Answer :

(i) 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 exists in the gaseous state only.

(ii) At the temperature T_{1} CO_{2} starts liquefying at the point ‘b’. Because at point ‘b’ the liquefication has just started or commences.

(iii) At the temperature T_{2,} CO_{2} will be completely liquefied at the point ‘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 ‘g’ all the gas has being converted to liquid.

(iv)As stated in the graph T_{3} > T_{C} > T_{2} > 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 a gas above which gas cannot be liquified howsoever high pressure is applied and T_{3} is greater than T_{C.}

(v)At the temperature T_{1} curve the equilibrium of liquid and gaseous state of CO_{2} is represented between the point’s ‘b’ and ‘c’. Because between the points ‘b’ and ‘c’ the pressure being constant volume of gas decreases till point ‘c’ so between these points CO_{2} gas partially exists as in liquid and in gaseous state i.e. existing in equilibrium.

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