Real gas compressibility isotherms(graphs)

[requal]

Homework Statement

Sketch the following compressibility isotherms(fancy word for graphs) (compressibility factor, PV/nRT versus pressure

:) for oxygen at a temperature just above the critical temperature
::) for oxygen at a temperature well above the critical temperature eg. 200 degrees.

Label all isotherms and explain the differences, if any, between the curves.

Homework Equations

The Attempt at a Solution

Well I thought that both graphs will start at 1, and will have a positive gradient, because of the molecular volume effect. The graph won't dip like it would if oxygen was below critical temperature, because at critical point the species will be moving too fast for the intermolecular forces to have effect. But what I'm not sure if the gradient of the curve for part :) will be higher than for part ::) because as temperature increases, PV/nRT will drop no?

 

[nate808]

Thank you for your post. I can provide some insights on the compressibility isotherms for oxygen at different temperatures.

Firstly, you are correct in stating that both graphs will start at a compressibility factor of 1 at low pressures, as this is the ideal gas behavior. However, as the pressure increases, the compressibility factor will deviate from 1 due to the intermolecular forces between oxygen molecules.

For the graph at a temperature just above the critical temperature, the curve will show a slight dip as the pressure increases, before rising again. This is because at this temperature, oxygen is in a state between gas and liquid, known as the critical point. The intermolecular forces are still present, but the molecules are moving too fast for them to have a significant effect. As the pressure increases, the molecules will come closer together and the intermolecular forces will start to have a larger impact, causing the dip in the curve.

On the other hand, for the graph at a temperature well above the critical temperature, the curve will show a more gradual increase with pressure. This is because at this temperature, oxygen is mostly in a gaseous state and the intermolecular forces have a smaller effect on the compressibility. Therefore, the curve will not dip as much as in the previous case.

In terms of the gradient of the curves, it is true that as temperature increases, the compressibility factor will decrease. However, this effect will be more significant in the graph at a temperature well above the critical temperature, as the molecules are moving faster and the intermolecular forces have less influence.

I hope this explanation helps you understand the differences between the two compressibility isotherms for oxygen at different temperatures. Keep up the good work in your studies!