# Why is the pressure correction added in the Van der Waals equation?

Since real gas particles experience attraction to each other, that means the pressure must be less from what we would expect. So why do we have to add a term to correct it? For example, in the correction for volume, we subtract a term to find the effective volume or the "real" volume. So, it follows that the "real" pressure must be less than the given one and so, should we not subtract a term?

This is mostly with regard to solving problems based on this law, where the values of pressure, volume, $$a$$ and $$b$$ are given

• The Van der Waals model was a crude method for adding some "fudge factors" into the ideal gas equation based on a theoretical justification. The Van der Waals model is by far not the best model for empirical data fitting. See Wikipedia article Real Gas – MaxW Jan 12 at 17:57
• So you basically say that the parameter would/should always be negative, right? Well, is it? – Karl Jan 12 at 17:57
• ".. real gas particles experience attraction to each other .." that is a very (too) simplistic statement. If you get the particles sufficiently close together, they always repel each other. – Karl Jan 12 at 17:59
• @Karl Interestingly, while atoms/molecules do repel each other at sufficiently close range, in gases that repulsion is entirely entropic. I.e., neutral gas atoms/molecules are always in the attractive range when it comes to the pair potential. You can squeeze atoms/molecules sufficiently close to reach the repulsive region of the pair potential, but if you get them that close they are no longer in the gas phase. – theorist Jan 13 at 8:16
• Doesn't the "a" term make the pressure less? – Chet Miller Jan 13 at 19:03

You may consider the vdW equation as the deviation from the ideal gas state equation $$pV_\mathrm{m} = RT$$.

If you look at the vdW equation

$$\left(P+\frac a{V_m^2}\right)(V_m-b)=R T$$

you can easily see why the corrections have their signs.

The pressure term correction has the positive sign, as the real gas has due molecule cohesion smaller volume than an ideal gas at the same pressure. Like if it was under bigger pressure than it really was.

Increasing $$p$$ term by a positive correction leads to lower predicted volume than for ideal gas, at the same pressure, because of cohesive force.

$$V_m=\frac {R T}{\left(P+\frac a{V_m^2}\right)}$$

The correction simulates this bigger pressure the ideal gas would need to get the same vol.

The volume term correction has the negative sign, as the gas has at the given volume higher pressure due molecule own volume.

Decreasing $$V$$ term by a negative correction leads to higher predicted pressure than for ideal gas, at the same pressure, because of own molecule volume.

$$P=\frac {R T}{(V_m-b)}$$

• Smaller volume would mean more pressure than expected so the term should be subtracted? – Micelle Jan 12 at 18:43
• @AnanyaPateriya Smaller volume here means smaller atomic/molecular volume, not smaller container volume. I.e., the particles themselves take up less of the space, so the "free volume" is greater, hence the pressure is lower. – theorist Jan 13 at 8:18