The gas described by van der Waals' equation in the limit of large molar volume, in the intermediate range of pressure experience balancing by two factors.
$$
\left(p+\frac{a}{V^2}\right)(V-b)=R T \Rightarrow p V=\frac{a}{V}-p b+\frac{a b}{V^2}=R T
$$
(We can neglect the product of two very small constants, $\frac{a b}{V^2}$ )
So,
$$
p V+\frac{a}{V}-p b=R T
$$
As
$$
\frac{a}{V}=p b
$$
The gas behaves ideally, $p V=R T$
- The van der Waals' constants ' $a$ ' and ' $b$ ' are characteristics of the gas. The value of these constants are determined by the critical constants of the gas
$$
a=3 p_C V_C^2=3 p_C\left(\frac{3 R T_C}{8 p_C}\right)^2=\frac{27\left(R T_C\right)^2}{64 p_C} ; b=\frac{1}{3}\left(\frac{3 R T_C}{8 p_C}\right)=\frac{R T_C}{8 p_C}
$$
Question is based on description by van der Waals' equation, that's why we have to ignore the effect of temperature on $a$ and $b$. Actually, the so-called constant varies to some extent with temperature and this shows that the van der Waals' equation is not a complete solution of the behaviour of real gas