Question

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 4.2 g of H2O, and the third, 0.40 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. What is the pressure? Express your answer to two significant figures and include the appropriate units.

Answer 1

First one needs to calculate partial pressures of the gases in the flasks when fully opened.

Accordingly

nitrogen partial pressure:

Initial N₂ pressure, P₁ = 1.5 atm

Initial volume of N₂, V₁ = 1 L

Final volume of N₂, V₂ = 3 L

Therefore final N_​2 pressure, P₂ = P₁V₁/V₂ = 1.5 x 1/3 = 0.5 atm

Vapor pressure of water = 42 mmHg/760 mmHg = 0.05526 atm

Final vapour pressure calculation of

water vapour after opening stopcocks = 0.05526 x 1/3 = 0.01842 atm

Vapor pressure of ethanol = 102 mmHg = 102/760 = 0.13421 atm

Final vapour pressure calculation

for ethanol after opening stopcocks = 0.13421/3 = 0.04474 atm

Final pressure of N₂, water vapor and ethanol in 3 flask(1L)

system after opening stopcocks = 0.5 atm + 0.01842 atm + 0.04474 atm = 0.56316 atm