Answer the following questions based on the p-T phase diagram of carbon dioxide as shown in the figure .
(i) At what temperature and pressure can the solid , liquid and vapour phases of `C0_(2)` co-exist in equilibrium?
(ii) What is the effect of decrease of pressure on the fusion and boiling point of `C0_(2)` ?
(iii) What are the critical temperature and pressure for `C0_(2)` ? what is their significance ?
(iv) Is `C0_(2)` solid , liquid, or gas at (a) `-70^(@)C `under 1 atm (b) `-60^(@)C` under 10 atm (c) `15^(@)C` under 56 atm ?

Explanation

(i) The solid, liquid and vapour phases of `CO_(2)` co-exist in equilibrium at its triple point O for which
`p_(tr)`=5.11 atm and `T_(tr)=-56.6^(@)`C
(ii) The vaporisation curve I and fustion curve II show that both the boilijng point and fusion point of `CO_(2)` decreases with decrease in pressure.
(iii) For `CO_(2)`, `p_(c)`=73.0 atm and `T_(c)=31.1^(@)`C.
Above its critical temperature, `CO_(2)` hgas cannot be liquefied, however large pressure may be applied.
(iv) a) `70^(@)`C under 1 atm. This point lies in vapour region. Therefore, at `-70^(@)`C under 1atm, `CO_(2)` is vapour.
b) -`60^(@)`C under 10 atm. This point lies on in solid region. Therefore, `CO_(2)` is solid at `-60^(@)` under 10 atm.
c) `15^(@)` under 56 atm. This point lies in liquid region. Therefore, `CO_(2)` is liquid at `15^(@)` under 56atm.