Calculate the volume occupied by 18g of water vapor at 100C at 1 atm. Show calculations
11.2L
Use PV = nRT. Don't forget to change 100 C to Kelvin.
23.57
Well, well, well, it seems like you need some help with vapor-ous calculations! Don't worry, I've got you covered.
First things first, let's convert grams to moles of water vapor. The molar mass of water (H₂O) is approximately 18 g/mol (we can ignore the slight difference with water vapor at this temperature). So, 18 g of water vapor is equal to 1 mole.
At this point, we need to know that the molar volume of any ideal gas at standard temperature and pressure (STP) is approximately 22.4 liters/mol.
However, the given conditions are not at STP but at 100°C and 1 atm. So, we need to make an adjustment to account for the changes in pressure and temperature.
Using the ideal gas law equation: PV = nRT, where:
P is the pressure (in atm),
V is the volume (in liters),
n is the number of moles,
R is the ideal gas constant (0.0821 L.atm/mol.K), and
T is the temperature (in Kelvin).
We know that:
P = 1 atm
n = 1 mole
R = 0.0821 L.atm/mol.K
T = 100°C = 373.15 K
Now, let's rearrange the equation to solve for V:
V = (nRT) / P
V = (1 mole * 0.0821 L.atm/mol.K * 373.15 K) / 1 atm
V ≈ 30.6 liters
Therefore, the volume occupied by 18 grams of water vapor at 100°C and 1 atm is approximately 30.6 liters. That's one steamy situation!
To calculate the volume occupied by 18g of water vapor at 100°C and 1 atm, we need to use the ideal gas law equation: PV = nRT.
The ideal gas law equation relates the pressure (P), volume (V), number of moles (n), and temperature (T) of a gas. In this case, we know the pressure and temperature, and our goal is to find the volume.
First, let's convert the mass of water vapor (18g) to moles. We'll use the molar mass of water (H2O) to do this.
The molar mass of H2O is 18.015 g/mol.
Number of moles = Mass / Molar mass = 18g / 18.015 g/mol ≈ 0.9982 mol
Now, we have the number of moles of water vapor (n = 0.9982 mol), pressure (P = 1 atm), and temperature (T = 100°C = 373.15 K).
We can rearrange the ideal gas law equation to solve for volume (V):
V = (nRT) / P
V = (0.9982 mol * 0.0821 L·atm/mol·K * 373.15 K) / 1 atm
Simplifying the equation:
V = 30.112 L
Therefore, the volume occupied by 18g of water vapor at 100°C and 1 atm is approximately 30.112 liters.