2
$\begingroup$

$$ \require{begingroup} \begingroup \newcommand{\md}[0]{\mathrm{d}} \md U = \pi_T\,\md V + C_V \,\md T$$ where $\pi_T$ is the internal pressure and is given by $\displaystyle {\left(\partial U \over \partial V\right)_T}$.

  • What does the internal pressure actually physically mean? I can see it is the slope of the $U$-$V$ graph at constant temperature but that does not clear things up.

  • Also why is this quantity called the internal pressure and not something else? $\endgroup$

Improve this question
$\endgroup$
10
  • $\begingroup$ I'd assume that the phrase is trying to acknowledge the wall effect. Pressure at the wall of the container isn't the same as pressure away from the wall. $\endgroup$ – MaxW Feb 11 '17 at 18:45
  • $\begingroup$ @MaxW Outside the container or inside the conatiner ? $\endgroup$ – A---B Feb 11 '17 at 19:08
  • $\begingroup$ Inside of course. what would the pressure outside of the container have to do with anything? $\endgroup$ – MaxW Feb 11 '17 at 19:12
  • $\begingroup$ I also thought that but I did not understand where that pressure will be acting ? on other molecules ? $\endgroup$ – A---B Feb 11 '17 at 19:14
  • 1
    $\begingroup$ @MaxW Yes sorry, my mistake I thought the force field is inside the sphere of gas, whereas it is the exact opposite. $\endgroup$ – A---B Feb 11 '17 at 19:47
3
$\begingroup$

See orthocresol's derivation of an internal pressure $\pi_T$ relation $(1)$ here.

$$\pi_T = T\left(\frac{\partial P}{\partial T}\right)_V - P\tag{1}$$

Internal pressure $\pi_T$ measures what you would expect from its definition:

$$\pi_T \equiv \left(\frac{\partial U}{\partial V}\right)_T.$$

It shows how internal energy changes when volume is changed and temperature is constant. For ideal gases the change is zero, something that also follows from the equipartition theorem.

$$\mathrm{d}U_{\text{ideal gas}}= \frac{\nu}{2}nR\mathrm{d}T \implies \left(\frac{\partial U}{\partial V}\right)_T = 0 \ \ \text{for an ideal gas}$$

In other words, the total energy an ideal gas has is not dependent on volume. There is no repulsion nor attraction amongst ideal gas particles.

As a first approximation to model real gases, the van der Waals equation $(2)$ is used.

$$\left(P+a\frac{n^2}{V^2}\right)\left(V-nb\right)=nRT\tag{2}$$

Hence,

$$P = \frac{nRT}{V-nb} - a\frac{n^2}{V^2}\tag{3}.$$

Using relations $(1)$ and $(3)$

$$\left(\frac{\partial P}{\partial T}\right)_V = \frac{nR}{V-nb} \overset{(3)}{=} \frac{1}{T}\left(P + a\frac{n^2}{V^2}\right).$$

Again via formula $(1)$

$$\pi_T = T \cdot \overbrace{\frac{1}{T}\left(P + a\frac{n^2}{V^2}\right)}^{(\partial P/\partial T)_V} - P = a\frac{n^2}{V^2}\tag{4}.$$

Result $(4)$ $\pi_T = a\frac{n^2}{V^2}$ demonstrates that for real gases internal energy will change when volume is decreased or increased. Real particles have interparticular forces that vary with distance. A hint as to why $\pi_T$ is called internal pressure will be given when $(4)$ is substituted into $(2)$:

$$\left(P+\pi_T\right)\left(V-nb\right)=nRT,$$

or even better, $(3)$:

$$P = \frac{nRT}{V-nb} - \pi_T.$$

  • If $\pi_T > 0,$ then $P$ will be smaller. Therefore, attractive forces are dominant.
  • If $\pi_T < 0,$ then $P$ will be bigger. Therefore, repulsive forces are dominant.
Improve this answer
$\endgroup$
5
  • 1
    $\begingroup$ I don't think this is what the OP is asking about. I think question is more about why "internal pressure" vs. just "pressure." In other words what does "internal" add? $\endgroup$ – MaxW Feb 11 '17 at 19:19
  • 1
    $\begingroup$ @MaxW I did always think about it as 'a correction to pressure that derives from intermolecular forces'. As opposed to traditional (ideal gas) pressure which can be modelled as collisions with container walls. To recap, real gases occupy a finite volume, and if one decreases the volume of the system (via an external pressure), the molecules are forced more close together. How these particles respond is what internal pressure measures. This is, in turn, to be differentiated from the traditional pressure (of which internal pressure is a component). $\endgroup$ – Linear Christmas Feb 11 '17 at 19:35
  • $\begingroup$ @MaxW Whether this is a good way of thinking about it, remains to be seen. :=) (Feel free to downvote accordingly if this is wrong.) $\endgroup$ – Linear Christmas Feb 11 '17 at 19:37
  • 1
    $\begingroup$ @LinearChristmas Wow, this is easily one of the most complete answers I have ever got. Crystal clear. Thanks for the answer. $\endgroup$ – A---B Feb 11 '17 at 19:47
  • 1
    $\begingroup$ That is pretty much the significance of $\pi_T$, it is a measure of intermolecular (IM) forces. As V changes, the average IM distance will change, and then the difference in the IM potential energy will affect U. For an ideal gas, there are no IM forces; hence there's no effect on U and $\pi_T = 0$. (And I realise I am just repeating what you said... whoops) $\endgroup$ – orthocresol♦ Feb 11 '17 at 20:13
1
$\begingroup$

Assume that the gas is enclosed in a cylinder with a movable piston but the piston is not moving now. If the piston - wall interface is frictionless then the internal pressure (the one you are asking about) is the same as the external pressure (the one that can be controlled by mechanical means, e.g., by pushing or pulling on it.) If there is no friction and the process is reversible then the internal pressure is equal to the eternal pressure (the control variable) and is a state variable one that can be described by some other equilibrium parameters, such as $U$ or $T$ or $V$. If there is friction between the piston and the wall then depending on whether one pushes the piston in or pulls it out the internal pressure maybe larger or smaller than the external one and there is no such functional relationship among these parameters.

Improve this answer
$\endgroup$

Your Answer

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

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