Instruction
1
Suppose you know the number of moles of oxygen (e.g. 5). You face the question: how much under normal conditions takes these 5 moles? The solution is: under normal conditions the volume of 1 mole of any gas is constant and is approximately equal to 22.4 liters. Therefore, 5 moles of oxygen at normal conditions will occupy a volume of 22.4*5 = 112 liters.
2
Well, if you know the mass of oxygen? Suppose 96 grams. What is the amount they take under normal conditions? First, find out how many moles of oxygen contains 96 grams of this substance. Molar mass of oxygen (from the formula O2) = 32 grams/mol. Therefore, 96 grams of 3 moles. Performing the multiplication, we get the following response: 22,4*3 = 67.2 per liter.
3
But if you need to determine the amount of oxygen at conditions other than normal? Then you come to the aid of the universal equation Mendeleev-Clapeyron describing the state of the so-called "ideal gas". It is written like this:
PV = RTM/m, where p is the gas pressure in Pascals, V is the volume in liters, R is the universal gas constant, T is the gas temperature in degrees Kelvin, M is the mass of gas, m is its molar mass.
PV = RTM/m, where p is the gas pressure in Pascals, V is the volume in liters, R is the universal gas constant, T is the gas temperature in degrees Kelvin, M is the mass of gas, m is its molar mass.
4
Transforming the equation, we get:
V = RTM/mP.
V = RTM/mP.
5
As you can see, if you have the necessary for calculation data (temperature, mass, and oxygen pressure), calculate its volume very simple. Because the values of R (8,31) and m (32) you know.
Useful advice
Of course, the oxygen (or any other gas) cannot be considered an ideal gas, so the equation Mendeleev-Clapeyron, strictly speaking, not quite accurately describes his condition. However, if the pressure and temperature of the gas is not very different from normal, it can be used in the calculations. The error will be negligible.
