(i) $\mathrm{CO}_2$ exists as in the gaseous state between the points ' $a$ ' and '$b$' at temperature $T_1$ because from point ' $a$ ' to ' $b$ ' volume starts decreasing and the pressure increases and the gaseous molecules start to come closer but exist in the gaseous state only.
(ii) At the temperaturte $\mathrm{T}_1, \mathrm{CO}_2$ starts liquefying at the point '$b$'. Because at point ' $b$ ' the liquefication has just started or commences.
(iii) At the temperature $\mathrm{T}_2, \mathrm{CO}_2$ will be completely liquefied at the point ' $\mathrm{g}$ '. Because in the curve at the temperature $T_2$ point ' $f$ ' to ' $g$ ' represents the phase where the gas is being converted to liquid and at point ' $\mathrm{g}$ ' all the gas has been converted to liquid.
(iv) As stated in the graph $\mathrm{T}_3>\mathrm{T}_{\mathrm{C}}>\mathrm{T}_2>\mathrm{T}_1$. Temperature $T_3>T_C$ i.e., the critical temperature so condensation will not take place when the temperature is $T_3$. Because critical temperature is the temperature of gas above which gas cannot be liquified howsoever high pressure is applied and $\mathrm{T}_3$ is greater than $\mathrm{T}_{\mathrm{C}}$.
(v) At the temperature $T_1$ curve the equilibrium of liquid and gaseous state of $\mathrm{CO}_2$ is represented between the point's ' $b$ ' and ' $c$ '. Because between the points ' $b$ ' and ' $\mathrm{c}$ ', the pressure being the constant volume of a gas decreases till point ' $\mathrm{c}$ ' so between these points $\mathrm{CO}_2$ gas partially exists as in liquid and the gaseous state i.e., existing in equilibrium.