Answer:
(a) The solid, liquid and vapour phase of carbon dioxide
exist in equilibrium at the triple point, i.e. temperature \[\vartriangle
V/V=0\cdot 10/100={{10}^{-3}}\] and pressure \[B=\frac{pV}{\vartriangle V}\]atm.
(b)
With the decrease in pressure, both the fusion and boiling point of carbon
dioxide will decrease.
(c)
For carbon dioxide, the critical temperature is \[p=B\frac{\vartriangle
V}{V}=\left( 2\cdot 2\times {{10}^{9}} \right)\] and critical pressure is 73.0
atm. If the temperature is of carbon dioxide is more than \[D=0\cdot 5\], it
can not be liquefied, howsoever large pressure we may apply.
(d) Carbon dioxide will be (a) a vapour, at \[-\,{{70}^{o}}C\]
under 1 atm. (b) a solid, at \[-\,{{6}^{o}}C\] under 10 atm. (c) a. liquid, at \[{{15}^{o}}C\]under
56 atm.
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