Isotherms of carbon dioxide at various temperatures are represented in Fig. 5.5. Answer the following questions based on this figure.

(i) In which state will CO₂ exist between the points a and b at temperature T₁?

(ii) At what point will CO₂ start be liquefying when temperature is T₁?

(iii) At what point will CO₂ be completely liquefied when temperature is T₂.

(iv) Will condensation take place when the temperature is T₃.

(v) What portion of the isotherm at T₁represent liquid and gaseous CO₂ at equilibrium?

(i) CO₂ exists as in the gaseous state between the points ‘a’ and ‘b ’at temperature T₁ 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 T₁ CO₂ starts liquefying at the point ‘b’. Because at point ‘b’ the liquefication has just started or commences.

(iii) At the temperature T_2, CO₂ will be completely liquefied at the point ‘g’. Because in the curve at the temperature T₂ 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 T₃ > T_C > T₂ > T₁. Temperature T₃ > T_C i.e. the critical temperature so condensation will not take place when the temperature is T₃. Because critical temperature is the temperature of a gas above which gas cannot be liquified howsoever high pressure is applied and T₃ is greater than T_C.

(v)At the temperature T₁ curve the equilibrium of liquid and gaseous state of CO₂ 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 CO₂ gas partially exists as in liquid and in gaseous state i.e. existing in equilibrium.