To solve the problem step by step, we need to determine how much water remains in the liquid state after some of it has evaporated in a closed cubic vessel.
### Step 1: Understand the given data
- Volume of water (liquid) = \(4 \times 10^{-3} \, m^3\)
- Vapor pressure of water at 400 K = \(0.0842 \, \text{atm}\)
- Density of water vapor = \(0.89 \, \text{g/cm}^3\) or \(890 \, \text{g/L}\)
- Edge length of the cubic vessel = \(1 \, m\)
### Step 2: Convert the volume of water from cubic meters to liters
1 cubic meter = 1000 liters, so:
\[
\text{Volume of water} = 4 \times 10^{-3} \, m^3 = 4 \times 10^{-3} \times 1000 \, L = 4 \, L
\]
### Step 3: Calculate the volume of the cubic vessel
The volume of the cubic vessel (V) is given by:
\[
V = \text{edge length}^3 = 1^3 = 1 \, m^3 = 1000 \, L
\]
### Step 4: Use the ideal gas law to find the amount of vapor produced
The ideal gas law is given by:
\[
PV = nRT
\]
Where:
- \(P\) = vapor pressure = \(0.0842 \, \text{atm}\)
- \(V\) = volume of the vessel = \(1000 \, L\)
- \(n\) = number of moles of vapor
- \(R\) = gas constant = \(0.0821 \, \text{L atm/(K mol)}\)
- \(T\) = temperature = \(400 \, K\)
Rearranging the ideal gas law gives:
\[
n = \frac{PV}{RT}
\]
### Step 5: Substitute the values into the equation
Substituting the known values:
\[
n = \frac{0.0842 \, \text{atm} \times 1000 \, L}{0.0821 \, \text{L atm/(K mol)} \times 400 \, K}
\]
Calculating the right side:
\[
n = \frac{84.2}{32.84} \approx 2.56 \, \text{moles}
\]
### Step 6: Calculate the mass of the vapor
Using the number of moles to find the mass:
\[
\text{Mass} = n \times \text{molar mass of } H_2O = 2.56 \, \text{moles} \times 18 \, \text{g/mol} = 46.08 \, g
\]
### Step 7: Convert the mass of vapor to volume
Using the density of water vapor:
\[
\text{Volume of vapor} = \frac{\text{Mass}}{\text{Density}} = \frac{46.08 \, g}{890 \, g/L} \approx 0.0518 \, L
\]
### Step 8: Calculate the remaining liquid volume
The initial volume of water was \(4 \, L\), so the volume of liquid left is:
\[
\text{Volume of liquid left} = 4 \, L - 0.0518 \, L \approx 3.9482 \, L
\]
### Final Answer
Approximately **3.9482 liters** of water remains in the liquid state.
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