Question

Consider a cylinder containing a mixture of liquid carbon dioxide in equilibrium with gaseous carbon dioxide at an initial pressure of \({\rm{65}}\) atm and a temperature of \({\rm{2}}{{\rm{0}}^{\rm{o}}}{\rm{C}}\). Sketch a plot depicting the change in the cylinder pressure with time as gaseous carbon dioxide is released at constant temperature.

Short answer

One of the states of matter is the liquid state. Because the particles in a liquid are free to move, it lacks a defined form even while having a defined volume.

Step-by-step solution

Definition of liquid

One of the states of matter is the liquid state. Because the particles in a liquid are free to move, it lacks a defined form even while having a defined volume.

 Sketching a plot

The reaction

- The temperature is \({20^\circ }{\rm{C}}\)

- The pressure of \({\rm{C}}{{\rm{O}}_{\rm{2}}}\)in a closed cylinder at equilibrium is \({\rm{65\;atm}}\)

Let us sketch a plot with the change in pressure over time, as gaseous \({\rm{C}}{{\rm{O}}_{\rm{2}}}\) is released (the temperature is constant).

In the beginning, the vapour and liquid phases of carbon dioxide are in balance. The pressure is kept at \({\rm{65}}\) atmospheres. The system's pressure drops after the gaseous carbon dioxide is discharged.

Hence, when some gaseous \({\rm{C}}{{\rm{O}}_{\rm{2}}}\)is released, the equilibrium of the system will move to the right, and so some liquid \({\rm{C}}{{\rm{O}}_{\rm{2}}}\)will convert into gas, to maintain the equilibrium pressure of \({\rm{65\;atm}}\)_.So, as long as we have liquid \({\rm{C}}{{\rm{O}}_{\rm{2}}}\),the pressure will be \({\rm{65\;atm}}\)_.As soon as we convert all liquid \({\rm{C}}{{\rm{O}}_{\rm{2}}}\)into gas, the pressure will begin to drop.