Assume you heat 30 g of water until it all turns into gas. At standard conditions, what volume will this water occupy?
Answer in units of liters.
30 grams is how many mols? That's mols = g/molar mass = 30/18 = estimated 2 mols but you need to do it more accurately.
Then you know that 1 mol will occupy 22.4 L at STP. Convert moles you have to L.
Oh, water turning into gas? That's a groundbreaking performance! Alright, now let's calculate the volume this disappearing act will create.
First, we need to know the molar mass of water, which is roughly 18 g/mol. Since you have 30 g of water, we can convert that to moles by dividing the mass by the molar mass:
30 g / 18 g/mol ≈ 1.67 mol
Now, the fun part. At standard conditions (STP) of 1 atmosphere of pressure and 0 degrees Celsius (273.15 Kelvin), one mole of any gas occupies 22.4 liters. So, using the power of math and the magic of conversion:
1.67 mol × 22.4 L/mol ≈ 37.41 L
Voila! The water will occupy approximately 37.41 liters when it turns into gas.
To determine the volume of water vaporized at standard conditions, we need to use the ideal gas law equation:
PV = nRT
Where:
P = Pressure (at standard conditions, the pressure is 1 atm)
V = Volume
n = Number of moles
R = Ideal gas constant (0.0821 L·atm/(mol·K))
T = Temperature (at standard conditions, the temperature is 273.15 K)
First, we need to calculate the number of moles of water vapor.
To do this, we will use the molar mass of water (H2O), which is approximately 18 g/mol.
Number of moles (n) = mass (m) / molar mass (M)
= 30 g / 18 g/mol
= 1.67 mol (rounded to two decimal places)
Now, we can substitute the values into the ideal gas law equation:
PV = nRT
(1 atm)(V) = (1.67 mol)(0.0821 L·atm/(mol·K))(273.15 K)
V = (1.67 mol)(0.0821 L·atm/(mol·K))(273.15 K) / (1 atm)
V = 37.33 L
Therefore, 30 grams of water vaporized at standard conditions will occupy a volume of 37.33 liters.
To determine the volume of water when it turns into a gas, we need to use the ideal gas law. The ideal gas law is given by the equation:
PV = nRT
Where:
P = pressure (in units of atmosphere, atm)
V = volume (in units of liters, L)
n = number of moles of gas
R = ideal gas constant (0.0821 L·atm/(mol·K) )
T = temperature (in units of Kelvin, K)
Since we are given the mass of water (30 g) but we want to find the volume, we need to convert mass to moles.
To do this, we need to use the molar mass of water (H2O). The molar mass of water is approximately 18 g/mol (2 hydrogen atoms x 1 g/mol + 1 oxygen atom x 16 g/mol).
First, let's calculate the number of moles:
moles = mass / molar mass
moles = 30 g / 18 g/mol
moles = 1.67 mol (rounded to two decimal places)
Next, we need to assume standard conditions for temperature and pressure. Standard temperature is 273.15 K (0 degrees Celsius) and standard pressure is 1 atm.
Now we can rearrange the ideal gas law equation to solve for volume:
V = (nRT) / P
Substituting the values:
V = (1.67 mol * 0.0821 L·atm/(mol·K) * 273.15 K) / 1 atm
Calculating this expression:
V ≈ 38.74 L
So, when all the water turns into gas at standard conditions, it will occupy approximately 38.74 liters of volume.