Answer:
0.6103 atm.
Explanation:
- We need to calculate the vapor pressure of each component after the stopcocks are opened.
- Volume after the stopcocks are opened = 3.0 L.
<u><em>1) For N₂:</em></u>
P₁V₁ = P₂V₂
P₁ = 1.5 atm & V₁ = 1.0 L & V₂ = 3.0 L.
P₂ of N₂ = P₁V₁ / V₂ = (1.5 atm) (1.0 L) / (3.0 L) = 0.5 atm.
<u><em>2) For H₂O:</em></u>
Pressure of water at 308 K is 42.0 mmHg.
we need to convert from mmHg to atm: <em>(1.0 atm = 760.0 mmHg)</em>.
P of H₂O = (1.0 atm x 42.0 mmHg) / (760.0 mmHg) = 0.0553 atm.
We must check if more 2.2 g of water is evaporated,
n = PV/RT = (0.0553 atm) (3.0 L) / (0.082 L.atm/mol.K) (308 K) = 0.00656 mole.
m = n x cmolar mass = (0.00656 mole) (18.0 g/mole) = 0.118 g.
It is lower than the mass of water in the flask (2.2 g).
<em><u>3) For C₂H₅OH:</u></em>
Pressure of C₂H₅OH at 308 K is 102.0 mmHg.
we need to convert from mmHg to atm: (1.0 atm = 760.0 mmHg).
P of C₂H₅OH = (1.0 atm x 102.0 mmHg) / (760.0 mmHg) = 0.13421 atm.
We must check if more 0.3 g of C₂H₅OH is evaporated,
n = PV/RT = (0.13421 atm) (3.0 L) / (0.082 L.atm/mol.K) (308 K) = 0.01594 mole.
m = n x molar mass = (0.01594 mole) (46.07 g/mole) = 0.7344 g.
<em>It is more than the amount in the flask (0.3 g), so the pressure should be less than 0.13421 atm.</em>
We have n = mass / molar mass = (0.30 g) / (46.07 g/mole) = 0.00651 mole.
So, P of C₂H₅OH = nRT / V = (0.00651 mole) (0.082 L.atm/mole.K) (308.0 K) / (3.0 L) = 0.055 atm.
- <em>So, </em><em>total pressure</em><em> = </em><em>P of N₂ + P of H₂O + P of C₂H₅OH</em><em> = 0.5 atm + 0.0553 atm + 0.055 atm = </em><em>0.6103 atm</em><em>.</em>
= 0.531 mol O₂