The molar volume It is an intensive property that indicates how much space a mole of a certain substance or compound occupies. It is represented by the symbol Vm, y is expressed in units of dm3/ mol for gases, and cm3/ mol for liquids and solids, because the latter are more confined by their higher intermolecular forces.
This property is recurrent when studying thermodynamic systems that involve gases; since, for liquids and solids the equations to determine Vm they become more complicated and inaccurate. Therefore, as far as basic courses are concerned, the molar volume is always associated with the ideal gas theory..
This is due to the fact that structural aspects are irrelevant for ideal or perfect gases; all its particles are visualized as spheres that elastically collide with each other and behave in the same way no matter what their masses or properties are.
This being the case, a mole of any ideal gas will occupy, at a given pressure and temperature, the same volume Vm. It is then said that under normal conditions of P and T, 1 atm and 0 ºC, respectively, one mole of an ideal gas will occupy a volume of 22.4 liters. This value is useful and approximate even when evaluating real gases.
Article index
The immediate formula to calculate the molar volume of a species is:
Vm = V / n
Where V is the volume it occupies, and n the amount of the species in moles. The problem is that Vm depends on the pressure and temperature that the molecules experience, and a mathematical expression that takes these variables into account is desired.
The ethylene in the image, HtwoC = CHtwo, it has an associated molecular volume limited by a green ellipsoid. This HtwoC = CHtwo It can rotate in multiple ways, which is as if said ellipsoid were moved in space to visualize how much volume it would occupy (obviously negligible).
However, if the volume of such a green ellipsoid we multiply it by NTO, Avogadro's number, we will then have one mole of ethylene molecules; a mole of ellipsoids interacting with each other. At higher temperatures, the molecules will separate from each other; while at higher pressure, they will contract and reduce their volume.
Therefore, Vm is dependent on P and T. Ethylene has a plane geometry, so it cannot be thought that its Vm is precisely and exactly the same as that of methane, CH4, of tetrahedral geometry and capable of being represented with a sphere and not an ellipsoid.
The molecules or atoms of liquids and solids also have their own Vm, which can be roughly related to its density:
Vm = m / (dn)
Temperature affects molar volume for liquids and solids more than pressure, as long as the latter does not vary sharply or is exorbitant (in the order of GPa). Likewise, as mentioned with ethylene, the geometries and molecular structures have a great influence on the values of Vm.
However, under normal conditions it is observed that the densities for different liquids or solids do not vary too much in their magnitudes; the same is true of its molar volumes. Note that the denser they are, the smaller will be Vm.
Regarding solids, their molar volume also depends on their crystalline structures (the volume of their unit cell)..
Unlike liquids and solids, for ideal gases there is an equation that allows us to calculate Vm as a function of P and T and their changes; this is, that of ideal gases:
P = nRT / V
Which is accommodating to express V / n:
V / n = RT / P
Vm = RT / P
If we use the gas constant R = 0.082 L atm K-1Mole-1, then the temperatures should be expressed in kelvin (K), and the pressures in atmospheres. Note that here we can see why Vm is an intensive property: T and P have nothing to do with the mass of the gas but with its volume.
These calculations are only valid under conditions where gases behave close to ideality. However, the values obtained through experimentation have a small margin of error in relation to the theoretical.
There is a gas Y whose density is 8.5 · 10-4 g / cm3. If we have 16 grams equivalent to 0.92 moles of Y, calculate its molar volume.
From the density formula we can calculate what volume of Y these 16 grams occupy:
V = 16 g / (8.5 10-4 g / cm3)
= 18.823.52 cm3 or 18.82 L
So Vm It is calculated directly by dividing this volume by the number of moles given:
Vm = 18.82 L / 0.92 mol
= 20.45 L / mol or L mol-1 or dm3Mole-1
In the previous example of Y, it was not specified at any time what was the temperature that the particles of said gas were experiencing. Assuming that Y was worked at atmospheric pressure, calculate the temperature needed to compress it to the determined molar volume.
The statement of the exercise is longer than its resolution. We use the equation:
Vm = RT / P
But we solve for T, and knowing that the atmospheric pressure is 1 atm, we solve:
T = VmP / R
= (20.45 L / mol) (1 atm) / (0.082 L atm / K mol)
= 249.39 K
That is, one mole of Y will occupy 20.45 liters at a temperature close to -23.76 ºC.
Following the previous results, determine Vm at 0 ºC, 25 ºC and at absolute zero at atmospheric pressure.
Transforming the temperatures to kelvin, we first have 273.17 K, 298.15 K and 0 K. We solve directly by substituting for the first and second temperatures:
Vm = RT / P
= (0.082 L atm / K mol) (273.15 K) / 1 atm
= 22.40 L / mol (0 ºC)
= (0.082 L atm / K mol) (298.15 K) / 1 atm
= 24.45 L / mol (25ºC)
The value of 22.4 liters was mentioned at the beginning. Note how Vm increases with temperature. When we want to do the same calculation with absolute zero, we stumble upon the third law of thermodynamics:
(0.082 L atm / K mol) (0 K) / 1 atm
= 0 L / mol (-273.15 ºC)
Gas Y cannot have a non-existent molar volume; that means that it has been transformed into a liquid and the previous equation is no longer valid.
On the other hand, the impossibility of calculating Vm at absolute zero it obeys the third law of thermodynamics, which says that it is impossible to cool any substance to the temperature of absolute zero.
Yet No Comments