(N/A) Boyle's Law states that at constant temperature,the pressure of a fixed amount of gas is inversely proportional to its volume ($p \propto 1/V$ or $pV = k$).
$(i)$ Plot of $V$ vs $p$: Shows an isothermal curve (hyperbola) representing $V \propto 1/p$.
(ii) Plot of $V$ vs $1/p$: Shows a straight line passing through the origin,confirming $V = k(1/p)$.
(iii) Plot of $p$ vs $1/V$: Shows a straight line passing through the origin,confirming $p = k(1/V)$.
(iv) Plot of $\log_{10} p$ vs $\log_{10} V$: Since $pV = k$,taking log on both sides gives $\log p + \log V = \log k$,or $\log p = -\log V + \log k$. This is a straight line with a slope of $-1$.
$(v)$ Plot of $p$ vs $V$ at different temperatures: Shows isotherms where $p$ decreases as $V$ increases. Higher temperature curves are further from the origin $(T_3 > T_2 > T_1)$.
(vi) Plot of $p$ vs $1/V$ at different temperatures: Shows straight lines passing through the origin. The slope $(k = nRT/V)$ increases with temperature,so $T_3 > T_2 > T_1$ (Note: The image label $T_3 < T_2 < T_1$ in the provided diagram is physically incorrect for the slope of $p$ vs $1/V$ at constant $n$).