a 1.00g sample of water is allowed to vaporize completely in a 10.0L container. What is the pressure of the water vapor at 150 degrees celsius?
Use PV = nRT.
n = 1.00g/molar mass H2O
Don't forget T must be in kelvin.
PV=nRT
P=?
V=10L
n=0.056mol
R=0.082 atm*l/mol*K
T=423K
P=nRT/V
P=(0.056mol)(0.082atm*l/mol*K)(423K)/(10L)
P=0.1942 atm
To find the pressure of the water vapor at 150 degrees Celsius, you can use the Ideal Gas Law equation:
PV = nRT
Where:
P = pressure
V = volume
n = moles of gas
R = ideal gas constant
T = temperature in Kelvin
First, let's convert the temperature from Celsius to Kelvin:
T(K) = T(C) + 273.15
T(K) = 150 + 273.15
T(K) = 423.15 K
Now, we need to determine the number of moles of water vapor. To do this, we can use the molar mass of water, which is approximately 18.01528 g/mol.
moles (n) = mass (m) / molar mass (M)
moles (n) = 1.00 g / 18.01528 g/mol
moles (n) ≈ 0.05549 mol
Now, let's substitute the values into the Ideal Gas Law equation:
PV = nRT
P * 10.0 L = (0.05549 mol) * (0.0821 L∙atm/(mol∙K)) * 423.15 K
P * 10.0 L = 1.9237
P = 1.9237 / 10.0
P ≈ 0.1924 atm
Therefore, the pressure of the water vapor at 150 degrees Celsius is approximately 0.1924 atm.
To find the pressure of the water vapor, we can use the Ideal Gas Law equation:
PV = nRT
Where:
P = pressure (in units of pressure, such as atm)
V = volume (in units of volume, such as liters)
n = number of moles
R = ideal gas constant (0.0821 L·atm/(mol·K))
T = temperature (in units of temperature, such as Kelvin)
First, we need to determine the number of moles of water vapor present. We can do this by using the molar mass of water, which is about 18.015 g/mol.
Number of moles (n) = mass (m) / molar mass (M)
n = 1.00 g / 18.015 g/mol ≈ 0.0555 mol
Next, we need to convert the temperature from Celsius to Kelvin. The formula is:
T (in Kelvin) = T (in Celsius) + 273.15
T = 150 °C + 273.15 = 423.15 K
Now, we can substitute the known values into the Ideal Gas Law equation:
(P)(V) = (n)(R)(T)
P = (n)(R)(T) / V
P = (0.0555 mol)(0.0821 L·atm/(mol·K))(423.15 K) / 10.0 L
Calculating this expression will give us the pressure of the water vapor.