Isotherms of carbon dioxide at various temperatures are repersented in figure. Answer the following questions based on this figures.

(i) In which state will `CO_(2)` exist between the points a and b at temperature `T_(1)` ?
(ii) At what point will `CO_(2)` start liquefyinh when temperature is `T_(1)` ?
(iii) At what point will `CO_(2)` be completely liquefued when temperature is `T_(2)` ?
(iv) Will condensation take place when the temperature is `T_(3)` ?
(v) What portion of the isotherm at `T_(1)` represent liquid and gaseous `CO_(2)` at equilibrium ?

Answer

(1682.5 k , 2142 k)

Explanation

`u_(av)`(average velocity) `=sqrt((8RT_(1))/(pi M)) rArr (9 xx 10^(4))/(100) ms^(-1) = sqrt((8 xx 8.314 T_(1))/(3.14 xx 44 xx 10^(-3))) rArr T_(1) = 1682.5 K`
Also, for the same gas `(u_(av))/(u_(m ps)) = sqrt((8RT_(1))/(pi M)): sqrt((2RT_(2))/(M)) = sqrt((8T_(1))/(pi) xx (1)/(2T_(2))) = sqrt((4T_(1))/(pi T_(2))) rArr 1 = sqrt((4T_(1))/(pi T_(2)))`
`rArr T_(2) = (4T_(1))/(pi) = (4 xx 1682.5)/(3.14) = 2124K`, Hence, `T_(1) = 1682.5 K and T_(2) = 2142K`