Question

The pressure of real gas is less than the pressure of ideal gas because:
A. No. of collisions increases
B. Definite shape of the molecule
C. K.E. of molecule increases
D. Intermolecular forces

Answer 1

Hint:Vander Waal equation is a thermodynamic equation of state based on the theory that liquids are composed of particles with non-zero volumes but have interparticle attractive force. It is for real gases. When in case the correction factor of pressure and volume becomes negligible, the pressure and volume of real gas become equal to an ideal gas. All real gases start to behave like ideal gases at low pressure and high temperature.

Complete step by step answer:
Vander Waals equation of state for real gases is obtained from the modification of the ideal gas equation or law.
The Vander Waals equation considers the molecular interaction forces i.e. both the attractive and repulsive forces and the molecular size. There comes a volume correction in the Vander Waal equation. As the particles have a definite volume, the volume available for the movement of molecules is not the entire volume of the container but less than that.
The volume of an ideal gas is considered as an overestimation and therefore it is reduced in the case of real gases.
 \[volume\,of\,gas({V_r}) = \dfrac{{volume\,of\,the\,container}}{{ideal\,gas({V_i}) - correction\,factor(b)}}\]
Volume correction for each particle is not the volume of the particle but in actual it is four times of it i.e. each particle has a sphere of influence 4 times its volume.
Volume correction for n particles (nb)= \[\dfrac{{4n \times 4}}{{3\pi {r^3}}}\]
Therefore, volume of real gases = \[{V_i}\] -nb
Now coming to the pressure correction of the Vander Waals equation:
For particles that are inside, the interactions cancel each other but the particles which are on the surface or near the wall of the container experience net pulling of the bulk molecules towards the bulk i.e. away from the surface. The molecules experiencing net interaction away from the walls will hit the walls with less pressure. That is why real gases exhibit lower pressure than shown by the ideal gas.
The pressure of real gas = \[P - \dfrac{{a{n^2}}}{{V_i^2}}\]
On substituting these pressure and volume correction in ideal gas law equation, we get
 \[\left( {P - \dfrac{{a{n^2}}}{{V_i^2}}} \right)({V_i} - nb) = nRT\]
Where ‘a’ and ‘b’ are Van Der Waals constants and have positive values. They are the characteristics of an individual gas.
So, due to the intermolecular attraction, the molecules cannot hit the walls with their full force.

So, the correct answer is, D.

Note:
When in case the correction factor of pressure and volume becomes negligible, the pressure and volume of real gas become equal to an ideal gas. All real gases start to behave like ideal gases at low pressure and high temperature.