question_answer 17)

Answer the following questions based on the P-T phase diagram of \[=0\cdot 5\times {{10}^{-3}}\] as given in Q.16. (a) \[=0\cdot 5\times {{10}^{-3}}\] at 1 atm. pressure and temperature \[-\,{{60}^{o}}C\] is compressed isothennally. Does it go through a liquid phase? (b) What happens when \[=5\times {{10}^{-4}}\] at 4 atm. pressure is cooled from room temperature at constant pressure? (c) Describe qualitatively the changes in a given mass of solid \[F=50,000N\] at 10 atm. pressure and temperature \[{{65}^{o}}C\] as it is heated up to room temperature at constant pressure. (d) \[5\times {{10}^{-4}}\] is heated to a temperature \[{{70}^{o}}C\] and compressed isothennally. What changes in its properties do you expect to observe?

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.