Question

For ${{H}_{2}}$ gas. the compressibility factor, $Z=\dfrac{PV}{nRT}$ is
A. Equal to 1
B. Equal to 0
C. Always greater than 1
D. Initially less than 1 and then becomes greater than 1 at high pressure.

Answer 1

Ideal gases are considered to be point masses. Therefore, there is no compressibility factor to be calculated since Z = 1 for an ideal gas. Compressibility factor or compression factor appears for only real gases because in real gases the intermolecular force of attraction is present. Thus the value of Z will not be equal to 1.

Complete step by step solution:
Let us first understand the fact that in nature, there is no ideal gas. Ideal gases are theoretical because all the properties of ideal gases are not possible for a gas to acquire in a normal environment. Only under certain conditions, real gases tend to behave like ideal gases.
But first, let us see some main properties of ideal gases:
- The molecules of ideal gases do not have mass and thus the volume occupied by ideal gas moles are considered insignificant with respect to the volume occupied by the gas.
- It is considered that there is no intermolecular force of attraction or any kind of repulsion between the molecules of an ideal gas. That is why the compressibility factor is equal to one. We will know more about the compressibility factor later.
- There is no loss of internal energy of the ideal gas molecules when they collide with each other i.e. their collisions are perfectly elastic.
On the other hand, real gas molecules have masses, occupy significant volume, have an intermolecular force of attraction and the collisions between molecules are not elastic.
The compressibility factor can be defined as the ratio of the actual volume of a real gas as a given temperature and pressure to the ideal volume at the same temperature and pressure.
If Z is the compressibility factor then,
$Z=\dfrac{PV}{nRT}$
We can write the Vander wall’s equation for real gas as:
$\left( P+\dfrac{a{{n}^{2}}}{{{V}^{2}}} \right)\left( V-nb \right)=nRT$
Here, a is the Vander wall’s correction constant for the intermolecular force of attraction and b is the Vander wall’s correction for the occupied volume by real gas.
The compressibility factor at a particular temperature and pressure depends on the intermolecular force of attraction. Since the mass of hydrogen is very small the intermolecular force of attraction is negligible. Hence we can say that the value $\dfrac{a{{n}^{2}}}{{{V}^{2}}}$ is also negligible. Thus the Vander wall’s equation becomes:

& P\times \left( V-nb \right)=nRT \\\ & \Rightarrow PV-nPb=nRT \\\ & \Rightarrow PV=n\left( RT+Pb \right) \\\ & \Rightarrow \dfrac{PV}{nRT}=\dfrac{\left( RT+Pb \right)}{RT} \\\ & \Rightarrow \dfrac{PV}{nRT}=1+\dfrac{Pb}{RT} \\\ & \Rightarrow Z=\dfrac{PV}{nRT}=1+\dfrac{Pb}{RT} \\\ \end{aligned}$$ Thus, we can say that the value of Z will always be greater than one. **Hence option (C) is the correct one.** **Note:** Always remember that the value of Z will always be greater than one only for hydrogen and helium gases because of their small masses. When the mass of the mass molecules is very small we can ignore Vander wall’s constant for the correction of the intermolecular force of attraction ‘a’. But in case of gases like ${{N}_{2}}$ and ${{O}_{2}}$, the mass of the gas molecules are large and we cannot ignore the Vanderwall constant ‘a’ for these gases. For them, the compressibility factor Z at a certain temperature will be less than 1 at lower pressures but it will increase at higher pressures.