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Background

Recently I have been studying about the states of matter and came to the topic of ideal gases and real gases, and the laws related to them. While studying it from my textbook, I saw the plot of pressure vs volume for real gas and ideal gas. The problem is that the interpretation I am getting from the graph is quite contrary to what I have learnt till now (this might be because of some conceptual misunderstanding of mine. Sorry if the question looks foolish).

Question Explained

Plot of pressure Vs volume for real and ideal gas

In the above image of the plot, there is a point where real gas behaves the same as ideal gas (where the curves intersect). From this point, as we decrease the volume in the graph we find that pressure exerted by the real gas is greater than that by the ideal gas. This seems to be contrary to what I have learnt, which is as the volume of a real gas decreases the pressure exerted by a real gas becomes lesser than ideal gas at the same volume (due to increased intermolecular attraction). In reality even this isn't the case as decreasing the volume much further causes the intermolecular repulsion to overcome the attraction at some point and hence increase the pressure. But as far as I know the graph has been made with the use of the van der Waals equation (I checked it myself via plotting the graph in Desmos and it looked the same as shown above). But the assumption with which the equation (and hence the graph) were derived say something different. The equation I'm talking about is $P_\mathrm{ideal} = P_\mathrm{real} + \dfrac {an^2}{V^2}$, and as far as I can comprehend it says pressure by an ideal gas is greater than pressure by a real gas at a given volume. Hence the question: Why does the equation $\left(P_\mathrm{ideal}=P_\mathrm{real}+\dfrac {an^2}{V^2}\right)$ and the graph (shown above) say totally different thing about the pressure at a given volume? Please do point out if I am going wrong anywhere. Thanks in advance! Here in this crash Course video the same logic as mine is given for decrease in pressure for real gas.

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  • $\begingroup$ The curves of real gases and ideal gases do not cross one another. Never ! $\endgroup$
    – Maurice
    Aug 4 at 19:38
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    $\begingroup$ Could you please provide a reference or link to the source of that figure. It is misleading and has appeared in numerous other posts. It would be great if we could track down the culprit. $\endgroup$
    – Buck Thorn
    Aug 7 at 10:31
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    $\begingroup$ @BuckThorn Please see pg. 14 ncert.nic.in/ncerts/l/kech105.pdf $\endgroup$
    – Lllt
    Aug 7 at 13:23
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    $\begingroup$ @Maurice Can you provide a reference of your statement? $\endgroup$
    – Apurvium
    Aug 8 at 6:09
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The lower right part of graph indicates that real gases exert less pressure than ideal gas for a fixed volume. This happens because the ideal gas molecules are not attracted by surrounding molecules (therefore if an ideal gas molecule move at velocity $V_1$, It will strike the wall with same velocity, provided it doesn't collide with another molecule) while real gas molecules face attraction from surrounding molecules as they move towards the wall ( thus if a real gas molecule moves with a velocity $V_1$, it will strike the wall with somewhat less speed as nearby molecules' attractive force will slow it down)

In the upper left part of graph, you can see that to exert same pressure , we need more of a real gas then an ideal gas or similarly at a certain volume of both , the pressure exerted by real gas is more than that of ideal gas. This can be explained by fact that molecular size of ideal gas particles is negligible while that of real gas particles is considerable. If you take equal volume $V$ of both, the ideal gas has occupied complete volume $V$ while real gas has occupied effectively somewhat less volume (which is $V'=V-nb$). Therefore even if we have same number molecules of real gas as that of ideal gas, real gas molecules have lower volume to move into thus they exert more pressure.

The Van der Waals equation of state is

\begin{align} p &= \frac{nRT}{V-nb} -\frac{an^2}{V^2}\\ &= \frac{\frac{nRT}{V}}{1-\frac{nb}{V}} - \frac{an^2}{V^2}\\ &= \frac{nRT}{V}\left({1-\frac{nb}{V}}\right)^{-1} -\frac{an^2}{V^2}\\ &= \frac{nRT}{V}\left({1+\frac{nb}{V}}\right) - \frac{an^2}{V^2} & \text{assuming } nb &\ll V\\ &= \frac{nRT}{V} + \frac{n^2bRT}{V^2} - \frac{an^2}{V^2}\\ &= p_\mathrm{ideal} + \frac{n^2}{V^2}(bRT-a) \end{align}

Depending on whether $bRT$ is less than or greater than $a$, you have $p$ less than or greater than $p_\mathrm{ideal}$.

Also note that when $V$ is very large the second term can be neglected and in that case real gas tends to ideal conditions.

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  • $\begingroup$ Please compare your answer with the equation also for both situations, i.e. large and small volumes. $\endgroup$
    – Apurvium
    Aug 5 at 5:07
  • $\begingroup$ @Apurvium If repulsive forces become dominant then why should pressure be less than $P_{ideal}$, it should be more $\endgroup$
    – Lllt
    Aug 5 at 6:17
  • $\begingroup$ Yes, you are right, I will repost. $\endgroup$
    – Apurvium
    Aug 5 at 6:26
  • $\begingroup$ The $T$ is constant & I think the $V$ is volume of container. The only way we can increase the $P$ is by reducing the volume of container by a piston or so. When we do so, the repulsive forces also become significant for real gas. Now, the pressure exerted at the walls of the container should be more than $P_{ideal}$. This is not according to the equation between $P_{ideal}$ and $P_{real}$ but yes, according to the graph. At high volume the situation is vice versa. $\endgroup$
    – Apurvium
    Aug 5 at 6:29
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    $\begingroup$ $\frac{1}{1-a} \ne 1+a$ // $\frac{1}{1-a} \approx 1+a$ for $|a| \ll 1$. So there is the condition $nb \ll V$ $\endgroup$
    – Poutnik
    Aug 5 at 7:34
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Summary

At sufficiently low volumes, the excluded volume effect causes the van der Waals (VDW) pressure of all gases to exceed the ideal gas (IG) pressure. At higher volumes, the VDW pressure may or may not cross below the IG pressure. This depends on the temperature, and the relative size of the VDW a and b terms. If the temperature is sufficiently low, and if the a term is sufficiently large relative to the b term, you will see a crossover. Otherwise, you won't.

Notes:

(1) The OP asks for a discussion of VDW gases, and thus most of this analysis concerns the predictions of the VDW model. However, the curves presented in the OP's screenshot are for real gases and thus, at the end, I discuss their behavior.

(2) Below the VDW critical temperature (which equals $\frac{8a}{27bR}$, and is thus gas-specific), the VDW isotherms (the curves on a $p-V$ graph) begin to show oscillations. Those oscillations are beyond the scope of this question, but they should be noted for completeness. This answer assumes we are above the VDW critical temperature for each gas.

The VDW model

The VDW model assumes the gas molecules are hard spheres that interact only attractively. Unlike the case with the attractive forces, there are no long-range repulsive forces in the VDW model. Indeed, there are no repulsive forces at all between the gas molecules (except insofar as they are hard spheres that can't interpenetrate).

Instead, the molecules are assumed to have a physical volume, and the effect of this volume is to reduce the free volume left for the gas. The volume of one mole of molecules (or atoms) is given by the VDW "b" term.

This should help illustrate how the physical volume of the gas particles affects the pressure:

Suppose you are at a volume where the hard spheres themselves take up half the container. If there were no attractive forces, the pressure of the VDW gas would be double that of the ideal gas, because only half as much free volume is left for the VDW gas, such that p will be a monotonically decreasing function of V.

Physical explanation of difference between pressure of VDW gas and ideal gas

At very low volume (low relative to the volume taken up by the hard spheres), the molecular volume term dominates, and thus p_VDW > p_IG (where IG = ideal gas).

What happens at higher volumes, however, is gas- and temperature-specific. And thus whether or not there is a crossover is gas- and temperature-specific.

Case I: Lower temperature

At a sufficiently high volume, the pressure-increasing effect of molecular volume becomes less than the pressure-reducing effect of the attractive force, and thus p_VDW < p_IG. Hence there is a point at which the VDW and IG curves cross over.

Again, let me emphasize that what constitutes "lower" vs. "higher" temperature is gas-specific. For instance, while $T = 273 \, \ce{K}$ is in the low-temperature regime for $\ce {N2(g)}$, it's in the high-temperature regime for Ne. Thus, at this temp, you would see a crossover for $\ce {N2(g)}$, but not for Ne.

More specifically, $\ce {Ne(g)}$, which has a small a value relative to b, loses its crossover behavior above $149.8 \, \ce{K}$, while $\ce {N2(g)}$ doesn't lose its crossover behavior until the temperature exceeds $425.8 \, \ce{K}$.

Case II: Higher temperature

As explained above, as the volume increases, the effect of molecular volume decreases. However, the intermolecular spacing also increases, lessening the average attractive force between the particles. Further at, high temperatures the kinetic energy of the gas is higher, thus making the effect on pressure of intermolecular attraction relatively less. If the attractive force is sufficiently low, and the temperature is sufficiently high, then the attractive force is never able to overcome the molecular volume effect. Hence, for these gases at these temperatures, p_VDW > p_IG at all volumes.

Note: the above discussion is referring to the VDW model, not necessarily to how real gases actually behave.

Mathematical explanation of difference between pressure of VDW gas and ideal gas

We can also see this from the mathematical structure of the VDW equation:

$$p=\frac{RT}{V_m-b}-\frac{a}{V_m^2}$$

In the limit as $V_m \rightarrow b$, $(V_m-b)\rightarrow 0$, causing the first term to dominate over the second.

As $V_m$ gets large, whether the attraction can win out over the excluded volume is determined by the magnitude of T, and the relative values of a and b.

In the limit as $V_m \rightarrow \infty$, for all gases, the VDW equation reduce to the ideal gas equation:

$$p=\frac{RT}{V_m}$$

Plots, Case I, lower T ($\ce{N2(g)}$ at $\pu{273K}$)

Comparative pressure vs. volume plots of VDW and ideal gas

Now let's take a look at what the actual $p-V$ graphs look like, using $\ce {N2(g)}$ as an example. The VDW a and b terms for $\ce {N2(g)}$ are:

$$a=1.37 \ce{ \frac{bar L^2}{mol^2}}$$ $$b=0.0387 \ce{ \frac{L}{mol}}$$

Assuming $T=273 \,\ce{K}$, we can determine crossover point by setting pIdeal = pN2VDW (the VDW pressure for $\ce{N2(g)}$), and solving for $p$. The crossover occurs at $V = 0.108 \,\ce{ L}$. But if we plot pIdeal and pN2VDW from about 0.1 L to 10.0 L, we find they are nearly on top of each other:

enter image description here

Thus to see the crossover, we need to focus on the region around $V = 0.108 \,\ce{ L}$: enter image description here

Plot of p_VDW–p_ideal vs volume, using $\ce{N2 (g)}$ as the VDW gas

We can further improve our understanding by plotting the difference between pIdeal and pN2VDW vs $V$. This allows us to see, directly, how the deviation between the VDW equation and the ideal gas equation changes with volume.

Let's start at high volume, and move to the left. At high volume, the attractive term dominates. As the volume decreases, the molecules get closer together, so the attractive term has even more effect, increasing the difference between pIdeal and pVDWN2. This continues until $V=0.154 \,\ce{L}$ [I determined this minimum by setting the derivative of (pN2VDW - pIdeal) equal to zero, and solving for $V$.] Beyond this, the increase in pressure due to the diminishing free volume becomes greater than the decrease in pressure due to the increasingly strong attractive forces. Eventually, at $V=0.108 \,\ce{L}$ the effect of the hard sphere volume just balances that of the attractive forces, and the curve crosses above the zero line.

enter image description here

Plots, Case II, higher T ($\ce{N2(g)}$ at $\pu{500 K}$)

Assuming $T = 500 \,\ce{K}$, if we solve for the crossover point as we did above, we get a non-physical answer of $V = -0.222 \,\ce{ L}$. This indicates the curves don't cross over. We can see this from the plots:

enter image description here

enter image description here

Real Gases

The easiest way to see how the isotherms for real gases compare with those for an ideal gas is to look at the measured compressibility factor, $Z$:

$$Z=\frac{p V_b}{R T}$$

$Z = 1$ for an ideal gas. Thus, for a given $V_b$ and $T$, $Z>1 \implies p_{real} >p_{ideal}$, and visa-versa.

From https://en.wikipedia.org/wiki/Compressibility_factor#/media/File:Z_Overview.png, we have the following diagram for $\ce{N2(g)}$. It appears to show behavior similar to what the VDW equation predicts— at lower temperatures, we see a crossover from $p_{real} > p_{ideal}$ at high $p$ (low $V$), to $p_{real} < p_{ideal}$ at low $p$ (high $V$).

However, as the temperature increases, $p_{real}$ reduces its tendency to dip below $p_{ideal}$ (see red curve). However, a direct examination of the numerical data would be needed to determine if, at high $T$, there is no crossover at all.

enter image description here

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  • $\begingroup$ Can even at very high pressure $V_m$ $\ce{->}$$b$ ? $\endgroup$
    – Apurvium
    Aug 8 at 6:48
  • $\begingroup$ +1 for your effort but I am confused as there are contradictory answers. Few confusions: (1) From the VDW equation you mentioned, when $V_m$$\ce{->}\infty$, it reduces to ideal gas expression but from the graph, $P_{real}<P_{ideal}$. $P_{real}=P_{ideal}$ point is at a much lower $V_m$. (2) Assuming the graph do cross, can we say that at very high $V_m$, a real gas is more $perfect$ than an $ideal$ gas? $\endgroup$
    – Apurvium
    Aug 8 at 7:53
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    $\begingroup$ @Apurvium (1) According to the VDW model, the pressure goes to infinity in the limit as Vm -> b. Physically, it's much more complicated, because gases aren't hard spheres, and they will typically become solids before extreme pressure pushes them into a repulsive regime. I.e., energetically, gas interactions are always attractive, and it's entropy that keeps them apart. $\endgroup$
    – theorist
    Aug 9 at 1:43
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    $\begingroup$ @Apurvium Yes, I agree with the expression dervied by Lllt for the high-volume regime ($V_b>>b$). Assuming you are above the VDW critical temperature ($\frac{8a}{27bR}$), if there is a $T$ below which $a<bRT$ then, at high volumes, the attractive forces will win, and you will see a crossover. Otherwise, you will not, because the VDW pressure will always be greater than the IG pressure. $\endgroup$
    – theorist
    Aug 9 at 19:06
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    $\begingroup$ @Apurvium See the note I added at the end about the behavior of real gases. They also show a crossover at lower temperatures. $\endgroup$
    – theorist
    Aug 10 at 6:03
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When the pressure tends to infinity, the volume of the ideal gas tends to zero, and the volume of the real gas does not. It tends to the proper volume of the molecules, which is greater than zero. This is exactly what you see on the upper part of the picture.

And when the pressure is extremely low, the volume increases but not as much as calculated with the ideal law, because there is some small attraction between the molecules. This is exactly what you see on your picture if you draw a horizontal line at its lower part

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    $\begingroup$ At very large volumes and low pressures, the behavior of a real gas should approach that of an ideal gas. $\endgroup$ Dec 11 '19 at 23:10
  • $\begingroup$ Sorry but I have a confusion. If that's the case then aren't the both graphs (real as well as ideal) showing the fact of non zero volume (reason being that both look like a rectangular hyperbola to me)? $\endgroup$
    – user84548
    Dec 12 '19 at 15:00
  • $\begingroup$ The ideal gas is a hyperbola. The real gas has a non analytical curve. $\endgroup$
    – Maurice
    Dec 12 '19 at 16:00
  • $\begingroup$ @maurice At a constant $V$, if we compare the pressures vis-a-vis the equation between $p_{ideal}$ and $p_{real}$? $\endgroup$
    – Apurvium
    Jul 29 at 6:59
  • $\begingroup$ Related $\endgroup$
    – Apurvium
    Aug 4 at 16:38
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The ideal gas pressure is always above the 'real' gas pressure as given by the van-der Waals eqation $$p=\frac{RT}{V_\mathrm{m}-b}-\frac{a}{V_\mathrm{m}^2},$$ where the volume is the molar volume $V_m$. The molar volume varies from say $0.1$ to $\pu{5 dm^3/mol}$ and by experiment constant $a$ in units $\pu{dm6 bar mol-2}$ is numerically greater than $b$ in units $\pu{dm3/mol}$ and the gas constant $R=\pu{0.08314 dm3 bar mol-1 K-1}$.

At large volume the second term tends to zero and the first to $RT/V$ the ideal gas value. At small volumes the second term becomes larger and negative and reduces the pressure from the ideal behaviour. Thus the ideal and real curves do not cross and the ideal is always greater than the real.

The interaction between molecules ensures that the real pressure is lower than in the ideal case. (You can imagine this in a hand waving sort of way by imagining that a small fraction of molecules form dimers so that the expected pressure is not measured because some molecules are paired up.)

Thus the figure in the question is wrong on a number of points.

vdw-pv

I'm adding some notes here about the vdw equation as there is confusion in the comments.

First it is clear that the van-der-Waals equation can produce negative pressures. It is an empirical equation that produces largely qualitative results. All that has to happen to get negative pressures is that the second term exceeds the first in the equation above. Negative pressures are, of course, unphysical and this shows that this equation has to be used with great care otherwise completely misleading results will be obtained.

Second, at large volumes the $V_m-b$ term is smaller than $V$ because $b$ is positive, so the pressure will exceed that given by the ideal gas but the effect is small. A typical molar volume of $0.25\;\mathrm{ dm^3/mol}$ vs a typical value of $b=0.02\;\mathrm{dm^3/mol}$, so small as to be physically meaningless but mathematically possible, what I mean is that we would normally take $V_m-b \to V_m$ at large $V_m$.

The important point about the two curves in the figure is that the interaction between molecules, (due to polarisability/ dipoles ) makes the experimental pressure less than that of the ideal, e.g. methane at 273.15 K pressure is 78.6 bar by experiment but 90.8 by ideal gas law and 72.9 by van-der-Waals. The van-der-Waals eqn. tries to model the experimental results and does so quite well but should not be pushed to the mathematical limits as here it fails.

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  • $\begingroup$ Your answer is definitely enlightening ,are you sure that those curves never cross each other (I am asking because the figure taken above is from the national book of India) $\endgroup$
    – Lllt
    Aug 6 at 13:39
  • $\begingroup$ They never cross, you can see that in the limit when V is very large the ideal gas law is produced to that they tend to the same value. Try this with some numbers, say at 300 K for methane a=2.3 , b=0.043 in the units I gave and V=1 vs V=10 etc. $\endgroup$
    – porphyrin
    Aug 6 at 14:15
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    $\begingroup$ I am afraid the national book of India is wrong ! $\endgroup$
    – Maurice
    Aug 6 at 20:44
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    $\begingroup$ Correcting my earlier comment: Sometimes the curves do cross over, sometimes they don't. As $V_m$ gets large, whether the attraction can win out over the excluded volume is determined by the magnitude of T, and the relative values of a and b. $\endgroup$
    – theorist
    Aug 9 at 0:44
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    $\begingroup$ @TooTea, quite so, I agree, but this is at an extreme. The figure in the original question must relate to the gas at sort of normal conditions one might meet as its a from a general textbook, it is very misleading for students otherwise, i.e. to show extreme case (as it does) and ignore more general case. My comments related to that general position, interactions between molecules lead to a reduced pressure vs ideal. From this position you can then go on to extremes. $\endgroup$
    – porphyrin
    Aug 10 at 11:12

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