Real gas

[m25]

My TBR book (page 12, chemistry II) states that when dealing with real gases, expected changes in the volume of gas is never as large as it is predicted by the ideal gas law. For example, if a pressure is halved while all the other conditions except for volume is constant, the volume of the real gas will be 2V - a little bit, and if a pressure is doubled while all the other conditions except for volume is constant, the volume of real gas will be 0.5V + a little bit. It then goes on to explain how the changes in real gas is never as large as you think.
Now, my question is, is the change being "never as large as you think" also apply when I change the volume and predict the pressure?
So for example, what will be the real pressure of a gas when the volume is doubled? Would it be 0.5P + a little(pressure changed less than predicted) bit, or 0.5P - a little(pressure changed more than predicted)?

Thank you!

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[Chrisz]

pv=nrt. this is the ideal gas law.
If you incorporate correction for volume, the equition will become p(v-nb)=nrt. where b is the volume intrinsic of gases, which we have to discount from the total volume of the container. In this case, i have assumed that the molecules do not exert attraction to make the case little bit easier, since no mention of attraction in your questions. If you double the total volume, the new effective volume would become 2v-nb. Comparing this to v-nb, we get (2v-nb)/(v-nb)=1+v/(v-nb)>2.
This would mean increase the container volume by twice, we increase the effective volume by a factor more than 2, so the pressure has to decrease by a factor more than 2. So, the answer to your question would be no, the opposite applies.