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 T2, 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 > TC > T2 > T1. Temperature T3 > TC 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 TC.
(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.
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