A rigid vessel containing a \(3: 1 \mathrm{~mol}\) ratio of carbon dioxide and water vapor is held at \(200^{\circ} \mathrm{C}\) where it has a total pressure of \(202.7 \mathrm{kPa}\). If the vessel is cooled to \(10^{\circ} \mathrm{C}\) so that all of the water vapor condenses, what is the pressure of carbon dioxide? Neglect the volume of the liquid water that forms on cooling.

Given the total pressure of the mixture is \(202.7 kPa\), and the mol ratio of carbon dioxide to water vapor is \(3:1\), we can find the partial pressures of each component. Let \(P_{CO_2}\) and \(P_{H_2O}\) be the partial pressures of carbon dioxide and water vapor, respectively. Then, we can write: \[P_{CO_2} + P_{H_2O} = 202.7\] Since the mol ratio of carbon dioxide to water vapor is \(3:1\), we can write: \[P_{CO_2} = 3P_{H_2O}\] Now, we can substitute this expression for \(P_{CO_2}\) back into the first equation: \[3P_{H_2O} + P_{H_2O} = 202.7\] \[4P_{H_2O} = 202.7\] Solving for \(P_{H_2O}\), we get: \[P_{H_2O} = 50.675 kPa\] Now, we can find the partial pressure of carbon dioxide: \[P_{CO_2} = 3(50.675) = 152.025 kPa\]

Now that we have the initial partial pressure of carbon dioxide, we can use the ideal gas law to find the initial number of moles of carbon dioxide in the mixture. The ideal gas law is given by: \[PV = nRT\] Where \(P\) is the pressure, \(V\) is the volume, \(n\) is the number of moles, \(R\) is the ideal gas constant, and \(T\) is the temperature. We can rearrange for the number of moles: \[n_{CO_2} = \frac{P_{CO_2}V}{RT}\] Since the vessel is rigid and the volume of liquid water is negligible, we can assume that the volume remains constant throughout the process. Since we're only interested in the pressure of carbon dioxide, we don't need to directly compute the volume. We'll only need the ratio of the initial and final temperatures. The initial and final temperatures in Kelvin are: \[T_{1} = 200 + 273.15 = 473.15 K\] \[T_{2} = 10 + 273.15 = 283.15 K\]

Now that we have the initial number of moles of carbon dioxide and the initial and final temperatures, we can apply the ideal gas law to determine the final pressure of carbon dioxide. Since the volume and the number of moles of carbon dioxide (water vapor has condensed) remain constant: \[\frac{P_{1}}{T_{1}} = \frac{P_{2}}{T_{2}}\] Where \(P_{1}\) is the initial partial pressure of carbon dioxide, and \(P_{2}\) is the final pressure of carbon dioxide. We can solve for the final pressure, \(P_{2}\): \[P_{2} = \frac{P_{1}T_{2}}{T_{1}}\] Substituting the values we found earlier: \[P_{2} = \frac{152.025 \times 283.15}{473.15}\] Evaluating the expression, we find the final pressure of carbon dioxide: \[P_{2} = 91.5 kPa\] Therefore, the pressure of carbon dioxide after cooling the rigid vessel to \(10^{\circ}C\) and condensing all the water vapor is approximately \(91.5 kPa\).