Intermolecular interaction Van Der Waals

Question

In the empirical demonstration of the Van der Waals equation at Khan Academy, they made a demonstration of the real volume, for example, it was said that it has to be larger than the ideal volume, since the size of the molecules is not disregarded, hence it would be:

$$ V_{\text{ideal}} = V_{\text{real}} + nb $$

Where, $ b $ was said to be related to the size of the molecule.

But why don't I also include intermolecular interaction? Depending on it, the volume may even be smaller.

How would interatomic interactions affect the volume? Can you please elaborate on how you think interatomic interactions affect the volume? — Nisarg Bhavsar, Apr 20 '21 at 15:31
Interatomic interactions will affect the pressure and not volume. — Nisarg Bhavsar, Apr 20 '21 at 16:07
@NisargBhavsar Intermolecular interactions can cause attraction or repulsion, changing the volume, right? — Lambert macuse, Apr 20 '21 at 16:13
Atomic interactions can't change the volume of a gas. Just assume if they did decrease the volume than wouldn't there be vacuum spaces inside a gas. — Nisarg Bhavsar, Apr 20 '21 at 16:18
If you see the full van der Waals equation, the parameter a is associated with the correction for intermolecular attraction. — S R Maiti, Apr 20 '21 at 16:21
@NisargBhavsar I disagree. In an attraction for example, there would still be vacuum spaces, they would just be a little bit smaller — Lambert macuse, Apr 20 '21 at 16:22
Let's say it is mobile, a piston, the kinetic energy (with the addition of heat, say) overcomes the interatomic attraction, and the volume increases. If you increase the attraction while increasing the kinetic energy, which is bizarre, you would even have that he said — Lambert macuse, Apr 20 '21 at 16:39
@NisargBhavsar To say that interaction does not change volume is absurd. If it were in this logic, the compressibility factor Z would always be greater than 1, and guess what, even if the real gas does not have a negligible volume, the compressibility factor is often less than 1, due to interactions — Lambert macuse, Apr 20 '21 at 17:03
Compressibility and Volume are two different quantities. Compressibility does depend on interatomic interactions. But that doesn't mean volume depends on interatomic interactions. — Nisarg Bhavsar, Apr 20 '21 at 17:11
*Volume ratio at the same pressure and to assure that you will have to include $a$. $Z= \frac{(P+ \frac{an}{V^2}) (V-nb) }{nRT}$. And I just noticed your equation you mentioned is also incorrect. — Nisarg Bhavsar, Apr 20 '21 at 17:24
$V_{\text{ideal}} = V_{\text{real}} - nb$ is correct. Refer this question — Nisarg Bhavsar, Apr 20 '21 at 17:28
I think you got something backward: the volume V is always "the real volume", no matter what gas you have. But to obtain an expression that looks like the ideal-gas you can make the substitution $V_{ideal}=V_{real}-nb$. — Buck Thorn, Apr 20 '21 at 18:31

score 3 · Accepted Answer · edited Apr 20 '21 at 20:13

Intermolecular attraction can reduce the volume, in the same way that increasing the pressure can lower the volume (Boyle's law).

One way to see the effect is to take the derivative of the volume, as given by the van der Waals (vdW) equation, with respect to the attractive parameter $a.$

The vdW equation is

$$\left(p+\frac{a}{\bar{V}^2}\right)(\bar{V}-b)=RT\tag{1}$$

Ignore for simplicity the excluded volume term $b,$ so that

$$p\bar{V}=RT-\frac{a}{\bar{V}}\tag{2}$$

Then the derivative of the volume wrt $a$ is

$$\left(\frac{\partial V}{\partial a} \right)_{p,T} = \frac{\bar{V}}{a-p\bar{V}^2}=-(RT)^{-1}\tag{3}$$

In essence what this result says is that the volume decreases as you increase the parameter $a.$

Intermolecular interaction Van Der Waals

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