Water has an equilibrium vapor pressure of 23.8 mm Hg at 25 C. If 1.00 g H2O is sealed in a 20 L flask, what mass will remain in the liquid phase when equilibrium is established? ( 760 mmHg = 1 atm, R =0.0821 L.atm/mol.K)

I would solve PV = nRT for n to determine moles of gas that can be accommodated in the 20L flask. Then compare that with the number of moles H2O available.

To solve this problem, we can use the ideal gas law and the concept of partial pressure.

Firstly, let's calculate the moles of water vapor in the equilibrium state:

1. Convert the equilibrium vapor pressure from mmHg to atm:
23.8 mmHg * (1 atm / 760 mmHg) = 0.0313 atm

2. Use the ideal gas law to find the number of moles of water vapor:
PV = nRT

Where:
P is the pressure (in atm)
V is the volume (in L)
n is the number of moles of gas
R is the ideal gas constant (0.0821 L·atm/mol·K)
T is the temperature (in K)

n = (PV) / (RT)
= (0.0313 atm * 20 L) / (0.0821 L·atm/mol·K * 298 K)
≈ 0.25 mol

Now, let's calculate the molecular weight of water (H2O):

1. The molecular weight of water (H2O) = 2 * Atomic weight of hydrogen + Atomic weight of oxygen
= (2 * 1.01 g/mol) + 16.00 g/mol
= 18.02 g/mol

2. The mass of water vapor in the equilibrium state = moles of water vapor * molecular weight of water
= 0.25 mol * 18.02 g/mol
≈ 4.50 g

Therefore, when equilibrium is established, approximately 4.50 grams of water will remain in the liquid phase.

To solve this problem, we need to use the Clausius-Clapeyron equation, which relates the vapor pressure of a substance to its temperature and molar mass.

The Clausius-Clapeyron equation is given by:
ln(P1/P2) = (ΔHvap/R) * (1/T2 - 1/T1)

where P1 and P2 are the initial and final pressures respectively, ΔHvap is the enthalpy of vaporization, R is the ideal gas constant, T1 and T2 are the initial and final temperatures respectively.

First, we need to find the initial pressure using the given equilibrium vapor pressure of water at 25°C. Convert 23.8 mmHg to atm by dividing by 760 mmHg/1 atm.

P1 = 23.8 mmHg / 760 mmHg/atm = 0.0313 atm

Next, we need to convert the temperature into Kelvin. Add 273.15 to the given temperature of 25°C.

T2 = 25°C + 273.15 = 298.15 K

The final temperature (T1) is not given, but since we are asked to find the mass that remains in the liquid phase when equilibrium is established, we can assume that it is the boiling point of water, which is 100°C or 373.15 K.

T1 = 373.15 K

Substituting the values into the Clausius-Clapeyron equation:

ln(P1/P2) = (ΔHvap/R) * (1/T2 - 1/T1)

ln(0.0313/P2) = (ΔHvap/0.0821) * (1/298.15 - 1/373.15)

To solve for ln(P2), let's rearrange the equation:

ln(P2) = ln(0.0313) - (ΔHvap/0.0821) * (1/298.15 - 1/373.15)

Now we need to find the value of ΔHvap, which is the enthalpy of vaporization of water. It is given as 40.7 kJ/mol, but we need to convert it to J/mol by multiplying by 1000.

ΔHvap = 40.7 kJ/mol * 1000 J/kJ = 40700 J/mol

Now we can substitute the values to solve for ln(P2):

ln(P2) = ln(0.0313) - (40700/0.0821) * (1/298.15 - 1/373.15)

Calculate the right side of the equation:

ln(P2) ≈ -3.4625

To solve for P2, take the antilog (e^x) of both sides of the equation:

P2 ≈ e^(-3.4625)

P2 ≈ 0.0317 atm

Finally, to find the mass of water remaining in the liquid phase, we use the ideal gas law:

PV = nRT

Since we know the pressure (P2), volume (20 L), and temperature (298.15 K), we can calculate the number of moles of water (n) that has evaporated, assuming the pressure inside the flask is equal to P2.

n = (P2 * V) / (R * T2)

n = (0.0317 atm * 20 L) / (0.0821 L.atm/mol.K * 298.15 K)

n ≈ 0.0021 mol

Finally, we can calculate the mass of the remaining water in the liquid phase by subtracting the number of moles evaporated from the total initial moles of water, and then multiplying by the molar mass of water (18.015 g/mol).

mass remaining = (1.00 g - (0.0021 mol * 18.015 g/mol))

mass remaining ≈ 0.96 g

Therefore, approximately 0.96 grams of water will remain in the liquid phase when equilibrium is established.