# Final pressure?

Three 1.0-L flasks, maintained at 308 K, are connected to each other with stopcocks. Initially the stopcocks are closed. One of the flasks contains 1.5 atm of N2, the second 5.0 g of H2O, and the third, 0.60 g of ethanol, C2H6O. The vapor pressure of H2O at 308 K is 42 mmHg and that of ethanol is 102 mmHg. The stopcocks are then opened and the contents mix freely.

Roger the Mole: The first questions to be addressed are

The first questions to be addressed are: Do the liquids vaporize completely, or is there some liquid(s) remaining in the final state? What are the three partial pressures? n = PV / RT = (42 mmHg) x (3 L H2O vapor) / ((62.36367 L mmHg/K mol) x (308 K)) = 0.00656 mol H2O (0.00656 mol H2O) x (18.01532 g H2O/mol) = 0.118 g H2O required to fill all three flasks with vapor So there will be liquid water left over, and the partial pressure of H2O will be 42 mmHg in the final state (102 mmHg) x (3 L ethanol vapor) / ((62.36367 L mmHg/K mol) x (308 K)) = 0.01593 mol ethanol (0.01593 mol ethanol) x (46.0684 g CH3CH2OH/mol) = 0.734 g ethanol to fill 3 L But there is not that much ethanol present, so all of the ethanol will evaporate, leaving no liquid ethanol. So the partial pressure of ethanol will be: P = nRT / V = ((0.60 g CH3CH2OH) / (46.0684 g CH3CH2OH/mol)) x ((62.36367 L mmHg/K mol) x (308 K)) / (3 L) = 83.4 mmHg (1.5 atm N2) x (1 L) / (3 L) x (760 mmHg/atm) = 380 mmHg of N2 So the total pressure when the stopcocks are opened (and enough time has elapsed for the necessary amount of H2O to evaporate and all the gases to mix) is: 42 mmHg + 83.4 mmHg + 380 mmHg = 505 mmHg total