(A) $-p, s(\mathrm{~B})-r$ (C) $-p, q$ (D) $-r$
Explanation van der Waals' equation
$$
\left(p+\frac{n^2 a}{V_m^2}\right)\left(V_m-b\right)=n R T
$$

For hydrogen gas $(p=200 \mathrm{~atm}, T=273 \mathrm{~K})$
As pressure is large $V_m$ can be assumed small, thus ' $b$ ' can not be ignored, while due to high pressure $a / V_m^2$ can be considered negligible in comparison to $p$.
$$
p\left(V_m-b\right)=R T \quad \text { and } \quad Z=1+\frac{p b}{R T}
$$
For hydrogen gas $(p \sim 0, T=273 \mathrm{~K})$
when pressure occurs of low about $1 \mathrm{~atm}$ or less and temperature is not very close to the point of liquification $\left[T_c\left(\mathrm{H}_2\right)=33.3 \mathrm{~K}\right]$ gas behaves ideally.
For
$$
P V=n R T
$$
Temperature is close to the point of liquification $\left[T_C\left(\mathrm{CO}_2\right)=304.2\right]$ thus, deviation from ideality appears very high (due to high attractive force of attraction).
For real gas with very large molar volume.
As molar volume is very large $a / V_m^2$ will be negligible and at the same time ' $b$ ' in comparison to $V_m$ is also considered negligible, thus, $p V_m=n R T$