1} = 0.0389\text{ atm}\)