Calculate Pressure at Equilibrium: Thermodynamics Problem

[mandangalo]

I am trying to calculate the pressure once a closed system reaches equilibrium at constant temperature and volume. I am to start with a tank filled with 1000L of at air 20^oC and 1 atm. It is then filled with 900L propane at the same temperature without venting. Assuming the air doesn't dissolve into the liquid propane, what is the pressure at equilibrium.

I thought this was a case of figuring out what the partial pressure of the air is once the liquid propane has been added. However, once equilibrium is reached the total volume for the vapor phase is going to be greater than 100L. This is where my issue lies: How do I know how much of the propane is going into the vapor phase? Once I know how to calculate that I should be able to find out each partial pressure and sum them up to get the total pressure (unless my ideas behind that are wrong too).

 

[Valkarie]

The answer is 1.85 atm. To calculate the pressure, you will need to calculate the total moles of the system (the air and the propane). The total moles of the system can be calculated using the ideal gas law: n = PV/RT Where n = total moles, P = pressure, V = volume, R = universal gas constant, and T = temperature. For the air, you can calculate the moles using the initial conditions: n_air = (1 atm)(1000 L)/(0.08206 L*atm/mole*K)(293 K) = 37.6 moles For the propane, you can use the same equation, but now you have 900L of propane at the same temperature, so the moles of propane is: n_propane = (1 atm)(900 L)/(0.08206 L*atm/mole*K)(293 K) = 33.8 moles Now that you have the total moles of air and propane, you can calculate the partial pressures of each gas using the ideal gas law. The total pressure of the system is equal to the sum of the partial pressures: P_total = P_air + P_propane To calculate the partial pressures, you will need to know the ratio of moles of each gas. The ratio of moles is equal to the ratio of the volumes of each gas, since the temperature and pressure are the same for both gases. So the ratio of moles is equal to: n_air/n_propane = 1000 L/900 L = 1.11 Therefore, the partial pressure of the air is equal to 1.11 times the total pressure, and the partial pressure of the propane is equal to the total pressure minus the partial pressure of the air. So, the partial pressure of the air is: P_air = (1.11)(P_total) And the partial pressure of the propane is: P_propane = P_total - P_air Now that you have the partial pressures, you can calculate the total pressure of the system. The total pressure is