2
$\begingroup$

This question has been troubling me for quite a while now. Which has greater pressure: Real gas or an ideal gas under similar conditions? What about their volumes?

I read this in many sources and each giving a different answer.

What I thought was $(p+an^2/V^2)(V-nb)=RT$ according to van der Waals equation where $p_\text{ideal} = p_\text{real} + \text{correction term}$.

I feel that $p_\text{ideal}$ is greater than $p_\text{real}$ as of there were no attraction forces between the molecules(as in ideal gas), they would strike the walls of the container with a greater impact.

I also feel that $V_\text{ideal}$ also greater than $V_\text{real}$ as $V_\text{ideal}$ is the volume of the container and $V_\text{real}$ is the effective/free volume which subtracts excluded volume from total volume of container.

$\endgroup$
  • 1
    $\begingroup$ Depending on the particular T&P a real gas may have a pressure above or below that predicted by ideal behavior. In fact a particular gas may be above ideal pressure for one T and below ideal pressure for a different T. $\endgroup$ – MaxW Sep 21 '18 at 18:08
  • $\begingroup$ Thanks for your quick reply. Is this in any way related to the compressibility factor? Does volume of real gas in comparison with ideal gas also depend on temperature/pressure? $\endgroup$ – thewitness Sep 21 '18 at 18:15
  • $\begingroup$ ?!? For a given number of moles, $n$, $$\dfrac{PV}{T}= nR$$ so pick whichever "variable" that you want from the three (P, V, and T) and the other two are constants. Typically the experimental P is used as the variable which is compared to the ideal pressure. So the comprehensibility factor $$Z=\dfrac{PV}{nRT}$$ shows deviation from ideal behavior. $\endgroup$ – MaxW Sep 21 '18 at 19:40
  • $\begingroup$ Have you tried determining the compressibility factor from the van der Waals equation? $\endgroup$ – Chet Miller Sep 21 '18 at 19:55
  • $\begingroup$ @ChesterMiller - Huh?!? the whole point of the comprehensibility factor is to quantify the deviation from "ideal" behavior. $\endgroup$ – MaxW Sep 22 '18 at 4:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.