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We know that ideal gas equation is $PV= nRT$ where P is the pressure of the ideal gas and V is the volume of the ideal gas. Thus, we can write :

$$ P_{\textrm{ideal}}V_{\textrm{ideal}} = nRT$$

Now from volume correction term of van der Waals' equation we know that

$$ V_{\textrm{real}} = V_{\textrm{ideal}} - nb $$ where $ V_{\textrm{real}} $ is the volume of the real gas and $ V_{\textrm{ideal}} $ is the volume of the ideal gas. Clearly, $ V_{\textrm{ideal}}$ and $ V_{\textrm{real}} $ are different.

Then how can we write :

$$P_{\textrm{ideal}} (V_{\textrm{ideal}} - nb) = nRT$$

How can we replace $ V_{\textrm{ideal}}$ with $ V_{\textrm{real}}$?

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    $\begingroup$ There is frequent confusion: The term $V - nb$ is not correction of ideal volume to real volume, but the opposite, it is the correction of the real volume to ideal volume. similarly for $p + an^2/V^2$ is the correction of the real pressure to the ideal pressure. The ideal volume would be for the same pressure smaller, not bigger. similarly, the ideal pressure would be at the same volume bigger, not smaller. $\endgroup$
    – Poutnik
    Apr 15, 2021 at 15:10
  • $\begingroup$ Also relevant is that more sophisticated equations come with sets of consistent factors that work together and tearing out a correction term into another set of equations with different assumptions will not go well. $\endgroup$
    – matt_black
    Apr 15, 2021 at 15:15
  • $\begingroup$ @matt_black Sure, but I have not wanted to go so far. :-) $\endgroup$
    – Poutnik
    Apr 15, 2021 at 15:22

4 Answers 4

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The actual volume occupied by the gas is less than the measured volume(volume of the container).

In reality gaseous particles are not point objects and don't occupy zero space, which is against one of the postulates of the Kinetic Theory of Gases.

Thus while dealing with real gases we have to account for the volume which has been occupied by the gaseous particles. We do so by assuming the volume of gaseous particles to be $nb$.

Now, the ideal volume(actual volume) occupied by the gas is the measured volume(volume of container) minus the volume occupied by the gaseous particles.

Thus, $V_{ideal} = V_{real} - nb$ and not the other way around.

Note, $V_{ideal} =V_{actual}$ and $V_{real} =V_{measured}$.

The mistake you are making is formula you are using and thus

$P_{ideal}(V_{ideal}-nb) ≠ nRT$, but

$P_{ideal}(V_{real}-nb) = nRT$

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    $\begingroup$ In $PV = nRT$, P represents the actual pressure exerted by the gas. Now, we know that the pressure exerted by a ideal gas is greater than the pressure exerted by a real gas since intermolecular forces of attraction come into play in real gas. Thus, $P_{ideal} > P_{real}$. So, $P_{real} = P_{ideal} - something$. From van der Waals, we know that that '$something$' is $an²/V²$. Thus, $P_{real} or P_{actual} = P_{ideal} - an²/V²$. So, $(P_{ideal} - an²/V²)(V) = nRT$. What is wrong here? $\endgroup$
    – user281837
    Apr 15, 2021 at 15:22
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    $\begingroup$ @user281837 You did the same mistake again. "real" pressure is not the "actual" pressure. Actual pressure is considered ideal pressure. The measured pressure is the real pressure. Real pressure means the pressure we observe. Now substitute the actual pressure(ideal pressure) in $PV=nRT$ and not the real pressure. $\endgroup$ Apr 15, 2021 at 16:08
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    $\begingroup$ If you understand what I mean than please consider accepting and upvoting the answer. $\endgroup$ Apr 15, 2021 at 16:09
  • $\begingroup$ How $V_{\text {actual}}=V_{\text {ideal}}$? In ideal gas, the molecules have zero volume (point mass). $\endgroup$
    – Apurvium
    Mar 11 at 12:20
  • $\begingroup$ @Apurvium The actual volume here doesn't mean the volume of the container. It means the volume of the container excluding the volume occupied by the molecules themselves. Thus the next line about real volume = measured volume. $\endgroup$ Mar 12 at 7:30
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Here I think you are considering a container and labelling it as that of containing a volume $V$ and then considering it to be ideal. Then you go on to derive the Van der waals equation and maybe you think that $V\mathrm{_{real}}$ would thus be equal to the $V\mathrm{_{ideal}} - nb$ as now that $V\mathrm{_{real}}$ shouldn’t have that volume accessible to the particles.(This is what I perceive might be your problem, otherwise you may see the other answers.)

This is actually incorrect.

Consider a container of volume $V$ given below:

Gas in a container`

Here thus we can see that the volume of the container cannot be ideal volume rather it is real volume as it also has the volume of the particles and the inaccessible region around them.

Therefore, $$V\mathrm{_{real}}=V\mathrm{_{ideal}}+nb$$ and thus $$V\mathrm{_{ideal}}=V\mathrm{_{real}}-nb$$ where $V\mathrm{_{real}}$ is volume $V$ of the container.

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The Volume of a gas filled in a container is defined as the space in which the gas particles can move freely.

In case of Ideal gas where the gas particles have negligible Volume, the whole container is available for ideal gas particles to move and due to this reason $V$of ideal gas=Volume of container.

In the equation $PV=nRT$ the $V$ here is the Space available for the motion of the gas particles freely and not the necessarily the volume of the whole container, while P is the pressure exerted by the particles on the walls of the close container. And I assume that the source of your confusion is assuming that the defination of $V$ in the above equation (i.e. the volume of the gas) is the Volume of the whole container

The actual thing is that the value of the above $V$ comes out to be equal to the volume of container in case of an Ideal gas as explained above

Now in question or whenever, the Volume of gas is measured it is always the volume of the container given, for example,

The measured volume of a real gas = $V^c$ (say)

This means that the volume of the container in which real gas is present is $V^c$ that is $V_{real}=V^c$ thus the actual volume of this real gas ($V$) i.e. the volume available of the motion of its particle, which must be less than $V_{real}$ and the difference must be equal to the the Volume the real gas particles so, $$V=V_{real}-V'$$ And thus the equation, $$PV=nRT\Rightarrow (P_{real}+P')(V_{real}-V')=nRT$$ Where $P'$ and $V'$ are Pressure and Volume correction terms respectively.

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I realise I am a bit late to answer this, but I found out that no one had given the perfect answer to this. So, here's my two cents.

First off, we should just leave out the words $V_{ideal}, V_{real}$. They're just confusing.

Next, the ideal gas equation taught to you isn't completely correct. The actual equation is:$$P_{ideal} * Free\, Space =nRT$$ where $Free\, space$ means space allowing molecular motion.

Now, when dealing with ideal gases, you must know that their volume can be neglected, and hence the $Free \, Space$ is simply the volume of the container $V$.

However, when dealing with real gases, they are not point objects and don't occupy zero space. So, the value of $Free \, Space$ becomes $V-nb$. Here, I'm assuming you know why and how we took $b$ to be $4*Molar\, Volume$, so I'm not going to show its derivation.

So, the Van der Waals equation then becomes $$(P+\frac{an^2}{V^2})(V-nb)=nRT$$

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    $\begingroup$ $V_\mathrm{real}$ is the (measurable) volume occupied by a real gas of given amount, at given p and T. $V_\mathrm{ideal}$ is the volume a real gas would have occupied at the same conditions, if it had been behaving like an ideal gas. For real and ideal pressure analogy, physical chemistry uses for the latter the term fugacity. $\endgroup$
    – Poutnik
    Dec 7, 2021 at 8:47
  • $\begingroup$ Instead of downvoting, maybe y'all could tell me whats wrong with my answer... $\endgroup$ Dec 7, 2021 at 14:17
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    $\begingroup$ V-nb is not Free Space. b is not molar volume of molecules but its quadruple. So abandoning V_ideal and V_real in favour of Free Space is more confusing than keeping them. $\endgroup$
    – Poutnik
    Dec 7, 2021 at 14:27
  • $\begingroup$ Of course b is not molar volume, but I assumed OP knew that. Explaining why we put in $4*molar \, volume$ would just unnecessarily lengthen the answer. But V-nb is not free space?? Could you explain why? $\endgroup$ Dec 7, 2021 at 16:28
  • $\begingroup$ Well, it depends what you consider as free space. In sense of space allowing molecular motion it is. In sense of space not being occupied it is not. // Regardless of all that, using V_ideal and V_real, or p_ideal and p_real is clear. $\endgroup$
    – Poutnik
    Dec 7, 2021 at 17:48

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