-2
$\begingroup$

The following equation is valid for ideal gases

$$pV=nRT$$

Please mind the following notation, any parameter that is related to an ideal gas would be denoted with a subscript $i$ and anything related to a real gas with a subscript $r$

For example, the pressure exerted by a real gas is denoted by $p_r$ and for an ideal gas, $p_i$

For real gases, when using the van der Waals equation, the following correction terms are introduced,

For the pressure correction,

$$p_i = p_r + \frac{an^2}{V^2} \tag{1}$$

For the volume correction:

$$V_r = V_i - nb\tag{2}$$

What I think is, since the relation holds for an ideal gas, we should multiply $p_i$ & $V_i$ with the necessary rearrangements.

But the equation taught to us as the van der Waals equation, if you think about it, is multiplying $p_i$ and $V_r$.

Which is correct and why?

$\endgroup$
8
  • 1
    $\begingroup$ Please see the following possible duplicate: chemistry.stackexchange.com/questions/37029/… as well as many other earlier posts on the subject $\endgroup$
    – Buck Thorn
    Sep 1, 2022 at 14:58
  • $\begingroup$ The site expects that you write explicit compact summary of your prior effort to answer the question, based on your knowledge and on searching for existing related info or answers. It helps preventing others to tell you what you already know or what you could easily find yourself what was provided many times in multiple ways. $\endgroup$
    – Poutnik
    Sep 2, 2022 at 5:38
  • $\begingroup$ @Poutnik the equation you have wrote above is wrong in my opinion specifically where you mention V(i) = V(r) - nb , how can the ideal volume be less than the real volume. $\endgroup$ Sep 2, 2022 at 7:04
  • $\begingroup$ @Poutnik From what i understand , the volume considered in the equations is the volume of free space in which gases can move, and not the volume of the gases itself. Now in ideal case it's almost equivalent to all of the space in the container since assumption is made that molecular volume is negligibly small. In the real case however the molecules carry significant volume hence the volume for movement is actually lesser than what we thought to be, therefore the subtractive term. And now the equation becomes V(r) = V(i) - nb , but you have wrote it other way around. $\endgroup$ Sep 2, 2022 at 7:20
  • $\begingroup$ @Poutnik That doesn't answer my question, to be specific i asked why are we multiplying P (i) with V (r) $\endgroup$ Sep 2, 2022 at 7:40

2 Answers 2

0
$\begingroup$

This was the way I understood the corrections in the van der Waals equation for real gasses.

Before which, a bit of history. How was it derived?


TL;DR: An empirical guide as to how the ideal gas law was proposed

The derivation of the gas equation

The ideal gas equation in its most simplified derivation is the combination of three of four different laws - Charles's Law ($V \propto T$), Boyle's Law ($p \propto V)$, Avogadro's Law ($n \propto V$) and Gay-Lusaac's Law ($p \propto T$).

Combining them, we get the following equation:

$$pV = nRT \tag{1}$$

This is the gas equation that follows. Notice that I haven't used the term ideal gas equation yet. This is because we now move on to the assumptions of an ideal gas to understand why this is the ideal gas scenario.

What makes a gas ideal?

A gas is considered to be ideal if it follows the following criteria as taken from this book

For a gas to be “ideal” there are four governing assumptions:

  1. The gas particles have negligible volume.
  2. The gas particles are equally sized and do not have intermolecular forces (attraction or repulsion) with other gas particles.
  3. The gas particles move randomly in agreement with Newton’s Laws of Motion.
  4. The gas particles have perfect elastic collisions with no energy loss.

Here, properties 1 and 2 are important for this discussion. An ideal gas occupies no volume and doesn't have any forces between particles

Now that we know what an ideal gas is, let's go back to the discussion at hand.

The ideal(?) gas equation

As mentioned in equation 1, the gas equation uses the formula $pV = nRT$. But what is $p$, $V$, $n$ and $T$ in terms of what we can measure? Well, $p$ is supposed to be the pressure exerted on the container by the gas itself. $V$, the volume of the container where the gas can move around in, $T$ is the temperature of the container and $n$ is the amount of gas within the container. But then why do we say equation 1 is the ideal gas equation?

If you look at it carefully and think about it from a molecular standpoint, the above gas equation only holds when the gas occupies no volume and has no inter molecular forces between particles. This means that the pressure observed is the same as the pressure exerted and the volume observed is the same as the volume where the gas can move around. Hence our gas equation has become the ideal gas equation


Time to make things real (or closer to real)

Our gas equation holds because they were derived from empirical relations. This means that it should work for ideal as well as real gasses. However, real gasses don't have the the same two properties that made the ideal gas law so simple to write down. They interact between molecules as well as take up space.

Going back to the definitions of the parameters of the gas equation as well as the above paragraph, we notice one thing. What we measure is not what is actually what we need. Intermolecular forces pull back on the molecules nearer to the edges of the container reducing the actual pressure observed and the molecules actually take up space reducing the volume that they can move around in

Van der Waals came up with the corrections whose proofs can be found online. They were,

$$p_\mathrm{correction} = \frac{an^2}{V^2} \tag{2}$$ $$V_\mathrm{correction} = nb\tag{3}$$

Applying this corrections using the discussion so far, we get:

$$\left(p + \frac{an^2}{V^2}\right)\left(V - nb\right) = nRT \tag{4}$$

Here $p$ and $V$ stand for the observed values. This is why the term for pressure is additive and the term for volume is subtractive.

$\endgroup$
0
$\begingroup$

$$nRT = p_\mathrm{i}V_\mathrm{i} = (p_\mathrm{r}+a\frac{n^2}{V^2})(V_\mathrm{r} - nb)$$

@Poutnik the equation you have wrote above is wrong in my opinion specifically where you mention V(i) = V(r) - nb , how can the ideal volume be less than the real volume.


The root cause of your confusion is that you consider V as the free volume between gas molecules and not as the total gas volume, as intended.


When real volume approaches the minimal volume of free molecule motion, it is like if an ideal gas volume approaches zero. In both cases, the pressure would shoot upwards.

By other words, if there are negligible cohesive forces, a real gas has bigger volume than an ideal gas would have at the same p and T.

By yet other words, ideal gases have infinite compressibility, real gases do not. At extreme pressures, real gas volume can be orders bigger than ideal gas volume.

@Poutnik From what i understand , the volume considered in the equations is the volume of free space in which gases can move, and not the volume of the gases itself. Now in ideal case it's almost equivalent to all of the space in the container since assumption is made that molecular volume is negligibly small. In the real case however the molecules carry significant volume hence the volume for movement is actually lesser than what we thought to be, therefore the subtractive term. And now the equation becomes V(r) = V(i) - nb , but you have wrote it other way around.

The volume in the van der Waals state equation for real gases and in the ideal gas state equation $pV=nRT$ means in both cases the volume of the container.

$V_\mathrm{r} > V_\mathrm{i} = \frac{nRT}{p}$ if cohesion has lower impact than molecular volume. Both terms in parenthases of van der Waals equation means corrections what would be p and V of an ideal gas, if there was no cohession and zero molecular volume.

@Poutnik That doesn't answer my question, to be specific i asked why are we multiplying P (i) with V (r).

We do not. You do.

If you isothermally compress 1 L of helium at 100 kPa by pressure 100 MPa, will be the final volume more or less than the expected 1 mL for ideal behaviour? Is then real volume more or less then ideal volume?

The own volume of gas molecules causes more frequent collisions of molecules with container walls, compared to an ideal gas at the same conditions. That means

  • either higher real then ideal gas pressure in the same volume,
  • either bigger real than ideal gas volume at the same pressure.

Van der Waals equation pretends a real gas is an ideal gas, following principles of Perturbation theory(*). It masks the real gas non-idealities by corrective factors:

  • Increases the real gas pressure by $+a\frac{n^2}{V^2}$
    (as real pressure at walls is decreased by cohesion).
  • Decreases the real gas volume by $-nb$
    (as real volume is increased by lower compressibility due molecular volume).

That leads to modification of $V_\mathrm{r}$,$p_\mathrm{r}$, pretending to be ideal gas values $V_\mathrm{i}$,$p_\mathrm{i}$, plugging the latter to the ideal gas equation:

$$nRT = p_\mathrm{i}V_\mathrm{i} = (p_\mathrm{r}+a\frac{n^2}{V^2})(V_\mathrm{r} - nb)$$

Real volume is decreased to the smaller ideal volume ($V_\mathrm{i}=V_\mathrm{r} - nb$), because an ideal gas is more compressible. The extrapolation of the van der Waals model to extreme conditions $V_\mathrm{r} = nb$ leads as well as for $V_\mathrm{i} = 0$ to infinite pressure for $n \gt 0$, $T \gt 0$.


(*) Perturbation theory offers an approximate solution of a complex scenario by corrections(perturbations) of the known, exact solution of a related, but simpler scenario.

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.