Answer:
0.6103 atm.
Explanation:
- We need to evaluate the vapor pressure for each substance once the stopcocks are opened.
- The volume after the stopcocks open is 3.0 L.
1) Concerning N₂:
P₁V₁ = P₂V₂
P₁ = 1.5 atm & V₁ = 1.0 L & V₂ = 3.0 L.
P₂ for N₂ = P₁V₁ / V₂ = (1.5 atm) (1.0 L) / (3.0 L) = 0.5 atm.
2) Concerning H₂O:
At 308 K, the vapor pressure of water is 42.0 mmHg.
We must convert mmHg to atm: (1.0 atm = 760.0 mmHg).
P of H₂O = (1.0 atm x 42.0 mmHg) / (760.0 mmHg) = 0.0553 atm.
We need to check whether an additional 2.2 g of water has 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.
This is less than the water mass in the flask (2.2 g).
3) For C₂H₅OH:
At 308 K, the vapor pressure of C₂H₅OH is 102.0 mmHg.
We need to convert 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 should check if an additional 0.3 g of C₂H₅OH has 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.
This exceeds the amount in the flask (0.3 g), indicating the pressure must be lower than 0.13421 atm.
Thus, n = mass / molar mass = (0.30 g) / (46.07 g/mole) = 0.00651 mole.
Consequently, P of C₂H₅OH = nRT / V = (0.00651 mole) (0.082 L.atm/mole.K) (308.0 K) / (3.0 L) = 0.055 atm.
- Therefore, total pressure = P of N₂ + P of H₂O + P of C₂H₅OH = 0.5 atm + 0.0553 atm + 0.055 atm = 0.6103 atm.
