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For the pressure exerted by compressed gas, its stated in my textbook: "The result is that at high pressures the molecules of a gas are so compressed that their volume becomes a significant fraction of the total volume of the gas. Since this reduces the volume available for molecular motion, collision occur more frequently.The pressure exerted by the real gas is higher than that predicted by Boyle's law for that particular volume of an ideal gas." What confuses me is the fact that the statement does not take in consideration for gravitational force of attraction between gas particles. Under the same temperature, if the volume is decreased by a significant amount, wouldn't the gravitational force of attraction between gas particles be no longer negligible?( following FG is inversely proportional to square of the distance between particles), hence this will cause the decrease in real pressure exerted to exterior surface( wall of the container).But, as stated in the textbook, its also true that there is a increase in real pressure due to smaller space for random motion of the particles( taking consideration for volume of the particles which is negligible within ideal gas's assumptions). So, doesn't these two make a contradiction? Or the increase in real pressure is larger than the decrease numerically so there is still an overall increase?

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You are right to think it through and an attractive force would tend to reduce pressure. However in this example the gravitational force is simply too weak to overturn the repulsive effect of the electrons of one molecule repelling those of another. So overall the force is repulsive when molecules are close.

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  • $\begingroup$ Thank you, so can you please explain this: " the intermolecular forces have more effect because of the slower speed of the molecules."? Can this be interpreted as in a particular instance the relative separation between molecules is greater hence a greater intermolecular force? And thats indeed the reason for the large deviation between ideal gas pressure and real gas pressure in low temperatures? $\endgroup$ – Wang Rui Mar 17 at 10:47
  • $\begingroup$ And won't the effect of intermolecular force of attraction be greater if the volume if decreased as well? ( constant temperature), so its again a decrease in pressure? and there is again a increase in pressure due to more rapid collision within smaller volume? $\endgroup$ – Wang Rui Mar 17 at 10:57
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Intermolecular/interatomic forces are much, much, much stronger than the gravitational force.

Consider for example an Argon gas. The interaction between Argon atoms is well described by the Lennard-Jones potential,

$$ U(r) = 4 \epsilon \left[\left(\frac \sigma r\right)^{12} -\left(\frac \sigma r\right)^6\right] $$

where $r$ is the distance between the centers of mass of the atoms and for Argon,

$$\epsilon/k_B \simeq 126 K$$ $$\sigma \simeq 0.335 nm$$

At short distances, only the repulsive part is relevant, and therefore

$$ U(r) \simeq 4 \epsilon \left(\frac \sigma r\right)^{12} $$

Now let's compare this with the gravitational interaction,

$$ U_G(r) = - \frac{G m m} {r} $$

For two Argon atoms,

$$G m m/k_B = 2.13 \cdot 10^{-38} m K$$

Comparing the absolute values of the potentials we obtain therefore

$$ \frac{U(r)}{|U_G(r)|}\simeq \frac{4 \epsilon \sigma}{Gmm} \left(\frac r \sigma\right)^{-11} = 7.92 \cdot 10^{30} \left(\frac r \sigma\right)^{-11} $$

or

$$ \frac{U(r)}{|U_G(r)|}\simeq 4.65 \cdot 10^{25} x^{-11} $$

where $x$ is the numerical value of $r$ in nm.

So you can see that the gravitational interaction is completely negligible compared to interatomic forces. This is true in general, and not only for ideal gases, even if finding an approximate form for the interatomic/intermolecular potential is quite difficult in general.

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  • $\begingroup$ So, in my comments above, the large deviation in pressure between real gas and ideal gas cause by the low temperature( greater effect of intermolecular force of attraction) is only true based on the temperature is not too low, and the intermolecular force stays attractive? $\endgroup$ – Wang Rui Mar 17 at 11:22

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