I got how the real pressure would be less than ideal pressure due to net force of attraction between gas molecules acting towards the center.
But, why is the real volume greater than ideal volume? I thought the real volume would be less than ideal one since the ideal volume is the volume of container (volume of gas particles negligible) while the real volume is the available volume for gases to move excluding their individual volumes i.e. Vi = Vr + volume of gas particles
In the ideal gas equation $$p_iV_i = nRT$$ $p_i$ is the pressure if there was no intermolecular forces between the molecules and $V_i$ is the free volume available for the molecules to do motion in, which is the volume of the container.
In real gases however, the pressure that you observe (real pressure) is less than the expected pressure. This is because of the intermolecular forces. So you convert the $p_r$ to a pressure which the gas would exert if there were no intermolecular forces. $$p + a\frac{n^2}{V^2}$$
The free volume available to real gas is less than the volume considered in ideal gas equation. That is because Kinetic theory of gases made an assumption that the volume occupied by molecules of an ideal gas is negligible. But in real gases, the molecules are of considerable size. Thus, the full container will not be free for motion, some volume of the container will be occupied by the molecules. Let the volume occupied by one mole gas molecules be $b$, for n mole gas this is $nb$, so free volume available to gas is less than the volume of container $$V_i = V_{\text container} - nb$$
Edit: In your question, the $V_r$ they're talking about is volume of the container. And $V_i$ is actual free volume available to the real gas.
