Respuesta :
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.
1) For N₂:
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.
2) For H₂O:
Pressure of water at 308 K is 42.0 mmHg.
we need to convert from 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 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).
3) For C₂H₅OH:
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.
It is more than the amount in the flask (0.3 g), so the pressure should be less than 0.13421 atm.
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.
- So, total pressure = P of N₂ + P of H₂O + P of C₂H₅OH = 0.5 atm + 0.0553 atm + 0.055 atm = 0.6103 atm.
Vapor pressure is defined as the pressure exerted by the vapors on the walls of the container. The total pressure of the solution is 0.6103 atm.
Given that,
For Nitrogen:
- P₁V₁ = P₂V₂
- P₁ = 1.5 atm
- V₁ = 1.0 L
- V₂ = 3.0 L.
P₂ of N₂ = P₁V₁ / V₂ = [tex]\dfrac{(1.5 \text {atm}) \times (1.0 \text L)} {(3.0 \text L)} = 0.5 \text{atm.}[/tex]
Similarly, for water molecule:
Pressure of water at 308 K = 42.0 mmHg
Converting from mmHg to atm: (1.0 atm = 760.0 mmHg)
P of H₂O = [tex]\dfrac{(1.0 \text{atm} \times 42.0 \text{mmHg)}}{ (760.0\text{ mmHg})} = 0.0553 \text{atm}[/tex]
Now, from the ideal gas equation:
PV = nRT
n = [tex]\dfrac{\text{PV}} {\text{RT}}[/tex]
Substituting the values,
n = [tex]\dfrac{\text{0.0553 atm}\times \text{3 L}} {\text{0.082} \times \text{308 K}}& = 0.000656[/tex] moles.
From the formula of number of moles:
m = n x molar mass = (0.00656 mole) x (18.0 g/mole) = 0.118 g
The mass of water is lower than the mass in the flask, that is 2.2 g.
Similarly, for ethanol:
Pressure of water at 308 K = 102.0 mmHg
Converting from mmHg to atm: (1.0 atm = 760.0 mmHg)
P of C₂H₅OH = [tex]\dfrac{(1.0 \text{atm} \times 102.0 \text{mmHg)}}{ (760.0\text{ mmHg})} = 0.13421 \text{atm}[/tex]
Now, from the equation of ideal gas, number of moles will be:
n = [tex]\dfrac{\text{PV}} {\text{RT}}[/tex]
n = [tex]\dfrac{\text{0.13421 atm}\times \text{3 L}} {\text{0.082} \times \text{308 K}}& = 0.01594[/tex]
From the formula of number of moles:
m = n x molar mass = (0.01594 mole) x (46.07 g/mole) = 0.7344 g.
The mass of ethanol is more than the mass in the flask, that is 0.3 g.
Now,
P of C₂H₅OH = nRT / V
[tex]\dfrac{\text{0.00651 atm}\times \text{0.082} \times \text{308 K}}{\text{3 L}}& = 0.055 atm.[/tex]
Now, the total pressure:
Total Pressure = P of N₂ + P of H₂O + P of C₂H₅OH = 0.5 atm + 0.0553 atm + 0.055 atm = 0.6103 atm.
Therefore, the total pressure of the solution is 0.6103 atm.
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