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, 2143.4 K

Explanation

The average velocity of a gas is given by
`baru =sqrt((8RT)/(piM))`
In the present case, `baru =9.0xx10^4 " cm " s^(-1)`,
`R=8.314xx10^7 " erg " K^(-1) mol^(-1)`
`T=T_1 (?) and M=44`
`:. " " 9.0xx10^4=sqrt((8xx8.314xx10^7xxT_1)/(3.14xx44))`
Hence, `T_1 = 1682.5 K`.
The most probable velocity is given by
`alpha = sqrt((2RT)/M)`
`:. " " 9.0xx10^4=sqrt((2xx8.314xx10^7xxT_2)/(44))`
Which gives `T_2 = 2143.3 K`.