A 0.0320-m3 container is initially evacuated. Then, 5.27 g of water is placed in the container, and, after some time, all of the water evaporates. If the temperature of the water vapor is 352 K, what is its pressure?
To find the pressure of the water vapor, we can use the ideal gas law equation:
PV = nRT
Where:
P = Pressure
V = Volume
n = number of moles of gas
R = ideal gas constant (8.314 J/(mol*K))
T = temperature in Kelvin
First, we need to determine the number of moles of water vapor. We can do this by using the molar mass of water.
1. Calculate the number of moles of water:
Number of moles = mass / molar mass
The molar mass of water (H2O) = 2(1.008 g/mol) + 16.00 g/mol = 18.02 g/mol
Number of moles of water = 5.27 g / 18.02 g/mol
Now we can calculate the pressure of the water vapor:
2. Rearrange the ideal gas law equation to solve for pressure:
P = nRT / V
3. Convert the volume from cubic meters to liters:
Volume = 0.0320 m^3 * 1000 L/m^3 = 32 L
4. Plug in the values into the equation:
P = (number of moles of water * ideal gas constant * temperature) / volume
P = (5.27 g / 18.02 g/mol) * (8.314 J/(mol*K)) * (352 K) / 32 L
Now, we can calculate the pressure.
To find the pressure of the water vapor, we can use the ideal gas law equation, which is as follows:
PV = nRT
Where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature in Kelvin.
To apply this equation, we need to calculate the number of moles of water vapor first.
1. Convert the mass of water (5.27 g) to moles:
To do this, divide the mass by the molar mass of water, which is approximately 18 g/mol.
Number of moles = 5.27 g / 18 g/mol ≈ 0.2923 mol
2. Now, let's calculate the pressure. Given that the volume is 0.0320 m3 and the temperature is 352 K, we need to rearrange the ideal gas law equation to:
P = nRT / V
Substituting the values:
P = (0.2923 mol) * (8.314 J/(mol·K)) * (352 K) / 0.0320 m3
Note that we used the value of the ideal gas constant, R, which is 8.314 J/(mol·K).
Now, let's calculate:
P ≈ 1006.8 J/(mol·K) / 0.0320 m3
P ≈ 31462.5 J/m3
The pressure of the water vapor is approximately 31.5 kPa (since 1 J/m3 ≈ 1 Pa and 1 kPa = 1000 Pa).
Therefore, the pressure of the water vapor is approximately 31.5 kPa.