How much volume in liters will 100.0 g of oxygen occupy at 100 0C and 100 atm pressure?
Calculate the number of moles of O2 in 100.0 g using
Number of moles = mass/mass of one mole
this then gives you n in
PV=nRT
You find n from the mass of oxygen (O2)
n = (100.0 g of O2) / (1 mol O2 /32.00 g O2) =3.1 mol of O2
1 mol of any gas at STP has a volume of 22.4 L
1 mole of O2 = 32.0 g of O2 1 mole of O2 = 22.4 L of O2
100.0 g of O2 / 32.00 g of O2 = 3.1 mol of O2
3.1 mol of O2 * 22.4 L/ mol = 69.4 L of O2
But note that 100 C and 100 atm are not STP conditions.
To calculate the volume of a gas, we can use the Ideal Gas Law equation, which states:
PV = nRT
Where:
P = Pressure (in atmospheres)
V = Volume (in liters)
n = Number of moles of gas
R = Ideal gas constant (0.0821 L·atm/(mol·K))
T = Temperature (in Kelvin)
First, let's convert the temperature from Celsius to Kelvin:
T(K) = T(C) + 273.15
So in this case, the temperature would be:
T(K) = 100°C + 273.15 = 373.15 K
Now we need to calculate the number of moles of oxygen. We can use the molar mass of oxygen to convert the given mass to moles:
Molar Mass of Oxygen (O2) = 32.00 g/mol
moles = mass / molar mass
moles = 100.0 g / 32.00 g/mol = 3.125 mol
Now we have all the values we need to plug into the Ideal Gas Law equation and solve for the volume (V):
PV = nRT
V = (nRT) / P
We can substitute the values into the equation:
V = (3.125 mol * 0.0821 L·atm/(mol·K) * 373.15 K) / 100 atm
After performing the calculations:
V ≈ 9.23 L
Therefore, 100.0 grams of oxygen will occupy approximately 9.23 liters at a temperature of 100°C and a pressure of 100 atm.