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Pressure of real gas is less that the pressure of ideal gas because:
A. Number of collisions increases
B. Define shape of molecule
C. Kinetic energy of molecules increases
D. Intermolecular forces

Last updated date: 21st Apr 2024
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Hint: To answer the question, we should know what real gas and ideal gas are.
Real gas: “A real gas is a gas that does not behave as an ideal gas because of interactions between gas molecules”.
Idea gas: “The gas in which all collisions between atoms or molecules are perfectly elastic and in which there is no intermolecular force of attraction is called ideal gas”.

Complete step by step answer:
We know that for real gases
\[\left( P+\frac{a{{n}^{2}}}{{{V}^{2}}} \right)\left( V-nb \right)=nRT\]
a = intermolecular forces
b = total volume per mole occupied by the gas
n = number of moles of gas
P = Pressure
V = Volume
R = gas constant
T = Temperature
For ideal gas
\[\left( PV \right)=nRT\]
From the above equations it is clear that for real gases there is a role of intermolecular forces, but in case of ideal gas there are no intermolecular forces.
On observing the formula of real gases we can say that there is some amount of attraction between the gas molecules namely the van der waal forces.
These van der waal forces don't let the molecules hit the wall of the container with full force and gas molecules held back by these attractive forces. Due to these forces of attraction (intermolecular forces), real gases tend to show slightly less pressure as compared to ideal gasses.

So, the correct option is D.

Note: Real gases don’t obey all the gas laws at all temperature and pressure but ideal gas obeys all the laws at all temperatures and pressures. Real gases only obey all Gas Laws at high temperature and low pressure.