Question

Three 1.0L flasks, maintained at 308K are connected to each other with stopcocks. Initially, the stopcocks are closed: One of the flasks contains 1.0atm N2, the second 2.0g of H2O and the third, 0.50g ethanol, C2H6O. The vapor pressure of H2O at 308K is 24mmHG and that of ethanol is 102mmHG. The stopcocks are then opened and the contents mix freely. What is the pressure of the system after mixing? Please give all work including a description of each step.

Answer 1

Answer:

Pt = 0.392 atm

Explanation:

First, we must separate the initial condition for each gas:

- N₂ :

V₁ = 1 Lt , T₁ = 308 K , P₁ = 1 atm

- H₂O :

V₂= 1 Lt , T₂= 308 K, P₂= 24 mm Hg = 0.032 atm

- C₂H₆O:

V₃= 1 Lt, T₂= 308 K, P₃= 102 mm Hg = 0.134 atm

For this case, we consider that 1 atm = 760 mm Hg

We then calculate the number of moles for each case (in gas phase), and we consider that al gasses behave like an ideal gas:

PV = nRT we consider R = 0.082 Lt*atm/K*mol

- N₂ :

n H₂ = PV/RT = 1x1/0.082x308 = 0.04

- H₂O:

n H₂O = PV/RT = 0.032x1/0.082x308 = 1.27x10⁻³

- C₂H₆O:

n C₂H₆O = PV/RT = 0.134x1/0.082x308 = 5.31x10⁻³

Finally, when the stopcocks are open and the 3 gases are mixed together, T will remain as 308 K, total volume will be the volume of the three flasks (3 Lt), so total pressure (Pt) after mixing the 3 gases will be:

Pt = (n N₂ + n H₂O + n C₂H₆O)xRT/V = (0.04658x0.082x308)/3

Pt = 0.392 atm