[confusedstud]

At low temperatures, the compressibility factor z falls below 1 and according to this link: http://www.chemguide.co.uk/physical/kt/realgases.html it is because when the molecules are about to collide with the walls they are pull back by the attractive forces of the other molecules. And so the pressure measured would be lower than the ideal pressure if it were an ideal gas. This effect is even more prominent at lower temperatures because the molecules are moving more slowly on average. Any pull they feel back into the gas will have relatively more effect on a slow moving particle than a faster one.

So when i was trying to relate this to the van der Waals equation, initially I thought that P(ideal)=P(real)+a(n/v)² so the actual pressure we measure is the P(real) and so P(real)=P(ideal)-a(n/v)² which causes the P(real) to be lesser and hence when we substitute it into the compressibility factor formula (PV/nRT) since P(real)<P(ideal) the z will fall below 1.

But after going through that again I realised that that did not make much sense because the a(n/v)² term does not contain any variable that changes with temperature. So at any temperature the same substraction of the a(n/v)² would occur so it could not be the reason why the P(real) is lesser than P(ideal) causing the z to fall below 1. 

I think I understand the 'theory' part but I can't relate it back to the van der Waals equation. Why does the z fall below 1 at low temperatures according to the van der Waal equation?

[confusedstud]

I think I have been viewing this problem in the wrong manner. Instead I should have used the full van der Waals equation to determine how each parameter affects the compressibility factor.

PV/nRT = z = V/V-nb - (an/RTV)
This means that if the volume is very small the -nb factor would be very prominent so ignoring the a term of the equation, z would be positive.
Similarly if we ignore the b term z would be negative. However, we would have to consider both terms together into to tell whether the gas would have a z that is less than or more than 1.

But if we compare low temperatures vs high temperatures at a specific pressure, only the second part of the equation (an/RTV) would be affected. And at lower temperatures, the an/RTV factor would be larger compared to at higher temperatures and as a result at lower temperatures z would tend to fall below zero compared to at higher temperature.

I think this is the better way to approach this question. Hope it helps any curious persons.

[Enthalpy]

z decreases as the molecules attract an other. This is more important when less heat is available to separate them. The extreme case is the liquid, where heat doesn't separate the molecules at all, and gets noticed in the gas when approaching the liquid conditions.

Van der Waals equation contains the volume in the term modifying the pressure, and this volume depends both on the pressure and the temperature.

[confusedstud]

z decreases as the molecules attract an other. This is more important when less heat is available to separate them. The extreme case is the liquid, where heat doesn't separate the molecules at all, and gets noticed in the gas when approaching the liquid conditions.

Van der Waals equation contains the volume in the term modifying the pressure, and this volume depends both on the pressure and the temperature.

Question 1:

I was trying to resolve why z<1 at low temperatures while z>1 at higher temperatures at a specific pressure using the van der Waal equation but this is really difficult for me. This is because at different temperatures the volume of the gas changes and trying to solve for V using the van der Waals equation is really difficult which is why I'm stuck.

According to this equation z = V/V-nb - (an/RTV) So assuming T=100K and T=200K. V(100K) should be lower than V(200K) just by using the ideal gas law's correlation. So the V/V-nb term at 100K is larger than the V/V-nb term at 200K. However, the an/RTV term at 100K is larger than the an/RTV term at 200K. So I'm guessing comparing 100K and 200K the second term is more significant than the first term so at lower temperatures z still falls below 1 despite the first term's opposite effect?


Question 2:

Actually I have one problem that I just thought of. We know that at very high pressures the z should always be greater than 1. However, when I was testing that out using this equation PV/nRT = z = V/V-nb - (an/RTV) I realised that the -(an/RTV) term also increases in magnitude as the pressure increases because the V decreases causing the an/RTV to increase. However, the V/V-nb term also increases as the pressure increases as the volume decreases. So either parts of the equation would have a opposite effect on z. But since we have already know that z increases as pressure increases that must mean that the V/V-nb factor must be greater than the -(an/RTV) factor. So now I'm substituting arbitrary values to confirm that that is true.

I set nb=5, an=5 and RT=3 and substituted V=8 so z=(8/8-5)-(5/3x8)=2.46
Next I substituted V=7 and z=3.26, next V=6 and z=5.72. So this proved to me that as the pressure increases the z will continue to increase. The thing to note here for me is that V>nb for it to be true. Initially I accidentally substituted V=0.5 which got me the opposite trend. But I was wondering, when V=nb the first term V/V-nb would give me an infinite number. So does this mean it is impossible to increase the pressure to the point where V=nb?