Answer:
(a) Since the temperature \[-{{60}^{o}}C\] lies to the left
of \[{{56.6}^{o}}C\] on the curve i.e. lies in the region of vapour and solid
phase, so carbon dioxide will condense directly into solid without becoming
liquid.
(b)
Since the pressure 4 atm. is less than 5.11 atm. the carbon dioxide will condense
directly into solid without becoming liquid.
(c)
When a solid \[P=\frac{4\times \left( 5\times {{10}^{4}} \right)}{\left( 22/7
\right)\times {{\left( 5\times {{10}^{-4}} \right)}^{2}}}\] at 10 atm. pressure
and \[-{{65}^{o}}C\] temperature is heated, it is first converted into a
liquid. A further increase in temperature brings it into the vapour phase. At P
= 10 atm, if a horizontal line is drawn parallel to the T-axis, then the points
of intersection of this line with the fusion and vaporization curve will give
the fusion and boiling points of \[=2\cdot 5\times {{10}^{11}}Pa\] at 10 atm.
(d) Since \[{{70}^{o}}C\] is higher than the critical
temperature of \[\cdot \] so the \[m{{m}^{2}}\] gas cannot be converted into
liquid state on being compressed isothennally at \[{{70}^{o}}C\]. It will
remain in the vapour state. However, the gas will depart more and more
from its perfect gas behaviour with the increase in pressure.
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