Ideal Gasses vs Real gasses

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Feb 17, 2014
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Getting a little mixed up with Real vs. Ideal gasses. Could someone let me know if im thinking about it correctly?

PV = nRT is the ideal gas law
At STP(1 atm, 273 K), 1 mole of an ideal gas occupies 22.4L of volume. We assume molecular volume is negligible, intermolecular forces are negligible.

[P-(an^2)/(V^2)][V-nb] = nRT for a real gas, which is van der waals equation.
Real gas has a larger volume than an ideal gas because molecular volume is NOT negligible.
Real gas has a lower pressure than an ideal gas because of intermolecular forces(attractive).
Real gasses act more like ideal gasses with higher temperature(to break any intermolecular forces) and at low pressure(to minimize interactions between gas molecules)

However, I was reading up on something about total volume vs free volume and that's where my confusion comes in. When we use the ideal gas law, the V is equal to the total volume which is equal to the free volume. When we use van der waals, does the V refer to the total volume or the free volume?

So, at STP, does a real gas have more volume or less volume than an ideal gas?

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Real gases have larger volume than ideal gases because you have to take into account the molecular volume. So total volume is equal to the volume occupied by the particles of the gas plus the volume of the empty space between them. In an ideal gas, you assume the first term is zero, making the total volume equal to just the volume of the empty space between particles. But in a real gas, the first term matters.

You can also see this in the van der Waals equation - you have to subtract a term (n*b) from the real volume in order to get it to conform to the "ideal" volume term. Thus, the real volume has to be larger than the ideal volume.

This applies at the same given pressure.
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So, just to clear up one more thing, could the real pressure ever be higher than the ideal pressure?

Generally not because of the intermolecular attractive forces between gas particles (van der Waals attractive forces). To see this, refer to this formulation of the van der Waals equation:

P = (nRT/(V-nb)) - a(n/V)^2

Repulsive forces come into play when you pack the molecules in with each other so that they get close enough to repel. So that's taken into account by the "b" term. So as b increases, you'll see that P also increases. The "a" term factors in the attractive forces. So as a increases, P decreases. But the term (n/V) is squared, which magnifies its effect, so that you generally don't see a P(real) that's higher than P(ideal). To compare, you can get P(ideal) by setting a=b=0.